We provide matrices practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. We focus on matrices skills mastery so, below you will get all questions that are also asking in the competition exam beside that classroom.
List of matrices Questions
| Question No | Questions | Class |
|---|---|---|
| 1 | ( fleft(begin{array}{ccc}-1 & 2 & 4 \ 3 & 6 & -5end{array}right] ) then find ( 3 A ) | 12 |
| 2 | The element in the first row and third column of the inverse of the matrix ( left[begin{array}{ccc}1 & 2 & -3 \ 0 & 1 & 2 \ 0 & 0 & 1end{array}right] ) is A . -2 B. ( c cdot 1 ) D. 7 | 12 |
| 3 | Trace of matrix ( boldsymbol{A}^{k} ) is ( mathbf{A} cdot 3^{k}+1+(-1)^{k} ) B . ( 2^{k}+3^{k}-2 ) c. ( 3^{k}-2^{k}+2 ) ( mathbf{D} cdot 2^{k}+1 ) | 12 |
| 4 | If ( A=left[begin{array}{ccc}1 & 2 & 3 \ 2 & 3 & 4 \ 0 & 5 & 6end{array}right], ) then ( 2 A= ) A. ( left[begin{array}{lll}2 & 4 & 6 \ 2 & 3 & 4 \ 0 & 5 & 6end{array}right] ) в. ( left[begin{array}{lll}1 & 2 & 3 \ 4 & 6 & 8 \ 0 & 5 & 6end{array}right] ) с. ( left[begin{array}{lll}1 & 2 & 3 \ 2 & 3 & 4 \ 0 & 10 & 12end{array}right] ) D. ( left[begin{array}{lll}2 & 4 & 6 \ 4 & 6 & 8 \ 0 & 10 & 12end{array}right] ) | 12 |
| 5 | If ( boldsymbol{A}=left[begin{array}{ll}2 & 3 \ 5 & 7end{array}right], B=left[begin{array}{cc}0 & 4 \ -1 & 7end{array}right], C= ) ( left[begin{array}{cc}1 & 0 \ -1 & 4end{array}right], ) find ( A C+B^{2}-10 C ) | 12 |
| 6 | Q Type your question ( left.begin{array}{lll}mathbf{0} & mathbf{0} & mathbf{1}end{array}right], mathbf{i} mathbf{z}- ) ( left[begin{array}{lll}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{0} & mathbf{1} \ mathbf{0} & mathbf{1} & mathbf{0}end{array}right], boldsymbol{P}_{mathbf{3}}=left[begin{array}{lll}mathbf{0} & mathbf{1} & mathbf{0} \ mathbf{1} & mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{0} & mathbf{1}end{array}right] ) ( boldsymbol{P}_{4}=left[begin{array}{ccc}0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 1end{array}right], P_{5}= ) ( left[begin{array}{lll}0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0end{array}right], P_{6}=left[begin{array}{lll}0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0end{array}right], ) and ( boldsymbol{X}=sum_{boldsymbol{k}=1}^{6} boldsymbol{P}_{boldsymbol{K}}left[begin{array}{lll}2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1end{array}right] boldsymbol{P}_{K}^{T} ) Where ( P_{K}^{T} ) denotes the transpose of matrix ( P_{k} ). Then which of the following option is / are correct? This question has multiple correct options A. ( X ) is symmetric matrix B. The sum of the diagonal entries of ( X ) is 18 c. If ( Xleft[begin{array}{l}1 \ 1 \ 1end{array}right]=alphaleft[begin{array}{l}1 \ 1 \ 1end{array}right] ) then ( alpha=30 ) D. ( X-30 I ) is an invertible matrix | 12 |
| 7 | Assertion ( mathbf{f}[boldsymbol{x} mathbf{1}]left[begin{array}{cc}mathbf{1} & mathbf{0} \ -mathbf{2} & mathbf{3}end{array}right]left[begin{array}{c}boldsymbol{x} \ -mathbf{5}end{array}right]=mathbf{0}, ) then value of ( x ) is either- 3 or 5 Reason Two matrices ( left[begin{array}{ll}boldsymbol{x} & boldsymbol{y} \ boldsymbol{u} & boldsymbol{v}end{array}right] ) & ( left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right] ) are equal if ( & ) only if their corresponding entries are equal & only if their corresponding entries are equal A. Both (A) & (R) are individually true & (R) is correct explanation of (A) B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A). c. (A)is true but (R) is false D. (A)is false but (R) is true | 12 |
| 8 | Suppose ( A ) and ( B ) are two square matrices of same order. If ( boldsymbol{A}, boldsymbol{B} ) are symmetric matrices, then ( A B-B A ) is A. A symmetric matrix B. A skew symmetric c. A scalar matrix D. A triangular matrix | 12 |
| 9 | (a) a = 2ab, B = a? +62 (©) a = 2? +B2, B= 2ab (b) a = a? +62, B = ab (d) a=a? +62, B=a? –62. | 12 |
| 10 | Find the inverse of ( boldsymbol{A}=left[begin{array}{lll}mathbf{0} & mathbf{1} & mathbf{2} \ mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{3} & mathbf{1} & mathbf{1}end{array}right] ) ( boldsymbol{A}^{-1}=left[begin{array}{ccc}1 / 2 & -1 / 2 & 1 / 2 \ a & 3 & b \ c & -3 / 2 & 1 / 2end{array}right] ) Find ( |boldsymbol{a} boldsymbol{b} boldsymbol{c}| ? ) | 12 |
| 11 | Find the order of the following matrices. (i) ( left.begin{array}{ccc}1 & -1 & 5 \ -2 & 3 & 4end{array}right) ) (ii) (iii) ( begin{array}{rl}3 & -26 \ 6 & -11 \ 2 & 4end{array} ) (iv) (345) ( (v)left[begin{array}{cc}1 & 2 \ -2 & 3 \ 9 & 7 \ 6 & 4end{array}right] ) | 12 |
| 12 | If ( boldsymbol{A}=left[begin{array}{cc}1 & 2 \ 3 & 4 \ 5 & 6end{array}right] ) and ( B=left[begin{array}{cc}-3 & -2 \ 1 & -5 \ 4 & 3end{array}right] ) then find ( D=left[begin{array}{ll}boldsymbol{p} & boldsymbol{q} \ boldsymbol{r} & boldsymbol{s} \ boldsymbol{t} & boldsymbol{u}end{array}right] ) such that ( boldsymbol{A}+ ) ( boldsymbol{B}-boldsymbol{D}=boldsymbol{O} ) ( mathbf{A} cdotleft[begin{array}{cc}-1 & 0 \ 4 & -1 \ 9 & 9end{array}right] ) ( mathbf{B} cdotleft[begin{array}{cc}-3 & 0 \ 4 & -1 \ 9 & 9end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-2 & 0 \ 4 & -1 \ 9 & 9end{array}right] ) D. ( left[begin{array}{cc}-2 & 0 \ 4 & -5 \ 9 & 9end{array}right] ) | 12 |
| 13 | Let ( quad A=left(begin{array}{ccc}x^{2} & 6 & 8 \ 3 & y^{2} & 9 \ 4 & 5 & z^{2}end{array}right) ) and ( B= ) ( left(begin{array}{ccc}2 x & 3 & 5 \ 2 & 2 y & 6 \ 1 & 4 & 2 z-3end{array}right) ) be two matrices and if ( operatorname{Tr}(boldsymbol{A})=boldsymbol{T} boldsymbol{r}(boldsymbol{B}), ) then the value of ( (x+y+z) ) is equal to (Note: ( operatorname{Tr}(P) ) denotes trace of matrix ( P ) ) ( A ) B. ( c ) ( D ) | 12 |
| 14 | Which one of the following is true for any two square matrices ( A ) and ( B ) of same order? ( mathbf{A} cdot(A B)^{T}=A^{T} B^{T} ) B . ( left(A^{T} Bright)^{T}=A^{T} B^{T} ) c. ( (A B)^{T}=B A ) D. ( (A B)^{T}=B^{T} A^{T} ) | 12 |
| 15 | A matrix has 8 elements. What are the possible orders it can have? | 12 |
| 16 | If ( boldsymbol{A} ) satisfies the equation ( boldsymbol{x}^{3}-mathbf{5} boldsymbol{x}^{2}+ ) ( 4 x+k I=0, ) then ( A^{-1} ) exists if A. ( k neq-1 ) в. ( k neq 0 ) c. ( k neq 1 ) D. none of these | 12 |
| 17 | If ( A ) is a skew-symmetric matrix, then trace of ( boldsymbol{A} ) is | 12 |
| 18 | If ( A=left[begin{array}{cr}2 & 5 a \ -3 & 1end{array}right] ) and ( A ) doesn’t have multiplicative inverse then find ( mathbf{A} ) | 12 |
| 19 | For what value of ( x, ) is the matrix ( A= ) ( left[begin{array}{ccc}mathbf{0} & mathbf{1} & mathbf{- 1} \ -mathbf{1} & mathbf{0} & mathbf{3} \ boldsymbol{x} & mathbf{- 3} & mathbf{0}end{array}right] ) a skew symmetric matrix? | 12 |
| 20 | matrix is a square matrix in which all the elements other than the principal diagonal elements are zero. A. scalar B. null c. diagonal D. unit | 12 |
| 21 | ( left|begin{array}{cc}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{4}end{array}right|+left|begin{array}{cc}boldsymbol{x} & mathbf{3} \ boldsymbol{y} & mathbf{1}end{array}right|=left|begin{array}{cc}mathbf{1 0} & mathbf{6} \ mathbf{8} & mathbf{5}end{array}right|, ) then ( (mathbf{x}, mathbf{y})= ) A ( cdot(4,8) ) ( B cdot(8,4) ) ( c cdot(1,2) ) ( D cdot(2,4) ) | 12 |
| 22 | If ( mathbf{A}=left[mathbf{a}_{mathbf{i} mathbf{j}}right] ) is a scalar matrix of order ( boldsymbol{n} times boldsymbol{n} ) such that ( mathbf{a}_{mathbf{i j}}=mathbf{k} ) for all ( mathbf{i}=boldsymbol{j}, ) then trace of ( mathbf{A}= ) ( A cdot ) nk B. ( n+k ) c. ( frac{n}{k} ) D. | 12 |
| 23 | Let ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & mathbf{3} \ mathbf{- 1} & mathbf{5}end{array}right] cdot ) If ( boldsymbol{A}^{-mathbf{1}}=boldsymbol{x} boldsymbol{A}+boldsymbol{y} boldsymbol{I} ) find ( boldsymbol{x}+mathbf{2} boldsymbol{y} ) | 12 |
| 24 | cos -sino 41. IfA= sino cose ] |, then the matrix A-50 when 0 = *, is equal to: [JEE M 2019-9 Jan (M)] (b) 1 V3 ISO-IN (d) v3 | 12 |
| 25 | Find the value of ( y-x ) from the following equation ( mathbf{2}left[begin{array}{cc}boldsymbol{x} & mathbf{5} \ mathbf{7} & boldsymbol{y}-mathbf{3}end{array}right]+left[begin{array}{cc}mathbf{3} & mathbf{- 4} \ mathbf{1} & mathbf{2}end{array}right]=left[begin{array}{cc}mathbf{7} & mathbf{6} \ mathbf{1 5} & mathbf{1 4}end{array}right] ) | 12 |
| 26 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{3} & -mathbf{4} \ mathbf{1} & -mathbf{1}end{array}right], ) then prove that ( boldsymbol{A}- ) ( A^{T} ) is a skew-symmetric matrix. | 12 |
| 27 | For what value of ( x, ) is the matrix ( A= ) ( left[begin{array}{ccc}mathbf{0} & mathbf{1} & -mathbf{2} \ -mathbf{1} & mathbf{0} & mathbf{3} \ boldsymbol{x} & mathbf{- 3} & mathbf{0}end{array}right] ) a skew-symmetric matrix? | 12 |
| 28 | ( mathbf{f}left[begin{array}{ccc}mathbf{1} & mathbf{3} & mathbf{0} \ mathbf{1} & mathbf{0} & -mathbf{2} \ mathbf{- 4} & mathbf{- 4} & mathbf{4}end{array}right]=mathbf{A}+mathbf{B} ) where ( mathbf{A} ) is symmetric matrix and B is skew- symmetric, then ( mathbf{A}-mathbf{B} ) is equal to ( mathbf{A} cdotleft[begin{array}{ccc}1 & 1 & -4 \ 3 & 0 & -4 \ 0 & -2 & 4end{array}right] ) B. ( left[begin{array}{lll}2 & 1 & 3 \ -1 & 2 & 4 \ 3 & -1 & 2end{array}right] ) ( mathbf{c} cdotleft[begin{array}{lll}0 & 1 & -1 \ 2 & 3 & 4 \ -4 & 1 & 2end{array}right] ) D. ( left[begin{array}{ccc}2 & -3 & 0 \ 0 & 1 & 2 \ 2 & 4 & 0end{array}right] ) | 12 |
| 29 | In the matrix, write: ( A=left[begin{array}{ccc}2 & 5 & 19-7 \ 35-2 & frac{5}{2} & 12 \ sqrt{3} & 1 & -517end{array}right] ) (i) The order of the matrix (ii) The number of elements (iii) Write the elements ( a_{13}, a_{21}, a_{33}, a_{24}, a_{23} ) | 12 |
| 30 | Assertion The possible dimension of a matrix consisting 27 elements is 4 Reason The number of ways of expressing 27 as a product of two positive integers is 4 A. Both Assertion & Reason are individually correct & Reason is correct explanation of Assertion, B. Both Assertion & Reason are individually true but Reason is Not the correct explanation of Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct. | 12 |
| 31 | If ( boldsymbol{A}=left|begin{array}{c}-mathbf{2} \ mathbf{4} \ mathbf{5}end{array}right|, boldsymbol{B}=|mathbf{1} quad mathbf{3}-mathbf{6}|, ) State whether ist is true or false ( (A B)^{1}= ) ( boldsymbol{B}^{1} boldsymbol{A}^{1} ) A. True B. False | 12 |
| 32 | If ( mathbf{A}=left[begin{array}{cc}mathbf{2} & mathbf{2} \ -mathbf{3} & mathbf{2}end{array}right], mathbf{B}=left[begin{array}{cc}mathbf{0} & mathbf{- 1} \ mathbf{1} & mathbf{0}end{array}right] ) then ( mathbf{B}^{-mathbf{1}} mathbf{A}^{-mathbf{1}} ) A ( cdot frac{1}{10}left[begin{array}{cc}2 & 2 \ -3 & 2end{array}right] ) в. ( frac{1}{10}left[begin{array}{cc}2 & -2 \ 2 & 3end{array}right] ) c. ( frac{1}{10}left[begin{array}{cc}2 & 2 \ -2 & 3end{array}right] ) D. ( cdot frac{1}{10}left[begin{array}{cc}-2 & 2 \ 2 & 3end{array}right] ) | 12 |
| 33 | If ( A ) is an ( m times n ) matrix and ( B ) is ( n times p ) matrix, then does ( A B ) exist? If yes, write its order. | 12 |
| 34 | Find the matrix ( A ), such that ( left[begin{array}{l}4 \ 1 \ 3end{array}right] A= ) ( left[begin{array}{ccc}-4 & 8 & 4 \ -1 & 2 & 1 \ -3 & 6 & 3end{array}right] ) | 12 |
| 35 | If ( boldsymbol{A}=left[begin{array}{lll}1 & 1 & 1end{array}right], ) then ( A ) is a A. Identity matrix B. Null matrix c. Diagonal matrix D. Row matrix | 12 |
| 36 | If the order of ( mathbf{A} ) is ( mathbf{4} times mathbf{3}, ) the order of ( mathbf{B} ) is ( 4 times 5 ) and the order of ( C ) is ( 7 times 3 ), then the order of ( left(mathbf{A}^{T} mathbf{B}right)^{T} mathbf{C}^{T} ) is A. ( 4 times 5 ) в. ( 3 times 7 ) c. ( 4 x 3 ) D. ( 5 times 7 ) | 12 |
| 37 | If ( boldsymbol{A}=left[begin{array}{ll}2 & 5 \ 4 & 4end{array}right], quad B=left[begin{array}{ll}0 & 5 \ 1 & 6end{array}right], ) find ( 5 boldsymbol{A}^{prime}+3 B^{prime} ) | 12 |
| 38 | Consider ( A ) and ( B ) two square matrices of same order. Select the correct alternative. ( mathbf{A} cdot|A B| ) must be greater than ( |A| ) B. ( left[begin{array}{ll}1 & 1 \ 1 & 1end{array}right] ) is not unit matrix C ( cdot|A+B| ) must be greater than ( |A| ) D. If ( A B=0 ), either ( A ) or ( B ) must be zero matrix | 12 |
| 39 | Find the adjoint of the matrix ( A= ) ( left[begin{array}{lll}1 & 4 & 3 \ 4 & 2 & 1 \ 3 & 2 & 2end{array}right] ) | 12 |
| 40 | If ( A=left|begin{array}{ccc}3 & -1 & 0 \ -1 & 2 & 3end{array}right| ) and ( B=left|begin{array}{ccc}-1 & 1 & 2 \ 0 & 2 & -1end{array}right| ) then find ( left(A B^{T}right)^{T} ) A ( cdotleft|begin{array}{cc}2 & -5 \ 1 & 6end{array}right| ) в. ( left|begin{array}{cc}3 & -4 \ 7 & -6end{array}right| ) с. ( left|begin{array}{cc}-4 & 9 \ -2 & 1end{array}right| ) D ( cdotleft|begin{array}{cc}-3 & 8 \ 4 & 2end{array}right| ) | 12 |
| 41 | If matrices ( A ) and ( B ) anticommute then ( mathbf{A} cdot A B=B A ) B. ( A B=-B A ) c ( cdot(A B)=(B A)^{-1} ) D. None of these | 12 |
| 42 | Using elementary tansormations, find the inverse of each of the matrices, if it exists in ( left[begin{array}{ll}2 & 1 \ 1 & 1end{array}right] ) | 12 |
| 43 | Assertion If ( boldsymbol{F}(boldsymbol{alpha})=left[begin{array}{ccc}cos boldsymbol{alpha} & -sin boldsymbol{alpha} & boldsymbol{0} \ boldsymbol{s i n} boldsymbol{alpha} & boldsymbol{operatorname { c o s } boldsymbol { alpha }} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{1}end{array}right], ) then ( [boldsymbol{F}(boldsymbol{alpha})]^{-1}=boldsymbol{F}(-boldsymbol{alpha}) ) Reason For matrix ( G(beta)=left[begin{array}{ccc}cos beta & 0 & sin beta \ 0 & 1 & 0 \ -sin beta & 0 & cos betaend{array}right] ) we have ( [G(beta)]^{-1}=G(-beta) ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 44 | If ( A ) is a ( 3 times 3 ) invertible matrix, then what will be the value of ( k ) if ( operatorname{det}left(A^{-1}right)=(operatorname{det} A)^{k} ) | 12 |
| 45 | Find ( x, y ) satisfying the matrix equations ( left[begin{array}{ccc}boldsymbol{x}-boldsymbol{y} & boldsymbol{2} & -boldsymbol{2} \ boldsymbol{4} & boldsymbol{x} & boldsymbol{6}end{array}right]+left[begin{array}{ccc}boldsymbol{3} & -boldsymbol{2} & boldsymbol{2} \ boldsymbol{1} & boldsymbol{0} & boldsymbol{-} boldsymbol{1}end{array}right]= ) ( left[begin{array}{ccc}mathbf{6} & mathbf{0} & mathbf{0} \ mathbf{5} & mathbf{2} boldsymbol{x}+boldsymbol{y} & mathbf{5}end{array}right] ) | 12 |
| 46 | The number of ( A ) in ( T_{p} ) such that the trace of ( A ) is not divisible by ( p ) but ( operatorname{det}(A) ) divisible by p is ?[Note: The trace of matrix is the sum of its diagonal entries]. A ( cdot(p-1)left(p^{2}-p+1right) ) B . ( p^{3}-(p-1)^{2} ) c. ( (p-1)^{2} ) D. ( (p-1)left(p^{2}-2right) ) | 12 |
| 47 | Assertion The matrix ( boldsymbol{A}=left(begin{array}{ccc}mathbf{0} & boldsymbol{a} & boldsymbol{b} \ -boldsymbol{a} & mathbf{0} & boldsymbol{c} \ -boldsymbol{b} & -boldsymbol{c} & mathbf{0}end{array}right) ) is a skew symmetric matrix. Reason A square matrix ( boldsymbol{A}=left(boldsymbol{a}_{boldsymbol{i} j}right) ) of order ( mathbf{m} ) is said to be skew symmetric if ( boldsymbol{A}^{boldsymbol{T}}=-boldsymbol{A} ) A. Both (A) & (R) are individually true & (R) is correct explanation of (A), B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A). C. (A)is true but (R) is false, D. (A)is false but (R) is true. | 12 |
| 48 | If ( boldsymbol{A} ) is a square matrix such that ( boldsymbol{A}^{2}= ) ( I, ) then find the simplified value of ( (A- ) ( boldsymbol{I})^{3}+(boldsymbol{A}+boldsymbol{I})^{3}-mathbf{7} boldsymbol{A} ) | 12 |
| 49 | Find the values of ( x ) and ( y, ) if ( left[begin{array}{cc}boldsymbol{x}+mathbf{1 0} & boldsymbol{y}^{2}+mathbf{2} boldsymbol{y} \ mathbf{0} & -mathbf{4} \ mathbf{3} boldsymbol{x}+mathbf{4} & mathbf{3} \ mathbf{0} & boldsymbol{y}^{mathbf{2}}-mathbf{5} boldsymbol{y}end{array}right]= ) | 12 |
| 50 | If ( boldsymbol{A}=left|begin{array}{cc}mathbf{5} & boldsymbol{x}-mathbf{2} \ mathbf{2} boldsymbol{x}+mathbf{3} & boldsymbol{x}+mathbf{1}end{array}right| ) is symmetric ( operatorname{then} x= ) A .4 B. 5 c. -5 D. – | 12 |
| 51 | ( operatorname{Let} boldsymbol{A}+mathbf{2} boldsymbol{B}=left[begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{0} \ mathbf{6} & -mathbf{3} & mathbf{3} \ -mathbf{5} & mathbf{3} & mathbf{1}end{array}right] ) and ( mathbf{2} boldsymbol{A}-boldsymbol{B}=left[begin{array}{ccc}mathbf{2} & mathbf{- 1} & mathbf{5} \ mathbf{2} & -mathbf{1} & mathbf{6} \ mathbf{0} & mathbf{1} & mathbf{2}end{array}right], ) then ( boldsymbol{t} boldsymbol{r}(boldsymbol{A}) ) ( t r(B) ) has the value equal to ( A ) B. ( c cdot 2 ) D. none of these | 12 |
| 52 | If ( boldsymbol{A}=left(begin{array}{c}-121 \ 1 & 23end{array}right), B=(1) ) and ( C=(21) ) verify ( (A B) C=A(B C) ) | 12 |
| 53 | Matrix A shows the weight of four boys and four girls in kg at the beginning of a diet programme to lose weight. Matrix B shows the corresponding weights after the diet programme. ( boldsymbol{A}=left[begin{array}{llll}mathbf{3 5} & mathbf{4 0} & mathbf{2 8} & mathbf{4 5} \ mathbf{4 2} & mathbf{3 8} & mathbf{4 1} & mathbf{3 0}end{array}right] underset{mathbf{G i r l s}}{mathbf{B o y s}}, mathbf{B}= ) ( left[begin{array}{cccc}mathbf{3 2} & mathbf{3 5} & mathbf{2 7} & mathbf{4 1} \ mathbf{4 0} & mathbf{3 0} & mathbf{3 4} & mathbf{2 7}end{array}right] begin{array}{c}boldsymbol{B} mathbf{o y s} \ text {Girls}end{array} ) Find the weight loss of the Boys and Girls. | 12 |
| 54 | Out of the following matrices, choose that matrix which is a scalar matrix. ( A cdotleft[begin{array}{ll}0 & 0 \ 0 & 0end{array}right] ) В. ( left[begin{array}{lll}0 & 0 & 0 \ 0 & 0 & 0end{array}right] ) c. ( left[begin{array}{ll}0 & 0 \ 0 & 0 \ 0 & 0end{array}right] ) D. ( left[begin{array}{l}0 \ 0 \ 0end{array}right] ) | 12 |
| 55 | If ( A=operatorname{diag}left[d_{1}, d_{2}, d_{3}right] ) then ( a^{n} ) is equal to A ( cdot operatorname{diag}left[d_{1}^{n-1}, d_{2}^{n-1}, d_{3}^{n-1}right] ) B. A c. ( operatorname{diag}left[d_{1}^{n}, d_{2}^{n}, d_{3}^{n}right] ) D. none | 12 |
| 56 | Write the element ( a_{21} ) of the matrix ( boldsymbol{A}=left[boldsymbol{a}_{i j}right]_{2 times 2} ) whose elements ( boldsymbol{a}_{i j} ) are ( operatorname{given} ) by ( a_{a j}=e^{2 i x} cos j x ) | 12 |
| 57 | [ operatorname{ftg}left[begin{array}{ccc} 3 & 2 & -1 \ 2 & -2 & 0 \ 1 & 3 & 1 end{array}right], Bleft[begin{array}{ccc} -3 & -1 & 0 \ 2 & 1 & 3 \ 4 & -1 & 2 end{array}right] ] and ( X=A+B ) then find ( X ) | 12 |
| 58 | If ( mathbf{A} ) is a non-singular square matrix of order ( 3 times 3, ) find ( |a d j A| ) | 12 |
| 59 | fthe matrix ( boldsymbol{A}=left[begin{array}{lll}2 & 0 & 0 \ 0 & 2 & 0 \ 2 & 0 & 2end{array}right], ) then ( boldsymbol{A}^{n}=left[begin{array}{lll}boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{0} \ boldsymbol{b} & boldsymbol{0} & boldsymbol{a}end{array}right] . boldsymbol{n} in boldsymbol{N} ) where A ( cdot a=2 n, b=2^{n} ) B . ( a=2^{n}, b=2 n ) c. ( a=2^{n}, b=n 2^{n-1} ) D. ( a=2^{n}, b=n 2^{n} ) | 12 |
| 60 | Three roots of ( n ) are A. 0,1,2 в. -1,1,3 ( mathrm{c} .-2,2,3 ) D. -3,1,5 | 12 |
| 61 | Find the value of ( x ) in ( left[begin{array}{cc}2 x-y & 5 \ 3 & yend{array}right]= ) ( left[begin{array}{cc}mathbf{6} & mathbf{5} \ mathbf{3} & -mathbf{2}end{array}right] ) | 12 |
| 62 | ( mathbf{f} boldsymbol{A}=left|begin{array}{cc}mathbf{2} & mathbf{0} \ mathbf{5} & -mathbf{3}end{array}right| boldsymbol{B}=left|begin{array}{cc}mathbf{- 2} & mathbf{1} \ mathbf{3} & mathbf{-} mathbf{1}end{array}right|, ) then the find the trace of ( left(A B^{T}right)^{T} ) A . 10 B. 12 ( c cdot 14 ) D. 16 | 12 |
| 63 | ( left[begin{array}{ll}boldsymbol{x}+boldsymbol{y} & boldsymbol{2} \ boldsymbol{5}+boldsymbol{z} & boldsymbol{x} boldsymbol{y}end{array}right]=left[begin{array}{ll}boldsymbol{6} & boldsymbol{2} \ boldsymbol{5} & boldsymbol{8}end{array}right] ) Find ( boldsymbol{x} boldsymbol{y} boldsymbol{z} ? ) | 12 |
| 64 | If ( boldsymbol{A}=left(begin{array}{c}cos boldsymbol{alpha} sin boldsymbol{alpha} \ -sin alpha cos boldsymbol{alpha}end{array}right), ) find ( boldsymbol{alpha} ) satisfying ( 0<alpha<frac{pi}{2} ) when ( A+A^{T}= ) ( sqrt{2} I_{2} ; ) where ( A^{T} ) is transpose of ( A ) | 12 |
| 65 | -b7 34. If A= and A adj A = A A”, then 5a +b is equal [JEE M 2016] to: (a) 4 (c) 1 (b) 13 (d) 5 | 12 |
| 66 | ( left.left[begin{array}{cc}mathbf{2 x}+boldsymbol{y} & boldsymbol{y} \ mathbf{1 – x} & mathbf{4 x}end{array}right]=mid begin{array}{cc}mathbf{1} & mathbf{- 1} \ mathbf{0} & mathbf{4}end{array}right], ) find the values of ( boldsymbol{x}+boldsymbol{y} ) | 12 |
| 67 | ( A_{n times n} ) and ( B_{n times n} ) are diagonal matrices then ( A B=_{-dots} dots dots dots . . . ) matrix This question has multiple correct options A. square B. diagonal c. scalar D. rectangular | 12 |
| 68 | ( left[begin{array}{lll}boldsymbol{x} & boldsymbol{4} & boldsymbol{1}end{array}right]left[begin{array}{ccc}boldsymbol{2} & boldsymbol{1} & boldsymbol{2} \ boldsymbol{1} & boldsymbol{0} & boldsymbol{2} \ boldsymbol{0} & boldsymbol{2} & boldsymbol{-} boldsymbol{4}end{array}right]left[begin{array}{c}boldsymbol{x} \ boldsymbol{4} \ -boldsymbol{1}end{array}right]=mathbf{0} ) then find ( x ) | 12 |
| 69 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}-mathbf{1} & mathbf{2} & mathbf{0} \ -mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{0} & mathbf{1} & mathbf{0}end{array}right], ) Show that ( boldsymbol{A}^{2}=boldsymbol{A}^{-1} ) | 12 |
| 70 | A skew-symmetric matrix ( M ) satisfies the relation ( M^{2}+I=0, ) where ( I ) is the unit matrix. Then, ( M M^{prime} ) is equal to A . ( I ) B . ( 2 I ) ( c .-I ) D. None of these | 12 |
| 71 | If ( boldsymbol{A}=left[begin{array}{lll}mathbf{6} & mathbf{1 0} & mathbf{1 0 0} \ mathbf{7} & mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{9} & mathbf{1 0}end{array}right], ) then ( operatorname{Tr}left(boldsymbol{A}^{T}right)= ) (Tr denotes trace of a matrix) A . -17 B. 17 ( c cdot-frac{1}{15} ) D. ( frac{1}{17} ) | 12 |
| 72 | Identify the order of matrix ( left[begin{array}{ccc}1 & 0 & -4 \ 2 & -1 & 3end{array}right] ) | 12 |
| 73 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{5} \ mathbf{6} & mathbf{7}end{array}right], ) then find ( boldsymbol{A}+boldsymbol{A}^{prime} ) | 12 |
| 74 | State whether the following statement is true or false. Enter 1 for true and 0 for false If ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right] ) then the value of ( mathrm{f} ) and ( mathrm{g} ) satisfying the matrix equation ( A^{2}+ ) ( boldsymbol{f} boldsymbol{A}+boldsymbol{g} boldsymbol{I}=boldsymbol{O} ) are equal to ( -boldsymbol{t}_{boldsymbol{r}}(boldsymbol{A}) ) and determinant of A respectively. Given a b, ( c, d ) are non zero reals and ( I= ) ( left[begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right] ; boldsymbol{O}=left[begin{array}{ll}mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{0}end{array}right] ) | 12 |
| 75 | If ( mathbf{A}-mathbf{2 B}=left[begin{array}{cc}mathbf{1} & -mathbf{2} \ mathbf{3} & mathbf{0}end{array}right] ) and ( mathbf{2} mathbf{A}-mathbf{3} mathbf{B}= ) ( left[begin{array}{cc}-mathbf{3} & mathbf{3} \ mathbf{1} & mathbf{- 1}end{array}right], ) then ( mathbf{B}= ) A ( cdotleft[begin{array}{cc}-5 & 7 \ 5 & 1end{array}right] ) B. ( left[begin{array}{cc}-5 & 7 \ -5 & -1end{array}right] ) c. ( left[begin{array}{cc}-5 & 7 \ 5 & -1end{array}right] ) D. ( left[begin{array}{cc}-5 & -7 \ -5 & -1end{array}right] ) | 12 |
| 76 | Find the inverse of the following matrix, using elementary transformations: ( boldsymbol{A}=left[begin{array}{lll}mathbf{2} & mathbf{3} & mathbf{1} \ mathbf{2} & mathbf{4} & mathbf{1} \ mathbf{3} & mathbf{7} & mathbf{2}end{array}right] ) | 12 |
| 77 | Number of real values of ( left|begin{array}{ccc}mathbf{3}-boldsymbol{x} & mathbf{2} & mathbf{2} \ mathbf{2} & mathbf{4}-boldsymbol{x} & mathbf{1} \ mathbf{- 2} & mathbf{- 4} & mathbf{- 1}-boldsymbol{x}end{array}right| ) is singular then ( A cdot 1 ) B. 3 ( c cdot 2 ) D. infinite | 12 |
| 78 | ff ( C=left[begin{array}{cc}3 & -6 \ 0 & 9end{array}right] ) find ¡) 2C ii) ( frac{1}{3} C ) iii) – | 12 |
| 79 | Find the inverse of matrices by elementary row transformation. | 12 |
| 80 | Let ( A ) be an invertible matrix then which of the following is/are true This question has multiple correct options A ( cdotleft|A^{-1}right|=|A|^{-1} ) – ( left.^{-1}|A|^{-1}|=| Aright|^{-1} ) B. ( left(A^{2}right)^{-1}=left(A^{-1}right)^{2} ) C ( cdotleft(A^{T}right)^{-1}=left(A^{-1}right)^{T} ) D. none of these | 12 |
| 81 | If ( boldsymbol{A}=left(begin{array}{cc}mathbf{7} & mathbf{2} \ mathbf{1} & mathbf{3}end{array}right) ) and ( boldsymbol{A}+boldsymbol{B}= ) ( left(begin{array}{cc}mathbf{- 1} & mathbf{0} \ mathbf{2} & mathbf{- 4}end{array}right) ) then matrix ( boldsymbol{B}=mathbf{?} ) ( mathbf{A} cdotleft(begin{array}{cc}1 & 0 \ 1 & 1end{array}right) ) B. ( left(begin{array}{cc}6 & 2 \ 3 & -1end{array}right) ) c. ( left(begin{array}{cc}-8 & -2 \ 1 & -7end{array}right) ) D. ( left(begin{array}{cc}8 & 2 \ -1 & 7end{array}right) ) | 12 |
| 82 | Find the inverse of the matrix ( left[begin{array}{ccc}1 & 2 & 1 \ -1 & 0 & 2 \ 2 & 1 & -3end{array}right] ) by elementary row transformation. Hence solve the system of equations ( boldsymbol{x}+mathbf{2} boldsymbol{y}+boldsymbol{z}=mathbf{4},-boldsymbol{x}+ ) ( mathbf{2} z=mathbf{0}, mathbf{2} boldsymbol{x}+boldsymbol{y}-mathbf{3} boldsymbol{z}=mathbf{0} ) | 12 |
| 83 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{c}-mathbf{1} \ mathbf{2} \ mathbf{3}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{lll}-mathbf{2} & -mathbf{1} & -mathbf{4}end{array}right] ) If trace of matrix ( A B ) is -12 , then the value of ( k ) ( A cdot 7 ) B. ( c cdot 2 ) D. none of these | 12 |
| 84 | ( operatorname{Let} boldsymbol{a}_{boldsymbol{k}}=boldsymbol{k}left(^{10} boldsymbol{C}_{boldsymbol{k}}right), boldsymbol{b}_{boldsymbol{k}}=(mathbf{1 0}-boldsymbol{k})left(^{10} boldsymbol{C}_{boldsymbol{k}}right) ) and ( boldsymbol{A}_{boldsymbol{k}}=left[begin{array}{ll}boldsymbol{a}_{boldsymbol{k}} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{b}_{boldsymbol{k}}end{array}right] ) If ( boldsymbol{A}=sum_{boldsymbol{k}=1}^{9} boldsymbol{A}_{boldsymbol{k}}=left[begin{array}{ll}boldsymbol{a} & mathbf{0} \ mathbf{0} & boldsymbol{b}end{array}right], ) find the value of ( a+b ) | 12 |
| 85 | ( boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5} & mathbf{6} \ mathbf{7} & mathbf{1} & mathbf{0}end{array}right], boldsymbol{B}=left[begin{array}{lll}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{3} & mathbf{0} \ mathbf{0} & mathbf{4} & mathbf{5}end{array}right] ) ( operatorname{Tr}(A B)=lambda operatorname{Tr}(mathrm{A}) . operatorname{Tr}(mathrm{B}), ) then ( lambda= ) ( A ) B. 0 ( c cdot frac{6}{5} ) ( frac{20}{27} ) | 12 |
| 86 | ( A ) and ( B ) are two square matrices of same order. If ( A B=B^{-1}, ) then ( A^{-1}= ) A. ( B A ) B . ( A^{2} ) ( c cdot B^{2} ) D. ( B ) | 12 |
| 87 | If ( P ) is a two-rowed matrix satisfying ( P^{T}=P^{-1}, ) then ( P ) can be ( mathbf{A} cdotleft[begin{array}{cc}cos theta & -sin theta \ -sin theta & cos thetaend{array}right] ) B. ( left[begin{array}{cc}cos theta & sin theta \ -sin theta & cos thetaend{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-cos theta & sin theta \ sin theta & -cos thetaend{array}right] ) D. none of these | 12 |
| 88 | ff ( A=left[begin{array}{ccc}1 & 3 & 1 \ 2 & 1 & -1 \ 3 & 0 & 1end{array}right], ) then ( operatorname{rank}(A) ) is equal to ( mathbf{A} cdot mathbf{4} ) B. ( c cdot 2 ) D. 3 | 12 |
| 89 | Elements of a matrix ( A ) of order ( 10 times 10 ) are defined as ( a_{i j}=w^{i+j} ) (where ( w ) is cube root of unity), then trace ( (A) ) of the matrix is ( A cdot omega ) B. ( c cdot omega^{2} ) D. | 12 |
| 90 | ( mathbf{A}=left[begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{- 3} \ mathbf{5} & mathbf{0} & mathbf{2} \ mathbf{1} & mathbf{- 1} & mathbf{1}end{array}right] ) and ( mathbf{B}= ) ( left[begin{array}{ccc}3 & -1 & 2 \ 4 & 2 & 5 \ 2 & 0 & 3end{array}right] ) Find the matrix ( C ) satisfying the relation ( A+2 C=B ) | 12 |
| 91 | If ( left(begin{array}{c}3 x+7 \ y+1 & 2-3 xend{array}right)=left(begin{array}{cc}1 & y-2 \ 8 & 8end{array}right) ) then the values of ( x ) and ( y ) respectively are A. -2,7 B. ( -frac{1}{3}, 7 ) ( c cdot-frac{1}{3},-frac{2}{3} ) D. 2,-7 | 12 |
| 92 | Using elementary row transformation, find the inverse of ( left[begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3end{array}right] ) | 12 |
| 93 | If ( 2left[begin{array}{ll}3 & 4 \ 5 & xend{array}right]+left[begin{array}{ll}1 & y \ 0 & 1end{array}right]=left[begin{array}{cc}7 & 0 \ 10 & 5end{array}right], ) find ( (x-y) ) | 12 |
| 94 | ff ( left[begin{array}{ll}boldsymbol{x} & mathbf{0} \ mathbf{1} & boldsymbol{y}end{array}right]+left[begin{array}{cc}-mathbf{2} & mathbf{1} \ mathbf{3} & mathbf{4}end{array}right]=left[begin{array}{ll}mathbf{3} & mathbf{5} \ mathbf{6} & mathbf{3}end{array}right]- ) ( left[begin{array}{ll}2 & 4 \ 2 & 1end{array}right], ) then A. ( x=-3, y=-2 ) в. ( x=3, y=-2 ) c. ( x=3, y=2 ) D. ( x=-3, y=2 ) | 12 |
| 95 | Let ( boldsymbol{A}=left[begin{array}{c}-mathbf{1} \ mathbf{2} \ mathbf{3}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{lll}-mathbf{2} & -mathbf{1} & -mathbf{4}end{array}right] ) The skew symmetric part of ( A B ) is? ( mathbf{A} cdotleft(begin{array}{ccc}0 & 5 / 2 & 5 \ -5 / 2 & 0 & -5 / 2 \ -5 & 5 / 2 & 0end{array}right) ) В. ( left(begin{array}{ccc}0 & 5 & 10 \ -5 & 0 & 11 \ -10 & -11 & 0end{array}right) ) ( mathbf{c} cdotleft(begin{array}{ccc}0 & 3 & 5 \ -3 & 0 & 11 \ -5 & -11 & 0end{array}right) ) D. None of these | 12 |
| 96 | If ( 2 A+3 B=left[begin{array}{ccc}2 & -1 & 4 \ 3 & 2 & 5end{array}right] ) and ( A+2 B= ) ( left[begin{array}{ccc}5 & 0 & 3 \ 1 & 6 & 2end{array}right] ) Then ( B ) is A ( cdotleft[begin{array}{ccc}8 & -1 & 2 \ -1 & 10 & -1end{array}right] ) B. ( left[begin{array}{ccc}8 & 1 & 2 \ -1 & 10 & -1end{array}right] ) c. ( left[begin{array}{ccc}8 & 1 & -2 \ -1 & 10 & -1end{array}right] ) D. ( left[begin{array}{lll}8 & 1 & 2 \ 1 & 10 & 1end{array}right] ) | 12 |
| 97 | B is said to be a skew symmetric matrix, if A. ( B^{T}=B ) B. B = 0 ( c cdot B=-B ) D. ( B^{T}=-B ) | 12 |
| 98 | If ( boldsymbol{A}=left(begin{array}{l}23 \ 4 \ 5end{array}right), ) then find the transpose of ( boldsymbol{A} ) | 12 |
| 99 | If ( [boldsymbol{x} quad boldsymbol{y}]=left[begin{array}{ll}1 & 5end{array}right], ) then ( 2 boldsymbol{x}+mathbf{5} boldsymbol{y}= ) A . 26 B. 27 c. 29 D. None of these | 12 |
| 100 | Evaluate ( left[begin{array}{lll}mathbf{3} & mathbf{4} & mathbf{1}end{array}right]left[begin{array}{c}mathbf{3} \ -mathbf{1} \ mathbf{3}end{array}right] ) | 12 |
| 101 | If ( boldsymbol{A}=left[begin{array}{ccc}1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & bend{array}right] ) is a matrix satisfying the equation ( boldsymbol{A} boldsymbol{A}^{boldsymbol{T}}=mathbf{9} boldsymbol{I} ) where ( I ) is ( 3 times 3 ) identity matrix, then the ordered pair ( (a, b) ) is equal: A ( .(2,-11) ) в. (-2,1) c. (2,1) () D. (2,-1) | 12 |
| 102 | ( mathbf{A}=left[begin{array}{cc}mathbf{1 1} & mathbf{1} \ mathbf{0} & mathbf{1 1}end{array}right], mathbf{B}=left[begin{array}{cc}mathbf{0} & -mathbf{2} \ -mathbf{3} & mathbf{4}end{array}right], mathbf{I}=left[begin{array}{cc}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right] ) Find ( boldsymbol{A}+mathbf{3} boldsymbol{B}+boldsymbol{4} boldsymbol{I} ) A ( cdotleft[begin{array}{cc}0 & 2 \ -3 & 0end{array}right] ) B ( cdotleft[begin{array}{cc}15 & 2 \ -13 & -16end{array}right] ) c. ( left[begin{array}{cc}15 & -5 \ -9 & 27end{array}right] ) D. ( left[begin{array}{cc}11 & 2 \ -4 & -5end{array}right] ) | 12 |
| 103 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) and ( boldsymbol{a}_{i j}=boldsymbol{i}(boldsymbol{i}+boldsymbol{j}) ) then trace of ( boldsymbol{A}= ) A ( cdot frac{n(n+1)(2 n+1)}{6} ) B. ( frac{n(n+1)(2 n+1)}{3} ) c. ( frac{n(n+1)}{2} ) D. ( frac{n^{2}(n+1)^{2}}{4} ) | 12 |
| 104 | ( left[begin{array}{ll}boldsymbol{x}-boldsymbol{y} & boldsymbol{4} \ boldsymbol{z}+boldsymbol{6} & boldsymbol{x}+boldsymbol{y}end{array}right]=left[begin{array}{ll}boldsymbol{8} & boldsymbol{w} \ boldsymbol{0} & boldsymbol{6}end{array}right], ) write the value of ( (boldsymbol{x}+boldsymbol{y}+boldsymbol{z}) ) | 12 |
| 105 | Consider the following relation ( R ) on the set of real square matrices of order 3 ( boldsymbol{R}={(boldsymbol{A}, boldsymbol{B}), mid boldsymbol{A} boldsymbol{B}=boldsymbol{B} boldsymbol{A}} ) STATEMENT-1: Relation R is equivalence. STATEMENT-2: Relation R is symmetric. A. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT- B. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 C. STATEMENT -1 is True, STATEMENT-2 is False D. STATEMENT-1 is False, STATEMENT-2 is True | 12 |
| 106 | ( boldsymbol{A}=left[begin{array}{ll}mathbf{5} & mathbf{7} \ mathbf{9} & mathbf{4}end{array}right] boldsymbol{B}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{5}end{array}right] ) find ( boldsymbol{A}-boldsymbol{B} ) | 12 |
| 107 | If ( boldsymbol{A}^{prime}=left[begin{array}{cc}-2 & 3 \ 1 & 2end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}-1 & 0 \ 1 & 2end{array}right] ) then find ( (boldsymbol{A}+mathbf{2} boldsymbol{B})^{prime} ) | 12 |
| 108 | If ( boldsymbol{A}=left[begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2end{array}right], ) verify that ( boldsymbol{A}^{3}-mathbf{6} boldsymbol{A}^{2}+mathbf{9} boldsymbol{A}-mathbf{4} boldsymbol{I}=mathbf{0} . ) Hence find ( boldsymbol{A}^{-1} ) | 12 |
| 109 | ( operatorname{Let} boldsymbol{A}+mathbf{2} boldsymbol{B}=left[begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{0} \ mathbf{6} & -mathbf{3} & mathbf{3} \ mathbf{- 5} & mathbf{3} & mathbf{1}end{array}right] ) and ( mathbf{2} boldsymbol{A}-boldsymbol{B}=left[begin{array}{ccc}mathbf{2} & mathbf{- 1} & mathbf{5} \ mathbf{2} & -mathbf{1} & mathbf{6} \ mathbf{0} & mathbf{1} & mathbf{2}end{array}right] ) then Det ( operatorname{Tr}(A)-operatorname{Tr}(B)) ) has the value equal to ( A ) B. ( c ) ( D ) | 12 |
| 110 | If ( boldsymbol{A}=left[begin{array}{cccc}mathbf{5} & mathbf{6} & mathbf{- 2} & mathbf{3} \ mathbf{1} & mathbf{0} & mathbf{4} & mathbf{2}end{array}right] ) and ( boldsymbol{B}= ) ( left[begin{array}{cccc}mathbf{3} & mathbf{- 1} & mathbf{4} & mathbf{7} \ mathbf{2} & mathbf{8} & mathbf{2} & mathbf{3}end{array}right], ) then find ( mathbf{A}+mathbf{B} ) | 12 |
| 111 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) then number of elements in ( A ) are A .4 B. 3 ( c cdot 2 ) D. None of these | 12 |
| 112 | ( operatorname{Let} A=left[begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1end{array}right] ) and ( 10 B= ) ( left[begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & alpha \ 1 & -2 & 3end{array}right] . ) If ( B ) is the inverse of matrix ( A ), then ( alpha ) is ( A ) в. ( c cdot 2 ) ( D ) | 12 |
| 113 | If ( boldsymbol{A}=left(begin{array}{ccc}4 & 1 & 2 \ 1 & -2 & 3 \ 0 & 3 & 2end{array}right), boldsymbol{B}=left(begin{array}{ccc}2 & 0 & 4 \ 6 & 2 & 8 \ 2 & 4 & 6end{array}right) ) and ( boldsymbol{C}= ) ( left(begin{array}{ccc}1 & 2 & -3 \ 5 & 0 & 2 \ 1 & -1 & 1end{array}right), ) then verify that ( boldsymbol{A}+(boldsymbol{B}+boldsymbol{C})=(boldsymbol{A}+boldsymbol{B})+ ) ( boldsymbol{C} ) | 12 |
| 114 | If (i) ( A=left[begin{array}{cc}cos alpha & sin alpha \ -sin alpha & cos alphaend{array}right], ) then verify that ( boldsymbol{A}^{prime} boldsymbol{A}=boldsymbol{I} ) (ii) ( A=left[begin{array}{cc}sin alpha & cos alpha \ -cos alpha & sin alphaend{array}right], ) then verify that ( boldsymbol{A}^{prime} boldsymbol{A}=boldsymbol{I} ) | 12 |
| 115 | If ( mathbf{3} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{2} \ mathbf{2} & mathbf{1} & -mathbf{2} \ boldsymbol{x} & mathbf{2} & boldsymbol{y}end{array}right], ) and ( boldsymbol{A} boldsymbol{A}^{prime}=boldsymbol{I} ) then ( -boldsymbol{x}-boldsymbol{y}=ldots ) | 12 |
| 116 | Using elementary tansormations, find the inverse of each of the matrices, if it exists in ( left[begin{array}{ll}2 & 3 \ 5 & 7end{array}right] ) | 12 |
| 117 | For ( k=frac{1}{sqrt{50}}, ) find ( a, b, c ) such that ( boldsymbol{P P}^{T}=boldsymbol{I} ) where, ( boldsymbol{P}=left[begin{array}{ccc}frac{2}{3} & boldsymbol{3} boldsymbol{k} & boldsymbol{a} \ -frac{1}{3} & -boldsymbol{4} boldsymbol{k} & boldsymbol{b} \ frac{2}{3} & -boldsymbol{5} boldsymbol{k} & boldsymbol{c}end{array}right] ) A ( cdot a=pm frac{13}{15 sqrt{2}}, b=pm frac{16}{15 sqrt{2}}, c=frac{1}{3 sqrt{2}} ) в. ( a=pm frac{13}{15 sqrt{2}}, b=pm frac{16}{15 sqrt{2}}, c=pm frac{1}{15 sqrt{2}} ) c. ( a=pm frac{13}{15 sqrt{2}}, b=pm frac{16}{15 sqrt{2}}, c=pm frac{1}{3 sqrt{2}} ) D. None of these | 12 |
| 118 | Prove that ( (A B)(A B)^{-1}=1 ) | 12 |
| 119 | If matrix ( mathbf{A}=[boldsymbol{a} boldsymbol{i} boldsymbol{j}]_{3 times 2}, ) and ( boldsymbol{a} boldsymbol{i} boldsymbol{j}=(boldsymbol{3} boldsymbol{i}- ) ( 2 j)^{2}, ) then find the matrix ( A ) | 12 |
| 120 | Find the values of a and ( b, ) if ( A=B, ) where [ begin{array}{c} boldsymbol{A}=left[begin{array}{cc} boldsymbol{a}+boldsymbol{4} & boldsymbol{3} boldsymbol{b} \ boldsymbol{8} & -boldsymbol{6} end{array}right] text { and } boldsymbol{B}= \ {left[begin{array}{cc} boldsymbol{2} boldsymbol{a}+boldsymbol{2} & boldsymbol{b}^{2}+boldsymbol{2} \ boldsymbol{8} & boldsymbol{b}^{2}-boldsymbol{5} boldsymbol{b} end{array}right]} end{array} ] | 12 |
| 121 | If ( A ) and ( B ) are symmetric matrices, then ( (A B-B A) ) is skew-symmetric. A. True B. False | 12 |
| 122 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & mathbf{1} \ -mathbf{1} & mathbf{2}end{array}right], boldsymbol{B}=left[begin{array}{cc}mathbf{1} & -mathbf{2} \ mathbf{2} & mathbf{1}end{array}right], boldsymbol{C}= ) ( left[begin{array}{cc}1 & -3 \ 2 & 1end{array}right], ) then which of the following is true A ( . A+B=B+A ) and ( A+(B+C)=(A+B)+C ) B. ( A+B=B+A ) and ( A C=B C ) c. ( A+B=B+A ) and ( A B=B C ) D. ( A C=B C ) and ( A=B C ) | 12 |
| 123 | If ( mathbf{A}+left|begin{array}{cc}mathbf{4} & mathbf{2} \ mathbf{1} & mathbf{3}end{array}right|=left|begin{array}{cc}mathbf{6} & mathbf{9} \ mathbf{1} & mathbf{4}end{array}right| ) then ( mathbf{A}= ) ( mathbf{A} cdotleft|begin{array}{ll}2 & 7 \ 0 & 1end{array}right| ) ( mathbf{B} cdotleft|begin{array}{ll}0 & 1 \ 2 & 7end{array}right| ) ( mathbf{C} cdotleft|begin{array}{ll}1 & 0 \ 2 & 7end{array}right| ) ( mathbf{D} cdotleft|begin{array}{ll}2 & 1 \ 0 & 7end{array}right| ) | 12 |
| 124 | If ( A^{T}=left[begin{array}{cc}3 & 4 \ -1 & 2 \ 0 & 1end{array}right] ) and ( B= ) ( left[begin{array}{ccc}-1 & 2 & 1 \ 1 & 2 & 3end{array}right], ) then find ( A^{T}-B^{T} ) | 12 |
| 125 | If ( boldsymbol{A}=left[begin{array}{c}-2 \ 4 \ 6end{array}right] ) and ( B=[14-6] ) then find ( boldsymbol{A B} ) | 12 |
| 126 | Find the inverse of the matrix ( left[begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & 1end{array}right] ) | 12 |
| 127 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & mathbf{1} \ -mathbf{1} & mathbf{1}end{array}right] ) and ( quad boldsymbol{n} boldsymbol{epsilon} boldsymbol{N}, ) then ( boldsymbol{A}^{n} ) is equal to ( mathbf{A} cdot 2^{n} A ) B ( cdot 2^{n-1} A ) ( c cdot n A ) D. None of these | 12 |
| 128 | ( fleft|begin{array}{ll}boldsymbol{x} & boldsymbol{y} \ mathbf{1} & boldsymbol{6}end{array}right|=left|begin{array}{ll}mathbf{1} & mathbf{8} \ mathbf{1} & mathbf{6}end{array}right| ) then ( mathbf{x}+2 mathbf{y}= ) ( A cdot 9 ) B. 17 c. 10 ( D ) | 12 |
| 129 | 21. The number of 3 x 3 non-singular matrices, with four entries as 1 and all other entries as 0, is [2010] (a) je (b) 6 (c) at least 7 (d) less than 4 | 12 |
| 130 | If ( a, b, c ) and ( d ) are real numbers such that and if ( boldsymbol{A}=left|begin{array}{cc}boldsymbol{a}+boldsymbol{i b} & boldsymbol{c}+boldsymbol{i d} \ -boldsymbol{c}+boldsymbol{i d} & boldsymbol{a}-boldsymbol{i b}end{array}right| ) then ( A^{-1}= ) ( mathbf{A} cdotleft|begin{array}{ll}a+i b & -c-i d \ c-i d & a-i bend{array}right| ) B. ( left|begin{array}{ll}a-i b & c+i d \ -c+i d & a+i bend{array}right| ) c. ( left|begin{array}{cc}a-i b & -c-i d \ c-i d & a+i bend{array}right| ) D. ( left|begin{array}{ll}a+i b & c+i d \ c-i d & a-i bend{array}right| ) | 12 |
| 131 | The total number of matrices formed with the help of 6 different numbers are ( mathbf{A} cdot mathbf{6} ) в. ( 6 ! ) c. ( 2(6 !) ) D. ( 4(6 !) ) | 12 |
| 132 | Find ( x, ) if the matrix ( left|begin{array}{ccc}-1 & 2 & 3 \ 2 & 5 & 6 \ 3 & x & 7end{array}right| ) is a symmetric matrix. | 12 |
| 133 | 12. If A and B are square matrices of size n x n such that AC – B4 = (A – B)(A+B), then which of the following will be always true? [2006] (a) A=B (b) AB = BA (c) either of A or B is a zero matrix (d) either of A or B is identity matrix | 12 |
| 134 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose element ( a_{i j} ) is ( a_{i j}=frac{(i-j)^{2}}{2} ) | 12 |
| 135 | Find the order of the matrix ( left[begin{array}{ccc}1 & 1 & 3 \ 5 & 2 & 6 \ -2 & -1 & -3end{array}right] ) | 12 |
| 136 | What must be the matrix ( boldsymbol{X} ) if ( 2 boldsymbol{X}+ ) ( left[begin{array}{l}12 \ 34end{array}right]=left[begin{array}{l}38 \ 72end{array}right] ? ) A ( cdotleft[begin{array}{l}13 \ 2-1end{array}right] ) в. ( left[begin{array}{l}1-3 \ 2-1end{array}right] ) c. ( left[begin{array}{l}26 \ 4-2end{array}right] ) D. ( left[begin{array}{l}2-6 \ 4-2end{array}right] ) | 12 |
| 137 | Let ( A ) be an invertible matrix. Which of the following is not true? A ( cdot A^{-1}=|A|^{-1} ) B. ( left(A^{2}right)^{-1}=left(A^{-1}right)^{2} ) c. ( left(A^{T}right)^{-1}=left(A^{-1}right)^{T} ) D. None of these | 12 |
| 138 | [ begin{array}{l} boldsymbol{A}_{mathbf{1}}=left[boldsymbol{a}_{1}right] \ boldsymbol{A}_{2}=left[begin{array}{lll} boldsymbol{a}_{2} & boldsymbol{a}_{3} \ boldsymbol{a}_{boldsymbol{4}} & boldsymbol{a}_{5} end{array}right] \ boldsymbol{A}_{boldsymbol{3}}=left[begin{array}{lll} boldsymbol{a}_{boldsymbol{6}} & boldsymbol{a}_{boldsymbol{7}} & boldsymbol{a}_{boldsymbol{8}} \ boldsymbol{a}_{boldsymbol{9}} & boldsymbol{a}_{boldsymbol{1 0}} & boldsymbol{a}_{11} \ boldsymbol{a}_{boldsymbol{1 2}} & boldsymbol{a}_{boldsymbol{1 3}} & boldsymbol{a}_{14} end{array}right] ldots ldots boldsymbol{A}_{boldsymbol{n}}=[ldots .] end{array} ] where ( quad a_{r}=left[log _{2} rright]([.] ) denotes greatest integer. Then trance of ( boldsymbol{A}_{mathbf{1 0}} ) | 12 |
| 139 | Find ( boldsymbol{A}, ) if ( boldsymbol{A}+boldsymbol{B}=left[begin{array}{ll}mathbf{5} & mathbf{2} \ mathbf{0} & mathbf{9}end{array}right] ) and ( boldsymbol{A}- ) ( boldsymbol{B}=left[begin{array}{cc}-mathbf{3} & -mathbf{6} \ mathbf{4} & -mathbf{1}end{array}right] ) | 12 |
| 140 | If the order of the matrix is ( 1 times 2, ) then it is a A. Row matrix B. Column matrix c. Square matrix D. None of these | 12 |
| 141 | ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{a}end{array}right] ) and ( boldsymbol{A}^{2}=left[begin{array}{ll}boldsymbol{alpha} & boldsymbol{beta} \ boldsymbol{beta} & boldsymbol{alpha}end{array}right] ) then ( mathbf{A} cdot alpha=a^{2}+b^{2}, beta=2 a b ) B . ( alpha=a^{2}+b^{2}, beta=a^{2}-b^{2} ) C ( cdot alpha=2 a b, beta=a^{2}+b^{2} ) D . ( alpha=a^{2}+b^{2}, beta=a b ) | 12 |
| 142 | Using elementary row transformations, find the inverse of the matrix ( left[begin{array}{ccc}1 & 2 & 3 \ 2 & 5 & 7 \ -2 & -4 & 5end{array}right] ) | 12 |
| 143 | Suppose ( A ) and ( B ) are two ( 3 times 3 ) non singular matrices such that ( (A B)^{k}= ) ( boldsymbol{A}^{k} boldsymbol{B}^{k} ) for ( k=2008,2009,2010, ) then This question has multiple correct options ( mathbf{A} cdot A B^{-1} A^{-1}=B^{-1} ) B. ( A^{-1} B^{-2} A=B^{2} ) ( mathbf{c} . A B=B A ) D. ( A^{-2} B^{2} A^{2}=left(A^{-1} B Aright)^{2} ) | 12 |
| 144 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}end{array}right], ) then ( boldsymbol{A}^{100} ) is equal to ( mathbf{A} cdot 2^{100} A ) в. ( 2^{99} ) А ( c cdot 100 A ) D . ( 299 A ) | 12 |
| 145 | If ( A ) be any ( m times n ) matrix and both ( A B ) and BA are defined prove that B should be ( boldsymbol{m} times boldsymbol{n} ) matrix | 12 |
| 146 | Suppose ( A ) is any ( 3 times 3 ) non-singular matrix and ( (boldsymbol{A}-mathbf{3} boldsymbol{I})(boldsymbol{A}-mathbf{5} boldsymbol{I})=boldsymbol{O} ) where ( boldsymbol{I}=boldsymbol{I}_{3} ) and ( boldsymbol{O}=boldsymbol{O}_{3}, ) If ( boldsymbol{alpha} boldsymbol{A}+ ) ( beta A^{-1}=4 I, ) then ( alpha+beta ) is equal to A . 8 B. 7 c. 13 D. 12 | 12 |
| 147 | If ( A ) and ( B ) are square matrices such that ( B=-A^{-1} B A, ) then ( (A+B)^{2} ) is equal to ( mathbf{A} cdot mathbf{0} ) B. ( A^{2}+B^{2} ) c. ( A^{2}+2 A B+B^{2} ) D. ( A+B ) | 12 |
| 148 | ( mathbf{f} A=left[begin{array}{cc}cos boldsymbol{x} & -sin boldsymbol{x} \ sin boldsymbol{x} & cos boldsymbol{x}end{array}right], ) then find ( boldsymbol{A} boldsymbol{A}^{boldsymbol{T}} ) | 12 |
| 149 | f matrix ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & mathbf{- 2} \ mathbf{- 2} & mathbf{2}end{array}right] ) and ( boldsymbol{A}^{mathbf{2}}=boldsymbol{p} boldsymbol{A} ) then write the value of ( p ) | 12 |
| 150 | Let ( A ) be a ( 3 times 3 ) matrix such that [ boldsymbol{A} timesleft[begin{array}{lll} mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{0} & mathbf{2} & mathbf{3} \ mathbf{0} & mathbf{1} & mathbf{1} end{array}right]=left[begin{array}{lll} mathbf{0} & mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{1} & mathbf{0} end{array}right] ] Then ( boldsymbol{A}^{-1} ) is : A. [ left[begin{array}{lll} 3 & 2 & 1 \ 3 & 2 & 0 \ 1 & 1 & 0 end{array}right] ] B. [ left[begin{array}{lll} 0 & 1 & 3 \ 0 & 2 & 3 \ 1 & 1 & 1 end{array}right] ] c. [ left[begin{array}{lll} 3 & 1 & 2 \ 3 & 0 & 2 \ 1 & 0 & 1 end{array}right] ] D. [ left[begin{array}{lll} 1 & 2 & 3 \ 0 & 1 & 1 \ 0 & 2 & 3 end{array}right] ] | 12 |
| 151 | Inverse of the matrix ( left[begin{array}{cc}cos 2 theta & -sin 2 theta \ sin 2 theta & cos 2 thetaend{array}right] ) is. ( mathbf{A} cdotleft[begin{array}{cc}cos 2 theta & -sin 2 theta \ sin 2 theta & cos 2 thetaend{array}right] ) B. ( left[begin{array}{cc}cos 2 theta & sin 2 theta \ sin 2 theta & -cos 2 thetaend{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}cos 2 theta & sin 2 theta \ sin 2 theta & cos 2 thetaend{array}right] ) D. ( left[begin{array}{cc}cos 2 theta & sin 2 theta \ -sin 2 theta & cos 2 thetaend{array}right] ) | 12 |
| 152 | If ( A ) is a square matrix then ( A-A^{prime} ) is a A. diagonal matrix B. skew symmetric matrix c. symmetric matrix D. None of these | 12 |
| 153 | If ( A ) is a skew-symmetric matrix of order ( n ) and ( C ) is a column matrix of order ( n times 1 ) then ( C^{T} A C ) is A. a identity matrix of order ( n ). B. a unit matrix of order c. a zero matrix of order 1 D. none of these | 12 |
| 154 | If ( alpha ) and ( beta ) differ by an odd multiply on ( pi / 2 ) prove that the product of the two matrices given below is a null matrix | 12 |
| 155 | If ( boldsymbol{A}=left[begin{array}{l}mathbf{1} \ mathbf{2} \ mathbf{3}end{array}right], ) then find ( boldsymbol{A} boldsymbol{A}^{boldsymbol{T}} ) | 12 |
| 156 | If ( A ) is ( n ) squared matrix then ( A A^{prime} ) and ( A^{prime} A ) are symmetric. | 12 |
| 157 | ( A 2 times 2 ) matrix whose elements ( a_{i j} ) are given by ( a_{i j}=i-j ) is ( mathbf{A} cdotleft[begin{array}{ll}0 & 1 \ 1 & 0end{array}right] ) B. ( left[begin{array}{cc}0 & -1 \ 1 & 0end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-1 & 0 \ 0 & 1end{array}right] ) D. ( left[begin{array}{cc}0 & 1 \ -1 & 0end{array}right] ) | 12 |
| 158 | Choose the correct answer A. Every scalar matrix is an identity matrix B. Every identity matrix is a scalar matrix C. Every diagonal matrix is an identity matrix D. A square matrix whose each element is 1 is an identity matrix | 12 |
| 159 | If ( boldsymbol{A}=left[begin{array}{lll}boldsymbol{a}^{2} & boldsymbol{a b} & boldsymbol{a c} \ boldsymbol{a b} & boldsymbol{b}^{2} & boldsymbol{b c} \ boldsymbol{a c} & boldsymbol{b c} & boldsymbol{c}^{2}end{array}right] ) and ( boldsymbol{a}^{2}+boldsymbol{b}^{2}+ ) ( c^{2}=1 ) then ( A^{2}= ) A ( .2 A ) в. ( c .3 A ) D. ( frac{1}{2} ) | 12 |
| 160 | Which of the following property is not always true for metrics but in numbers? A. ( A+B=0 ) B. AB = BA ( mathrm{c} cdot mathrm{AB}=0 ) D. None of the | 12 |
| 161 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) is a skew-symmetric matrix then write a value of ( sum_{i} sum_{j} a_{i j} ) | 12 |
| 162 | Write a ( 2 times 2 ) matrix which is both symmetric and skew-symmetric. | 12 |
| 163 | 7. Let M be a 2 x 2 symmetric matrix with integer entries. Then Mis invertible if (JEE Adv. 2014) (a) The first column of M is the transpose of the second row of M (b) The second row of Mis the transpose of the first column of M (c) Mis a diagonal matrix with non-zero entries in the main diagonal The product of entries in the main diagonal of M is not the square of an integer (d) The | 12 |
| 164 | If ( A ) and ( B ) are two square matrices such that ( B=-A^{-1} B A, ) then ( (A+B)^{2} ) is equal to ( mathbf{A} cdot A^{2}+B^{2} ) в. ( O ) c. ( A^{2}+2 A B+B^{2} ) D. ( A+B ) | 12 |
| 165 | If ( A ) and ( B ) are invertible matrices, which one of the following statement is/are correct This question has multiple correct options A ( . A d j(A)=|A| A^{-1} ) B. ( operatorname{det}left(A^{-1}right)=|operatorname{det}(A)|^{-1} ) c. ( (A+B)^{-1}=B^{-1}+A^{-1} ) D. ( (A B)^{-1}=B^{-1} A^{-1} ) | 12 |
| 166 | If ( mathbf{A} ) and ( mathbf{B} ) are two square matrices such that ( mathbf{B}=-mathbf{A}^{-1} mathbf{B} mathbf{A} ) then ( (mathbf{A}+ ) ( mathbf{B})^{2}= ) A . 0 B. ( A^{2}+B^{2} ) c. ( A^{2}+2 A B+B^{2} ) D. ( A+B ) | 12 |
| 167 | ( operatorname{Let} A=left[begin{array}{c}3 x^{2} \ 1 \ 6 xend{array}right], B=[a b c], ) and ( C= ) ( left[begin{array}{ccc}(x+2)^{2} & 5 x^{2} & 2 x \ 5 x^{2} & 2 x & (x+2)^{2} \ 2 x & (x+2)^{2} & 5 x^{2}end{array}right] ) three given matrices, where ( a, b, c ) and ( x ) ( in mathrm{R} . ) Given that ( boldsymbol{t r}(boldsymbol{A B})=boldsymbol{t r}(boldsymbol{C}) boldsymbol{x} in boldsymbol{R} ) where ( operatorname{tr}(mathrm{A}) ) denotes trace of ( mathrm{A} ). If ( f(x)= ) ( a x^{2}+b x+c, ) then the value of ( f(1) ) is | 12 |
| 168 | ff ( left[begin{array}{cc}boldsymbol{x} & mathbf{2} \ mathbf{1 8} & boldsymbol{x}end{array}right]=left[begin{array}{cc}mathbf{6} & mathbf{2} \ mathbf{1 8} & mathbf{6}end{array}right] ) then ( mathbf{x}= ) ( A cdot pm 6 ) B. 6 ( c .-5 ) D. | 12 |
| 169 | If ( A^{prime} ) is the transpose of a square matrix A, then A ( cdot|A| neq mid A^{prime} ) B ( cdot|A|=left|A^{prime}right| ) c ( cdot|A|+left|A^{prime}right|=0 ) D cdot ( |A|=left|A^{prime}right| ) only when A is symmetric | 12 |
| 170 | Find the inverse of the following matrix by using elementary row transformation ( left[begin{array}{lll}0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1end{array}right] ) | 12 |
| 171 | If ( boldsymbol{A}=left[begin{array}{c}mathbf{2} \ -mathbf{4} \ mathbf{1}end{array}right], boldsymbol{B}=left[begin{array}{lll}mathbf{5} & mathbf{3} & -mathbf{1}end{array}right] ) then verify that ( (boldsymbol{A} boldsymbol{B})^{prime}=boldsymbol{B}^{prime} boldsymbol{A}^{prime} ) | 12 |
| 172 | If ( boldsymbol{A}=left[begin{array}{ccc}boldsymbol{4} & mathbf{1} & boldsymbol{0} \ mathbf{1} & -mathbf{2} & mathbf{2}end{array}right], boldsymbol{B}= ) ( left[begin{array}{ccc}mathbf{2} & mathbf{0} & -mathbf{1} \ mathbf{3} & mathbf{1} & mathbf{4}end{array}right], boldsymbol{C}=left[begin{array}{c}mathbf{1} \ mathbf{2} \ -mathbf{1}end{array}right] ) and ( (mathbf{3} boldsymbol{B}-mathbf{2} boldsymbol{A}) boldsymbol{C}+mathbf{2} boldsymbol{X}=mathbf{0} ) then ( boldsymbol{X}= ) A ( cdot frac{1}{2}left[begin{array}{c}3 \ 13end{array}right] ) в. ( frac{1}{2}left[begin{array}{c}3 \ -13end{array}right] ) c. ( frac{1}{2}left[begin{array}{c}-3 \ 13end{array}right] ) D. ( left[begin{array}{c}3 \ -13end{array}right] ) | 12 |
| 173 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{ccc}mathbf{2} & mathbf{0} & mathbf{7} \ mathbf{0} & mathbf{1} & mathbf{0} \ mathbf{1} & mathbf{- 2} & mathbf{1}end{array}right] ) and ( boldsymbol{B}= ) ( left[begin{array}{ccc}-boldsymbol{x} & mathbf{1 4 x} & mathbf{7 x} \ mathbf{0} & mathbf{1} & mathbf{0} \ boldsymbol{x} & mathbf{- 4 x} & mathbf{- 2 x}end{array}right] ) are two matrices such that ( A B=(A B)^{-1} ) and ( A B neq I ) (where ( I ) is an identity matrix of order ( 3 times 3) ) Find the value of ( boldsymbol{T} boldsymbol{r} cdotleft(boldsymbol{A} boldsymbol{B}+(boldsymbol{A} boldsymbol{B})^{2}+(boldsymbol{A} boldsymbol{B})^{3}+ldots+(boldsymbol{A} boldsymbol{B}right. ) where ( T r .(A) ) denotes the trace of matrix ( boldsymbol{A} ) A. 98 в. 99 ( c cdot 100 ) D. 10 | 12 |
| 174 | ( fleft[begin{array}{ll}mathbf{4} & mathbf{3} \ boldsymbol{x} & mathbf{5}end{array}right]=left[begin{array}{ll}boldsymbol{y} & boldsymbol{z} \ mathbf{1} & mathbf{5}end{array}right] ) then find the value of ( x, y ) and ( z ) | 12 |
| 175 | ( mathbf{A B A}^{-1}=mathbf{X} ) then ( mathbf{B}^{mathbf{2}}= ) A ( cdot x^{2} ) B. AxA ( ^{-1} ) c. ( A x^{2} A^{-1} ) D. ( A^{-1} x^{2} A ) | 12 |
| 176 | Assertion If the matrices ( A, B,(A+B) ) are nonsingular, then ( left[boldsymbol{A}(boldsymbol{A}+boldsymbol{B})^{-1} boldsymbol{B}right]^{-1}= ) ( boldsymbol{B}^{-1}+boldsymbol{A}^{-1} ) Reason ( left[boldsymbol{A}(boldsymbol{A}+boldsymbol{B})^{-1} boldsymbol{B}right]^{-1}=left[boldsymbol{A}left(boldsymbol{A}^{-1}+right.right. ) ( left.left.boldsymbol{B}^{-1}right) boldsymbol{B}right]^{-1}=left[left(boldsymbol{I}+boldsymbol{A} boldsymbol{B}^{-1}right) boldsymbol{B}right]^{-1}= ) ( left[left(boldsymbol{B}+boldsymbol{A B}^{-1} boldsymbol{B}right)right]^{-1}=[(boldsymbol{B}+ ) ( boldsymbol{A I})]^{-1}=[(boldsymbol{B}+mathbf{1})]^{-1}=boldsymbol{B}^{-1}+boldsymbol{A}^{-1} ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 177 | If ( A B=A ) and ( B A=B, ) then This question has multiple correct options ( mathbf{A} cdot A^{2} B=A^{2} ) B. ( B^{2} A=B^{2} ) c. ( A B A=A ) D. ( B A B=B ) | 12 |
| 178 | Find the rank of the matrix ( boldsymbol{A}= ) ( left[begin{array}{ccc}1 & 5 & 9 \ 4 & 8 & 12 \ 7 & 11 & 15end{array}right] ) | 12 |
| 179 | A square matrix A has 9 elements. What is the possible order of ( A ? ) ( mathbf{A} cdot 1 times 9 ) B. ( 9 times 9 ) c. ( 3 times 3 ) D. ( 2 times 7 ) | 12 |
| 180 | If ( A ) satisfies the equation ( x^{3}-5 x^{2}+ ) ( 4 x+lambda=0, ) then ( A^{-1} ) exists if A. ( lambda neq 1 ) B. ( lambda neq 2 ) c. ( lambda neq-1 ) D. ( lambda neq 0 ) | 12 |
| 181 | If ( mathrm{A}=left[a_{i j}right]_{2 times 2} ) such that ( a_{i j}=i-j+3 ) then find ( boldsymbol{A} ) ( mathbf{A} cdotleft[begin{array}{ll}2 & 3 \ 4 & 2end{array}right] ) в. ( left[begin{array}{ll}3 & 4 \ 2 & 3end{array}right] ) c. ( left[begin{array}{ll}4 & 2 \ 2 & 3end{array}right] ) D. ( left[begin{array}{ll}3 & 2 \ 4 & 3end{array}right] ) | 12 |
| 182 | Construct a ( 2 times 3 ) matrix ( A=left[a_{i j}right] ) whose element ( a_{i j} ) is ( a_{i j}=2 i-j ) | 12 |
| 183 | If ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{x} \ boldsymbol{y} & boldsymbol{a}end{array}right] ) and if ( boldsymbol{x} boldsymbol{y}=mathbf{1}, ) then ( operatorname{det}left(A A^{T}right) ) is equal to: ( mathbf{A} cdot a^{2}-1 ) B ( cdotleft(a^{2}+1right)^{2} ) c. ( 1-a^{2} ) D. ( left(a^{2}-1right)^{2} ) E ( cdot(a-1)^{2} ) | 12 |
| 184 | Multiply the given matrices: [ left[begin{array}{cc} 1 & -2 \ 2 & 3 end{array}right]left[begin{array}{ccc} 1 & 2 & 3 \ -3 & 2 & -1 end{array}right] ] | 12 |
| 185 | ( fleft(frac{2}{3} 1 frac{5}{3}right) quadleft[frac{2}{3} frac{2}{3} frac{4}{3}right] ) and ( B=left[frac{2}{5} frac{3}{5} frac{4}{5} frac{4}{5}right], ) then compute ( 3 A-5 B ) | 12 |
| 186 | If ( A ) and ( B ) are two non singular matrices of the same order such that ( B^{r}=I, ) for some positive integer ( r>1 ) ( operatorname{then} boldsymbol{A}^{-1} boldsymbol{B}^{r-1} boldsymbol{A}-boldsymbol{A}^{-1} boldsymbol{B}^{-1} boldsymbol{A}= ) A . ( I ) B. ( 2 I ) ( c cdot O ) D. ( -I ) | 12 |
| 187 | ( boldsymbol{A}=left(begin{array}{ccc}1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0end{array}right) Rightarrow A^{2}-2 A= ) ( A cdot A^{-1} ) B ( .-A^{-1} ) ( c ) D. -1 | 12 |
| 188 | Solve by matrix method ( boldsymbol{x}+mathbf{2} boldsymbol{y}+mathbf{3} boldsymbol{z}= ) ( mathbf{2}, mathbf{2} boldsymbol{x}+mathbf{3} boldsymbol{y}+boldsymbol{z}=-mathbf{1}, boldsymbol{x}-boldsymbol{y}-boldsymbol{z}=-mathbf{2} ) | 12 |
| 189 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{4} & mathbf{2} \ mathbf{- 1} & mathbf{1}end{array}right], ) then ( (boldsymbol{A}-mathbf{2} boldsymbol{I})(boldsymbol{A}-mathbf{3} boldsymbol{I}) ) equals- ( mathbf{A} cdot mathbf{0} ) в. c. ( I ) D. ( 5 I ) | 12 |
| 190 | To find the inverse of the matrix ( boldsymbol{A}= ) ( [121 ; 011 ; 311] ) by elementary transformation method | 12 |
| 191 | ff ( boldsymbol{P}=left[begin{array}{ccc}2 & 3 & 1 \ 0 & -1 & 5end{array}right] ) and ( Q= ) ( left[begin{array}{ccc}1 & 2 & -6 \ 0 & -1 & 3end{array}right], ) Evaluate ( 3 P-4 Q ) ( mathbf{3} P-4 Q=left[begin{array}{lll}a & b & c \ d & e & fend{array}right], ) find sum of ( a, b, c, d, e, f ) | 12 |
| 192 | A matrix consists of 30 elements. What are the possible orders it can have? | 12 |
| 193 | ff ( left[begin{array}{cc}boldsymbol{alpha} & boldsymbol{beta} \ boldsymbol{gamma} & -boldsymbol{alpha}end{array}right] ) is to be the square root of ( mathbf{a} ) two -rowed unit matrix, then ( alpha, beta ) and ( gamma ) should satisfy the relation. A. ( 1+alpha^{2}+beta gamma=0 ) B . ( 1-alpha^{2}-beta gamma=0 ) C ( cdot 1-alpha^{2}+beta gamma=0 ) D. ( 1+alpha^{2}-beta gamma=0 ) | 12 |
| 194 | Let ( A, B, C, D ) be (not necessarily square) real matrices such that ( boldsymbol{A}^{boldsymbol{T}}= ) ( boldsymbol{B} boldsymbol{C} boldsymbol{D} ; boldsymbol{B}^{T}=boldsymbol{C} boldsymbol{D} boldsymbol{A} ; boldsymbol{C}^{T}=boldsymbol{D} boldsymbol{A} boldsymbol{B} ) and ( D^{T}=A B C ) for the matrix ( S=A B C D ), consider the two statements. ( boldsymbol{S}^{3}=boldsymbol{S} ) ( | S^{2}=S^{4} ) A. II is true but not B. I is true but not I c. Both I and II are true D. Both I and II are false. | 12 |
| 195 | ( mathrm{ff} mathrm{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{0} \ mathbf{4} & mathbf{1}end{array}right] ) and ( mathrm{B}=left[begin{array}{lll}mathbf{0} & mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{2} & mathbf{1} \ mathbf{2} & mathbf{3} & mathbf{1}end{array}right] ) find BA. Can we find AB also? | 12 |
| 196 | If ( A ) and ( B ) are square matrices of same order and ( mathrm{B} ) is a skew-symmetric matrix, show that ( A^{prime} B A ) is a skew symmetric matrix. | 12 |
| 197 | If ( A ) and ( B ) are square matrices such that ( A B=I ) and ( B A=I, ) then ( B ) is A. Unit matrix B. Null matrix c. Multiplicative inverse matrix of ( A ) D. ( -A ) | 12 |
| 198 | f matrix ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{4} & mathbf{3}end{array}right] ) such that ( boldsymbol{A} boldsymbol{x}=boldsymbol{I} ) then ( boldsymbol{x}= ) begin{tabular}{r} ( mathbf{A} cdot frac{1}{5}left[begin{array}{rr}1 & 3 \ 2 & -1end{array}right] ) \ hline end{tabular} в. 1 [ begin{array}{ll}overline{overline{5}} & {left[begin{array}{ll}4 & -1end{array}right]}end{array} ] c. ( 1left[begin{array}{ll}-3 & 2 \ 5 & -1end{array}right] ) D. ( frac{1}{5}left[begin{array}{ll}-1 & 2 \ -1 & 4end{array}right] ) | 12 |
| 199 | toppr LoGin Joln Now Q Type your question two classes of alphabets as in the two matrices given below. The columns and rows of Matrix I are numbered from 0 to 4 and that of Matrix II are numbered from 5 to ( 9 . ) A letter from these matrices can be represented first by its row and next by its column, e.g., ( ^{prime} boldsymbol{R}^{prime} ) can be represented by 04,42 etc., and ( ^{prime} D^{prime} ) can be represented by 57,76 etc. Similarly, you have to identify the set for the word ( mathrm{ROAD}^{prime} ) Matrix 1 begin{tabular}{cc|c|c|c|c} & 0 & 1 & 2 & 3 & 4 \ hline 0 & ( F ) & ( O ) & ( M ) & ( S ) & ( R ) \ 1 & ( S ) & ( R ) & ( F ) & ( O ) & ( M ) \ 2 & ( O ) & ( M ) & ( S ) & ( R ) & ( F ) \ 3 & ( R ) & ( F ) & ( O ) & ( M ) & ( S ) \ 4 & ( M ) & ( S ) & ( R ) & ( F ) & ( O ) end{tabular} Matrix I begin{tabular}{cccc|cc} & 5 & 6 & 7 & 8 & 9 \ 5 & ( A ) & ( T ) & ( D ) & ( I ) & ( P ) \ 6 & ( I ) & ( P ) & ( A ) & ( T ) & ( D ) \ 7 & ( T ) & ( D ) & ( I ) & ( P ) & ( A ) \ 8 & ( P ) & ( A ) & ( T ) & ( D ) & ( I ) \ 9 & ( D ) & ( I ) & ( P ) & ( A ) & ( T ) end{tabular} ( A ) 42,32,49,58 B. 23,32,98,99 ( c ) 11,13,67,69 13,67,69 00 ( D ) | 12 |
| 200 | Assertion For a singular square matrix ( mathbf{A}, boldsymbol{A} boldsymbol{B}= ) ( boldsymbol{A C} Rightarrow boldsymbol{B}=boldsymbol{C} ) Reason If ( |A|=0, ) then ( A^{-1} ) does not exist. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 201 | If ( boldsymbol{A}=left[begin{array}{c}-2 \ 4 \ 5end{array}right] ) and ( B=left[begin{array}{lll}1 & 3 & -6end{array}right], ) then verify that ( (A B)^{T}=B^{T} A^{T} ) | 12 |
| 202 | IF ( A ) be a square matrix, then ( boldsymbol{A}+boldsymbol{A}^{boldsymbol{T}} ) is a symmetric matrix? A. True B. False | 12 |
| 203 | If ( A ) and ( B ) are symmetric matrices of the same order, write whether ( A B- ) ( B A ) is symmetric or skew-symmetric or neither of the two. | 12 |
| 204 | Let ( boldsymbol{P} ) be an ( boldsymbol{m} times boldsymbol{m} ) matrix such that ( boldsymbol{P}^{2}=boldsymbol{P} . ) Then ( (boldsymbol{I}+boldsymbol{P})^{n} ) equals. ( mathbf{A} cdot I+P ) B. ( I+n P ) c. ( I+2^{n} P ) D. ( I+left(2^{n}-1right) P ) | 12 |
| 205 | ( fleft[begin{array}{cc}cos ^{2} alpha & cos alpha sin alpha \ cos alpha sin alpha & sin ^{2} alphaend{array}right] ) and ( B= ) ( left[begin{array}{cc}cos ^{2} beta & cos beta sin beta \ cos beta sin beta & sin ^{2} betaend{array}right] ) are two matrices such that the product ( A B ) is null matirx, then ( alpha-beta ) is ( mathbf{A} cdot mathbf{0} ) B. Multiple of ( pi ) c. An odd number of ( frac{pi}{2} ) D. None of the above | 12 |
| 206 | If ( boldsymbol{A}^{prime}left[begin{array}{cc}-mathbf{2} & mathbf{3} \ mathbf{1} & mathbf{2}end{array}right], boldsymbol{B}=left[begin{array}{cc}-mathbf{1} & mathbf{0} \ mathbf{1} & mathbf{2}end{array}right] ) Find ( [boldsymbol{A}+mathbf{2} boldsymbol{B}] ) | 12 |
| 207 | ( boldsymbol{A}=left[begin{array}{lll}2 & 3 & 1 \ 4 & 1 & 5 \ 3 & 9 & 7end{array}right] . ) Then the additive inverse of ( boldsymbol{A} ) is: ( A ) [ left[begin{array}{ccc} -2 & -3 & 1 \ 4 & -1 & -5 \ -3 & 9 & -7 end{array}right] ] B. [ left[begin{array}{ccc} -2 & -3 & -1 \ -4 & -1 & -5 \ -3 & -9 & -7 end{array}right] ] ( c ) [ left[begin{array}{ccc} 2 & -3 & -1 \ -4 & 1 & -5 \ -3 & -9 & 7 end{array}right] ] D. [ left[begin{array}{ccc} -2 & -3 & -1 \ -4 & -1 & -5 \ -3 & 9 & -7 end{array}right] ] | 12 |
| 208 | If ( S=left[begin{array}{ll}6 & -8 \ 2 & 10end{array}right]=P+Q, ) where ( P ) is ( a ) symmetric & Q is a skew -symmetric matrix, then ( Q= ) A. ( left[begin{array}{cc}0 & 5 \ -5 & 0end{array}right] ) В. ( left[begin{array}{cc}0 & -5 \ 5 & 0end{array}right] ) C ( cdotleft[begin{array}{cc}0 & 8 \ -8 & 0end{array}right] ) D. ( left[begin{array}{cc}0 & 6 \ -6 & 0end{array}right] ) | 12 |
| 209 | For any square matrix ( boldsymbol{A}, boldsymbol{A}+boldsymbol{A}^{T} ) is A. unit matrix B. symmetric matrix c. skew symmetric matrix D. zero matrix | 12 |
| 210 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 2} & mathbf{0} \ mathbf{2} & mathbf{1} & mathbf{3} \ mathbf{0} & mathbf{- 2} & mathbf{1}end{array}right], ) then ( boldsymbol{A}^{-1}= ) | 12 |
| 211 | Express the matrix ( left[begin{array}{ccc}mathbf{3} & -mathbf{2} & mathbf{- 4} \ mathbf{3} & -mathbf{2} & -mathbf{5} \ -mathbf{1} & mathbf{1} & mathbf{2}end{array}right] ) as the sum of a symmetric and skew- symmetric matrix | 12 |
| 212 | If the matrix ( left[begin{array}{ccc}mathbf{0} & boldsymbol{a} & mathbf{3} \ mathbf{2} & boldsymbol{b} & mathbf{- 1} \ boldsymbol{c} & mathbf{1} & mathbf{0}end{array}right] ) is a skew symmetric matrix, find ( a, b ) and ( c ) | 12 |
| 213 | If ( boldsymbol{A}=operatorname{diag}(mathbf{2}-mathbf{5} mathbf{9}), boldsymbol{B}=operatorname{diag}(mathbf{1} mathbf{1}- ) 4) and ( C=operatorname{diag}(-634), ) then find ( A ) ( 2 B ) | 12 |
| 214 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{ll}mathbf{1} & frac{mathbf{3}}{2} \ mathbf{1} & mathbf{2}end{array}right], boldsymbol{B}= ) ( left[begin{array}{cc}mathbf{4} & -mathbf{3} \ -mathbf{2} & mathbf{2}end{array}right] ) and ( boldsymbol{C}_{boldsymbol{r}}= ) ( left[begin{array}{cc}r .3^{r} & 2^{r} \ 0 & (r-1) 3^{r}end{array}right] ) be 3 given matrices. Compute the value of ( sum_{r=1}^{50} t r cdotleft((A B)^{r} C_{r}right) .(text { where } t r .(A) ) denotes trace of matrix ( mathbf{A} ) ) ( mathbf{A} cdot 3left(49.3^{50}+1right) ) B ( cdot 3left(49.3^{49}+1right) ) C ( .3left(49.3^{48}+1right) ) D. None of these | 12 |
| 215 | Three school ( A, B ) and ( C ) organized a mela for collecting funds for helping the rehabilitation of flood victims. They sold hand made fans, mats and plates from recycled material at a cost of Rs ( 25, ) Rs 100 and ( R s 50 ) each. The number of articles sold are given below: school ( begin{array}{lll}text { School } & text { School } & text { Sc } \ text { A } & text { B } & text { c }end{array} ) Articles Hand40 25 fans Mats 50 40 Plates 20 30 40 Find the fund collected by each school separately by selling the above articles. Also, find the total funds collected for the purpose. | 12 |
| 216 | ( mathbf{f} mathbf{A}=left[begin{array}{ccc}mathbf{1} & mathbf{0} & -mathbf{2} \ mathbf{2} & mathbf{- 3} & mathbf{4}end{array}right], ) then the matrix ( X ) for which ( 2 X+3 A=0 ) holds true is ( mathbf{A} cdotleft[begin{array}{ccc}-frac{3}{2} & 0 & -3 \ -3 & -frac{9}{2} & -6end{array}right] ) ( mathbf{B} cdotleft[begin{array}{ccc}frac{3}{2} & 0 & -3 \ 3 & -frac{9}{2} & -6end{array}right] ) ( mathbf{C} cdotleft[begin{array}{lll}frac{3}{2} & 0 & 3 \ 3 & frac{9}{2} & 6end{array}right] ) ( ^{mathrm{D}} cdotleft[begin{array}{ccc}-frac{3}{2} & 0 & 3 \ -3 & frac{9}{2} & -6end{array}right] ) | 12 |
| 217 | Prove that the matrix ( B^{prime} A B ) is symmetric or skew symmetric according as ( A ) is symmetric or skew symmetric. | 12 |
| 218 | If ( boldsymbol{A}^{-1}=left[begin{array}{cc}1 & -2 \ -2 & 2end{array}right], ) then what is ( operatorname{det}(A) ) equal to ( ? ) ( A cdot 2 ) B. -2 c. ( 1 / 2 ) D. ( -1 / 2 ) | 12 |
| 219 | If matrix ( A ) is an circulant matrix whose elements of first row are ( mathbf{a}, mathbf{b}, mathbf{c} ) all ( >mathbf{0} ) such that abc ( =mathbf{1} ) and ( A^{T} A=I ) then ( a^{3}+b^{3}+c^{3} ) equals ( A cdot 0 ) B. 3 ( c .1 ) D. 4 | 12 |
| 220 | ( left[begin{array}{lll}0 & 0 & 0end{array}right] ) is A. Identity matrix B. diagonal matrix c. scalar matrix D. null matrix | 12 |
| 221 | Determine of ( U ) is A . 13 B. 15 ( c .3 ) D. 2 | 12 |
| 222 | ( |f| A mid=47, ) then find ( left|A^{T}right| ) | 12 |
| 223 | Express the following matrices as the sum of a symmetric and a skew symmetric matrix : ( left[begin{array}{ccc}6 & -2 & 2 \ -2 & 3 & -1 \ 2 & -1 & 3end{array}right] ) | 12 |
| 224 | The transpose of a rectangular matrix is ( mathbf{a} ) A. rectangular matrix B. diagonal matrix c. square matrix D. scalar matrix | 12 |
| 225 | Which of the following is correct? A. Determinant is a square matrix B. Determinant is a number associated to a matrix C. Determinant is a number associated to a square matrix D. None of these | 12 |
| 226 | ( boldsymbol{A}=left[boldsymbol{a}_{i j}right]_{m times n} ) is a square matrix , if ( mathbf{A} cdot mn ) c. ( m=n ) D. None of these | 12 |
| 227 | Let ( A ) and ( B ) are two matrices of same order ( 3 times 3 ) given by ( A= ) ( left[begin{array}{ccc}1 & 3 & lambda+2 \ 2 & 4 & 6 \ 3 & 5 & 8end{array}right] B=left[begin{array}{ccc}3 & 2 & 4 \ 3 & 2 & 5 \ 2 & 1 & 4end{array}right] ) If ( lambda=4 ) ( operatorname{then} frac{1}{6}{operatorname{tr}(A B)+operatorname{tr}(B A)} ) is equal to ( A cdot 42 ) B. 37 ( c .35 ) D. None of thes | 12 |
| 228 | Find the symmetric and skew symmetric parts of the matrix ( boldsymbol{A}=left[begin{array}{lll}1 & 2 & 4 \ 6 & 8 & 1 \ 3 & 5 & 7end{array}right] ) | 12 |
| 229 | ( fleft(begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0end{array}right], ) then solve is ( A^{3}= ) I? If correct state true else false. A. True B. False | 12 |
| 230 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose element ( a_{i j} ) is ( a_{i j}=frac{|-3 i+j|}{2} ) | 12 |
| 231 | If ( A ) and ( B ) are square matrices of same order and ( A A^{T}=I ) then ( left(A^{T} B Aright)^{10} ) is equal to A. ( A B^{10} A^{T} ) B . ( A^{T} B^{10} A ) c. ( A^{10} B^{10}left(A^{T}right)^{10} ) D. ( 10 A^{T} B A ) | 12 |
| 232 | ( boldsymbol{A}=left(begin{array}{cc}mathbf{3} & mathbf{2} \ -mathbf{1} & mathbf{4}end{array}right), boldsymbol{B}=left(begin{array}{cc}-mathbf{2} & mathbf{5} \ mathbf{6} & mathbf{7}end{array}right) ) and ( C=left(begin{array}{cc}1 & 1 \ -5 & 3end{array}right), ) then verify that ( boldsymbol{A}(boldsymbol{B}+boldsymbol{C})=boldsymbol{A} boldsymbol{B}+boldsymbol{A} boldsymbol{C} ) | 12 |
| 233 | The maximum number of different possible non-zero entries in a skew- symmetric matrix of order ‘n’ is A ( cdot frac{1}{2}left(n^{2}-nright) ) B ( cdot frac{1}{2}left(n^{2}+nright) ) ( c cdot n^{2} ) D. ( left(n^{2}-nright) ) | 12 |
| 234 | Find the value of ( x ) for which the matrix product [ left[begin{array}{ccc} 2 & 0 & 7 \ 0 & 1 & 0 \ 1 & -2 & 1 end{array}right]left[begin{array}{ccc} -x & 14 x & 7 x \ 0 & 1 & 0 \ x & -4 x & -2 x end{array}right] text { equa } ] to an identity matrix. | 12 |
| 235 | If ( A=[123] ) and ( B=left[begin{array}{l}1 \ 2 \ 3end{array}right], ) then find ( (A B)^{prime} ) | 12 |
| 236 | Let ( A ) be a ( 3 times 3 ) matrix such that ( a_{11}= ) ( a_{33}=2 ) and all the other ( a_{i j}=1 . ) Let ( boldsymbol{A}^{-1}=boldsymbol{x} boldsymbol{A}^{2}+boldsymbol{y} boldsymbol{A}-boldsymbol{z} boldsymbol{I}, ) then find the value of ( (x+y+z) ) where ( I ) is a unit of matrix of order 3 A . -9 B. 9 c. 1 D. – | 12 |
| 237 | Assertion ( boldsymbol{operatorname { T r }}(boldsymbol{A})=mathbf{0} ) Reason ( |A|=0 ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 12 |
| 238 | For matrices ( A ) and ( B ; ) if ( A B=4 I, ) then ( A^{-1} ) is ( = ) A ( .4 B ) в. ( 4 B^{-1} ) c. ( frac{1}{4} ) D. ( frac{1}{4} B^{-1} ) | 12 |
| 239 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) such that ( boldsymbol{A} boldsymbol{X}=boldsymbol{I}, ) then find ( boldsymbol{X} ) | 12 |
| 240 | Find matrix ( boldsymbol{X} ) so that ( boldsymbol{X}left[begin{array}{lll}mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5} & mathbf{6}end{array}right]= ) ( left[begin{array}{ccc}-mathbf{7} & -mathbf{8} & -mathbf{9} \ mathbf{2} & mathbf{4} & mathbf{6}end{array}right] ) | 12 |
| 241 | If ( A ) is a matrix of order ( 2 times 3, B ) is a matrix of order ( 3 times 5 ), then what in the order of matrix ( (A B) ) or ( (A B)^{t} ) | 12 |
| 242 | If ( mathbf{A} ) is non-singular matrix such that ( A^{2}=A^{-1} ) then ( a d j A= ) A. B. ( A^{-1} ) c. ( A^{3} ) D. ( left(mathrm{A}^{-1}right)^{2} ) | 12 |
| 243 | Let ( A ) and ( B ) be matrices of orders ( 3 times 2 ) and ( 2 times 4 ) respectively. Write the order of matrix ( boldsymbol{A} boldsymbol{B} ) | 12 |
| 244 | If ( A=operatorname{diag}(1,-1,2) ) and ( B= ) ( operatorname{diag}(2,3,-1) operatorname{then} 3 A+4 B= ) ( operatorname{diag}(a, b, c) . ) Then ( a-b-c= ) | 12 |
| 245 | If ( boldsymbol{A}_{1}, boldsymbol{A}_{3}, dots dots A_{2 n-1} ) are ( boldsymbol{n} ) skew symmetric matrices of same order, then ( boldsymbol{B}=sum_{r=1}^{n}(2 r-1)left(A_{2 r-1}right)^{2 r-1} ) will be A. symmetric B. skew-symmetricç c. neither symmetric nor skew-symmetric D. data not adequate | 12 |
| 246 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{0} & mathbf{- 1} & mathbf{2} \ mathbf{1} & mathbf{0} & mathbf{3} \ -mathbf{2} & mathbf{- 3} & mathbf{0}end{array}right], ) then ( boldsymbol{A}+mathbf{2} boldsymbol{A}^{prime} ) equals A. ( A ) B . ( A^{prime} ) ( c cdot-A ) D. ( 2 A ) | 12 |
| 247 | f ( A ) and ( B ) are two matrices such that ( A B=B ) and ( B A=A ) and ( left(A^{2}+B^{2}right)=lambda(A+B) . ) Considering ( boldsymbol{f}(boldsymbol{x})=|[sin boldsymbol{x}]+[cos boldsymbol{x}]| ; ) where [] is greatest integer function find ( boldsymbol{f}(mathbf{4} boldsymbol{lambda}) ) | 12 |
| 248 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{3} & mathbf{5} \ mathbf{7} & mathbf{- 9}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}mathbf{6} & mathbf{- 4} \ mathbf{2} & mathbf{3}end{array}right], ) find ( (4 A-3 B) ) | 12 |
| 249 | In a skew-symmetric matrix, the diagonal elements are all A. one B. zero c. different from each other D. non-zero | 12 |
| 250 | ff ( boldsymbol{I}=left[begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right] . boldsymbol{J}= ) ( left[begin{array}{cc}mathbf{0} & mathbf{1} \ -mathbf{1} & mathbf{0}end{array}right] ) and ( quad boldsymbol{B}= ) ( left[begin{array}{cc}cos theta & sin theta \ -sin theta & cos thetaend{array}right], quad ) then ( quad B= ) A ( . I cos theta+J sin theta ) B . ( operatorname{Isin} theta+J cos theta ) c. ( operatorname{Icos} theta-J sin theta ) D. ( -I cos theta-J sin theta ) | 12 |
| 251 | ( operatorname{Matrix} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{3} & mathbf{3} \ mathbf{2} & mathbf{4} & mathbf{1 0} \ mathbf{3} & mathbf{8} & mathbf{4}end{array}right] ) is similar to ( A ) [ left[begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{array}right] ] в. [ left[begin{array}{ccc} 1 & 0 & 9 \ 0 & 1 & -2 \ 0 & 0 & -7 end{array}right] ] ( c ) [ left[begin{array}{lll} 1 & 0 & 5 \ 0 & 1 & 2 \ 0 & 4 & 1 end{array}right] ] D. Both A and B | 12 |
| 252 | If ( boldsymbol{A}=left[begin{array}{cc}-boldsymbol{i} & mathbf{0} \ mathbf{0} & boldsymbol{i}end{array}right], ) then ( boldsymbol{A}^{prime} boldsymbol{A} ) is equal to A . ( I ) B. ( -i A ) ( c .-I ) D. ( i A ) | 12 |
| 253 | Find the inverse of the following ( operatorname{matrices} boldsymbol{A}=left(begin{array}{ccc}2 & mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{3} & mathbf{1} & mathbf{1}end{array}right) ) | 12 |
| 254 | Match the expression/statements on the left with the expression on the right. | 12 |
| 255 | ( mathrm{f} mathrm{A}=left[begin{array}{cc}1 & 3 \ 3 & 2 \ 2 & 5end{array}right], mathrm{B}=left[begin{array}{cc}-1 & -2 \ 0 & 5 \ 3 & 1end{array}right] ).Find the matrices D such that ( boldsymbol{A}+boldsymbol{B}-boldsymbol{D}=boldsymbol{O} ) e. zero matrices | 12 |
| 256 | Express the following matrices as the sum of a symmetric and a skew symmetric matrix: (i) ( left[begin{array}{cc}mathbf{3} & mathbf{5} \ mathbf{1} & -mathbf{1}end{array}right] ) (ii) ( left[begin{array}{ccc}6 & -2 & 2 \ -2 & 3 & -1 \ 2 & -1 & 3end{array}right] ) (iii) ( left[begin{array}{ccc}mathbf{3} & mathbf{3} & -mathbf{1} \ -mathbf{2}-mathbf{2} & mathbf{1} \ -mathbf{4}-mathbf{5} & mathbf{2}end{array}right] ) (iv) ( left[begin{array}{cc}mathbf{1} & mathbf{5} \ -mathbf{1} & mathbf{2}end{array}right] ) | 12 |
| 257 | ( operatorname{Let} A=left[begin{array}{ccc}1 & -1 & -1 \ 2 & 1 & -3 \ 1 & 1 & 1end{array}right] ) and ( 10 B= ) ( left[begin{array}{ccc}mathbf{4} & mathbf{2} & mathbf{2} \ -mathbf{5} & mathbf{0} & boldsymbol{alpha} \ mathbf{1} & mathbf{- 2} & mathbf{3}end{array}right], ) if ( boldsymbol{B} ) is the inverse of matrix ( A ), then ( alpha ) is ( A ) B. ( c cdot 2 ) ( D ) | 12 |
| 258 | If ( boldsymbol{A}=left[begin{array}{ccc}2 & 4 & -1 \ -1 & 0 & 2end{array}right], B=left[begin{array}{cc}3 & 4 \ -1 & 2 \ 2 & 1end{array}right] ) find ( (A B)^{T} ) | 12 |
| 259 | If ( A ) and ( B ) are two matrices of same order, then ( A+B ) is equal to ( mathbf{A} cdot B+A ) в. ( B A ) c. ( (A+B) T ) D. ( A-B ) | 12 |
| 260 | If ( A ) and ( B ) are symmetric matrices of the same order and ( X=A B+B A ) and ( boldsymbol{Y}=boldsymbol{A} boldsymbol{B}-boldsymbol{B} boldsymbol{A}, ) then ( boldsymbol{X} boldsymbol{Y}^{boldsymbol{T}} ) is equal to A . ( X Y ) в. ( Y X ) ( mathrm{c} cdot-Y X ) D. none of these | 12 |
| 261 | ( mathrm{f}[mathbf{2} mathbf{1} mathbf{3}]left[begin{array}{ccc}-mathbf{1} & mathbf{0} & mathbf{1} \ -mathbf{1} & mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1} & mathbf{1}end{array}right]left[begin{array}{l}mathbf{1} \ mathbf{0} \ mathbf{1}end{array}right]=boldsymbol{A}, ) then write the order of matrix ( boldsymbol{A} ) | 12 |
| 262 | In a upper triangular matrix ( n times n ) minimum number of zeroes is A ( cdot frac{n(n-1)}{2} ) в. ( frac{n(n+1)}{2} ) c. ( frac{2 n(n-1)}{2} ) D. None of these | 12 |
| 263 | A fruit vendor sells fruits from his shop. Selling prices of Apple, Mango and Orange are Rs. ( 20, ) Rs. 10 and Rs. 5 each respectively. The sales in three days are given below ( begin{array}{llll}text { Day } & text { Apples } & text { Mangoes } & text { Oranges } \ 1 & 50 & 60 & 30 \ 2 & 40 & 70 & 20 \ 3 & 60 & 40 & 10end{array} ) Write the matrix indicating the total amount collected on each day and hence find the total amount collected from selling of all three fruits combined. | 12 |
| 264 | If ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{b}end{array}right], boldsymbol{B}=left[begin{array}{ll}-boldsymbol{b} & -boldsymbol{a}end{array}right] ) and ( boldsymbol{C}= ) ( left[begin{array}{c}boldsymbol{a} \ -boldsymbol{a}end{array}right], ) then the correct statement is A. ( A=-B ) В. ( A+B=A-B ) c. ( A C=B C ) D. ( C A=C B ) | 12 |
| 265 | ( mathrm{f}left[begin{array}{l}mathbf{4} \ mathbf{1} \ mathbf{3}end{array}right] boldsymbol{A}=left[begin{array}{lll}-mathbf{4} & mathbf{8} & mathbf{4} \ -mathbf{1} & mathbf{2} & mathbf{1} \ -mathbf{3} & mathbf{6} & mathbf{3}end{array}right], ) if ( boldsymbol{A}= ) ( boldsymbol{a} quad boldsymbol{b} quad boldsymbol{c}] ) Find ( a+b+c ) | 12 |
| 266 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) is a scalar matrix of order ( boldsymbol{n} times boldsymbol{n} ) such that ( boldsymbol{a}_{boldsymbol{i} i}=boldsymbol{k} ) for all ( boldsymbol{i}, ) then trace of ( boldsymbol{A} ) is equal to ( mathbf{A} cdot n k ) ( mathbf{B} cdot n+k ) c. ( frac{n}{k} ) D. None of these | 12 |
| 267 | If ( boldsymbol{P}=left[begin{array}{ll}sqrt{mathbf{3}} / mathbf{2} & mathbf{1} / mathbf{2} \ -mathbf{1} / mathbf{2} & sqrt{mathbf{3}} / mathbf{2}end{array}right], boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{0} & mathbf{1}end{array}right] ) ( boldsymbol{Q}=boldsymbol{P} boldsymbol{A} boldsymbol{P}^{prime}, operatorname{then} boldsymbol{P}^{prime} boldsymbol{Q}^{2005} boldsymbol{P} ) is A. ( left[begin{array}{cc}1 & 1 \ 2005 & 1end{array}right] ) B. ( left[begin{array}{cc}1 & 2005 \ 0 & 1end{array}right] ) c. ( left[begin{array}{cc}1 & 0 \ 0 & 1end{array}right] ) D. ( left[begin{array}{cc}1 & 2005 \ 2005 & 1end{array}right] ) | 12 |
| 268 | If a matrix has 13 elements, then the possible dimensions (orders) of the matrix are ( mathbf{A} cdot 1 times 13 ) or ( 13 times 1 ) B. ( 1 times 26 ) or ( 26 times 1 ) c. ( 2 times 13 ) or ( 13 times 2 ) D. ( 13 times 13 ) | 12 |
| 269 | The matrix ( left[begin{array}{ll}0 & 1 \ 1 & 0end{array}right] ) is the matrix reflection in the line ( mathbf{A} cdot x=1 ) B . ( x+y=1 ) c. ( y=1 ) D. ( x=y ) | 12 |
| 270 | If ( C ) is skew-symmetric matrix of order ( n ) and ( X ) in ( n times 1 ) column matrix, then ( X^{T} ) ( mathrm{CX} ) is This question has multiple correct options A. singular B. non-singular c. invertible D. non-invertible | 12 |
| 271 | Assertion ( operatorname{As} A=left[begin{array}{lll}2 & 1 & 1 \ 0 & 1 & 1 \ 1 & 1 & 2end{array}right] ) satisfies the equation ( x^{3}-5 x^{2}+7 x-3=0 ) therefore ( A ) is invertible. Reason If a square matrix ( A ) satisfies the equation ( a_{0} x^{n}+a_{1} x^{n-1}+ldots a_{n-1} x+ ) ( a_{n}=0, ) and ( a_{n} neq 0, ) then ( A ) is invertible. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 272 | If ( P=left[begin{array}{cc}frac{sqrt{3}}{2} & frac{1}{2} \ -frac{1}{2} & frac{sqrt{3}}{2}end{array}right], A=left[begin{array}{cc}1 & 1 \ 0 & 1end{array}right] ) and ( Q=P A P^{prime}, ) then ( P^{prime} Q^{2015} P ) is: A. ( A=left[begin{array}{cc}0 & 2015 \ 0 & 0end{array}right] ) B. ( A=left[begin{array}{cc}2015 & 0 \ 1 & 2015end{array}right] ) c. ( A=left[begin{array}{cc}1 & 2015 \ 0 & 1end{array}right] ) D. ( A=left[begin{array}{cc}2015 & 1 \ 0 & 2015end{array}right] ) | 12 |
| 273 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) find ( boldsymbol{A}^{-1} ) by elementary transformation | 12 |
| 274 | If the value of a third order determinant is ( 11, ) then the value of the determinant of ( A^{-1}= ) A . 1 B. 121 ( c cdot 1 / 11 ) D. ( 1 / 121 ) | 12 |
| 275 | The transpose of a column matrix is A. zero matrix B. diagonal matrix c. column matrix D. row matrix | 12 |
| 276 | Using elementary transformations find the inverse of ( left[begin{array}{ccc}3 & 2 & 1 \ 2 & 4 & 3 \ 2 & -1 & 2end{array}right] ) | 12 |
| 277 | If ( A ) is a ( 3 times 3 ) invertible matrix, then what will be the value of ( k ) if ( operatorname{det}left(A^{-1}right)=(operatorname{det} A)^{k} ) | 12 |
| 278 | If ( boldsymbol{A}=frac{mathbf{1}}{sqrt{mathbf{3}}}left[begin{array}{cc}mathbf{1} & boldsymbol{i}+mathbf{1} \ boldsymbol{i}-mathbf{1} & mathbf{1}end{array}right], ) then ( boldsymbol{A}left(boldsymbol{A}^{boldsymbol{T}}right) ) equals ( mathbf{A} cdot mathbf{0} ) B. ( I ) ( c .-I ) D. ( 2 I ) | 12 |
| 279 | If ( A^{2}+2 A+10=0 ) ( A=left[begin{array}{lll}1 & 12 & 15 \ 2 & 10 & 4 \ 3 & 9 & 5end{array}right] ) find ( A^{-1} ) | 12 |
| 280 | Write ( boldsymbol{A}=left[begin{array}{cc}mathbf{3} & mathbf{5} \ mathbf{1} & mathbf{-} mathbf{1}end{array}right] ) as the sum of ( mathbf{a} ) symmetric and a skew-symmetric matrix. | 12 |
| 281 | ( a^{-1}+b^{-1}+c^{-1}=0 ) such that ( left|begin{array}{ccc}1+a & 1 & 1 \ 1 & 1+b & 1 \ 1 & 1 & 1+cend{array}right|=triangle ) then the value of ( triangle ) is ( mathbf{A} cdot mathbf{0} ) B. abco ( mathrm{c} .-a b c ) D. None of these | 12 |
| 282 | ( boldsymbol{A B}=mathbf{0} ) where ( boldsymbol{A}=left[begin{array}{cc}cos ^{2} boldsymbol{theta} & cos boldsymbol{theta} sin boldsymbol{theta} \ cos boldsymbol{theta} sin boldsymbol{theta} & sin ^{2} boldsymbol{theta}end{array}right] ) ( boldsymbol{B}=left[begin{array}{cc}cos ^{2} boldsymbol{phi} & cos phi sin phi \ cos phi sin phi & sin ^{2} phiend{array}right] ) then find ( |boldsymbol{theta}-boldsymbol{phi}|=? ) | 12 |
| 283 | Using elementary tansormations, find the inverse of each of the matrices, if it exists in ( left[begin{array}{ll}2 & 5 \ 1 & 3end{array}right] ) | 12 |
| 284 | ( f_{f} A=left[begin{array}{lll}0 & 1 & 2 \ 2 & 3 & 4 \ 4 & 5 & 6end{array}right], B=left[begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1end{array}right] ) Find ( 3 A-4 B ) | 12 |
| 285 | Express the matrices as the sum of systemmetric ( & ) a skew- symmetric [ text { matrices }left[begin{array}{ccc} mathbf{6} & -mathbf{2} & mathbf{2} \ -mathbf{2} & mathbf{3} & -mathbf{1} \ mathbf{2} & -mathbf{1} & mathbf{3} end{array}right] ] | 12 |
| 286 | ( A ) is of order ( m times n ) and ( B ) is of order ( p times ) ( q, ) addition of ( A ) and ( B ) is possible only if A ( . m=p ) B . ( n=q ) c. ( n=p ) D. ( m=p, n=q ) | 12 |
| 287 | If ( boldsymbol{A}= ) ( frac{1}{pi}left[begin{array}{cc}sin ^{-1}(pi x) & tan ^{-1}left(frac{pi}{pi}right) \ sin ^{-1}left(frac{x}{pi}right) & cot ^{-1}(pi x)end{array}right], B= ) ( frac{1}{pi}left[begin{array}{cc}-cos ^{-1}(pi x) & tan ^{-1}left(frac{x}{pi}right) \ sin ^{-1}left(frac{x}{pi}right) & -tan ^{-1}(pi x)end{array}right], ) then ( A-B ) is equal to ( A ) B. ( c .2 ) ( D cdot underline{1} ) | 12 |
| 288 | Suppose ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{5} \ mathbf{4} & mathbf{3} & mathbf{7}end{array}right] ) find ( mathbf{4} boldsymbol{A} ) | 12 |
| 289 | ( left[begin{array}{cc}2 & 1 \ 1 & 0 \ -3 & 4end{array}right]+left[begin{array}{cc}-1 & -8 \ 1 & -2 \ 9 & 22end{array}right] ) | 12 |
| 290 | If ( boldsymbol{A}=left(begin{array}{rr}mathbf{4} & -mathbf{2} \ mathbf{5} & -mathbf{9}end{array}right) ) and ( boldsymbol{B}=left(begin{array}{cc}mathbf{8} & mathbf{2} \ mathbf{- 1} & -mathbf{3}end{array}right) ) find ( mathbf{6} boldsymbol{A}-mathbf{3} boldsymbol{B} ) | 12 |
| 291 | If ( A ) and ( B ) are square matrices of order 3 such that ( |boldsymbol{A}|=-mathbf{1},|boldsymbol{B}|=mathbf{3}, ) then the determinant of ( 3 A B ) is equal to A . -9 B. -27 c. -81 D. 81 | 12 |
| 292 | Find ( p, q, r ) and ( s, ) if ( left[begin{array}{cc}boldsymbol{p}+boldsymbol{4} & boldsymbol{2} boldsymbol{q}-boldsymbol{7} \ boldsymbol{s}-boldsymbol{3} & boldsymbol{r}+boldsymbol{2} boldsymbol{s}end{array}right]=left[begin{array}{cc}boldsymbol{6} & -boldsymbol{3} \ boldsymbol{2} & boldsymbol{1} boldsymbol{4}end{array}right] ) | 12 |
| 293 | If ( boldsymbol{A}=left(begin{array}{c}4-2 \ 5-9end{array}right) ) and ( boldsymbol{B}=left(begin{array}{cc}8 & 2 \ -1 & -3end{array}right) ) find ( 6 A ) 3B. | 12 |
| 294 | Construct a ( 2 times 3 ) matrix ( A=left[a_{i j}right] ) whose elements are given by ( a_{i j}= ) ( |2 i-3 j| ) | 12 |
| 295 | If ( left[begin{array}{ccc}3 & 2 & -1 \ 4 & 9 & 2 \ 5 & 0 & -2end{array}right]left[begin{array}{l}x \ y \ zend{array}right]=left[begin{array}{l}0 \ 7 \ 2end{array}right], ) then ( (x, y, z)= ) В. (2,-1,-4) c. (3,0,6) ( mathbf{D} cdot(2,-1,4) ) | 12 |
| 296 | Find matrix ( boldsymbol{X}, ) if ( left[begin{array}{ccc}mathbf{3} & mathbf{5} & -mathbf{9} \ -mathbf{1} & mathbf{4} & -mathbf{7}end{array}right]+boldsymbol{X}= ) ( left[begin{array}{lll}mathbf{6} & mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{8} & mathbf{6}end{array}right] ) | 12 |
| 297 | The number of ( boldsymbol{A} ) in ( boldsymbol{T}_{boldsymbol{p}} ) such that ( boldsymbol{A} ) is either symmetric or skew-symmetric or both, and det(A) divisible by ( p ) is: A ( cdot(p-1)^{2} ) в. ( 2(p-1) ) c. ( (p-1)^{2}+1 ) D. ( 2 p-1 ) | 12 |
| 298 | If ( A=left[begin{array}{cc}-3 & 5 \ 5 & 0 \ -7 & 4end{array}right] ) and ( B=left[begin{array}{ccc}3 & -5 & 7 \ -5 & 0 & -4end{array}right] ) then find ( boldsymbol{A}+boldsymbol{B}^{boldsymbol{T}} ) ( A cdot 0 ) в. ( 2 B ) ( c cdot 2 A^{T} ) D. ( 2 B^{T} ) | 12 |
| 299 | Construct a ( 2 times 2 ) matrix, ( A=left[a_{i j}right] ) whose elements are given by: (i) ( a_{i j}=frac{(i+j)^{2}}{2} ) (ii) ( a_{i j}=frac{i}{j} ) (iii) ( a_{i j}=frac{(i+2 j)^{2}}{2} ) | 12 |
| 300 | ( operatorname{Let} A=left[begin{array}{ll}a & b \ c & dend{array}right], a, b, c, d neq 0, ) then ( boldsymbol{B}=boldsymbol{A} boldsymbol{A}^{prime}-boldsymbol{A}^{prime} boldsymbol{A} ) equals A. ( (a d-b c) I ) в. ( (a c-b d) I ) c. D. none of these | 12 |
| 301 | A matrix having ( m ) rows and ( n ) columns with ( m neq n ) is said to be a A. rectangular matrix B. square matrix c. identity matrix D. scalar matrix | 12 |
| 302 | Solve the following system of equations by using Matrix inversion method. ( mathbf{2} boldsymbol{x}-boldsymbol{y}+mathbf{3} boldsymbol{z}=mathbf{9}, boldsymbol{x}+boldsymbol{y}+boldsymbol{z}=mathbf{6}, boldsymbol{x}- ) ( boldsymbol{y}+boldsymbol{z}=mathbf{2} ) | 12 |
| 303 | Assertion The matrix ( left(begin{array}{cccc}.1 & 0 & 0 & 0 \ 0 & 2 & 0 & 0 \ 0 & 0 & 3 & 0end{array}right) ) is a diagonal matrix Reason ( A=left(a_{i j}right)_{m times m} ) is a square matrix such that entry ( a_{i j}=0 forall i neq j, ) then ( A ) is called diagonal matrix. A. Both (A) & (R) are individually true & (R) is correct explanation of (A), B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A). C. (A) is true but (R)is false, D. (A) is false but (R ) is true. | 12 |
| 304 | Choose the correct statement or statements: This question has multiple correct options | 12 |
| 305 | ( left[begin{array}{lll}boldsymbol{x} & mathbf{1} & -mathbf{1}end{array}right]left[begin{array}{lll}mathbf{0} & mathbf{1} & -mathbf{1} \ mathbf{2} & mathbf{1} & mathbf{3} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right]left[begin{array}{l}boldsymbol{x} \ -mathbf{1} \ mathbf{1}end{array}right]=mathbf{0} ) then find ( x ) | 12 |
| 306 | If ( A ) is an invertible matrix of order ( 3 times 3 ) such that ( |boldsymbol{A}|=mathbf{5}, ) find the value of ( left|boldsymbol{A}^{-1}right| ) | 12 |
| 307 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right]_{2 times 2} ) and ( boldsymbol{a}_{i j}=boldsymbol{i}+boldsymbol{j}, ) then ( boldsymbol{A}= ) ( A cdotleft(begin{array}{l}12 \ 3end{array}right) ) в. ( left(begin{array}{c}23 \ 3end{array}right) ) ( c cdotleft(begin{array}{cc}2 & 3 \ 4 & 5end{array}right) ) D. ( left(_{6}^{4} begin{array}{r}5 \ end{array}right) ) | 12 |
| 308 | ( operatorname{Let} C_{k}=^{n} C_{k} ) for ( 0 leq k leq n ) and ( boldsymbol{A}_{boldsymbol{k}}=left[begin{array}{cc}boldsymbol{C}_{boldsymbol{k}-1}^{2} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{C}_{boldsymbol{k}}^{2}end{array}right] ) for ( boldsymbol{k} geq 1, ) and ( boldsymbol{A}_{1}+ ) ( boldsymbol{A}_{2}+ldots+boldsymbol{A}_{n}=left[begin{array}{cc}boldsymbol{k}_{1} & mathbf{0} \ mathbf{0} & boldsymbol{k}_{2}end{array}right], ) then This question has multiple correct options A ( . k_{1}=k_{2} ) B . ( k_{1}+k_{2}=^{2 n} C_{2 n}+1 ) c. ( k_{1}=^{2 n} C_{n}-1 ) D. ( k_{2}=^{2 n} C_{n+1} ) | 12 |
| 309 | Let ( A, B, C, D ) be (not necessarily square) real matrices such that ( boldsymbol{A}^{boldsymbol{T}}= ) ( boldsymbol{B C D} ; boldsymbol{B}^{boldsymbol{T}}=boldsymbol{C D A} ; boldsymbol{C}^{boldsymbol{T}}=boldsymbol{D A B} ) and ( D^{T}=A B C ) for the matrix ( S=A B C D ) then which of the following is/are true This question has multiple correct options A ( cdot S^{3}=S ) B. ( S^{2}=S^{4} ) c. ( S=S^{2} ) D. none of these | 12 |
| 310 | ( mathbf{f}left[begin{array}{cc}boldsymbol{x}-boldsymbol{y} & boldsymbol{z} \ mathbf{2} boldsymbol{x}-boldsymbol{y} & boldsymbol{w}end{array}right]=left[begin{array}{cc}-mathbf{1} & mathbf{4} \ mathbf{0} & mathbf{5}end{array}right], ) find the value of ( boldsymbol{x}+boldsymbol{y} ) | 12 |
| 311 | Multiplication of 3 with the matrix ( left[begin{array}{lll}mathbf{4} & mathbf{1} & mathbf{1}end{array}right] ) gives ( mathbf{A} cdotleft[begin{array}{lll}12 & 3 & 3end{array}right] ) B cdot ( left[begin{array}{lll}12 & 1 & 1end{array}right] ) ( mathbf{c} cdotleft[begin{array}{lll}4 & 3 & 1end{array}right] ) D. ( left[begin{array}{lll}4 & 1 & 3end{array}right] ) | 12 |
| 312 | Let ( M ) and ( N ) be two ( 3 times 3 ) non-singular skew-symmetric matrices such that ( M N=N M . ) If ( P^{T} ) denotes the transpose of ( P, ) then ( M^{2} N^{2}left(M^{T} Nright)^{-1}left(M N^{-1}right)^{T} ) is equal to A ( . M^{2} ) B. ( -N^{2} ) c. ( -M^{2} ) D. ( M N ) | 12 |
| 313 | Let ( a ) denote the element of the ( i^{t h} ) row and ( j^{t h} ) column in a ( 3 times 3 ) matrix and let ( boldsymbol{a}_{i j}=-boldsymbol{a}_{j i} ) for every i and j then this matrix is an – A. Orthogonal matrix B. singular matrix c. matrix whose principal diagonal elements are all zero D. skew-symmetric matrix | 12 |
| 314 | Let ( A ) be a real matrix such that ( A^{67}= ) ( A^{-1}, ) then ( mathbf{A} cdot|A|=pm 1 ) B . ( |A|=1 ) C. ( A=I, I ) being unit matrix D. ( A ) is diagonal matrix | 12 |
| 315 | Find the values of ( x ), if ( left|begin{array}{ll}2 & 4 \ 5 & 1end{array}right|=left|begin{array}{cc}2 x & 4 \ 6 & xend{array}right| ) | 12 |
| 316 | If ( boldsymbol{P}=left[begin{array}{ccc}boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{b} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{c}end{array}right] ) then, ( operatorname{det}left(boldsymbol{P}^{-1}right) ) A ( . a b c ) B. ( a^{2} b^{2} c^{2} ) c. ( frac{1}{a b} ) D. ( frac{1}{a^{2} b^{2} c^{2}} ) | 12 |
| 317 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{ll}mathbf{3} & mathbf{7} \ mathbf{2} & mathbf{5}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{ll}mathbf{6} & mathbf{8} \ mathbf{7} & mathbf{9}end{array}right] . ) Verify ( operatorname{that}(A B)^{-1}=B^{-1} A^{-1} ) | 12 |
| 318 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{3} & mathbf{- 2} \ mathbf{5} & mathbf{4}end{array}right], boldsymbol{B}=left[begin{array}{cc}mathbf{1} & mathbf{4} \ mathbf{6} & -mathbf{7}end{array}right], ) then find the matrix ( A-4 B+7 I ), where lis the unit matrix of order 2 ( =left(begin{array}{rr}3 & -2 \ 5 & 4end{array}right], mathbf{B}=left[begin{array}{rr}1 & 4 \ 6 & -7end{array}right] ) | 12 |
| 319 | If order of a matrix is ( 3 times 3, ) then it is a A. square matrix B. rectangular matrix c. unit matrix D. None of these | 12 |
| 320 | ( mathbf{A}=left[begin{array}{cc}cos alpha & sin alpha \ -sin alpha & cos alphaend{array}right] ) then ( mathbf{A} . mathbf{A}^{mathbf{T}} ) A. Null matrix в. А ( c cdot I ) D. A | 12 |
| 321 | If ( boldsymbol{A}=left[begin{array}{cc}-mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}mathbf{3} & -mathbf{2} \ mathbf{1} & mathbf{5}end{array}right] ) ( mathbf{2} boldsymbol{A}+boldsymbol{B}+boldsymbol{X}=mathbf{0}, ) then the matrix ( boldsymbol{X} ) is A. ( left[begin{array}{ll}1 & 2 \ 7 & 13end{array}right] ) B. ( left[begin{array}{cc}-1 & 2 \ 7 & -13end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-1 & -2 \ 7 & 13end{array}right] ) D. ( left[begin{array}{cc}-1 & -2 \ -7 & -13end{array}right] ) | 12 |
| 322 | ( mathrm{ff} mathbf{A}=left[begin{array}{lll}1 & 2 & x \ 0 & 1 & 0 \ 0 & 0 & 1end{array}right] ) and ( B= ) ( left[begin{array}{ccc}1 & -2 & y \ 0 & 1 & 0 \ 0 & 0 & 1end{array}right] quad ) and ( A B=I_{3}, quad ) then ( quad x+ ) ( boldsymbol{y} quad boldsymbol{i s} quad boldsymbol{e q u a l} quad boldsymbol{t o} ) | 12 |
| 323 | If the matrix ( left|begin{array}{ccc}mathbf{1} & mathbf{3} & boldsymbol{lambda}+mathbf{2} \ mathbf{2} & mathbf{4} & mathbf{8} \ mathbf{3} & mathbf{5} & mathbf{1 0}end{array}right| ) is singular ( operatorname{then} lambda= ) ( A cdot-2 ) B. 4 ( c cdot 2 ) D. – – | 12 |
| 324 | Let ( A ) be a square matrix all of whose entries are integers. Then, which one of the following is true? ( mathbf{A} cdot ) If ( operatorname{det} A=pm 1, ) then ( A^{-1} ) exists and all its entries are integers B. If ( operatorname{det} A=pm 1, ) then ( A^{-1} ) need not exist C. If ( operatorname{det} A=pm 1, ) then ( A^{-1} ) exists but all its entries are not necessarily integers D. If det ( A neqpm 1 ), then ( A^{-1} ) exists and all its entries are non-integers | 12 |
| 325 | Is it possible to define the matrix AB and BA when : A has 4 rows and ( mathrm{B} ) has 4 columns | 12 |
| 326 | Using elementary operations, find the inverse of the following matrix: ( left(begin{array}{ccc}-1 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1end{array}right) ) | 12 |
| 327 | In the set of all ( 3 times 3 ) real matrices a relation is defined as follows. A matrix ( A ) is related to a matrix ( B ), if and only there is a non-singular ( 3 times 3 ) matrix ( P ) such that ( B=P^{-1} A P . ) This relation is A. reflexive,symmetric but not transitive B. reflexive, transitive but not symmetric c. symmetric, transitive but not reflexive D. an equivalence relation | 12 |
| 328 | Matrices ( A ) and ( B ) satisfy ( A B=B^{-1} ) where ( boldsymbol{B}=left[begin{array}{cc}2 & -1 \ 2 & 0end{array}right], ) then find without finding ( A^{-1}, ) the matrix ( X ) satisfying ( boldsymbol{A}^{-1} boldsymbol{X} boldsymbol{A}=? ) A. ( B ) в. ( B^{2} ) c. ( A ) D. None of these | 12 |
| 329 | If ( boldsymbol{m}left[begin{array}{ll}-mathbf{3} & mathbf{4}end{array}right]+boldsymbol{n}left[begin{array}{ll}mathbf{4} & -mathbf{3}end{array}right]=left[begin{array}{ll}mathbf{1 0} & -mathbf{1 1}end{array}right] ) then ( 3 m+7 n= ) ( A cdot 3 ) B. 5 c. 10 ( D ) | 12 |
| 330 | If ( boldsymbol{A}=left(left[begin{array}{llll}1 & 2 & 3 & 4end{array}right] text { and } A B=left[begin{array}{lll}3 & 4 & -1end{array}right]right. ) then the order of matrix B is ( A cdot 2 times 3 ) B. 3×3 ( c cdot 4 times 3 ) D. ( 1 times 3 ) | 12 |
| 331 | Assertion ( operatorname{Let} A=left[begin{array}{ll}a & b \ b & aend{array}right] ) and ( B=left[begin{array}{ll}p & q \ r & send{array}right] ) If ( b=0, ) then ( A B=B A ) Reason If ( b neq 0, ) then ( A B=B A ) | 12 |
| 332 | By using the elementary transformation, find the inverse of the ( operatorname{matrix} boldsymbol{A}=left[begin{array}{cc}mathbf{1} & -mathbf{2} \ mathbf{2} & mathbf{1}end{array}right] ) | 12 |
| 333 | If ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{4} & mathbf{7} \ mathbf{2} & mathbf{6} & mathbf{5} \ mathbf{3} & mathbf{- 1} & mathbf{2}end{array}right] ) and ( mathbf{B}=operatorname{diag}(12 ) 5), then trace of matrix ( boldsymbol{A B}^{mathbf{2}} ) is A . 74 B. 75 c. 529 D. 23 | 12 |
| 334 | By using elementary transformation find the inverse of the matrix ( boldsymbol{A}= ) ( left[begin{array}{ll}1 & 2 \ 2 & 1end{array}right] ) | 12 |
| 335 | If ( [123] mathrm{B}=[34], ) the order of the matrix B is ( mathbf{A} cdot 3 times 1 ) B. ( 1 times 3 ) c. ( 2 times 3 ) D. ( 3 times 2 ) | 12 |
| 336 | The management committee of a residential colony decided to award some of its members (say ( x ) ) for honesty, some (say ( y ) ) for helping others and some others (say ( z ) ) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12 Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is ( 33 . ) If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others using matrix category. Apart from these values namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards. | 12 |
| 337 | Find the value of ( y, ) if ( left[begin{array}{cc}x-y & 2 \ x & 5end{array}right]= ) ( left[begin{array}{ll}2 & 2 \ 3 & 5end{array}right] ) | 12 |
| 338 | Solve the following matrix equation for ( x ) ( [boldsymbol{x}-mathbf{5}-mathbf{1}]left[begin{array}{lll}mathbf{1} & mathbf{0} & mathbf{2} \ mathbf{0} & mathbf{2} & mathbf{1} \ mathbf{2} & mathbf{0} & mathbf{3}end{array}right]left[begin{array}{l}boldsymbol{x} \ mathbf{4} \ mathbf{1}end{array}right]=mathbf{0} ) | 12 |
| 339 | Solve for ( x ) and ( y, ) if ( left(begin{array}{l}x^{2} \ y^{2}end{array}right)+3left(begin{array}{c}2 x \ -yend{array}right)= ) ( left(begin{array}{c}-mathbf{9} \ mathbf{4}end{array}right) ) | 12 |
| 340 | If the matrix ( A B ) is a zero matrix, then which one of the following is correct? A. ( A ) must be equal to zero matrix or ( B ) must be equal to zero matrix B. ( A ) must be equal to zero matrix and ( B ) must be equal to zero matrix c. It is not necessary that either ( A ) is zero matrix or ( B ) is zero matrix D. None of the above | 12 |
| 341 | ( boldsymbol{A}=left[begin{array}{ll}mathbf{0} & mathbf{1} \ mathbf{0} & mathbf{0}end{array}right], ) show that ( (boldsymbol{a} boldsymbol{I}+boldsymbol{b} boldsymbol{A})^{n}= ) ( a^{n} I+n a^{n-1} b A, ) where ( I ) is the identity matrix of order 2 and ( n in N ) | 12 |
| 342 | Using elementary transformations, find the inverse of matrix, ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{3} \ mathbf{2} & mathbf{7}end{array}right] ) | 12 |
| 343 | If ( A ) is an invertible matrix of order 2 then ( operatorname{det}left(A^{-1}right) ) is equal to ( mathbf{A} cdot operatorname{det}(A) ) B. ( frac{1}{operatorname{det}(A)} ) ( c .1 ) D. 0 | 12 |
| 344 | The order of the matrix ( left[begin{array}{c}-1 \ 3 \ 4end{array}right] ) is : ( mathbf{A} cdot 1 times 3 ) B. ( 3 times 1 ) c. ( 1 times 1 ) D. ( 3 times 3 ) | 12 |
| 345 | If for a matrix ( boldsymbol{A}, boldsymbol{A}^{2}+boldsymbol{I}=boldsymbol{O} ) where ( boldsymbol{I} ) is the identity matrix, then ( A ) equals ( mathbf{A} cdotleft[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) В. ( left[begin{array}{cc}i & 0 \ i & -iend{array}right] ) c. ( left[begin{array}{cc}1 & 2 \ -1 & 1end{array}right] ) D. ( left[begin{array}{cc}-1 & 0 \ 0 & -1end{array}right] ) | 12 |
| 346 | If ( A ) is non-singular and ( (A+I)(A- ) ( mathbf{3} boldsymbol{I})=mathbf{0} ) then ( mathbf{3} boldsymbol{A}^{-1}-boldsymbol{A}+mathbf{2} boldsymbol{I} ) A . ( I ) B. c. ( 2 I ) D. ( 6 I ) | 12 |
| 347 | Prove that ( left|begin{array}{ccc}boldsymbol{y} boldsymbol{z}-boldsymbol{x}^{2} & boldsymbol{z} boldsymbol{x}-boldsymbol{y}^{2} & boldsymbol{x} boldsymbol{y}-boldsymbol{z}^{2} \ boldsymbol{z} boldsymbol{x}-boldsymbol{y}^{2} & boldsymbol{x} boldsymbol{y}-boldsymbol{z}^{2} & boldsymbol{y} boldsymbol{z}-boldsymbol{x}^{2} \ boldsymbol{x} boldsymbol{y}-boldsymbol{z}^{2} & boldsymbol{y} boldsymbol{z}-boldsymbol{x}^{2} & boldsymbol{z} boldsymbol{x}-boldsymbol{y}^{2}end{array}right| ) is divisible by ( (x+y+z), ) and hence find the quotient OR Using elementary transformations, find the inverse of the matrix ( boldsymbol{A}= ) ( left(begin{array}{lll}8 & 4 & 3 \ 2 & 1 & 1 \ 1 & 2 & 2end{array}right) ) and use it to solve the following system of linear equations: ( 8 x+4 y+3 z=19 ) ( 2 x+y+z=5 ) ( x+2 y+2 z=7 ) | 12 |
| 348 | If ( boldsymbol{A}=[mathbf{1}], ) then the order of the matrix is ( mathbf{A} cdot 1 times 1 ) в. ( 2 times 1 ) c. ( 1 times 2 ) D. None of these | 12 |
| 349 | Convert ( [1 quad-12 ) 3] into an identity matrix by suitable row transformations. | 12 |
| 350 | If ( A ) and ( B ) are skew symmetric matrices of order ( n ) then ( A+B ) is A. skew symmetric B. a diagonal matrix c. a null matrix D. symmetric matrix | 12 |
| 351 | If ( x ) and ( Y ) are the matrices of order ( 2 x ) 2 each and ( 2 X-3 Y=left|begin{array}{cc}-7 & 0 \ 7 & -13end{array}right| ) and ( mathbf{3} boldsymbol{X}+mathbf{2} boldsymbol{Y}=left|begin{array}{cc}mathbf{9} & mathbf{1 3} \ mathbf{4} & mathbf{1 3}end{array}right|, ) then what is ( boldsymbol{Y} ) equal to? A. ( left|begin{array}{cc}1 & 3 \ -2 & 1end{array}right| ) B. ( left|begin{array}{ll}1 & 3 \ 2 & 1end{array}right| ) begin{tabular}{l|ll|l} & 3 & 2 \ -1 & 5 end{tabular} D. ( left|begin{array}{cc}3 & 2 \ 1 & -5end{array}right| ) | 12 |
| 352 | If ( A=left|begin{array}{l}1 \ 3end{array}right| B=left|begin{array}{c}-1 \ 4end{array}right| ) then ( 2 A+B= ) ( A cdot mid 10 ) 9 B. ( mid 10 ) 1 ( c cdotleft|begin{array}{l}1 \ 10end{array}right| ) D. ( mid 1 ) 9 | 12 |
| 353 | For square matrix ( boldsymbol{A}, boldsymbol{A} boldsymbol{A}^{boldsymbol{T}} ) is – A. unit matrix B. symmetric matrix c. skew symmetric matrix D. diagonal matrix | 12 |
| 354 | If ( mathbf{A} ) is ( 3 x 4 ) matrix ( B^{T} ) is a matrix such that ( A^{T} B ) and ( B A^{T} ) are both defined then ( B ) is of the type ( A cdot 3 times 4 ) B. ( 3 times 3 ) c. ( 4 times 3 ) D. ( 4 times 4 ) | 12 |
| 355 | type of entertainment device sold at three of their branch stores so that they can monitor their purchases of supplies. The sales in two weeks are shown in the following spreadsheets. Find the sum of the items sold out in two weeks using matrix addition. | 12 |
| 356 | Construct a ( 4 times 3 ) matrix ( A=left[a_{i j}right] ) whose element ( a_{i j} ) is ( a_{i j}=2 i+frac{i}{j} ) | 12 |
| 357 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{1} & mathbf{1} \ mathbf{0} & -mathbf{2} & mathbf{4}end{array}right], boldsymbol{I}=left[begin{array}{ccc}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{0} & mathbf{1}end{array}right] ) and ( boldsymbol{A}^{-1}=frac{1}{boldsymbol{6}}left(boldsymbol{A}^{2}+boldsymbol{alpha} boldsymbol{A}+boldsymbol{beta} boldsymbol{I}right), ) find ( boldsymbol{beta} / mathbf{1 1} ) | 12 |
| 358 | Find ( boldsymbol{A}+boldsymbol{B} ) ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{3} & mathbf{- 1} \ mathbf{2} & mathbf{- 5} & mathbf{4} \ mathbf{3} & mathbf{2} & mathbf{- 6}end{array}right] ) and ( boldsymbol{B}= ) ( left[begin{array}{ccc}mathbf{2} & mathbf{4} & mathbf{0} \ mathbf{3} & mathbf{- 4} & mathbf{5} \ mathbf{2} & mathbf{3} & mathbf{- 5}end{array}right] ) | 12 |
| 359 | ( boldsymbol{A}=left[begin{array}{cc}boldsymbol{t} & boldsymbol{t}+mathbf{1} \ boldsymbol{t}-mathbf{1} & boldsymbol{t}end{array}right] ) is a matrix such that ( A A^{T}=I_{2} ) then trace of the matrix is A . 2 B. 0 ( c cdot 4 ) ( D ) | 12 |
| 360 | Prove the following ( left[begin{array}{llll}1 & 3 & 2 & 0 \ 4 & 1 & 5 & 9 \ 3 & 2 & 1 & 3end{array}right] pm ) ( left[begin{array}{cccc}1 & 0 & 5 & 8 \ 2 & 3 & 5 & 8 \ 1 & -5 & 2 & 3end{array}right] ) ( =left[begin{array}{cccc}2 & 3 & 7 & 8 \ 6 & 4 & 9 & 14 \ 4 & -3 & 3 & 6end{array}right] ) for plus ( =left[begin{array}{cccc}0 & 3 & -3 & -8 \ 4 & 2 & 1 & 4 \ 0 & 7 & -1 & 0end{array}right] ) for minus | 12 |
| 361 | By using elementary transformation find the inverse of ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{2}-mathbf{1}end{array}right] ) | 12 |
| 362 | For a matrix ( boldsymbol{A}left(begin{array}{ccc}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{2} & mathbf{1} & mathbf{0} \ mathbf{3} & mathbf{2} & mathbf{1}end{array}right), ) if ( boldsymbol{U}_{1}, boldsymbol{U}_{2} ) and ( U_{3} ) are ( 3 times 1 ) column matrices satisfying ( boldsymbol{A} boldsymbol{U}_{mathbf{1}}= ) ( left(begin{array}{l}1 \ 0 \ 0end{array}right), A U_{2}left(begin{array}{l}2 \ 3 \ 0end{array}right), A U_{3}=left(begin{array}{l}2 \ 3 \ 1end{array}right) ) and ( U ) is ( 3 times 3 ) matrix whose columns are ( U_{1}, U_{2} ) and ( U_{3} ) Then sum of the elements of ( U^{-1} ) is A . 6 B. ( 0(z e r o) ) ( c . ) D. ( 2 / 3 ) | 12 |
| 363 | If ( A ) is non-singular and ( (A-2 I)(A- ) ( 4 I)=0 ) then ( frac{1}{6} mathbf{A}+frac{4}{3} mathbf{A}^{-1}= ) ( A cdot I ) B. c. ( 2 I ) D. ( 6 I ) | 12 |
| 364 | If ( A ) is a diagonal matrix of order ( 3 times 3 ) is a commutative with every square matrix of order ( 3 times 3 ) under multiplication and ( operatorname{tr}(A)=12, ) then the value of ( |boldsymbol{A}| ) is | 12 |
| 365 | ( A=left[begin{array}{ll}1 & 2 \ 3 & 4end{array}right], B=left[begin{array}{ll}-1 & -1 \ -1 & -1end{array}right], C=left[begin{array}{ll}x & y \ z & rend{array}right] ) If ( boldsymbol{A}+mathbf{3} boldsymbol{B}=boldsymbol{C}, ) then ( boldsymbol{x}+boldsymbol{y}+boldsymbol{z}+boldsymbol{r} ) is ( A ) B. – 2 ( c .-1 ) ( D ) | 12 |
| 366 | If ( boldsymbol{P}=left[begin{array}{cc}frac{sqrt{mathbf{3}}}{mathbf{2}} & frac{mathbf{1}}{2} \ -frac{mathbf{1}}{mathbf{2}} & frac{sqrt{mathbf{3}}}{mathbf{2}}end{array}right], boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{0} & mathbf{1}end{array}right] ) and ( Q=P A P^{T} ) then ( Pleft(Q^{2005}right) P^{T} ) equal to begin{tabular}{l} A. ( left[begin{array}{cc}1 & 2005 \ 0 & 1end{array}right] ) \ hline end{tabular} B. ( left[begin{array}{cc}sqrt{3} / 2 & 2005 \ 1 & 0end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}1 & 2005 \ sqrt{3} / 2 & 1end{array}right] ) D. ( left[begin{array}{ll}1 & sqrt{3} / 2 \ 0 & 2005end{array}right] ) | 12 |
| 367 | If ( A ) and ( B ) are two non-zero square matrices of the same order such that the product ( A B=0, ) then A. both A and B must be singular B. exactly one of them must be singular c. atleast one of them must be non-singular D. none of these | 12 |
| 368 | ( left[begin{array}{lll}0 & 0 & 0end{array}right] ) is an example of A. Scalar matrix B. Diagonal matrix c. Identity matrix D. Null matrix | 12 |
| 369 | If ( mathbf{3} boldsymbol{A}-mathbf{2 B}=left(begin{array}{cc}mathbf{1} & mathbf{- 2} \ mathbf{3} & mathbf{0}end{array}right) ) and ( mathbf{2} boldsymbol{A}- ) ( mathbf{3} boldsymbol{B}=left(begin{array}{cc}mathbf{- 3} & mathbf{3} \ mathbf{1} & mathbf{- 1}end{array}right) ) then find ( boldsymbol{B} ) | 12 |
| 370 | ( operatorname{Let} P=left(begin{array}{cc}cos frac{pi}{4} & -sin frac{pi}{4} \ sin frac{pi}{4} & cos frac{pi}{4}end{array}right) ) and ( x= ) ( left(begin{array}{c}frac{1}{sqrt{2}} \ frac{1}{sqrt{2}}end{array}right) cdot operatorname{Then} P^{3} X ) is equal to A ( cdotleft(begin{array}{l}0 \ 1end{array}right) ) B. ( left(frac{-1}{sqrt{2}}right) ) ( left(begin{array}{c}-1 \ 0end{array}right) ) D ( left(begin{array}{l}-frac{1}{sqrt{2}} \ -frac{1}{sqrt{2}}end{array}right) ) | 12 |
| 371 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{1} & mathbf{1}end{array}right], ) and ( boldsymbol{I}=left[begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right] ) then prove that ( boldsymbol{A}^{n}=boldsymbol{n} boldsymbol{A}-(boldsymbol{n}-mathbf{1}) boldsymbol{I}, boldsymbol{n} geq 1 ) | 12 |
| 372 | Find the inverse of the matrix ( boldsymbol{A}= ) ( left[begin{array}{ll}1 & 2 \ 1 & 3end{array}right] ) using elementry transformations. | 12 |
| 373 | ( operatorname{Given} boldsymbol{A}=left[begin{array}{cc}mathbf{3} & mathbf{6} \ -mathbf{2} & -mathbf{8}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{ll}mathbf{2} & mathbf{1 6}end{array}right] ) find the matrix ( boldsymbol{X} ) such that ( boldsymbol{X} boldsymbol{A}=boldsymbol{B} ) A ( cdotleft[-frac{4}{3}-3right] ) в. ( left[frac{4}{3} 3right] ) ( ^{mathbf{c}} cdotleft[begin{array}{ll}frac{4}{3} & -3end{array}right] ) D. ( left[-frac{4}{3} quad 3right] ) | 12 |
| 374 | If ( boldsymbol{A} ) is symmetric matrix and ( boldsymbol{n} in boldsymbol{N} ) write whether ( A^{n} ) is symmetric or skewsymmetric or neither of these two | 12 |
| 375 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{2} & mathbf{2}end{array}right], boldsymbol{B}=left[begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{3} & mathbf{4}end{array}right] ) then find ( boldsymbol{A B} ) | 12 |
| 376 | Matrices ( A ) and ( B ) satisfy ( A B=B^{-1} ) where ( boldsymbol{B}=left[begin{array}{cc}2 & -1 \ 2 & 0end{array}right], ) then find without finding ( B^{-1}, ) the value of ( K ) for which ( boldsymbol{K} boldsymbol{A}-mathbf{2} boldsymbol{B}^{-mathbf{1}}+boldsymbol{I}=boldsymbol{O} ) | 12 |
| 377 | ( left[begin{array}{ccc}10 & 20 & 30 \ 20 & 45 & 80 \ 30 & 80 & 171end{array}right]= ) ( left[begin{array}{lll}1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 4 & 1end{array}right]left[begin{array}{lll}x & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 1end{array}right]left[begin{array}{lll}1 & 2 & 3 \ 0 & 1 & 4 \ 0 & 0 & 1end{array}right] ) then ( boldsymbol{x}= ) ( A cdot 10 ) B. 20 ( c cdot 30 ) D. 40 | 12 |
| 378 | If ( A=left[begin{array}{ll}alpha & 0 \ 1 & 1end{array}right], B=left[begin{array}{ll}1 & 0 \ 5 & 1end{array}right] ) whenever ( boldsymbol{A}^{2}=boldsymbol{B} ) then values of ( boldsymbol{alpha} ) is A . B. – ( c cdot 4 ) D. no real value of ( alpha ) | 12 |
| 379 | Find the value of ( x+y ) from the following equation: ( mathbf{2}left[begin{array}{cc}boldsymbol{x} & mathbf{5} \ mathbf{7} & boldsymbol{y}-mathbf{3}end{array}right]+left[begin{array}{cc}mathbf{3} & mathbf{- 4} \ mathbf{1} & mathbf{2}end{array}right]=left[begin{array}{cc}mathbf{7} & mathbf{6} \ mathbf{1 5} & mathbf{1 4}end{array}right] ) | 12 |
| 380 | Let ( A ) be the set of all ( 3 times 3 ) skew symmetric matrices whose entries are either ( -1,0, ) or ( 1 . ) If there are exactly three 0 s, three 1 s, and three (-1) s, then the number of such matrices is | 12 |
| 381 | Assertion If ( A ) and ( B ) are two ( 3 times 3 ) matrices such that ( A B=0, ) then ( A=0 ) or ( B=0 ) Reason If ( A, B ) and ( X ) are three ( 3 times 3 ) matrices such that ( boldsymbol{A} boldsymbol{X}=boldsymbol{B},|boldsymbol{A}| neq mathbf{0}, ) then ( boldsymbol{X}= ) ( A^{-1} B ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 382 | If the matrices has 13 elements , then the possible dimension (order) it can have are A . ( 1 times 13 ) or ( 13 times 1 ) B. ( 1 times 26 ) or ( 26 times 1 ) c. ( 2 times 13 ) or ( 13 times 2 ) D. None of these | 12 |
| 383 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right] ; boldsymbol{B}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{d}end{array}right], ) then show that ( (A . B)^{-1}=B^{-1} . A^{-1} ) | 12 |
| 384 | 11. 13 -1 -2] Let P= 2 0 a [3 -5 o matrix such that PQ = kl, where k ER,k te a ER. Suppose Q=[q;] is a u wherek eRk+ 0 and I is the k2 and det(Q) = identity matrix of order 3. If 423 = -3 and then (JEE Adv. 2016) (a) a= 0, k=8 (b) 4a-k+8=0 © det (Padj (Q))=29 (d) det (Q adj (P))=2 | 12 |
| 385 | Find the inverse of ( boldsymbol{A}= ) ( left[begin{array}{ccc}cos alpha & -sin alpha & 0 \ sin alpha & cos alpha & 0 \ 0 & 0 & 1end{array}right] ) using elementary transformation. | 12 |
| 386 | For the matrix ( boldsymbol{A}=left[begin{array}{ccc}mathbf{4} & -mathbf{4} & mathbf{5} \ -mathbf{2} & mathbf{3} & -mathbf{3} \ mathbf{3} & -mathbf{3} & mathbf{4}end{array}right] ) | 12 |
| 387 | State true or false: The determinant of a skew-symmetric matrix is a perfect square if it’s elements are integers. A. True B. False | 12 |
| 388 | Find the output order for the following matrix multiplication ( boldsymbol{A}_{mathbf{4} times mathbf{2}} times boldsymbol{B}_{mathbf{2} times mathbf{4}} ) ? ( A cdot 2 times 4 ) B. ( 4 times 4 ) ( c cdot 4 times 2 ) D. Multiplication not possible | 12 |
| 389 | Let ( A ) be the ( 2 times 2 ) matrices given by ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) where ( boldsymbol{a}_{i j}={mathbf{0}, mathbf{1}, mathbf{2}, mathbf{3}, mathbf{4}} ) such that ( boldsymbol{a}_{11}+boldsymbol{a}_{12}+boldsymbol{a}_{21}+boldsymbol{a}_{22}=boldsymbol{4} ) Find the number of matrices ( A ) such that the trace of ( A ) is equal to 4 ( A cdot 3 ) B. 4 ( c cdot 5 ) D. 6 | 12 |
| 390 | What is the result when you add the ( operatorname{matrix}left[begin{array}{ll}mathbf{4} & mathbf{5}end{array}right] ) to the matrix ( left[begin{array}{ll}mathbf{7} & -mathbf{3}end{array}right] ) and multiply the result by ( 2 ? ) ( mathbf{A} cdotleft[begin{array}{ll}2 & 6end{array}right] ) в. ( left[begin{array}{ll}11 & 2end{array}right] ) c. ( left[begin{array}{ll}22 & 4end{array}right] ) D. ( [28 quad-15] ) E . ( left[begin{array}{ll}54 & -30end{array}right] ) | 12 |
| 391 | The value of ( x ) satisfying the equation 2 ( left|begin{array}{cc}mathbf{3} & mathbf{1} \ mathbf{1} & mathbf{2}end{array}right|+left|begin{array}{cc}boldsymbol{x}^{mathbf{2}} & mathbf{9} \ -mathbf{1} & mathbf{0}end{array}right|=left|begin{array}{cc}mathbf{5} boldsymbol{x} & mathbf{6} \ mathbf{0} & mathbf{1}end{array}right|+left|begin{array}{cc}mathbf{0} & mathbf{5} \ mathbf{1} & mathbf{3}end{array}right| ) are ( A cdot 1,2 ) B. 2,3 ( c .pm 2 ) ( mathrm{D} cdot pm 3 ) | 12 |
| 392 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{1} & mathbf{1} \ mathbf{0} & -mathbf{2} & mathbf{4}end{array}right], mathbf{6} boldsymbol{A}^{-mathbf{1}}=boldsymbol{A}^{mathbf{2}}+ ) ( c A+d I, ) then ( (c, d) ) is equal to A ( cdot(-6,11) ) в. (-11,6) c. (11,6) ( D ) | 12 |
| 393 | ( left[begin{array}{cc}mathbf{1} & mathbf{1} \ mathbf{0} & mathbf{1}end{array}right]left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{0} & mathbf{1}end{array}right]left[begin{array}{ll}mathbf{1} & mathbf{3} \ mathbf{0} & mathbf{1}end{array}right] cdotleft[begin{array}{cc}mathbf{1} & boldsymbol{n}-mathbf{1} \ mathbf{0} & mathbf{1}end{array}right]= ) ( left[begin{array}{cc}mathbf{1} & mathbf{7 8} \ mathbf{0} & mathbf{1}end{array}right] ) If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & boldsymbol{n} \ mathbf{0} & mathbf{1}end{array}right] operatorname{then} boldsymbol{A}^{-mathbf{1}}=? ) A ( cdotleft[begin{array}{cc}1 & 12 \ 0 & 1end{array}right] ) B ( cdotleft[begin{array}{cc}1 & -13 \ 0 & 1end{array}right] ) c. ( left[begin{array}{cc}1 & -12 \ 0 & 1end{array}right] ) D. ( left[begin{array}{cc}1 & 0 \ -13 & 1end{array}right] ) | 12 |
| 394 | Let ( mathbf{A} ) be a square matrix. Consider ( left.left.text { 1) }left.mathbf{A}+mathbf{A}^{mathbf{T}} text { 2 }right) mathbf{A} mathbf{A}^{mathbf{T}} mathbf{3}right) mathbf{A}^{mathbf{T}} mathbf{A} mathbf{4}right) mathbf{A}^{mathbf{T}}+mathbf{A} ) 5) ( left.mathbf{A}-mathbf{A}^{mathbf{T}} mathbf{6}right) mathbf{A}^{mathbf{T}}-mathbf{A}, ) Then A. all are symmetric matrices B. (2),(4), (6) are symmetric matrices c. (1),(2),(3),(4) are symmetric matrices & (5),(6) are skew symmetric matrices D. 5,6 are symmetric | 12 |
| 395 | Find ( : A^{2} ) if ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{3} & mathbf{2} \ -mathbf{5} & mathbf{1} & mathbf{0} \ mathbf{1} & mathbf{2} & mathbf{5}end{array}right] ) | 12 |
| 396 | If ( X, Y ) are two matrices given by the equations ( boldsymbol{X}+boldsymbol{Y}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], boldsymbol{X}-boldsymbol{Y}= ) ( left[begin{array}{cc}mathbf{3} & mathbf{2} \ -mathbf{1} & mathbf{0}end{array}right] ) ( operatorname{then} boldsymbol{X} boldsymbol{Y}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right], ) find ( boldsymbol{a}-boldsymbol{b}+boldsymbol{c}+boldsymbol{d} ) | 12 |
| 397 | IF ( A, B, C ) are non-singular ( n times n ) matrices, then ( (A B C)^{-1}= ) ( mathbf{A} cdot A^{-1} C^{-1} B^{-1} ) B. ( C^{-1} B^{-1} A^{-1} ) c. ( C^{-1} A^{-1} B^{-1} ) D. ( B^{-1} C^{-1} A^{-1} ) | 12 |
| 398 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{5} boldsymbol{a} & -boldsymbol{b} \ mathbf{3} & mathbf{2}end{array}right] ) and ( boldsymbol{A} ) adj ( boldsymbol{A}=boldsymbol{A} boldsymbol{A}^{boldsymbol{T}} ) then ( 5 a+b ) is equal to: A . 13 B. – c. 5 D. | 12 |
| 399 | If ( boldsymbol{A}=left[begin{array}{cc}2 & 3 \ mathbf{5} & -2end{array}right] ) be such that ( boldsymbol{A}^{-1}=boldsymbol{k} boldsymbol{A} ) then find the value of ( k ) | 12 |
| 400 | Write a ( 2 times 2 ) matrix which is both symmetric and skew symmetric. | 12 |
| 401 | If ( boldsymbol{A}=left[begin{array}{cc}2 & 0 \ 5 & -3end{array}right] ) and ( B=left[begin{array}{cc}-2 & 1 \ 3 & -1end{array}right] ) then find the trace of ( left(A B^{T}right)^{T} ) | 12 |
| 402 | Matrices obtained by changing rows and columns is called A. rectangular matrix B. transpose c. symetric D. None of the Above | 12 |
| 403 | A is a skew symmetric matrix such that ( A^{T} A=I, ) then ( A^{4 n-1}(n in N) ) is equal to A . ( -A^{T} ) B. ( I ) ( c .-I ) D. ( A^{T} ) | 12 |
| 404 | Without expanding, show that the value of the following determinant is zero: ( left|begin{array}{ccc}mathbf{0} & boldsymbol{x} & boldsymbol{y} \ -boldsymbol{x} & boldsymbol{0} & boldsymbol{z} \ -boldsymbol{y} & -boldsymbol{z} & boldsymbol{0}end{array}right| ) | 12 |
| 405 | Let ( A ) be a square matrix. Which of the following is/are not skew-symmetric matrix/ces? A . ( A-A^{T} ) B. ( A^{T}-A ) c. ( A A^{T}-A^{T} A ) D. ( A+A^{T} ), when A is skew-symmetric | 12 |
| 406 | ff ( left[begin{array}{cc}boldsymbol{alpha} & boldsymbol{beta} \ boldsymbol{gamma} & -boldsymbol{alpha}end{array}right] ) is square root of ( boldsymbol{I}_{2}, ) then ( boldsymbol{alpha} ) ( beta ) and ( gamma ) will satisfy the relation A. ( 1+alpha^{2}+beta gamma=0 ) В. ( 1-alpha^{2}+beta gamma=0 ) C ( cdot 1+alpha^{2}-beta gamma=0 ) D. ( -1+alpha^{2}+beta gamma=0 ) | 12 |
| 407 | f ( a_{i j}=0(i neq j) ) and ( a_{i j}=1(i=j) ) then the matrix ( A=left[a_{i j}right]_{n times n} ) is a matrix A. Null B. Identity c. Scalar D. Triangular | 12 |
| 408 | ( mathbf{f}left[begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{5} & mathbf{7}end{array}right]left[begin{array}{cc}mathbf{1} & mathbf{- 3} \ -mathbf{2} & mathbf{4}end{array}right]=left[begin{array}{ll}-mathbf{4} & mathbf{6} \ -mathbf{9} & boldsymbol{x}end{array}right], ) write the value of ( x ) | 12 |
| 409 | ( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2}-boldsymbol{b} boldsymbol{c} \ mathbf{1} & boldsymbol{b} & boldsymbol{b}^{2}-boldsymbol{c} boldsymbol{a} \ mathbf{1} & boldsymbol{c} & boldsymbol{c}^{2}-boldsymbol{a} boldsymbol{b}end{array}right|=? ) ( mathbf{A} cdot mathbf{5} ) B. abc ( c cdot 1 ) ( D ) | 12 |
| 410 | Solve: ( left[begin{array}{ccc}mathbf{5} & mathbf{1 0} & mathbf{8} \ mathbf{3} & mathbf{2} & mathbf{6} \ mathbf{8} & boldsymbol{x}+mathbf{9} & mathbf{9}end{array}right]=mathbf{1 0 0} ) | 12 |
| 411 | If ( A=left[begin{array}{ccc}1 & 2 & 3 \ 2 & -3 & 0end{array}right] ) and ( B=left[begin{array}{ccc}3 & 4 & -2 \ 1 & 0 & 0end{array}right] ) then the order of ( A B^{T} ) is ( A cdot 2 times 3 ) B. ( 3 times 3 ) ( c cdot 3 times 2 ) D. ( 2 times 2 ) | 12 |
| 412 | Let ( A ) and ( B ) are two matrices of same order ( 3 times 3 ) given by ( A= ) ( left[begin{array}{ccc}1 & 3 & lambda+2 \ 2 & 4 & 6 \ 3 & 5 & 8end{array}right] B=left[begin{array}{lll}3 & 2 & 4 \ 3 & 2 & 5 \ 2 & 1 & 4end{array}right] ) If ( t r(A B)^{t}+t r(B A)^{t}=t r(A B) ) then the value of ( 2 lambda ) equals A . 103 в. 206 ( c .-103 ) D. -206 | 12 |
| 413 | Find inverse of the matrix ( left[begin{array}{ccc}1 & -1 & 1 \ 0 & 1 & 1 \ 3 & 2 & -4end{array}right] ) by elementary transformation. | 12 |
| 414 | If ( mathbf{5} boldsymbol{A}=left[begin{array}{cc}mathbf{3} & -mathbf{4} \ mathbf{4} & boldsymbol{x}end{array}right] ) and ( boldsymbol{A} boldsymbol{A}^{boldsymbol{T}}=boldsymbol{A}^{boldsymbol{T}} boldsymbol{A}=boldsymbol{I} ) then ( x=? ) ( mathbf{A} cdot mathbf{3} ) B. -3 c. 2 D. -2 | 12 |
| 415 | For ( boldsymbol{alpha}, boldsymbol{beta}, boldsymbol{gamma} in boldsymbol{R}, ) let ( boldsymbol{A}=left[begin{array}{ccc}boldsymbol{alpha}^{2} & boldsymbol{6} & boldsymbol{8} \ boldsymbol{3} & boldsymbol{beta}^{2} & boldsymbol{9} \ boldsymbol{4} & boldsymbol{5} & boldsymbol{gamma}^{2}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{ccc}mathbf{2} boldsymbol{alpha} & boldsymbol{3} & boldsymbol{5} \ boldsymbol{2} & boldsymbol{2} boldsymbol{beta} & boldsymbol{6} \ boldsymbol{1} & boldsymbol{4} & boldsymbol{2} boldsymbol{gamma}-boldsymbol{3}end{array}right] ) ( boldsymbol{T}_{boldsymbol{r}}(boldsymbol{A})=boldsymbol{T}_{boldsymbol{r}}(boldsymbol{B}) ) then the value of ( left(frac{1}{alpha}+frac{1}{beta}+frac{1}{gamma}right) ) is ( boldsymbol{T}_{r}(boldsymbol{A}) ) is a Trace ( (boldsymbol{A}) ) of a matrix ( A ) B. 2 ( c .3 ) D. | 12 |
| 416 | The number of ( A ) in ( T_{p} ) such that ( operatorname{det}(A) ) is not divisible by p is? ( A cdot 2 p^{2} ) B . ( p^{3}-5 p ) c. ( p^{3}-3 p ) D. ( p^{3}-p^{2} ) | 12 |
| 417 | Construct a ( 2 times 3 ) matrix ( A=left[a_{i j}right] ) whose elements are given by ( boldsymbol{a}_{boldsymbol{i} j}= ) ( mathbf{2}(boldsymbol{i}-boldsymbol{j}) ) В. ( left[begin{array}{ccc}0 & -2 & 4 \ 2 & 0 & -2end{array}right] ) ( begin{array}{lll}text { c. } & {left[begin{array}{ccc}0 & -2 & -4 \ 2 & 0 & 2end{array}right]} \ & text { I }end{array} ) D. ( left[begin{array}{ccc}0 & -2 & 4 \ 2 & 0 & 2end{array}right] ) | 12 |
| 418 | If ( boldsymbol{A}=left(begin{array}{ccc}1 & -1 & 3 & 2 \ 5 & -4 & 7 & 4 \ 6 & 0 & 9 & 8end{array}right), ) Write down the elements ( a_{24} ) and ( a_{32} ) | 12 |
| 419 | If ( A ) is an ( m times n ) matrix such that ( A B ) and ( B A ) are both defined, then order of ( B ) is ( mathbf{A} cdot m times n ) в. ( n times m ) c. ( n times n ) D. ( m times m ) | 12 |
| 420 | If ( boldsymbol{A}=left(begin{array}{ccc}1 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3end{array}right) ) and ( A^{3}-6 A^{2}+ ) ( 7 A+k I^{3}=0, ) find ( k ) | 12 |
| 421 | Assertion ( operatorname{Let} boldsymbol{A}=left[begin{array}{ll}boldsymbol{a}_{11} & boldsymbol{a}_{12} \ boldsymbol{a}_{21} & boldsymbol{a}_{22}end{array}right], boldsymbol{X}=left[begin{array}{l}boldsymbol{x}_{1} \ boldsymbol{x}_{2}end{array}right], boldsymbol{y}= ) ( left[begin{array}{l}boldsymbol{y}_{1} \ boldsymbol{y}_{2}end{array}right. ) If ( A ) is symmetric, then ( X^{prime} A Y=Y^{prime} A X ) for each pair of ( X ) and ( Y ) Reason If ( boldsymbol{X}^{prime} boldsymbol{A} boldsymbol{Y}=boldsymbol{Y}^{prime} boldsymbol{A} boldsymbol{X} ) for each pair of ( boldsymbol{X} ) and ( Y, ) then ( A ) is symmetric. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 422 | Let ( p ) be a nonsingular matrix, and ( I+ ) ( boldsymbol{p}+boldsymbol{p}^{2}+ldots . .+boldsymbol{p}^{n}=mathbf{0}, ) then find ( boldsymbol{p}^{-1} ) A . ( I ) B . ( p^{n+1} ) c. ( p^{n} ) D ( cdotleft(p^{n+1}-Iright)(p-I) ) | 12 |
| 423 | For matrix ( mathbf{A} ) ( (boldsymbol{alpha}+boldsymbol{beta}) boldsymbol{A}= ) A ( . alpha A ) B. ( alpha A+beta B ) ( mathbf{c} cdot alpha A+beta A ) D. ( alpha^{2} A+beta^{2} A ) | 12 |
| 424 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{3} & mathbf{7} \ mathbf{2} & mathbf{5}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}-mathbf{3} & mathbf{2} \ mathbf{4} & -mathbf{1}end{array}right] ) find the ( boldsymbol{A}+boldsymbol{B} ) | 12 |
| 425 | ( mathbf{f} boldsymbol{A}=left[begin{array}{lll}mathbf{0} & mathbf{1} & mathbf{2} \ mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{3} & boldsymbol{a} & mathbf{1}end{array}right] ) and ( boldsymbol{A}^{-mathbf{1}}= ) ( left[begin{array}{ccc}1 / 2 & -1 / 2 & 1 / 2 \ -4 & 3 & c \ 5 / 2 & -3 / 2 & 1 / 2end{array}right] ) then A ( a=2, c=1 / 2 ) 2 В. ( a=1, c=-1 ) c. ( a=-1, c=1 ) D. ( a=1 / 2, c=1 / 2 ) | 12 |
| 426 | If the matrix ( A ) is both symmetric and skew symmetric, then A. ( A ) is a diagonal matrix B. ( A ) is a zero matrix c. ( A ) is a square matrix D. None of these | 12 |
| 427 | 10. Let X and Y be two arbitrary, 3 x 3, non-zero, ric matrices and Z be an arbitrary bitrary, 3 x 3, non-zero, skew-symmet- ces and Z be an arbitrary 3 x 3, non zero, symmetric matrix. Then which of the following matrices is (ar) symmetric? (JEE Adv. 2015) (a) Y24-243 (b) X44+ y44 (C) X473 -2284 (d) X23 + y23 | 12 |
| 428 | A square matrix ( left(a_{i j}right) ) in which ( a_{i j}=0 ) for ( i neq j ) and ( a_{i j}=k(text { constant }) ) for ( i= ) ( j ) is a A. Unit matrix B. Scalar matrix c. Null matrix D. none | 12 |
| 429 | If ( boldsymbol{A}=left[begin{array}{rr}4 & boldsymbol{x}+mathbf{2} \ mathbf{2} boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{1}end{array}right] ) is symmetric then ( x= ) ( A cdot 3 ) B. 5 ( c cdot 2 ) D. 4 | 12 |
| 430 | f ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right] boldsymbol{B}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{5}end{array}right] ) then ( boldsymbol{A}+boldsymbol{B} ) | 12 |
| 431 | Simplify: ( operatorname{os} theta cdotleft[begin{array}{cc}cos theta & sin theta \ -sin theta & cos thetaend{array}right]+ ) ( operatorname{in} theta cdotleft[begin{array}{cc}sin theta & -cos theta \ cos theta & sin thetaend{array}right] ) | 12 |
| 432 | Compute ( left[begin{array}{cc}cos ^{2} x & sin ^{2} x \ sin ^{2} x & cos ^{2} xend{array}right]+ ) ( left[begin{array}{cc}sin ^{2} x & cos ^{2} x \ cos ^{2} x & sin ^{2} xend{array}right] ) | 12 |
| 433 | If ( A ) is a square matrix of order 5 and ( left.9 A^{-1}=4 A^{T} text { then ladj (adj }(operatorname{adj} A)right) ) (where ( A^{-T} ) and adj ( (A) ) denotes the inverse, transpose and adjoint of matrix A respectively) contains: ( (log 3= ) ( mathbf{0 . 4 7 7}, log mathbf{2}=mathbf{0 . 3 0 3}) ) A. 56 digits B. 60 digits c. 58 digits D. 53 digits | 12 |
| 434 | ff ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & mathbf{- 1} \ mathbf{- 1} & mathbf{2}end{array}right] ) Find ( boldsymbol{A}^{2} ) | 12 |
| 435 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & mathbf{3} \ mathbf{1} & -mathbf{4}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}mathbf{1} & -mathbf{2} \ -mathbf{1} & mathbf{3}end{array}right] ) then verify that ( (A B)^{-1}=B^{-1} A^{-1} ) | 12 |
| 436 | If ( A^{-1}=left[begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2end{array}right] ) and ( B= ) ( left[begin{array}{ccc}1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1end{array}right] ), then ( (A B)^{-1}=? ) | 12 |
| 437 | Find the inverse of the following ( operatorname{matrices} boldsymbol{A}=left(begin{array}{ccc}mathbf{2} & mathbf{1} & mathbf{3} \ mathbf{5} & mathbf{3} & mathbf{1} \ mathbf{3} & mathbf{2} & mathbf{3}end{array}right) ) | 12 |
| 438 | If ( boldsymbol{A}=left(begin{array}{ll}2 & 2 \ 9 & 4end{array}right) ; I=left(begin{array}{ll}1 & 0 \ 0 & 1end{array}right), ) then ( 10 A^{-1} ) is equal to A ( .4 I-A ) B. ( 6 I-A ) c. ( A-4 I ) D. ( A-6 I ) | 12 |
| 439 | Identify the matrix given below: ( left[begin{array}{lll}1 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 2end{array}right] ) | 12 |
| 440 | ( A=left[begin{array}{lll}1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1end{array}right], U_{1}, U_{2} ) and ( U_{3} ) are columns matrices satisfying ( boldsymbol{A} boldsymbol{U}_{1}= ) ( left[begin{array}{l}1 \ 0 \ 0end{array}right], A U_{2}=left[begin{array}{l}2 \ 3 \ 0end{array}right], A U_{3}=left[begin{array}{l}2 \ 3 \ 1end{array}right] ) and ( U ) is ( 3 times 3 ) matrix whose columns are ( U_{1}, U_{2}, U_{3} ) then answer the following question The value of ( left[begin{array}{lll}mathbf{3} & mathbf{2} & mathbf{0}end{array}right] boldsymbol{U}left[begin{array}{l}mathbf{3} \ mathbf{2} \ mathbf{0}end{array}right] ) is ( A cdot[5] ) в. ( left[frac{5}{2}right] ) ( c cdot[4] ) D. ( left[frac{3}{2}right. ) | 12 |
| 441 | Identify the matrix given below: ( left[begin{array}{lll}1 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 2end{array}right] ) A. unit matrix B. scalar matrix c. zero matrix D. diagonal matrix | 12 |
| 442 | covert it into an upper triangular matrix Convert ( left[begin{array}{l}1-1 \ 23end{array}right] ) into an identity matrix by suitable row transformations. | 12 |
| 443 | If ( boldsymbol{A}=left[begin{array}{ccc}1 & -2 & 3 \ -4 & 2 & 5end{array}right] ) and ( B= ) ( left[begin{array}{cc}1 & 3 \ -1 & 0 \ 2 & 4end{array}right], ) then show that ( (A B)^{prime}= ) ( boldsymbol{B}^{prime} boldsymbol{A}^{prime} ) | 12 |
| 444 | If ( A ) and ( B ) are skew symmetric matrices of same order then A. ( A B ) is skew symmetric B. ( A B+B A ) is symmetic C. ( A B-B A ) is symmetric D. none of these | 12 |
| 445 | If the matrix is a square matrix and it contains 36 elements, then the order of the matrix is: ( mathbf{A} cdot 4 times 4 ) B. ( 8 times 8 ) ( mathbf{c} cdot 6 times 6 ) D. ( 3 times 3 ) | 12 |
| 446 | Find ( boldsymbol{x}+boldsymbol{y} ) if ( left[begin{array}{cc}-2 & 0 \ 3 & 1end{array}right]left[begin{array}{l}-1 \ 2 xend{array}right]+ ) ( left[begin{array}{c}-2 \ 1end{array}right]=2left[begin{array}{l}y \ 3end{array}right] ) | 12 |
| 447 | ( fleft(begin{array}{cc}a & b \ c & -aend{array}right) ) such that ( A^{2}-I ) then | 12 |
| 448 | fthe matrix ( A ) is such that ( left[begin{array}{ll}1 & 3 \ 0 & 1end{array}right] A= ) ( left[begin{array}{cc}mathbf{1} & mathbf{1} \ mathbf{0} & -mathbf{1}end{array}right], ) then what is equal to ( mathbf{A} ? ) A. ( left[begin{array}{cc}1 & 4 \ 0 & -1end{array}right] ) в. ( left[begin{array}{ll}1 & 4 \ 0 & 1end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-1 & 4 \ 0 & -1end{array}right] ) D. ( left[begin{array}{cc}1 & -4 \ 0 & -1end{array}right] ) | 12 |
| 449 | Compute the following: (i) ( left[begin{array}{cc}boldsymbol{a} & boldsymbol{b} \ -boldsymbol{b} & boldsymbol{a}end{array}right]+left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{a}end{array}right] ) (ii) ( left[begin{array}{ll}boldsymbol{a}^{2}+boldsymbol{b}^{2} & boldsymbol{b}^{2}+boldsymbol{c}^{2} \ boldsymbol{a}^{2}+boldsymbol{c}^{2} & boldsymbol{a}^{2}+boldsymbol{b}^{2}end{array}right]+ ) ( left[begin{array}{cc}2 a b & 2 b c \ -2 a c & -2 a bend{array}right] ) (iii) ( left[begin{array}{cc}-14-6 \ 8516 \ 285end{array}right]+left[begin{array}{c}1276 \ 805 \ 324end{array}right] ) ( (operatorname{iv})left[begin{array}{cc}cos ^{2} x & sin ^{2} x \ sin ^{2} x & cos ^{2} xend{array}right]+left[begin{array}{cc}sin ^{2} x & cos ^{2} x \ cos ^{2} x & sin ^{2} xend{array}right] ) | 12 |
| 450 | ( mathbf{I f A}=left{begin{array}{ccc}mathbf{2} & boldsymbol{x}-mathbf{3} & boldsymbol{x}-mathbf{2} \ mathbf{3} & mathbf{- 2} & mathbf{- 1} \ mathbf{4} & mathbf{- 1} & mathbf{- 5}end{array}right} ) is a symmetric matrix then ( A cdot 0 ) B. 3 ( c .6 ) ( D ) | 12 |
| 451 | ff ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right] ) such that ( boldsymbol{A} ) satisfies the relation ( boldsymbol{A}^{2}-(boldsymbol{a}+boldsymbol{d}) boldsymbol{A}=mathbf{0}, ) then inverse of ( boldsymbol{A} ) is A . ( I ) в. c. ( (a+d) A ) D. none of these | 12 |
| 452 | fthe matrix ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) then ( boldsymbol{I}+boldsymbol{A}+ ) ( A^{2}+ldots ldots ldots . ) upto ( A^{infty}=ldots ) ( mathbf{A} cdotleft[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) В. ( left[begin{array}{cc}frac{1}{2} & frac{1}{3} \ frac{1}{-2} & 0end{array}right] ) C. ( left[begin{array}{ll}frac{-1}{2} & frac{1}{-3} \ frac{1}{-2} & 0end{array}right] ) D. ( left[begin{array}{cc}frac{1}{2} & frac{1}{-3} \ frac{1}{-2} & 0end{array}right] ) | 12 |
| 453 | If ( boldsymbol{x}left[begin{array}{l}2 \ mathbf{3}end{array}right]+boldsymbol{y}left[begin{array}{c}-mathbf{1} \ mathbf{1}end{array}right]=left[begin{array}{c}mathbf{1 0} \ mathbf{5}end{array}right] ) Find values of ( x ) and ( y ) | 12 |
| 454 | ( mathbf{f} mathbf{Delta}=left|begin{array}{ccc}mathbf{1} & mathbf{5} & mathbf{6} \ mathbf{0} & mathbf{1} & mathbf{7} \ mathbf{0} & mathbf{0} & mathbf{1}end{array}right| ) and ( Delta^{prime}=left|begin{array}{ccc}mathbf{1} & mathbf{0} & mathbf{1} \ mathbf{3} & mathbf{0} & mathbf{3} \ mathbf{4} & mathbf{6} & mathbf{1 0 0}end{array}right| ) then A ( cdot Delta^{2}-3 Delta^{prime}=0 ) B. ( left(Delta+Delta^{prime}right)^{2}-3left(Delta+Delta^{prime}right)+2=0 ) c. ( left(Delta+Delta^{prime}right)^{2}+3left(Delta+Delta^{prime}right)+5=0 ) D. ( Delta+3 Delta^{prime}+1=0 ) | 12 |
| 455 | If ( boldsymbol{A} ) is a ( mathbf{3} times mathbf{3} ) matrix ( |mathbf{3} boldsymbol{A}|=boldsymbol{k}|boldsymbol{A}| ), then write the value of ( k ) | 12 |
| 456 | Trace of ( A^{50} ) equals ( mathbf{A} cdot mathbf{0} ) B. ( c cdot 2 ) D. 3 | 12 |
| 457 | Define a scalar matrix. | 12 |
| 458 | Find the matrix ( boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{1} & mathbf{0} \ mathbf{1} & mathbf{2} & mathbf{1} \ mathbf{2} & mathbf{1} & mathbf{0}end{array}right], ) which of the following is correct A. ( A^{3}+3 A^{2}-I=0 ) B. ( A^{3}-3 A^{2}-I=0 ) c. ( A^{3}+2 A^{2}-I=0 ) D. ( A^{3}-A^{2}+I=0 ) | 12 |
| 459 | A matrix ( A=left(A_{i j}right)_{m times n} ) is said to be a square matrix if ( mathbf{A} cdot m=n ) в. ( m leq n ) c. ( m geq n ) D. ( m<n ) | 12 |
| 460 | If ( boldsymbol{A}=left[begin{array}{ccc}1 & -2 & 3 \ -4 & 2 & 5end{array}right] ) and ( B= ) ( left[begin{array}{cc}1 & 3 \ -1 & 0 \ 2 & 4end{array}right] . ) Show that ( (A B)^{prime}=? ) A ( cdot B^{prime} A^{prime} ) в. ( A^{prime} B^{prime} ) c. ( A B^{prime} ) D. ( A^{prime} B ) | 12 |
| 461 | If ( A ) and ( B ) are square matrices of order ( n ) ( x ) n such that ( A^{2}-B^{2}= ) ( (A-B)(A+B), ) then of the following will always be true? ( A cdot A=B ) c. either of A or B is a zero matrix D. either of A or B is an identify matrix | 12 |
| 462 | If ( A ) is square matrix such that ( boldsymbol{A}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A})=left(begin{array}{ccc}mathbf{4} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{4} & mathbf{0} \ mathbf{0} & mathbf{0} & mathbf{4}end{array}right) ) then det ( (mathbf{A} mathrm{d} ) ( A)= ) ( A cdot 4 ) B. 16 ( c cdot 64 ) D. 256 | 12 |
| 463 | ( $ $ text { lbegin{array }}{mid} A=mid ) left [ lbegin{array ( }[1332-4115 & 2 ) ( mathrm{~ l e n d { a r r a y } ~ | r i g h t ] ~ | , ~ | , ~ B = | l e f t [ ~} ) [ begin{array}{cc} 1 & -3 \ 5 & 8 end{array} ] |right] ( |=3 ) left Then find ( 3 A-5 B+ ) ( 4 I ) by using matrix | 12 |
| 464 | Find ( X, ) if ( Y=left[begin{array}{ll}3 & 2 \ 1 & 4end{array}right] ) and ( 2 X+Y= ) ( left[begin{array}{cc}1 & 0 \ -3 & 2end{array}right] ) | 12 |
| 465 | ff ( boldsymbol{A}=left(begin{array}{cc}mathbf{3} & mathbf{1} \ -mathbf{9} & -mathbf{3}end{array}right) ) then ( left(1+2 A+3 A^{2}+ldots . inftyright)^{-1} ) equals ( ^{A} cdotleft(begin{array}{cc}-5 & -2 \ 18 & 7end{array}right) ) B. ( left(begin{array}{rr}-5 & 18 \ -2 & 7end{array}right) ) c. ( left(begin{array}{cc}7 & -2 \ 18 & -5end{array}right) ) D. None of these | 12 |
| 466 | If ( A ) is skew-symmetric, then ( A^{n} ) for ( boldsymbol{n} in boldsymbol{N} ) is This question has multiple correct options A. Symmetric B. Skew-symmetric c. Diagonal D. None of these | 12 |
| 467 | Construct a ( 3 times 2 ) matrix ( A=left[a_{i j}right] ) whose elements are given by ( a_{i j}=frac{i}{j} ) | 12 |
| 468 | ( boldsymbol{A}=left[begin{array}{rrr}mathbf{1} & mathbf{- 2} & mathbf{3} \ mathbf{7} & -mathbf{8} & mathbf{9} \ mathbf{4} & mathbf{- 5} & mathbf{6}end{array}right] ) the new matrix formed by adding ( 2^{n d} ) row to ( 1^{s t} ) row will be A. ( left[begin{array}{ccc}8 & -10 & 12 \ 7 & -8 & 9 \ 4 & -5 & 6end{array}right] ) В. ( left[begin{array}{lll}6 & 6 & 6 \ 7 & 8 & 9 \ 4 & 5 & 6end{array}right] ) c. ( left[begin{array}{ccc}1 & 2 & 3 \ 7 & 8 & 9 \ 11 & -13 & 14end{array}right] ) D. ( left[begin{array}{ccc}1 & -2 & 3 \ 7 & 8 & -29 \ 4 & -2 & 6end{array}right] ) | 12 |
| 469 | By row transformation find ( boldsymbol{A}^{-1} ) if: ( boldsymbol{A}=left[begin{array}{ll}mathbf{2} & mathbf{1} \ mathbf{4} & mathbf{2}end{array}right] ) | 12 |
| 470 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{- 3} \ mathbf{5} & mathbf{0} & mathbf{2} \ mathbf{1} & mathbf{- 1} & mathbf{1}end{array}right], boldsymbol{B}= ) ( left[begin{array}{ccc}mathbf{3} & mathbf{- 1} & mathbf{2} \ mathbf{4} & mathbf{2} & mathbf{5} \ mathbf{2} & mathbf{0} & mathbf{3}end{array}right], boldsymbol{C}=left[begin{array}{ccc}mathbf{4} & mathbf{1} & mathbf{2} \ mathbf{0} & mathbf{3} & mathbf{2} \ mathbf{1} & mathbf{- 2} & mathbf{3}end{array}right], ) Then Compute ( (boldsymbol{A}+boldsymbol{B}) ) and ( (boldsymbol{B}-boldsymbol{C}) . ) Also verify that ( boldsymbol{A}+(boldsymbol{B}-boldsymbol{C})=(boldsymbol{A}+boldsymbol{B})- ) ( C ) | 12 |
| 471 | If ( A B=0, ) then for the matrices ( A= ) ( left[begin{array}{cc}cos ^{2} theta & cos theta sin theta \ cos theta sin theta & sin ^{2} thetaend{array}right] ) and ( B= ) ( left[begin{array}{cc}cos ^{2} phi & cos phi sin phi \ cos phi sin phi & sin ^{2} phiend{array}right], theta-phi ) is A ( cdot ) an odd muliple of ( frac{pi}{2} ) B. an odd multiple of ( pi ) C . an even multiple of ( frac{pi}{2} ) D. | 12 |
| 472 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose element ( a_{i j} ) is ( a_{i j}=frac{(i-2 j)^{2}}{2} ) | 12 |
| 473 | Find the matrices ( A ) and ( B ) such that ( boldsymbol{A}+boldsymbol{B}=left[begin{array}{ll}mathbf{5} & mathbf{4} \ mathbf{7} & mathbf{3}end{array}right] ) and ( boldsymbol{A}-boldsymbol{B}= ) ( left[begin{array}{cc}11 & 2 \ -1 & 7end{array}right] ) | 12 |
| 474 | If ( A ) is a ( 2 times 3 ) matrix and ( B ) is ( 3 times 2 ) matrix then the order of ( (A B)^{T} ) is equal to the order of ( mathbf{A} cdot A B ) в. ( A^{T} B^{T} ) ( c . ) ВА D. All of these | 12 |
| 475 | ff ( boldsymbol{A}=left[begin{array}{cc}boldsymbol{6} & boldsymbol{2} \ boldsymbol{5} & boldsymbol{-} boldsymbol{4}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}mathbf{1} & boldsymbol{2} \ -boldsymbol{5} & boldsymbol{1}end{array}right], ) find a matrix ( X ) such that ( 2 A+3 B- ) ( mathbf{5} boldsymbol{X}=mathbf{0} ) | 12 |
| 476 | ( operatorname{Let} A=left[begin{array}{lll}1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 1 & 1end{array}right] ) and ( B=A^{10} ) Then the sum of elements of the first column of ( boldsymbol{B} ) is | 12 |
| 477 | 3. Let M and N be two 3 x 3 non-singular skew- symmetric matrices such that MN = NM. If PT denotes the transpose of P, then M²N2 (MTN)-1 (MN-1)T is equal to (2011) (a) M2 (b) N2 (C) – M2 (d) MN | 12 |
| 478 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & -mathbf{3} \ mathbf{4} & mathbf{1}end{array}right], boldsymbol{B}=left[begin{array}{cc}mathbf{2} & mathbf{3} \ mathbf{5} & mathbf{0}end{array}right] ) and ( boldsymbol{C}=left[begin{array}{cc}-mathbf{1} & mathbf{2} \ mathbf{0} & mathbf{5}end{array}right], ) then find ( boldsymbol{A}(boldsymbol{B}+boldsymbol{C}) ) | 12 |
| 479 | What is meant by transposing of a matrix? Give an example. | 12 |
| 480 | Solve for ( x ) and ( boldsymbol{y} ) ( mathbf{2}left[begin{array}{cc}boldsymbol{x} & mathbf{7} \ mathbf{9} & boldsymbol{y}-mathbf{5}end{array}right]+left[begin{array}{cc}mathbf{6} & -mathbf{7} \ mathbf{4} & mathbf{5}end{array}right]=left[begin{array}{cc}mathbf{1 0} & mathbf{7} \ mathbf{2 2} & mathbf{1 5}end{array}right] ) | 12 |
| 481 | If ( boldsymbol{A}^{T}=left[begin{array}{cc}mathbf{4} & mathbf{5} \ -mathbf{1} & mathbf{0} \ mathbf{2} & mathbf{3}end{array}right] ) and ( boldsymbol{B}= ) ( left[begin{array}{ccc}2 & -1 & 1 \ 7 & 5 & -2end{array}right], ) verify the following ( mathbf{A} cdot(A+B)^{T}=A^{T}+B^{T}=B^{T}+A^{T} ) В ( cdot(A+B)^{T}=A^{T}-B^{T} ) c. ( left(B^{T}right)^{T}=B ) D. none of these | 12 |
| 482 | ff ( boldsymbol{A}=left[begin{array}{cc}cos boldsymbol{x} & sin boldsymbol{x} \ -sin boldsymbol{x} & cos boldsymbol{x}end{array}right], ) then find ( boldsymbol{x} ) satisfying ( mathbf{0}<boldsymbol{x}<frac{boldsymbol{pi}}{mathbf{2}} ) when ( boldsymbol{A}+boldsymbol{A}^{boldsymbol{T}}=boldsymbol{I} ) | 12 |
| 483 | ( mathrm{IF} mathrm{A}=left|begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{1} & mathbf{0}end{array}right| ) And ( mathrm{B}=left|begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right| ) then ( mathbf{A}+mathbf{B}= ) ( A cdot A ) B. B c. ( mid begin{array}{ll}2 & 0 \ 1 & 1end{array} ) D. ( left|begin{array}{ll}0 & 2 \ 2 & 2end{array}right| ) | 12 |
| 484 | If ( boldsymbol{A}=left[begin{array}{ccc}1 & 1 & -1 \ 2 & -3 & 4 \ 3 & -2 & 3end{array}right] ) and ( B= ) ( left[begin{array}{ccc}-1 & -2 & -1 \ 6 & 12 & 6 \ 5 & 10 & 5end{array}right], ) then which of the following is/are correct? 1. ( A ) and ( B ) commute. 2. AB is null matrix. Select the correct answer using the code given below: A. 1 only B. 2 only c. Both 1 and 2 D. Neither 1 nor 2 | 12 |
| 485 | If ( A ) is a matrix of order ( 3 times 4 ), then both ( A B^{T} ) and ( B^{T} A ) are defined if order of ( B ) is ( A cdot 3 times 3 ) B. ( 4 times 4 ) c. ( 4 times 3 ) D. ( 3 times 4 ) | 12 |
| 486 | ( mathrm{f} A=left[begin{array}{lll}1 & 3 & 3 \ 1 & 4 & 3 \ 1 & 3 & 4end{array}right] ) then find ( A^{-1} ) | 12 |
| 487 | Using elementary tansormations, find the inverse of each of the matrices, if it exists in ( left[begin{array}{ll}2 & 1 \ 7 & 4end{array}right] ) | 12 |
| 488 | Let ( A ) be a matrix of order ( 3 times 4 . ) If ( R_{1} ) denotes the first row of ( A ) and ( C_{2} ) denotes its second column, then determine the orders of matrices ( boldsymbol{R}_{1} ) and ( C_{2} ) | 12 |
| 489 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{3} & mathbf{1} \ -mathbf{1} & mathbf{2} \ mathbf{0} & mathbf{6}end{array}right] ) and ( boldsymbol{B}= ) ( left[begin{array}{ccc}5 & 4 & 6 \ 4 & 1 & 2 \ -5 & -1 & 1end{array}right], ) then A. ( A+B ) exists B. ( A B ) exists c. ( B A ) exists D. none of these | 12 |
| 490 | If ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 2} & mathbf{3} \ -mathbf{4} & mathbf{2} & mathbf{5}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5} \ mathbf{2} & mathbf{1}end{array}right] ) Check commutativity of the two matrices | 12 |
| 491 | not the square of a 3 x 3 (JEE Adv. 2017) a=-3. 13. Which of the following is(are) not the squ matrix with real entries? (1 o o7 To ol (a) Tolol Tolo Lo 0 1 (b) Lo 0 -1] 1007 Hool (c) 0 -1 0 (d) 1 0 -1 0 Lo o 1 To 0 -1 17 | 12 |
| 492 | If ( boldsymbol{a}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{4} & mathbf{2}end{array}right], ) than show that ( |mathbf{2} boldsymbol{A}|= ) ( mathbf{4}|boldsymbol{A}| ) | 12 |
| 493 | Convert ( left[begin{array}{cc}1 & -1 \ 2 & 3end{array}right] ) into on identity matrix by suitable raw transformation. | 12 |
| 494 | Let ( A ) be a symmetric matrix such that ( boldsymbol{A}^{5}=mathbf{0} ) and ( boldsymbol{B}=boldsymbol{I}+boldsymbol{A}+boldsymbol{A}^{2}+boldsymbol{A}^{3}+boldsymbol{A}^{4} ) then ( B ) is This question has multiple correct options A. symmetric B. singular c. non-singular D. skew symmetric | 12 |
| 495 | Matrix ( boldsymbol{A}=left[boldsymbol{a}_{i j}right]_{m times n} ) is a square matrix if ( A cdot mn ) ( c cdot m=1 ) ( D cdot m=n ) | 12 |
| 496 | Using elementary transformation, find the inverse of the matrix ( left[begin{array}{ccc}2 & -3 & 3 \ 2 & 2 & 3 \ 3 & -2 & 2end{array}right] ) | 12 |
| 497 | Let ( A=left(begin{array}{ll}1 & 2 \ 3 & 4end{array}right) ) and ( B= ) ( left(begin{array}{ll}a & 0 \ 0 & bend{array}right), a, b in N . ) Then: A. there exists exactly one B such that ( A B=B A ) B. there exist exactly infinitely many B’s such that ( A B= ) ( B A ) C. there cannot exist any B such that ( A B=B A ) D. there exist more than one but finite number of B’s such that ( A B=B A ) | 12 |
| 498 | If ( A=left|begin{array}{cc}0 & 1 \ 2 & 4end{array}right|, B=left|begin{array}{cc}-1 & 1 \ 2 & 2end{array}right| ) ( c=left|begin{array}{cc}1 & 0 \ 1 & 0end{array}right|, ) then ( 2 A+3 B-C= ) A ( cdotleft|begin{array}{cc}-4 & 5 \ 9 & 14end{array}right| ) в. ( left|begin{array}{cc}4 & 3 \ 9 & 10end{array}right| ) с. ( left|begin{array}{cc}4 & -5 \ 9 & 14end{array}right| ) О ( cdotleft|begin{array}{cc}-4 & 5 \ 14 & 9end{array}right| ) | 12 |
| 499 | The order the matrix is ( left[begin{array}{lll}2 & 3 & 4 \ 9 & 8 & 7end{array}right] ) is ( mathbf{A} cdot 4 times 3 ) B. ( 3 times 2 ) ( c cdot 2 times 3 ) D. ( 3 times 1 ) | 12 |
| 500 | ( left[begin{array}{lll}mathbf{1} & mathbf{1} & boldsymbol{x}end{array}right]left[begin{array}{lll}mathbf{1} & mathbf{0} & mathbf{2} \ mathbf{0} & mathbf{2} & mathbf{1} \ mathbf{2} & mathbf{1} & mathbf{0}end{array}right]left[begin{array}{l}mathbf{1} \ mathbf{1} \ mathbf{1}end{array}right]=mathbf{0}, ) then find ( x ) | 12 |
| 501 | Inverse of a diagonal non-singular matrix is A. Scalar matrix B. Skew symmetric matrix c. zero matrix D. Diagonal matrix | 12 |
| 502 | If ( boldsymbol{m}left[begin{array}{ll}-mathbf{3} & mathbf{4}end{array}right]+boldsymbol{n}left[begin{array}{ll}mathbf{4} & -mathbf{3}end{array}right]=left[begin{array}{ll}mathbf{1 0} & -mathbf{1 1}end{array}right] ) then ( 3 m+7 n= ) ( A cdot 3 ) B. 5 c. 10 ( D ) | 12 |
| 503 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose element ( a_{i j} ) is ( a_{i j}=frac{(i-j)^{2}}{2} ) | 12 |
| 504 | ( boldsymbol{A}=left[begin{array}{cc}-mathbf{3} & -mathbf{5} \ -mathbf{6} & mathbf{0}end{array}right], boldsymbol{A}-boldsymbol{B}=mathbf{2} boldsymbol{I} . ) Find ( boldsymbol{B} ) ( mathbf{A} cdotleft[begin{array}{cc}-5 & -5 \ -6 & -2end{array}right] ) в. ( left[begin{array}{cc}1 & 2 \ -3 & -6end{array}right] ) с. ( left[begin{array}{cc}5 & 5 \ -3 & -6end{array}right] ) О ( cdotleft[begin{array}{ll}1 & 2 \ 3 & 6end{array}right] ) | 12 |
| 505 | If ( A ) is ( 3 times 4 ) matrix and ( B ) is a matrix such that ( A^{prime} B ) and ( B^{prime} A ) are both defined, then the order of ( B ) is ( mathbf{A} cdot 4 times 4 ) B. ( 3 times 3 ) c. ( 3 times 4 ) D. ( 4 times 3 ) | 12 |
| 506 | Find the inverse of the following matrix by using elementary row transformation ( left[begin{array}{ll}2 & 5 \ 1 & 3end{array}right] ) | 12 |
| 507 | ff ( left[begin{array}{cc}1 & 2 \ 3 & -5end{array}right], ) then ( A^{-1} ) is equal to ( ^{mathbf{A}} cdotleft[begin{array}{cc}frac{5}{11} & frac{2}{11} \ frac{3}{11} & -frac{1}{11}end{array}right] ) в. ( left[begin{array}{rr}-frac{5}{11} & -frac{2}{11} \ -frac{3}{11} & -frac{1}{11}end{array}right] ) ( ^{mathbf{C}} cdotleft[begin{array}{cc}frac{5}{11} & frac{2}{11} \ frac{3}{11} & frac{1}{11}end{array}right] ) D. ( left[begin{array}{ll}5 & 2 \ 3 & -1end{array}right] ) | 12 |
| 508 | 25. Let A and B be two symmetric matrices of order 3. Statement-1: A(BA) and (AB)A are symmetric matrices Statement-2: AB is symmetric matrix ifmatrix multiplicat of A with B is commutative. [2011] (a) Statement-1 is true, Statement-2 is true; Statement- not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (C) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. | 12 |
| 509 | If ( A ) and ( B ) are ( 3 times 3 ) matrices and ( |boldsymbol{A}| neq mathbf{0}, ) then This question has multiple correct options A ( cdot|A B|=0 Rightarrow|B|=0 ) B . ( |A B| neq 0 Rightarrow|B| neq 0 ) C ( cdotleft|A^{-1}right|=|A|^{-1} ) D. ( |2 A|=2|A| ) | 12 |
| 510 | If ( Delta_{r}=left|begin{array}{ccc}r-1 & n & 6 \ (r-1)^{2} & 2 n^{2} & 4 n-2 \ (r-1)^{3} & 3 n^{3} & 3 n^{2}-3 nend{array}right| ) | 12 |
| 511 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{ll}mathbf{0} & boldsymbol{a} \ mathbf{0} & mathbf{0}end{array}right] ) and ( (boldsymbol{A}+boldsymbol{I})^{50} ) ( mathbf{5} mathbf{0} boldsymbol{A}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right] . ) Then the value of ( boldsymbol{a}+ ) ( boldsymbol{b}+boldsymbol{c}+boldsymbol{d} ) is A . 2 B. ( c cdot 4 ) D. none of these | 12 |
| 512 | ff ( Delta=left|begin{array}{lll}a r g z_{1} & a r g z_{2} & a r g z_{3} \ a r g z_{2} & a r g z_{3} & a r g z_{1} \ a r g z_{3} & a r g z_{1} & a r g z_{2}end{array}right|, ) the ( , Delta ) is divided by: A ( cdot arg left(z_{1}+z_{2}+z_{3}right) ) B ( cdot arg left(z_{1} cdot z_{2} cdot z_{3}right) ) C ( cdotleft(a r g z_{1}+a r g z_{2}+arg z_{3}right) ) D. N.O.T | 12 |
| 513 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{cc}sin boldsymbol{theta} & mathbf{0} \ mathbf{0} & -boldsymbol{operatorname { s i n } boldsymbol { theta }}end{array}right] cdot ). If ( boldsymbol{A}+boldsymbol{A}^{boldsymbol{T}} ) is a null matrix, then the number of values of ( boldsymbol{theta} ) in ( [mathbf{0}, mathbf{2} boldsymbol{pi}] ) is A . 4 B. 3 ( c cdot 2 ) D. | 12 |
| 514 | If ( boldsymbol{A}=left[begin{array}{cc}cos boldsymbol{theta} & -sin boldsymbol{theta} \ sin boldsymbol{theta} & cos boldsymbol{theta}end{array}right], ) then ( boldsymbol{A} boldsymbol{A}^{boldsymbol{T}} ) equals ( A cdotleft[begin{array}{cc}cos 2 theta & -sin 2 theta \ sin 2 theta & cos 2 thetaend{array}right] ) B. ( left[begin{array}{cc}cos ^{2} theta & sin ^{2} theta \ sin ^{2} theta & cos ^{2} thetaend{array}right] ) c. ( left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) D. ( left[begin{array}{ll}0 & 0 \ 0 & 0end{array}right] ) | 12 |
| 515 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) is a ( 2 times 2 ) matrix such that ( a_{i j}=i+2 j, ) then find ( A ) | 12 |
| 516 | ( left[begin{array}{cc}boldsymbol{a}+boldsymbol{b} & boldsymbol{2} \ mathbf{5} & boldsymbol{b}end{array}right]=left[begin{array}{ll}mathbf{6} & mathbf{5} \ mathbf{2} & mathbf{2}end{array}right], ) then find ( boldsymbol{a} ) | 12 |
| 517 | If matrix ( boldsymbol{A}=[mathbf{1} mathbf{2} mathbf{3}], ) then find ( boldsymbol{A} boldsymbol{A}^{boldsymbol{T}} ) | 12 |
| 518 | Given ( boldsymbol{x}-boldsymbol{y}+mathbf{3} boldsymbol{z}=mathbf{5} ; mathbf{4} boldsymbol{x}+mathbf{2} boldsymbol{y}-boldsymbol{z}=mathbf{0} ) ( ;-boldsymbol{x}+mathbf{3} boldsymbol{y}+boldsymbol{z}=mathbf{5} ) If ( mathbf{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{3} \ mathbf{4} & mathbf{2} & -mathbf{1} \ -mathbf{1} & mathbf{3} & mathbf{1}end{array}right], boldsymbol{X}=left[begin{array}{l}boldsymbol{x} \ boldsymbol{y} \ boldsymbol{z}end{array}right], boldsymbol{D}= ) ( left.begin{array}{l}mathbf{5} \ mathbf{0} \ mathbf{5}end{array}right} ) such that ( mathbf{A X}=mathbf{D} ) Show that ( A ) is non singular and the cofactor elements of a matrix ( boldsymbol{A} ) is ( left[begin{array}{ccc}+(2+3) & -(4-1) & +(12+2) \ -(-1-9) & +(1+3) & -(3-1) \ +(1-6) & -(-1-12) & +(2+4)end{array}right) ) | 12 |
| 519 | If ( 2 A+B=left[begin{array}{cc}3 & -1 \ 2 & 4end{array}right] ) and ( B= ) ( left[begin{array}{cc}-1 & -5 \ 0 & 2end{array}right], ) then find ( A ) | 12 |
| 520 | ( operatorname{Given} mathbf{3}left[begin{array}{cc}boldsymbol{x} & boldsymbol{y} \ boldsymbol{z} & boldsymbol{w}end{array}right]=left[begin{array}{cc}boldsymbol{x} & boldsymbol{6} \ -mathbf{1} & boldsymbol{2} boldsymbol{w}end{array}right]+ ) ( left[begin{array}{cc}4 & x+y \ z+w & 3end{array}right], ) find the values of ( boldsymbol{x}, boldsymbol{y}, boldsymbol{z} ) and ( boldsymbol{w} ) | 12 |
| 521 | If ( A=left(begin{array}{ccc}1 & -1 & 3 \ 5 & -4 & 7 \ 6 & 0 & 9 & 8end{array}right), ) Find the order of the matrix | 12 |
| 522 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) | 12 |
| 523 | If ( left(begin{array}{cc}1 & -tan theta \ tan theta & 1end{array}right)left(begin{array}{cc}1 & tan theta \ -tan theta & 1end{array}right)^{-1}= ) ( left[begin{array}{cc}boldsymbol{a} & -boldsymbol{b} \ boldsymbol{b} & boldsymbol{a}end{array}right], ) then This question has multiple correct options ( mathbf{A} cdot a=cos 2 theta ) В . ( a=1 ) c. ( b=sin 2 theta ) D. ( b=-1 ) | 12 |
| 524 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose element ( boldsymbol{a}_{boldsymbol{i} j} ) is ( boldsymbol{a}_{boldsymbol{i} j}=frac{|mathbf{2} boldsymbol{i}-boldsymbol{3} boldsymbol{j}|}{boldsymbol{2}} ) | 12 |
| 525 | ( A ) and ( B ) are symmetric matrices of the same order. ( boldsymbol{X}=boldsymbol{A B}+boldsymbol{B A} ) and ( boldsymbol{Y}= ) ( boldsymbol{A B}-boldsymbol{B A} ) ( (boldsymbol{X} boldsymbol{Y})^{boldsymbol{T}}= ) A . ( X Y ) в. ( Y X ) ( c cdot-Y X ) D. ( X+Y ) | 12 |
| 526 | Select the missing number from the given matrix: ( begin{array}{ccc}5 & 2 & 4 \ 4 & 4 & 7 \ 2 & 5 & 3 \ 18 & 30 & ?end{array} ) A . 43 B. 42 ( c .33 ) D. 32 | 12 |
| 527 | Prove that: [ begin{array}{l} {left[begin{array}{lll} mathbf{x} mathbf{y} & mathbf{z} end{array}right]left[begin{array}{lll} boldsymbol{a} & boldsymbol{h} & boldsymbol{g} \ boldsymbol{h} & boldsymbol{b} & boldsymbol{f} \ boldsymbol{g} & boldsymbol{f} & boldsymbol{c} end{array}right]left[begin{array}{l} boldsymbol{x} \ boldsymbol{y} \ boldsymbol{z} end{array}right]} \ = & {left[boldsymbol{a x}^{2}+boldsymbol{b} boldsymbol{y}^{2}+boldsymbol{c} boldsymbol{z}^{2}+boldsymbol{c f} boldsymbol{z}+boldsymbol{2 g} boldsymbol{z} boldsymbol{x}+boldsymbol{t}right.} end{array} ] ( 2 h x y ) | 12 |
| 528 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) is a scalar matrix of order ( boldsymbol{n} times boldsymbol{n} ) such that ( boldsymbol{a}_{boldsymbol{i} j}=boldsymbol{k} ) for all then trace of ( A ) is equal to ( A cdot n k ) в. ( n+k ) c. ( n / k ) D. none of these | 12 |
| 529 | f ( boldsymbol{A}=left(begin{array}{ll}1 & -1 \ 2 & -2end{array}right) ) Find ( 5 boldsymbol{I}-8 boldsymbol{A} ) | 12 |
| 530 | ( (A B)^{-1}= ) ( A cdot B A ) B ( cdot A^{-1} B^{-1} ) c. ( B^{-1} A^{-1} ) D. All of these | 12 |
| 531 | fthe matrix ( left(begin{array}{cc}mathbf{6} & -boldsymbol{x}^{2} \ mathbf{2} boldsymbol{x}-mathbf{1 5} & mathbf{1 0}end{array}right) ) symmetric, find the value of ( x ) | 12 |
| 532 | Find ( frac{1}{2}left(A+A^{T}right) ) and ( frac{1}{2}left(A-A^{T}right) ) when ( A=left[begin{array}{ccc}0 & a b \ -a & 0 & c \ -b-c 0end{array}right] ) | 12 |
| 533 | If ( A ) is a square of order 3 , then ( left|boldsymbol{A} boldsymbol{d} boldsymbol{j}left(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A}^{2}right)right|= ) ( mathbf{A} cdot|A|^{2} ) B . ( |A|^{4} ) c. ( |A|^{8} ) D・ ( |A|^{1} ) | 12 |
| 534 | If ( A ) is a square matrix, ( B ) is a singular matrix of same order, then for a positive integer ( n,left(A^{-1} B Aright)^{n} ) equals ( mathbf{A} cdot A^{-n} B^{n} A^{n} ) B ( cdot A^{n} B^{n} A^{-n} ) c. ( A^{-1} B^{n} A ) D. ( nleft(A^{-1} B Aright) ) | 12 |
| 535 | If ( A ) is square matrix of order ( 3, ) then ( left|A d jleft(A d j A^{2}right)right|= ) A ( cdot|A|^{2} ) B ( cdot|A|^{4} ) c. ( |A|^{8} ) D・ ( |A|^{16} ) | 12 |
| 536 | ( mathbf{1} ) ( left[begin{array}{ll}1 & 1 \ 0 & 1end{array}right]left[begin{array}{ll}1 & 2 \ 0 & 1end{array}right]left[begin{array}{ll}1 & 3 \ 0 & 1end{array}right] cdotleft[begin{array}{cc}1 & n-1 \ 0 & 1end{array}right]= ) ( left[begin{array}{ll}1 & 78 \ 0 & 1end{array}right], ) then the inverse of ( left[begin{array}{ll}1 & n \ 0 & 1end{array}right] ) is? A. ( left[begin{array}{cc}1 & -13 \ 0 & 1end{array}right] ) в. ( left[begin{array}{ll}1 & 0 \ 12 & 1end{array}right] ) c. ( left[begin{array}{cc}1 & -12 \ 0 & 1end{array}right] ) D. ( left[begin{array}{ll}1 & 0 \ 13 & 1end{array}right] ) | 12 |
| 537 | ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{8} \ mathbf{9} & mathbf{4}end{array}right] boldsymbol{B}=left[begin{array}{ll}mathbf{6} & mathbf{7} \ mathbf{3} & mathbf{5}end{array}right] ) Find ( boldsymbol{A}+boldsymbol{B} ) | 12 |
| 538 | If a matrix has equal number of columns and rows then it is said to be a A. row matrix B. identical matrix c. square matrix D. rectangular matrix | 12 |
| 539 | Evaluate ( left[begin{array}{l}3 \ 4 \ 1end{array}right]left[begin{array}{lll}2 & -1 & 3end{array}right] ) | 12 |
| 540 | The inverse of a symmetric matrix is A. symmetric B. skew-symmetric c. diagonal matrix D. singular matrix | 12 |
| 541 | Show that matrix ( A+B ) is symmetric or skew symmetric according as ( A ) and ( B ) are symmetric of skew symmetric. | 12 |
| 542 | ( A ) is a ( 3 times 3 ) diagonal matrix having integral entries such that ( operatorname{det}(A)=120 ) number of such matrices is ( 10 n ), then ( boldsymbol{n} ) is A . 36 B . 38 ( c cdot 34 ) D. 30 | 12 |
| 543 | Identify a matrix В. ( A={1,2} ) ( mathbf{c} cdot A=left[begin{array}{ll}1 & 2end{array}right] ) D. None of these | 12 |
| 544 | Let ( boldsymbol{alpha}=boldsymbol{pi} / mathbf{5} ) and [ begin{array}{c} boldsymbol{A}=left[begin{array}{cc} cos boldsymbol{alpha} & sin boldsymbol{alpha} \ -sin boldsymbol{alpha} & cos boldsymbol{alpha} end{array}right] text { and } boldsymbol{B}=boldsymbol{A}+ \ boldsymbol{A}^{2}+boldsymbol{A}^{3}+boldsymbol{A}^{4}, text { then } end{array} ] This question has multiple correct options A. singular B. non-singular c. skew-symmetric D. ( |B|=1 ) | 12 |
| 545 | Ti 007 p= 4 4 21. Let P- 1 0 and I be the identity matrix of order 3. 16 4 1 If O= [9] is a matrix such that P50 – Q =I, then 931+ 932 921 equals (a) 52 (6) 103 (c) 201 (JEE Adv. 2016) (2) 205 How many 3 x 3 matrico , | 12 |
| 546 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{3} & -mathbf{2} \ mathbf{4} & -mathbf{2}end{array}right], ) find ( boldsymbol{K} ) such that ( boldsymbol{A}^{mathbf{2}}= ) ( boldsymbol{K} boldsymbol{A}-boldsymbol{2} boldsymbol{I}, ) where ( boldsymbol{I} ) is the identity element. | 12 |
| 547 | If ( A ) is a square matrix, then ( A-A^{T} ) is A. unit matrix B. null matrix ( c . A ) D. a skew symmetric matrix | 12 |
| 548 | If ( A ) is a ( 3 times 3 ) skew-symmetric matrix, then the trace of ( A ) is equal to A . -1 B. c. ( |A| ) D. 0 | 12 |
| 549 | If ( left(A+B^{T}right)^{T} ) is a matrix of order ( 4 times 3 ) then the order of matrix B is ( A cdot 3 times 4 ) B. ( 4 times 3 ) ( c cdot 3 times 3 ) D. ( 4 times 4 ) | 12 |
| 550 | If ( boldsymbol{A}=left[begin{array}{cc}1 & tan x \ -tan x & 1end{array}right], ) then ( A^{T} A^{-1} ) is ( mathbf{A} cdotleft[begin{array}{ll}-cos 2 x & sin 2 x \ -sin 2 x & cos 2 xend{array}right] ) В. ( left[begin{array}{cc}cos 2 x & -sin 2 x \ sin 2 x & cos 2 xend{array}right] ) c. ( left[begin{array}{cc}cos 2 x & cos 2 x \ cos 2 x & sin 2 xend{array}right] ) D. none of these | 12 |
| 551 | f ( A+2 B=left[begin{array}{cc}2 & -4 \ 1 & 6end{array}right], A^{prime}+B^{prime}= ) ( left[begin{array}{cc}1 & 2 \ 0 & -1end{array}right], ) then ( A= ) | 12 |
| 552 | If ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{i} & mathbf{0} \ mathbf{0} & boldsymbol{i}end{array}right], boldsymbol{n} in boldsymbol{N}, ) then ( boldsymbol{A}^{4 n} ) equals ( A cdotleft[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) в. ( left[begin{array}{ll}i & 0 \ 0 & iend{array}right] ) c. ( left[begin{array}{ll}0 & i \ i & 0end{array}right] ) D. ( left[begin{array}{ll}0 & 0 \ 0 & 0end{array}right] ) | 12 |
| 553 | Given, matrix ( boldsymbol{A}=left[begin{array}{l}mathbf{3} \ mathbf{2}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{l}-mathbf{2} \ -mathbf{1}end{array}right] ) find the matrix ( X ) such that ( X-A=B ) ( A cdotleft[begin{array}{l}0 \ 0end{array}right] ) в. ( left[begin{array}{l}1 \ 1end{array}right] ) ( c cdotleft[begin{array}{l}4 \ 0end{array}right] ) D. ( left[begin{array}{l}1 \ -1end{array}right. ) | 12 |
| 554 | Find the values of ( x, y, a ) and ( b ) if ( left[begin{array}{cccc}3 x+4 y & 2 & x-2 y \ a+b & 2 a-b & -1end{array}right]= ) | 12 |
| 555 | For ( mathbf{3} times mathbf{3} ) matrices ( boldsymbol{A} ) and ( boldsymbol{B}, ) if ( |boldsymbol{B}|=mathbf{1} ) and ( A=2 B ) then find ( |A| ) A . B. 4 ( c cdot 2 ) D. 8 | 12 |
| 556 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) find ( |mathbf{2} boldsymbol{A}| ) | 12 |
| 557 | The construction of ( 3 times 4 ) matrix ( A ) whose elements ( a_{i j} ) is given by ( frac{(i+j)^{2}}{2} ) is ( mathbf{A} cdotleft[begin{array}{llll}2 & 9 / 2 & 8 & 25 \ 9 & 4 & 5 & 18 \ 8 & 25 & 18 & 49end{array}right] ) ( mathbf{B} cdotleft[begin{array}{cccc}2 & 9 / 2 & 25 / 2 & 9 \ 9 / 2 & 5 / 2 & 5 & 45 / 2 \ 25 & 18 & 25 & 9 / 2end{array}right] ) ( mathbf{C} cdotleft[begin{array}{cccc}2 & 9 / 2 & 8 & 25 / 2 \ 9 / 2 & 8 & 25 / 2 & 18 \ 8 & 25 / 2 & 18 & 49 / 2end{array}right] ) D. None of these | 12 |
| 558 | ( left(left[begin{array}{ll}mathbf{8} & mathbf{4} \ boldsymbol{x} & mathbf{8}end{array}right]right)=mathbf{4}left(left[begin{array}{ll}mathbf{2} & mathbf{1} \ mathbf{1} & mathbf{2}end{array}right]right) ) then the value of ( x ) is A . B. 2 ( c cdot frac{1}{4} ) ( D ) | 12 |
| 559 | Assertion ( operatorname{Let} a, b in R, ) and ( I=left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) and ( J= ) ( left[begin{array}{cc}mathbf{0} & mathbf{1} \ -mathbf{1} & mathbf{0}end{array}right] ) Inverse of ( a I+b J ) is ( c I+d J ) if and only if ( boldsymbol{a c}-boldsymbol{b d} neq mathbf{0} ) and ( boldsymbol{a} boldsymbol{d}+boldsymbol{b} boldsymbol{c}=mathbf{0} ) Reason ( (boldsymbol{a} boldsymbol{I}+boldsymbol{b} boldsymbol{J})(boldsymbol{c} boldsymbol{I}+boldsymbol{d} boldsymbol{J})=(boldsymbol{a} boldsymbol{c}-boldsymbol{b} boldsymbol{d}) boldsymbol{I}+ ) ( (a d+b c) J ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 560 | ( fleft(begin{array}{lll}1 & 5 & 2 \ 1 & -3 & 4end{array}right), ) then find ( A^{T} ) and ( left(A^{T}right)^{T} ) | 12 |
| 561 | Identify the incorrect statement in respect of two square matrices ( A ) and ( B ) conformable for sum and product A ( cdot t_{r}(A+B)=t_{r}(A)+t_{r}(B) ) B . ( t_{r}(alpha A)=alpha t_{r}(A), quad alpha epsilon R ) ( mathbf{c} cdot t_{r}left(A^{T}right)=t_{r}(A) ) D. none of these | 12 |
| 562 | A and B are two square matrices of same order and ( A^{prime} ) denotes the transpose of ( A ), then ( mathbf{A} cdot(A B)^{prime}=B^{prime} A^{prime} ) B. ( (A B)^{prime}=A^{prime} B^{prime} ) ( mathbf{C} cdot A B=0 Rightarrow|A|=0 ) or ( |B|=0 ) D . ( A B=0 Rightarrow A=0 ) or ( B=0 ) | 12 |
| 563 | Let three matrices ( boldsymbol{A}=left[begin{array}{ll}mathbf{2} & mathbf{1} \ mathbf{4} & mathbf{1}end{array}right] ; boldsymbol{B}= ) ( left[begin{array}{ll}mathbf{3} & mathbf{4} \ mathbf{2} & mathbf{3}end{array}right] ) and ( boldsymbol{C}=left[begin{array}{cc}mathbf{3} & -mathbf{4} \ -mathbf{2} & mathbf{3}end{array}right] ) then find ( operatorname{tr}(A)+operatorname{tr}left(frac{A B C}{2}right) operatorname{tr}left(frac{A(B C)^{2}}{4}right)+ ) ( operatorname{tr}left(frac{A(B C)^{3}}{8}right)+ldots+infty, ) where ( t r(A) ) represents trace of matrix ( boldsymbol{A} ) ( mathbf{A} cdot mathbf{6} ) B. c. 12 D. 15 | 12 |
| 564 | [ mathbf{A}=left[begin{array}{lll} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 end{array}right] ] The new matrix formed after interchanging ( 2^{n d} ) and ( 3^{r d} ) rows will be ( A ) [ -left[begin{array}{lll} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 end{array}right] ] в. [ left[begin{array}{lll} 4 & 5 & 6 \ 1 & 2 & 3 \ 7 & 8 & 9 end{array}right] ] c. [ -left[begin{array}{lll} 1 & 2 & 3 \ 7 & 8 & 9 \ 4 & 5 & 6 end{array}right] ] D. [ left[begin{array}{lll} 1 & 2 & 3 \ 7 & 8 & 9 \ 4 & 5 & 6 end{array}right] ] | 12 |
| 565 | If the matrices ( A, B,(A+B) ) are non singular then ( left[boldsymbol{A}(boldsymbol{A}+boldsymbol{B})^{-1} boldsymbol{B}right]^{-1} ) is equal to- ( mathbf{A} cdot A+B ) B. ( A^{-1}+B^{-1} ) c. ( A(A+B)^{-1} ) D. None | 12 |
| 566 | If ( boldsymbol{A}=left[begin{array}{lll}mathbf{0} & mathbf{2} & mathbf{3} \ mathbf{3} & mathbf{5} & mathbf{7}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{lll}mathbf{1} & mathbf{3} & mathbf{7} \ mathbf{2} & mathbf{4} & mathbf{1}end{array}right] ) if ( boldsymbol{A}+boldsymbol{B}=left[begin{array}{ccc}mathbf{1} & mathbf{5} & mathbf{1 0} \ mathbf{5} & boldsymbol{k} & mathbf{8}end{array}right] ) Find the value of ( mathbf{k} ) ( A cdot 9 ) B. 4 ( c cdot 5 ) ( D ) | 12 |
| 567 | ( boldsymbol{B}=boldsymbol{A}+boldsymbol{A}^{2}+boldsymbol{A}^{3}+boldsymbol{A}^{4} ) If order of ( A ) is 3 then order of ( B ) is ( A cdot 3 ) B. 6 ( c cdot 2 ) D. 9 | 12 |
| 568 | ( mathbf{f} boldsymbol{A}=left[begin{array}{cc}mathbf{0} & mathbf{3} \ mathbf{2} & -mathbf{5}end{array}right] ) ( & boldsymbol{K} boldsymbol{A}=left[begin{array}{cc}mathbf{0} & mathbf{4} boldsymbol{a} \ -mathbf{8} & mathbf{5} boldsymbol{b}end{array}right] ) then find the value of a and | 12 |
| 569 | Find the inverse of the matrix ( mathbf{A}=left[begin{array}{ccc}mathbf{0} & mathbf{1} & mathbf{2} \ {[mathbf{0 . 3 e m}] mathbf{1}} & mathbf{2} & mathbf{3} \ {[mathbf{0 . 3 e m}] mathbf{3}} & mathbf{1} & mathbf{1}end{array}right] ) | 12 |
| 570 | If ( A B=A ) and ( B A=B ) then ( B^{2} ) is equal to A. ( B ) в. ( A ) ( c .-B ) D. ( B^{2} ) | 12 |
| 571 | Which of the following is not true, if ( mathbf{A} ) and ( B ) are two matrices each of order ( boldsymbol{n} times boldsymbol{n}, ) then ( mathbf{A} cdot(A+B)^{T}=B^{T}+A^{T} ) ( mathbf{B} cdot(A-B)^{T}=A^{T}-B^{T} ) ( mathbf{C} cdot(A B)^{T}=A^{T} B^{T} ) ( mathbf{D} cdot(A B C)^{T}=C^{T} B^{T} A^{T} ) | 12 |
| 572 | The inverse of a diagonal matrix is a : This question has multiple correct options A. Symmetric matrix B. Skew-symmetric matrix c. Diagonal matrix D. None of the above | 12 |
| 573 | ( fleft(begin{array}{cc}cos alpha & sin alpha \ -sin alpha & cos alphaend{array}right], ) then find ( A^{2} ) | 12 |
| 574 | Unit matrix is a diagonal matrix in which all the diagonal elements are unity. Unit matrix of order ‘n’ is denoted by ( I_{n}(text { or } I) ) i.e. ( A=left[a_{i j}right]_{n} ) is a unit matrix when ( boldsymbol{a}_{boldsymbol{i} j}=mathbf{0} ) for ( boldsymbol{i} neq ) ( boldsymbol{j} boldsymbol{a} boldsymbol{n} boldsymbol{d} boldsymbol{a}_{i j}=mathbf{1} ) A . True B. False | 12 |
| 575 | The scalar matrixis ( mathbf{A} cdotleft[begin{array}{cc}-1 & 3 \ 2 & 4end{array}right] ) ( mathbf{B} cdotleft[begin{array}{ll}0 & 3 \ 2 & 0end{array}right] ) ( mathbf{C} cdotleft[begin{array}{ll}4 & 0 \ 0 & 4end{array}right] ) D. None of these | 12 |
| 576 | If ( A ) is symmetric and ( B ) is skew symmetric matrix, then which of the following is/are correct? This question has multiple correct options A ( cdot A B A^{T} ) is skew symmetric matrix B. ( A B^{T}+B A^{T} ) is symmetric matrix c. ( (A+B)(A-B) ) is symmetric D. ( (A+I)(B-I) ) is skew symmetric | 12 |
| 577 | The inverse of a symmetric matrix is A. Symmetric B. Skew-symmetric c. Diagonal D. None of these | 12 |
| 578 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{0} & boldsymbol{a}+mathbf{1} & boldsymbol{b}-mathbf{2} \ mathbf{2} boldsymbol{a}-mathbf{1} & mathbf{0} & boldsymbol{c}-mathbf{2} \ mathbf{2} boldsymbol{b}+mathbf{1} & mathbf{2}+boldsymbol{c} & mathbf{0}end{array}right] ) is skew symmetric then ( a+b+c= ) ( A cdot 3 ) B. -3 ( c cdot frac{1}{3} ) ( D ) | 12 |
| 579 | Let ( boldsymbol{A}=left[begin{array}{c}mathbf{3} boldsymbol{x}^{2} \ mathbf{1} \ mathbf{6} boldsymbol{x}end{array}right], boldsymbol{B}=[boldsymbol{a}, boldsymbol{b}, boldsymbol{c}] ) and ( boldsymbol{C}= ) ( left[begin{array}{ccc}(x+2)^{2} & 5 x^{2} & 2 x \ 5 x^{2} & 2 x & (x+2)^{2} \ 2 x & (x+2)^{2} & 5 x^{2}end{array}right] ) three given matrices, where ( a, b, c ) and ( boldsymbol{x} in boldsymbol{R}, ) Given that ( boldsymbol{t r} .(boldsymbol{A B})=boldsymbol{t r} .(boldsymbol{C}) vee ) ( boldsymbol{x} in boldsymbol{R}, ) where ( boldsymbol{t r} .(boldsymbol{A}) ) denotes trace of ( boldsymbol{A} ) Find the value of ( (boldsymbol{a}+boldsymbol{b}+boldsymbol{c}) ) ( A cdot 6 ) B. ( c cdot 8 ) D. | 12 |
| 580 | If ( A ) is a square matrix of order ( 3, ) then ( left|A d jleft(A d j A^{2}right)right|= ) A ( cdotleft|A^{2}right| ) B cdot ( left|A^{4}right| ) c. ( left|A^{8}right| ) ( mathbf{D} cdotleft|A^{16}right| ) | 12 |
| 581 | For any square matrix ( boldsymbol{A} ) with real numbers, ( A+A^{prime} ) is a symmetric and ( A-A^{prime} ) is a skew-symmetric | 12 |
| 582 | 16. Let lo 07 [ 1007 To 107 0 0 P = I=/0 1 0 P = 0 0 1, P₂ = 1 Lo 0 1 0 1 1 0 0 1 To 107 To o 17 T0 0 17 Pa = 0 0 1 , Pg = 1 0 0 P6= 0 1 0 1 0 0 0 1 0 1 0 0 6 [2 1 37. and X = P 1 0 2 PTS k=1 [3 2 1 Where P denotes the transpose of the matrix P… Then which of the following options is/are correct? (a) X is a symmetric matrix (JEE Adv. 2019) (b) The sum of diagonal entries of X is 18 (C) X-301 is an invertible matrix If X 1 = a 1, then a = 30 | 12 |
| 583 | sin -1-sin? 23. Let M= 1+ cose cose = a1 +BM-1 Where a= a(O) and B = B(O) are real numbers, and I is the 2 x 2 identity matrix. If a* is the minimum of the set Sale): 0 E 10, 2TT)} and B* is the minimum of the set SB(A):0 € 0, 21)}. Then the value of a*+b* is (JEE Adv. 2019) 17 37 (a) (b) 16 (©) 1 (d) 16 16 | 12 |
| 584 | (10 o 26. Let A= 2 1 0 . Ifu, and u, are column matrices such (3 2 1 | 12 |
| 585 | ( left[begin{array}{cc}1 & 1 \ 0 & 1end{array}right]left[begin{array}{cc}1 & 2 \ 0 & 1end{array}right]left[begin{array}{cc}1 & 3 \ 0 & 1end{array}right] ldots ldotsleft[begin{array}{cc}1 & n \ 0 & 1end{array}right]= ) ( A cdot 27 ) 3.26 ( c .376 ) D. 378 | 12 |
| 586 | Find the inverse of the following matrix using elementary row transformation. ( left[begin{array}{cc}mathbf{1} & mathbf{2} \ mathbf{2} & -mathbf{1}end{array}right] ) | 12 |
| 587 | ( operatorname{Let} A=left[begin{array}{lll}1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1end{array}right] . ) If ( u_{1} ) and ( u_{2} ) are column matrices such that ( boldsymbol{A} boldsymbol{u}_{1}=left[begin{array}{l}mathbf{1} \ mathbf{0} \ mathbf{0}end{array}right] ) and ( A u_{2}=left[begin{array}{l}0 \ 1 \ 0end{array}right] ) then ( u_{1}+u_{2} ) is equal to ( mathbf{A} cdotleft[begin{array}{c}-1 \ 1 \ 0end{array}right. ) B. ( left[begin{array}{c}1 \ -1 \ -1end{array}right. ) ( mathbf{c} cdotleft[begin{array}{c}-1 \ -1 \ 0end{array}right. ) D. ( left[begin{array}{c}-1 \ 1 \ -1end{array}right. ) | 12 |
| 588 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{1} & mathbf{1}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}mathbf{0} & -mathbf{1} \ mathbf{1} & mathbf{2}end{array}right], ) then what is ( B^{-1} A^{-1} ) equal to? begin{tabular}{ll} A. ( left[begin{array}{cc}1 & -3 \ -3 & 2end{array}right] ) \ hline end{tabular} B. ( left[begin{array}{cc}-1 & 3 \ 1 & -2end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-1 & 3 \ -1 & -2end{array}right] ) D. ( left[begin{array}{cc}-1 & 1 \ -3 & -2end{array}right] ) | 12 |
| 589 | If ( A ) is a ( 3 times 3 ) skew symmetric matrix, then trace of ( A ) is equal to A . -1 B. 7 c. ( |A| ) D. None of these | 12 |
| 590 | If ( P ) is a ( 3 times 3 ) matrix such that ( P^{T}= ) ( 2 P+I, ) where ( P^{T^{prime}} ) is the transpose of ( P ) and ( I ) is the ( 3 times 3 ) identity matrix, then there exists a column matrix ( boldsymbol{X}= ) ( left[begin{array}{l}boldsymbol{x} \ boldsymbol{y} \ boldsymbol{z}end{array}right] neqleft[begin{array}{l}mathbf{0} \ mathbf{0} \ mathbf{0}end{array}right] ) such that ( ^{mathbf{A}} cdot operatorname{PX}=left[begin{array}{l}0 \ 0 \ 0end{array}right] ) в. ( P X=X ) c. ( P X=2 X ) D. ( P X=-X ) E. [ P X=left[begin{array}{l}2 \ 0 \ 0end{array}right] ] F. ( P X=3 X ) G. ( P X=5 X ) H. ( P X=-3 X ) | 12 |
| 591 | ( boldsymbol{A}=left[begin{array}{cc}boldsymbol{x} & -mathbf{7} \ mathbf{7} & boldsymbol{y}end{array}right] ) is a skew-symmetric matrix, then ( (x, y)= ) A ( .(1,-1) ) B. (7,-7) ( c cdot(0,0) ) D. (14,-14) | 12 |
| 592 | A matrix consisting of a single column of m elements is know as A. Column matrix B. Row matrix c. Square matrix D. Null matrix | 12 |
| 593 | Let o be a complex cube root of unity with o # 1 and P=[P] be a nx n matrix with p;;= mitj. Then p2 +0, when n= (JEE Adv. 2013) (a) 57 (6) 55 (c) 58 (d) 56 | 12 |
| 594 | If ( A ) is a ( 3- ) rowed square matrix and ( |mathbf{3} boldsymbol{A}|=boldsymbol{k}|boldsymbol{A}| ) then ( boldsymbol{k}=? ) ( A cdot 3 ) B. 9 ( c cdot 27 ) D. | 12 |
| 595 | The table shows a five-day forecast indicating high (H) and Low(L) temperatures in Fahrenheit. Organise the temperatures in a matrix where the first and second rows represent the High and Low temperatures respectively and identify which day will be the warmest? begin{tabular}{|c|c|c|c|c|} hline Mon & Tue & wed & Thu & Fri \ hline( sum_{i}^{m} ) & ( sum_{i}^{m} ) & ( sum_{i}^{m} ) & ( sum_{i}^{m} ) & ( sum_{j}^{m} ) \ hline ( mathrm{H} 88 ) & ( mathrm{H} 90 ) & ( mathrm{H} 86 ) & ( mathrm{H} 84 ) & ( mathrm{H} 85 ) \ hline ( mathrm{L} 54 ) & ( mathrm{L} 56 ) & ( mathrm{L} 53 ) & ( mathrm{L} 52 ) & ( mathrm{L} 52 ) \ hline end{tabular} | 12 |
| 596 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose element ( a_{i j} ) is ( a_{i j}=frac{(2 i+j)^{2}}{2} ) | 12 |
| 597 | If ( boldsymbol{A}=left[begin{array}{ll}2 & 3 \ 5 & 7end{array}right], ) then find ( A+A^{T} ) | 12 |
| 598 | ff ( boldsymbol{A}=left[begin{array}{cc}cos boldsymbol{x} & sin boldsymbol{x} \ -sin boldsymbol{x} & cos boldsymbol{x}end{array}right], ) then find ( boldsymbol{x} ) satisfying ( mathbf{0}<boldsymbol{x}<frac{boldsymbol{pi}}{mathbf{2}} ) when ( boldsymbol{A}+boldsymbol{A}^{boldsymbol{T}}=boldsymbol{I} ) | 12 |
| 599 | 18. Let 0 +1 be a cube root of unity and S be the set of all [i a b non-singular matrices of the form @ 1 c 02 @ 1 where each of a, b and c is either o or 02. Then the number of distinct matrices in the set Sis (2011) (a) 2 (6) 6 (c) 4 (d) 8 | 12 |
| 600 | If ( A ) is a ( 3 x 3 ) non singular matrix and ( |boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}|=|boldsymbol{A}|^{x},|boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A})|= ) ( |boldsymbol{A}|^{y},left|boldsymbol{A}^{-1}right|=|boldsymbol{A}|^{z} ) then the values of ( mathbf{x}, mathbf{y}, mathbf{z}, ) in descending order A. ( mathrm{x}, mathrm{Y}, mathrm{z} ) в. Z, Y, х c. ( mathrm{z}, mathrm{x}, mathrm{y} ) D. Y, x, z | 12 |
| 601 | f ( boldsymbol{y}=left[begin{array}{cc}1 & 2 \ -1 & 5end{array}right], ) find a matrix ( X ) such that ( 2 boldsymbol{X}+boldsymbol{Y}=left[begin{array}{cc}mathbf{5} & mathbf{0} \ -mathbf{3} & mathbf{3}end{array}right] ) | 12 |
| 602 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{0}end{array}right], ) then ( boldsymbol{A}^{2} ) is equal to ( A cdotleft[begin{array}{ll}0 & 1 \ 1 & 0end{array}right] ) B. ( left[begin{array}{ll}1 & 0 \ 1 & 0end{array}right] ) c. ( left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) D. ( left[begin{array}{ll}0 & 1 \ 0 & 1end{array}right] ) | 12 |
| 603 | If ( A_{3} x_{3} ) and ( d e t A=2 ) then ( d e t A^{-1}= ) A ( cdot frac{1}{2} ) B. – ( c cdot frac{1}{4} ) D. – | 12 |
| 604 | If ( mathbf{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{0} & mathbf{3}end{array}right] ) and ( mathbf{B}=[mathbf{3}-mathbf{1}], ) then ( mathbf{B A}= ) A. ( left[begin{array}{ll}3 & 0 \ 0 & 3end{array}right] ) B. ( left[begin{array}{ll}3 & 0end{array}right] ) ( begin{array}{ll}text { c. }[3 & 3]end{array} ) D. ( [0-3] ) | 12 |
| 605 | If ( boldsymbol{A}=left[begin{array}{ccc}1 & -2 & 4 \ 2 & 3 & 2 \ 3 & 1 & 5end{array}right] ) and ( B= ) ( left[begin{array}{ccc}mathbf{0} & -mathbf{2} & mathbf{4} \ mathbf{1} & mathbf{3} & mathbf{2} \ mathbf{- 1} & mathbf{1} & mathbf{5}end{array}right], ) then ( boldsymbol{A}+boldsymbol{B} ) is ( ^{A} cdotleft[begin{array}{ccc}1 & -2 & 4 \ 3 & 3 & 2 \ 2 & 1 & 5end{array}right] ) B. ( left[begin{array}{lll}1 & -2 & 8 \ 3 & 3 & 4 \ 2 & 1 & 10end{array}right] ) C ( cdotleft[begin{array}{lll}1 & -4 & 8 \ 3 & 6 & 4 \ 2 & 2 & 10end{array}right] ) D. none of these | 12 |
| 606 | If ( boldsymbol{A}+boldsymbol{B}+boldsymbol{C}=boldsymbol{pi}, ) then ( left|begin{array}{ccc}sin (boldsymbol{A}+boldsymbol{B}+boldsymbol{C}) & sin boldsymbol{B} & cos boldsymbol{C} \ sin boldsymbol{theta} & tan boldsymbol{A} \ cos (boldsymbol{A}+boldsymbol{B}) & -tan boldsymbol{A} & boldsymbol{0}end{array}right|= ) | 12 |
| 607 | The order of ( [mathbf{x} mathbf{y} mathbf{z}]left[begin{array}{lll}mathbf{a} & mathbf{h} & mathbf{g} \ mathbf{h} & mathbf{b} & mathbf{f} \ mathbf{g} & mathbf{f} & mathbf{c}end{array}right]left[begin{array}{l}mathbf{x} \ mathbf{y} \ mathbf{z}end{array}right] ) is A . ( 3 times 1 ) B. 1x ( c .1 times 3 ) D. 3×3 | 12 |
| 608 | ( mathbf{f} boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{4} & mathbf{0} \ mathbf{2} & mathbf{5} & mathbf{0} \ mathbf{3} & mathbf{6} & mathbf{0}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{lll}mathbf{3} & mathbf{2} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5} & mathbf{6}end{array}right] ) and ( C=left[begin{array}{lll}3 & 2 & 1 \ 1 & 2 & 3 \ 7 & 8 & 9end{array}right], ) Then evaluate ( boldsymbol{A B}-boldsymbol{B C} ) | 12 |
| 609 | Which of the following matrix is inverse of itself A. ( left[begin{array}{lll}1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1end{array}right] ) В. ( left[begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1end{array}right] ) C. ( left[begin{array}{lll}1 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 1end{array}right] ) D. ( left[begin{array}{lll}0 & 1 & 0 \ 1 & 1 & 1 \ 0 & 1 & 0end{array}right] ) | 12 |
| 610 | If ( A ) is a skew-symmetric matrix of order ( 3, ) then prove that det ( A=0 ) | 12 |
| 611 | Show that all the diagonal elements of a skew-symmetric matrix are zero. | 12 |
| 612 | If ( 2left[begin{array}{ll}1 & 3 \ 0 & xend{array}right]+left[begin{array}{ll}y & 0 \ 1 & 2end{array}right]=left[begin{array}{ll}5 & 6 \ 1 & 8end{array}right], ) then the value of ( x ) and ( y ) are A ( . x=3, y=3 ) в. ( x=-3, y=3 ) c. ( x=3, y=-3 ) D. ( x=-3, y=-3 ) | 12 |
| 613 | Find ( boldsymbol{x}, boldsymbol{y} ) if ( left[begin{array}{cc}mathbf{0} & mathbf{4} \ boldsymbol{x}^{2} & boldsymbol{y}^{2}end{array}right]=left[begin{array}{ll}mathbf{0} & mathbf{4} \ boldsymbol{4} & mathbf{9}end{array}right] ) | 12 |
| 614 | ff ( P(x)=left[begin{array}{cc}cos x & sin x \ -sin x & cos xend{array}right] ) then ( boldsymbol{P}(boldsymbol{x}) cdot boldsymbol{P}(boldsymbol{y})= ) This question has multiple correct options A ( . P(x) . P(y)=P(x+y) ) в. ( P(x) . P(y)=P(x y) ) c. ( P(x) . P(y)=P(y) . P(x) ) | 12 |
| 615 | For the matrices ( A ) and ( B ), verify that ( (A B)^{prime}=B^{prime} A^{prime} ) where (i) ( boldsymbol{A}=left[begin{array}{c}mathbf{1} \ -mathbf{4} \ mathbf{3}end{array}right], boldsymbol{B}=left[begin{array}{lll}-mathbf{1} & mathbf{2} & mathbf{1}end{array}right] ) (ii) ( boldsymbol{A}=left[begin{array}{l}mathbf{0} \ mathbf{1} \ mathbf{2}end{array}right], boldsymbol{B}=left[begin{array}{lll}mathbf{1} & mathbf{5} & mathbf{7}end{array}right] ) | 12 |
| 616 | If ( boldsymbol{P}=left[begin{array}{cc}frac{sqrt{mathbf{3}}}{mathbf{2}} & frac{mathbf{1}}{mathbf{2}} \ -frac{mathbf{1}}{mathbf{2}} & frac{sqrt{mathbf{3}}}{mathbf{2}}end{array}right], boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{0} & mathbf{1}end{array}right] ) and ( boldsymbol{Q}=boldsymbol{P} boldsymbol{A} boldsymbol{P}^{T}, ) then ( boldsymbol{P}^{T} boldsymbol{Q}^{2015} boldsymbol{P} ) is: A. ( left[begin{array}{cc}0 & 2015 \ 0 & 0end{array}right] ) B. ( left[begin{array}{cc}2015 & 0 \ 1 & 2015end{array}right] ) c. ( left[begin{array}{cc}2015 & 1 \ 0 & 2015end{array}right] ) D. ( left[begin{array}{cc}1 & 2015 \ 0 & 1end{array}right] ) | 12 |
| 617 | fthe matrix ( left[begin{array}{cc}mathbf{2} & mathbf{3} \ mathbf{5} & -mathbf{1}end{array}right]=boldsymbol{A}+boldsymbol{B}, ) where A is symmetric and B is skew symmetric, then ( B= ) A. ( left[begin{array}{cc}2 & 4 \ 4 & -1end{array}right] ) B. ( left[begin{array}{cc}0 & -2 \ 2 & 0end{array}right] ) C ( cdotleft[begin{array}{cc}0 & 1 \ -1 & 0end{array}right] ) D. ( left[begin{array}{cc}0 & -1 \ 1 & 0end{array}right] ) | 12 |
| 618 | Simplify ( : cos Qleft[begin{array}{cc}cos Q & sin Q \ -sin Q & cos Qend{array}right]+ ) ( sin Qleft[begin{array}{cc}sin Q & -cos Q \ cos Q & sin Qend{array}right] ) | 12 |
| 619 | Given the matrices [ begin{array}{l} boldsymbol{A}=left[begin{array}{lll} 2 & 1 & 1 \ 3 & -1 & 0 \ 0 & 2 & 4 end{array}right], B=left[begin{array}{lll} 9 & 7 & -1 \ 3 & 5 & 4 \ 2 & 1 & 6 end{array}right] \ text { and } boldsymbol{C}=left[begin{array}{lll} 2 & -4 & 3 \ 1 & -1 & 0 \ 9 & 4 & 5 end{array}right] end{array} ] Verify that ( (boldsymbol{A}+boldsymbol{B})+boldsymbol{C}=boldsymbol{A}+(boldsymbol{B}+ ) ( C) ) | 12 |
| 620 | If ( 2 A-left[begin{array}{ll}1 & 2 \ 7 & 4end{array}right]=left[begin{array}{cc}3 & 0 \ 0 & -2end{array}right], ) then ( A ) is equal to ( mathbf{A} cdotleft[begin{array}{cc}2 & 1 \ 7 / 2 & 1end{array}right] ) В. ( left[begin{array}{cc}4 & 4 \ 7 / 2 & 1end{array}right] ) c. ( left[begin{array}{ll}3 & -1 \ 7 & 2end{array}right] ) D. None of these | 12 |
| 621 | If ( boldsymbol{A}=left[begin{array}{cc}cos boldsymbol{x} & sin boldsymbol{x} \ -sin boldsymbol{x} & cos boldsymbol{x}end{array}right] ) and ( boldsymbol{A}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A})=boldsymbol{k}left[begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right] ) then the value of ( k ) is ( mathbf{A} cdot sin x cos x ) B. c. D. | 12 |
| 622 | Find the inverse of the following matrix using transformation method. ( left[begin{array}{cc}2 & -3 \ -1 & 2end{array}right] ) | 12 |
| 623 | By using elementary transformation find the inverce of ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{2}-mathbf{1}end{array}right] ) | 12 |
| 624 | If the matrix ( left[begin{array}{ccc}mathbf{0} & mathbf{2} boldsymbol{beta} & mathbf{Upsilon} \ boldsymbol{alpha} & boldsymbol{beta} & -mathbf{Upsilon} \ boldsymbol{alpha} & -boldsymbol{beta} & mathbf{Upsilon}end{array}right] ) is orthogonal, then A ( cdot alpha=pm frac{1}{sqrt{2}} ) в. ( beta=pm frac{1}{sqrt{6}} ) c. ( _{gamma}=pm frac{1}{sqrt{3}} ) D. all of these | 12 |
| 625 | If the system of equations ( 3 x-2 y+ ) ( z=0, lambda x-14 y+15 z=0, x+2 y- ) ( mathbf{3} z=mathbf{0} ) have non zero solution, then find ( lambda ) | 12 |
| 626 | ff ( mathbf{A}=left[begin{array}{cccc}cos & boldsymbol{theta} & sin & boldsymbol{theta} \ boldsymbol{s i n} & boldsymbol{theta} & -boldsymbol{operatorname { c o s }} & boldsymbol{theta}end{array}right], mathbf{B}=left[begin{array}{cc}mathbf{1} & mathbf{0} \ -mathbf{1} & mathbf{1}end{array}right] ) ( mathbf{C}=boldsymbol{A} boldsymbol{B} boldsymbol{A}^{boldsymbol{T}} ) then ( A^{T} C A ) equals to ( (n quad in quad N) ) A. ( left[begin{array}{cc}-n & 1 \ 1 & 0end{array}right] ) B. ( left[begin{array}{cc}1 & -n \ 0 & 1end{array}right] ) C. ( left[begin{array}{cc}0 & 1 \ 1 & -nend{array}right] ) D. ( left[begin{array}{cc}1 & 0 \ -n & 1end{array}right] ) | 12 |
| 627 | Let ( boldsymbol{x} in boldsymbol{R} ) and let ( boldsymbol{P}=left[begin{array}{lll}1 & 1 & 1 \ 0 & 2 & 2 \ 0 & 0 & 3end{array}right], Q=left[begin{array}{lll}2 & x & x \ 0 & 4 & 0 \ x & x & 6end{array}right] ) and ( boldsymbol{R}=boldsymbol{P Q P}^{-1} ) Then which of the following is are correct This question has multiple correct options | 12 |
| 628 | ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{2} \ mathbf{- 1} & mathbf{0}end{array}right], boldsymbol{B}=left[begin{array}{ccc}mathbf{1} & mathbf{3} & mathbf{2} \ mathbf{4} & mathbf{- 1} & mathbf{3}end{array}right], ) then order of ( A B ) is A. ( 2 times 2 ) в. ( 3 times 3 ) ( c cdot 1 times 3 ) D. ( 3 times 2 ) | 12 |
| 629 | Find the inverse of the following matrices by the adjoining method ( left[begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1end{array}right] ) | 12 |
| 630 | ( mathbf{f} mathbf{P}=left{begin{array}{cc}frac{sqrt{3}}{2} & frac{1}{2} \ -frac{1}{2} & frac{sqrt{3}}{2}end{array}right} mathbf{A}=left{begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{0} & mathbf{1}end{array}right} ) and ( mathbf{Q}=mathbf{P} mathbf{A} mathbf{P}^{mathbf{T}} ) and ( mathbf{x}=mathbf{P}^{mathbf{T}} mathbf{Q}^{2005} mathbf{P} ) then ( x ) is equal to A. ( left{begin{array}{ll}1 & 2005 \ 0 & 1end{array}right} ) в. ( left{begin{array}{ll}4+2005 sqrt{3} & 6015 \ 2005 & 4-2005 sqrt{3}end{array}right} ) c. ( _{frac{1}{4}}left[begin{array}{cc}2+sqrt{3} & 1 \ -1 & 2-sqrt{3}end{array}right] ) D. ( frac{1}{4}left{begin{array}{ll}2005 & 2-sqrt{3} \ 2+sqrt{3} & 2005end{array}right} ) | 12 |
| 631 | If ( boldsymbol{A}=left(begin{array}{lll}1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1end{array}right) ) If ( A^{2}-4 A=p I ) where ( I ) and ( O ) are the unit matrix and the null matrix of order 3 respectively. Find the value of ( p ) A. ( p=2 ) в. ( p=3 ) ( mathbf{c} cdot p=4 ) D. ( p=5 ) | 12 |
| 632 | If ( boldsymbol{A}=left[begin{array}{cc}2 & boldsymbol{4} \ -mathbf{1} & boldsymbol{k}end{array}right] ) and ( boldsymbol{A}^{2}=mathbf{0}, ) find ( boldsymbol{k} ) | 12 |
| 633 | If ( boldsymbol{A}=left[begin{array}{cr}mathbf{2} & mathbf{3} \ -mathbf{1} & mathbf{4}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}-mathbf{3} & mathbf{1} \ mathbf{4} & -mathbf{2}end{array}right] ) then find ( boldsymbol{A}-boldsymbol{B} ) | 12 |
| 634 | Q Type your question Model 18 el ( x quad 20 ) Model 5 el Model ( z ) 19 11 The table above shows the number of TV sets that were sold during a three-day sale. The prices of models ( X, Y ) and ( Z ) ( operatorname{are} $ 99, $ 199, ) and ( $ 299, ) respectively. Which of the following matrix representations gives the total income, in dollars, received from the sale of the TV sets for each of the three days? ( mathbf{A} ) ( begin{array}{ll}199 & 299end{array} ) ( left[begin{array}{ccc}20 & 18 & 3 \ 16 & 5 & 8 \ 19 & 11 & 10end{array}right][99 ) B. ( left[begin{array}{ccc}20 & 18 & 3 \ 16 & 5 & 8 \ 19 & 11 & 10end{array}right]left[begin{array}{c}99 \ 199 \ 299end{array}right] ) ( c ) ( 199 quad 2 ) 99 299]( left[begin{array}{ccc}20 & 18 & 3 \ 16 & 5 & 8 \ 19 & 11 & 10end{array}right] ) D. ( left[begin{array}{c}99 \ 199 \ 299end{array}right] quadleft[begin{array}{ccc}20 & 18 & 3 \ 16 & 5 & 8 \ 19 & 11 & 10end{array}right] ) E . ( bullet cdot 199left[begin{array}{ccc}20 & 18 & 3 \ 16 & 5 & 8 \ 19 & 11 & 10end{array}right]+299left[begin{array}{ccc}20 & 18 & 3 \ 16 & 5 & 8 \ 19 & 11 & 10end{array}right] ) | 12 |
| 635 | Let ( A ) and ( B ) be two symmetic matrices of order 3 Statement ( -1: boldsymbol{A}(boldsymbol{B} boldsymbol{A}) ) and ( (boldsymbol{A B}) boldsymbol{A} ) are symmetric matrices. Statement – ( 2: A B ) is symmetric matrix if matrix multiplication of ( boldsymbol{A} ) and ( boldsymbol{B} ) is commutative. A. Statement-1 is True, Statement-2 is True, Statementis a correct explanation for Statement- B. Statement-1 is True, Statement-2 is True, Statementis NOT a correct explanation for Statement- c. Statement-1 is True, Statement-2 is False D. Statement-1 is False, Statement-2 is True | 12 |
| 636 | If ( A ) and ( B ) are matrices of the same order, then ( A B^{T}-B A^{T} ) is a A. Skew-symmetric matrix B. Null matrix c. Unit matrix D. symmetric matrix | 12 |
| 637 | Two ( n times n ) square matrices ( A ) and ( B ) are said to be similar if there exists a non- singular matrix ( boldsymbol{P} ) such that ( boldsymbol{P}^{-1} boldsymbol{A} boldsymbol{P}=boldsymbol{B} ) If ( A ) and ( B ) are similar matrices such that ( operatorname{det}(A)=1, ) then A. ( operatorname{det}(B)=1 ) B. ( operatorname{det}(A)+operatorname{det}(B)=0 ) c. ( operatorname{det}(B)=-1 ) D. none of these | 12 |
| 638 | Two ( n times n ) square matrices ( A ) and ( B ) are said to be similar if there exists a non- singular matrix ( boldsymbol{P} ) such that ( boldsymbol{P}^{-1} boldsymbol{A} boldsymbol{P}=boldsymbol{B} ) If ( A ) and ( B ) are similar and ( B ) and ( C ) are similar, then A. ( A B ) and ( B C ) are similar B. ( A ) and ( C ) are similar c. ( A+C ) and ( B ) are similar D. none of these | 12 |
| 639 | If ( A ) be a skew symmetric matrix of order ( m ) than ( A+A^{prime} ) is a A. Nilpotent matrix B. Orthogonal matrix c. Null matrix D. Skew symmetric | 12 |
| 640 | Matrices ( A ) and ( B ) will be inverse of each other only if ( mathbf{A} cdot A B=B A ) в. ( A B=0, B A=I ) c. ( A B=B A=0 ) D. ( A B=B A=I ) | 12 |
| 641 | What is the order of the product ( left[begin{array}{lll}boldsymbol{x} & boldsymbol{y} & boldsymbol{z}end{array}right]left[begin{array}{lll}boldsymbol{a} & boldsymbol{h} & boldsymbol{g} \ boldsymbol{h} & boldsymbol{b} & boldsymbol{f} \ boldsymbol{g} & boldsymbol{f} & boldsymbol{c}end{array}right]left[begin{array}{l}boldsymbol{x} \ boldsymbol{y} \ boldsymbol{z}end{array}right] ) ( A cdot 3 times 1 ) в. ( 1 times 1 ) ( c cdot 1 times 3 ) D. ( 3 times 3 ) | 12 |
| 642 | If ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{9} & mathbf{- 7} \ boldsymbol{i} & boldsymbol{omega}^{boldsymbol{n}} & mathbf{8} \ mathbf{1} & boldsymbol{6} & boldsymbol{omega}^{mathbf{2 n}}end{array}right] ) where ( boldsymbol{i}=sqrt{-mathbf{1}} ) and ( omega ) is complex cube root of unity, then ( operatorname{tr}(A) ) will be This question has multiple correct options A. ( 1, ) if ( n=3 k, k in N ) B. ( 3, ) if ( n=3 k, k in N ) c. ( 0, ) if ( n neq 3 k, k epsilon in N ) D. ( -1, ) if ( n neq 3 k, k epsilon in N ) | 12 |
| 643 | Express the square matrix ( A ) as the sum of a symmetric and a skewsymmetric matrix. | 12 |
| 644 | Find the values of ( x, y ) and ( z ) if ( left[begin{array}{lll}boldsymbol{x} & mathbf{5} & mathbf{4} \ mathbf{5} & mathbf{9} & mathbf{1}end{array}right]=left[begin{array}{lll}mathbf{3} & mathbf{5} & boldsymbol{z} \ mathbf{5} & boldsymbol{y} & mathbf{1}end{array}right] ) | 12 |
| 645 | Using elementary tansormations, find the inverse of each of the matrices, if it exists in ( left[begin{array}{ll}1 & 3 \ 2 & 7end{array}right] ) | 12 |
| 646 | If ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{2} \ mathbf{3} & mathbf{0} & -mathbf{2} \ mathbf{2} & mathbf{0} & mathbf{3}end{array}right] ) then ( (boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}) boldsymbol{A}= ) ( mathbf{A} cdotleft[begin{array}{ccc}13 & 0 & 0 \ 0 & 13 & 0 \ 0 & 0 & 13end{array}right] ) ( mathbf{B} cdotleft[begin{array}{ccc}7 & 0 & 0 \ 0 & 7 & 0 \ 0 & 0 & 7end{array}right] ) ( mathbf{C} cdotleft[begin{array}{ccc}-7 & 0 & 0 \ 0 & 0 & -7 \ 0 & 0 & -7end{array}right] ) D. ( left[begin{array}{ccc}11 & 0 & 0 \ 0 & 11 & 0 \ 0 & 0 & 11end{array}right] ) | 12 |
| 647 | There are 6 Higher Secondary Schools, 8 High Schools and 13 Primary Schools in a town. Represent these data in the form of ( 3 times 1 ) and ( 1 times 3 ) matrices. | 12 |
| 648 | f ( boldsymbol{A}=operatorname{diag}(mathbf{2}-mathbf{5} mathbf{9}), boldsymbol{B}=operatorname{diag}(mathbf{1} mathbf{1}- ) 4) and ( C=operatorname{diag}(-634), ) then find ( B+ ) ( C-2 A ) | 12 |
| 649 | If ( mathbf{2} boldsymbol{A}+boldsymbol{B}=left[begin{array}{cc}mathbf{1} & mathbf{-} mathbf{1} \ mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{-} mathbf{2}end{array}right] boldsymbol{A}- ) ( mathbf{2} boldsymbol{B}=left[begin{array}{cc}mathbf{0} & mathbf{1} \ mathbf{- 2} & mathbf{0} \ mathbf{1} & mathbf{- 1}end{array}right] ) find ( boldsymbol{A} ) and ( boldsymbol{B} ) | 12 |
| 650 | Find the inverse of the matrix ( $ $ mathrm{~ l l e f t [ l b e g i n ~ { a r r a y } ~} mid 5 & 11 backslash 14 & 9 ) ( mathrm{~ l e n d ~ { a r r a y } l r i g h t ] $ $ ~ b y ~ e l e m e n t a r y ~} ) transformations. | 12 |
| 651 | The set of natural numbers is divided into arrays of rows and columns in the form of matrices as ( boldsymbol{A}_{1}=(1), boldsymbol{A}_{2}=left(begin{array}{ll}2 & 3 \ 4 & 5end{array}right), boldsymbol{A}_{3}= ) ( left(begin{array}{ccc}mathbf{6} & mathbf{7} & mathbf{8} \ mathbf{9} & mathbf{1 0} & mathbf{1 1} \ mathbf{1 2} & mathbf{1 3} & mathbf{1 4}end{array}right) ldots mathbf{s o} ) on Find the value of ( boldsymbol{T}_{boldsymbol{r}}left(boldsymbol{A}_{mathbf{1 0}}right) ) ( left[mathrm{Note}: T_{r}(A) ) denotes sum of diagonal right. elements of ( boldsymbol{A} . ) A . 3355 в. 3434 ( c .5533 ) D. None of these | 12 |
| 652 | Find the inverse of the matrix ( left[begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & 1end{array}right] ) | 12 |
| 653 | 42. If [ 1 17.51 27 51 37 If ſo 1] [o 13o 1… [1 n-1] [1 787 Lo 1 )=l6 78). then the inverse of o 1 is: [JEE M 2019 –9 April (M] 1 -17 | 12 |
| 654 | Let ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & mathbf{4} \ mathbf{3} & mathbf{2}end{array}right], boldsymbol{B}=left[begin{array}{cc}mathbf{1} & mathbf{3} \ -mathbf{2} & mathbf{5}end{array}right] ) and ( boldsymbol{C}=left[begin{array}{cc}mathbf{- 2} & mathbf{5} \ mathbf{3} & mathbf{4}end{array}right] . ) Find: ( boldsymbol{A}-mathbf{2} boldsymbol{B}+mathbf{3} boldsymbol{C} ) | 12 |
| 655 | If ( A ) is ( 2 times 2 ) matrix such that ( A^{2}=0 ) then ( t r(A) ) is A B. ( c cdot-1 ) D. none of these | 12 |
| 656 | If the number of elements in a matrix is 60 then how many different order of matrix are possible A . 12 B. 6 ( c cdot 24 ) D. none of these | 12 |
| 657 | If ( A ) is ( 2 times 3 ) matrix and ( A B ) is a ( 2 times 5 ) matrix, then ( B ) must be a A. ( 3 times 5 ) matrix B. ( 5 times 3 ) matrix c. ( 3 times 2 ) matrix D. ( 5 times 2 ) matrix | 12 |
| 658 | If ( P=left[begin{array}{lll}4 & 3 & 2end{array}right] ) and ( Q=left[begin{array}{lll}-1 & 2 & 3end{array}right] ) then ( P-Q= ) ( left.begin{array}{lll}text { A } cdot[6 & -1 & -4end{array}right] ) B . ( left[begin{array}{lll}2 & -1 & -4end{array}right] ) ( mathbf{c} cdotleft[begin{array}{lll}6 & 1 & 4end{array}right] ) D・[-4 -1 6] | 12 |
| 659 | Consider three matrices ( A= ) ( left[begin{array}{ll}mathbf{2} & mathbf{1} \ mathbf{4} & mathbf{1}end{array}right] quad boldsymbol{B}=left[begin{array}{ll}mathbf{3} & mathbf{4} \ mathbf{2} & mathbf{3}end{array}right] quad ) and ( quad boldsymbol{C}= ) ( left[begin{array}{cc}mathbf{3} & -mathbf{4} \ -mathbf{2} & mathbf{3}end{array}right] ) The value of the sum ( t r(A)+ ) ( operatorname{tr}left(frac{A B C}{2}right)+operatorname{tr}left(frac{A(B C)^{2}}{4}right)+ ) ( operatorname{tr}left(frac{boldsymbol{A}(boldsymbol{B C})^{3}}{boldsymbol{8}}right)+ldots ldots ldots .+infty ) is ( (operatorname{tr}(A) text { denotes trace of a matrix } A) ) A . 6 B. 9 c. 12 D. none of these | 12 |
| 660 | If ( boldsymbol{A}=left(begin{array}{ll}mathbf{3} & mathbf{5} \ mathbf{7} & mathbf{9}end{array}right) ) is written as ( boldsymbol{A}=boldsymbol{P}+boldsymbol{Q} ) where ( boldsymbol{P} ) is a symmetric matrix and ( boldsymbol{Q} ) is skew-symmetric matrix, then write the ( operatorname{matrix} boldsymbol{P} ) | 12 |
| 661 | If ( boldsymbol{A}=left[begin{array}{ccc}boldsymbol{4} & boldsymbol{1} & boldsymbol{0} \ boldsymbol{1} & boldsymbol{-} boldsymbol{2} & boldsymbol{2}end{array}right], boldsymbol{B}= ) ( left[begin{array}{ccc}boldsymbol{2} & boldsymbol{0} & -boldsymbol{1} \ boldsymbol{3} & boldsymbol{1} & boldsymbol{4}end{array}right], boldsymbol{C}=left[begin{array}{c}boldsymbol{1} \ boldsymbol{2} \ -mathbf{1}end{array}right] ) and ( left(boldsymbol{3} boldsymbol{B}-boldsymbol{t}_{boldsymbol{}}right) ) ( boldsymbol{2} boldsymbol{A}) boldsymbol{C}+boldsymbol{2} boldsymbol{X}=boldsymbol{0} operatorname{then} boldsymbol{X}= ) A ( cdot frac{1}{2}left[begin{array}{c}3 \ 13end{array}right] ) B ( cdot frac{1}{2}left[begin{array}{c}3 \ -13end{array}right] ) c. ( frac{1}{2}left[begin{array}{c}-3 \ 13end{array}right] ) D. ( left[begin{array}{c}3 \ -13end{array}right] ) | 12 |
| 662 | The number of different possible orders of matrices having 18 identical elements is ( A cdot 3 ) B. ( c cdot 6 ) D. 4 | 12 |
| 663 | If ( A ) is a square matrix with ( |A|=8 ) Find the value of ( left|boldsymbol{A} boldsymbol{A}^{-1}right| ) | 12 |
| 664 | A square non-singular matrix A satisfies ( boldsymbol{A}^{2}-boldsymbol{A}+boldsymbol{2} boldsymbol{I}=boldsymbol{0}, ) then ( boldsymbol{A}^{-1}= ) ( mathbf{A} cdot I-A ) B ( cdot frac{1}{2}(I-A) ) c. ( I+A ) D ( cdot frac{1}{2}(I+A) ) | 12 |
| 665 | ( mathbf{f} A^{prime}=left[begin{array}{cc}mathbf{3} & mathbf{4} \ -mathbf{1} & mathbf{2} \ mathbf{0} & mathbf{1}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{ccc}-mathbf{1} & mathbf{2} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3}end{array}right] ) then verify that: ( (i)(A+B)^{prime}=A^{prime}+B^{prime} ) ( (i i)(A-B)^{prime}=A^{prime}-B^{prime} ) | 12 |
| 666 | If ( boldsymbol{A} ) is a skew symmetric matrix of even order then ( operatorname{det}(A) ) is ( mathbf{A} cdot mathbf{0} ) B ( . neq 0 ) c. non zero perfect square D. None of these | 12 |
| 667 | The number of ( A ) in ( T_{p} ) such that the trace of ( A ) is not divisible by ( p ) but ( operatorname{det}(A) ) is divisible by ( p ) is? [Note: The trace of a matrix is the sum of its diagonal entries. A ( cdot(p-1)left(p^{2}-p+1right) ) В . ( p^{3}-(p-1)^{2} ) c. ( (p-1)^{2} ) D. ( (p-1)left(p^{2}-2right) ) | 12 |
| 668 | [ text { If } boldsymbol{A}=left[begin{array}{cc} mathbf{2} & mathbf{2} \ mathbf{- 3} & mathbf{1} \ mathbf{4} & mathbf{0} end{array}right], boldsymbol{B}=left[begin{array}{cc} mathbf{6} & mathbf{2} \ mathbf{1} & mathbf{3} \ mathbf{0} & mathbf{4} end{array}right], text { find } ] matrix ( C ) such that ( A+B+C=0 ) where ( boldsymbol{O} ) is the zero matrix. | 12 |
| 669 | If ( X ) is a ( 2 times 3 ) matrix such that ( left|boldsymbol{X}^{boldsymbol{T}} boldsymbol{X}right| neq mathbf{0} ) and ( boldsymbol{A}=boldsymbol{I}_{2}- ) ( Xleft(X^{T} Xright)^{-1} X^{T} ) then ( A^{2} ) is equal to: ( left(X^{T} text { denotes transpose of matrix } Xright) ) A. ( A ) B. ( I ) ( c cdot A^{-1} ) D. ( A X ) | 12 |
| 670 | If ( A ) and ( B ) are square matrices of same order, then which of the following is correct – A. ( A+B=B+A ) в. ( A+B=A B ) c. ( A B=B A ) D. ( A B=B+A ) | 12 |
| 671 | ff ( D=left|begin{array}{ccc}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}end{array}right| ) and ( D_{0}= ) ( left|begin{array}{ccc}boldsymbol{k} boldsymbol{a}_{1} & boldsymbol{k} boldsymbol{b}_{1} & boldsymbol{k} boldsymbol{c}_{1} \ boldsymbol{k} boldsymbol{a}_{2} & boldsymbol{k} boldsymbol{b}_{2} & boldsymbol{k} boldsymbol{c}_{2} \ boldsymbol{k} boldsymbol{a}_{3} & boldsymbol{k} boldsymbol{b}_{3} & boldsymbol{k} boldsymbol{c}_{3}end{array}right| ) then show that ( boldsymbol{D}_{0}=boldsymbol{k}^{3} boldsymbol{D} ) | 12 |
| 672 | Assertion ( operatorname{Let} boldsymbol{A}=left[begin{array}{ll}boldsymbol{a}_{11} & boldsymbol{a}_{12} \ boldsymbol{a}_{21} & boldsymbol{a}_{22}end{array}right], boldsymbol{X}=left[begin{array}{l}boldsymbol{x}_{1} \ boldsymbol{x}_{2}end{array}right], boldsymbol{Y}= ) ( left[begin{array}{l}boldsymbol{y}_{1} \ boldsymbol{y}_{2}end{array}right] ) If ( boldsymbol{X}^{prime} boldsymbol{A} boldsymbol{X}=boldsymbol{0} ) for each ( boldsymbol{X}, ) then ( boldsymbol{A} ) must be a symmetric matrix. Reason If ( boldsymbol{A} ) is symmetric and ( boldsymbol{X}^{prime} boldsymbol{A} boldsymbol{X}=mathbf{0} ) for ( operatorname{each} X, ) then ( A=0 ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 673 | The matrix A satisfies the matrix equation if ( A=left[begin{array}{lll}1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1end{array}right] ) ( mathbf{A} cdot A^{2}-4 A-5 I=0 ) B ( cdot A^{2}-4 A-5=0 ) C ( cdot A^{2}+4 A-5 I=0 ) D ( cdot A^{2}+4 A-5=0 ) | 12 |
| 674 | Construct a ( 3 times 2 ) matrix ( A=left[a_{i j}right] ) whose elements are given by ( boldsymbol{a}_{boldsymbol{i} j}= ) ( frac{(i-2 j)^{2}}{2} ) | 12 |
| 675 | ( boldsymbol{A}=left(begin{array}{cc}mathbf{5} & mathbf{2} \ mathbf{7} & mathbf{3}end{array}right) ) and ( boldsymbol{B}=left(begin{array}{cc}mathbf{2} & mathbf{- 1} \ mathbf{- 1} & mathbf{2}end{array}right) ) verify that ( (boldsymbol{A B})^{boldsymbol{T}}=boldsymbol{B}^{boldsymbol{T}} boldsymbol{A}^{boldsymbol{T}} ) | 12 |
| 676 | If ( boldsymbol{A}=left[begin{array}{cc}-mathbf{3} & mathbf{5} \ mathbf{5} & mathbf{0} \ -mathbf{7} & mathbf{4}end{array}right] ) and ( boldsymbol{B}= ) ( left[begin{array}{ccc}mathbf{3} & mathbf{- 5} & mathbf{7} \ mathbf{- 5} & mathbf{0} & mathbf{- 4}end{array}right], ) then find ( boldsymbol{A}+boldsymbol{B}^{boldsymbol{T}} ) A . 0 в. ( 2 B ) c. ( 2 A^{text {न }} ) D. ( 2 B^{text {T }} ) | 12 |
| 677 | Let ( k ) be a positive real number and let [ begin{array}{l} boldsymbol{A}=left[begin{array}{ccc} 2 boldsymbol{k}-mathbf{1} & mathbf{2} sqrt{boldsymbol{k}} & mathbf{2} sqrt{boldsymbol{k}} \ mathbf{2} sqrt{boldsymbol{k}} & mathbf{1} & -mathbf{2} boldsymbol{k} \ -mathbf{2} sqrt{boldsymbol{k}} & mathbf{2} boldsymbol{k} & mathbf{1} end{array}right] \ boldsymbol{B}=left[begin{array}{ccc} mathbf{0} & mathbf{2} boldsymbol{k}-mathbf{1} & sqrt{boldsymbol{k}} \ mathbf{1}-mathbf{2} boldsymbol{k} & mathbf{0} & mathbf{2} sqrt{boldsymbol{k}} \ -sqrt{boldsymbol{k}} & -mathbf{2} sqrt{boldsymbol{k}} & mathbf{0} end{array}right] \ text { If det }(boldsymbol{A} boldsymbol{d} boldsymbol{j}(boldsymbol{A}))+operatorname{det}(boldsymbol{A} boldsymbol{d} boldsymbol{j}(boldsymbol{B}))=mathbf{1 0}^{boldsymbol{o}} end{array} ] then ( [k] ) is equal to 4 3.6 ( c ) ( D ) | 12 |
| 678 | If a matrix is of order ( 2 times 3 ), then the number of elements in the matrix is A. 5 B. 6 ( c cdot 2 ) D. 3 | 12 |
| 679 | Find the inverse of the following matrices by the adjoining method ( left[begin{array}{cc}-1 & 5 \ -3 & 2end{array}right] ) | 12 |
| 680 | Construct a ( 2 times 3 ) matrix ( A=left[a_{i j}right] ) whose elements are given by ( a_{i j}= ) ( left{begin{array}{l}i-j i geq j \ i+j i<jend{array}right. ) | 12 |
| 681 | Two matrices are equal if and only if they have the and corresponding elements are A. rows, equal B. order, equal C . columns, equal D. order, unequal | 12 |
| 682 | ( f(alpha, beta, gamma ) are three real numbers and ( boldsymbol{A}= ) ( left[begin{array}{ccc}mathbf{1} & cos (boldsymbol{alpha}-boldsymbol{beta}) & cos (boldsymbol{alpha}-gamma) \ cos (boldsymbol{beta}-boldsymbol{alpha}) & mathbf{1} & cos (boldsymbol{beta}-gamma) \ cos (gamma-boldsymbol{alpha}) & cos (gamma-boldsymbol{beta}) & mathbf{1}end{array}right] ) then This question has multiple correct options A. ( A ) is symmetric B. ( A ) is orthogona c. ( A ) is singular D. ( A ) is not invertible | 12 |
| 683 | ( fleft(begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4end{array}right], ) then show that [ |mathbf{3} boldsymbol{A}|=mathbf{2 7}|boldsymbol{A}| ] | 12 |
| 684 | 5. Let A= 0 0 (-1 0 -1 0 -1) 0 . The only correct 0 statement about the matrix A is (a) A2 = 1 (b) A=(-1)1, where I is a unit matrix (C) A-1 does not exist (d) Ais a zero matrix | 12 |
| 685 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) then find ( boldsymbol{A}+boldsymbol{A}^{boldsymbol{T}} ) | 12 |
| 686 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{0} & mathbf{5} \ mathbf{0} & mathbf{0}end{array}right] ) and ( boldsymbol{f}(boldsymbol{x})=mathbf{1}+boldsymbol{x}+ ) ( boldsymbol{x}^{2}+ldots ldots+boldsymbol{x}^{6}, operatorname{then} boldsymbol{f}(boldsymbol{A})= ) A . 0 в. ( left[begin{array}{ll}1 & 5 \ 0 & 1end{array}right] ) c. ( left[begin{array}{ll}1 & 5 \ 0 & 0end{array}right] ) D. ( left[begin{array}{ll}0 & 5 \ 1 & 1end{array}right] ) | 12 |
| 687 | A matrix having ( m ) rows and ( n ) columns with ( m=n ) is said to be a A. rectangular matrix B. square matrix c. identity matrix D. scalar matrix | 12 |
| 688 | a O 1 0 10. If A = and B= , then value of a for which 5 1 | (2003) A2 =B, is (a) 1 (c) 4 (b) -1 (d) no real values | 12 |
| 689 | If ( boldsymbol{A}=left[begin{array}{ll}1 & -2 \ 5 & -3end{array}right], ) then ( A+A^{T} ) equals A. ( left[begin{array}{cc}2 & 3 \ 3 & -6end{array}right] ) В. ( left[begin{array}{cc}2 & -4 \ 10 & -6end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}2 & 4 \ -10 & 6end{array}right] ) D. None of these | 12 |
| 690 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose elements are given by ( a_{i j}=2 i- ) ( boldsymbol{j} ) | 12 |
| 691 | Find the values of ( x, y ) and ( z ) if ( left[begin{array}{lll}boldsymbol{x} & mathbf{5} & mathbf{4} \ mathbf{5} & mathbf{9} & mathbf{1}end{array}right]=left[begin{array}{lll}mathbf{3} & mathbf{5} & boldsymbol{z} \ mathbf{5} & boldsymbol{y} & mathbf{1}end{array}right] ) | 12 |
| 692 | If ( A ) is a non-singular matrix, then This question has multiple correct options ( mathbf{A} cdot A^{-1} ) is symmetric if ( A ) is symmeteric B. ( A^{-1} ) is skew-symmetric if ( A ) is symmeteric ( mathbf{C} cdotleft|A^{-1}right|=|A| ) D ( cdotleft|A^{-1}right|=|A|^{-1} ) | 12 |
| 693 | If ( A ) is a square matrix such that ( A^{2}= ) ( I, ) then ( A^{-1} ) is equal to A . ( I ) в. 0 ( c . A ) ( mathbf{D} cdot I+A ) | 12 |
| 694 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) | 12 |
| 695 | If ( A=left[begin{array}{ll}1 & 2 \ 3 & 4end{array}right], B=left[begin{array}{ll}2 & 3 \ 4 & 5end{array}right], ) and ( 4 A-3 B+C ) ( =0, ) then ( C= ) A. ( left[begin{array}{cc}2 & -1 \ 0 & 1end{array}right] ) В. ( left[begin{array}{cc}2 & 1 \ 0 & -1end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-2 & 1 \ 0 & -1end{array}right] ) D. None | 12 |
| 696 | If ( boldsymbol{A}=left[begin{array}{ccc}1 & 2 & -3 \ 5 & 0 & 2 \ 1-1 & 1end{array}right], B=left[begin{array}{ccc}3 & -12 \ 4 & 2 & 5 \ 2 & 0 & 3end{array}right] ) and ( boldsymbol{C}=left[begin{array}{ccc}mathbf{4} & mathbf{1} & mathbf{2} \ mathbf{0} & mathbf{3} & mathbf{2} \ mathbf{1}-mathbf{2 3}end{array}right], ) then compute ( (boldsymbol{A}+boldsymbol{B}) ) and ( (B-C) ) Also, verify that ( boldsymbol{A}+(boldsymbol{B}-boldsymbol{C})= ) ( (boldsymbol{A}+boldsymbol{B})-boldsymbol{C} ) | 12 |
| 697 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 5} & mathbf{7} \ mathbf{0} & mathbf{7} & mathbf{9} \ mathbf{1 1} & mathbf{8} & mathbf{9}end{array}right], ) then trace of matrix A is. A . 17 B. 25 ( c .3 ) D. 12 | 12 |
| 698 | ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{- 1} & mathbf{- 2} & mathbf{- 3}end{array}right], ) then A is a nilpotent matrix of index ( A ) B. 3 ( c cdot 4 ) D. 5 | 12 |
| 699 | f ( boldsymbol{T} boldsymbol{r}(boldsymbol{A})=boldsymbol{6}, ) then ( boldsymbol{T} boldsymbol{r}(boldsymbol{4} boldsymbol{A})= ) ( A cdot frac{3}{2} ) B. 2 c. 12 D. 24 | 12 |
| 700 | Define a diagonal matrix. | 12 |
| 701 | Let ( M ) and ( N ) be two ( 3 times 3 ) be two non- singular skew symmetric matrices such that ( M N=N M ) further, if ( M neq ) ( n^{2} ). If ( P T ) denotes the transpose of ( P ) then ( M^{2} N^{2}left(M^{T} Nright)^{-1}(M N)^{-1} T ) is equal to | 12 |
| 702 | ff ( 2left[begin{array}{ll}3 & 4 \ 5 & xend{array}right]+left[begin{array}{ll}1 & y \ 0 & 1end{array}right]=left[begin{array}{cc}7 & 0 \ 10 & 5end{array}right] ) matrices. Find the value of ( x & y ) | 12 |
| 703 | If ( A ) and ( B ) are invertible matrices of order 3. ( |boldsymbol{A}|=mathbf{2} ) and ( left|(boldsymbol{A B})^{-1}right|=-frac{mathbf{1}}{mathbf{6}} ) Find ( |boldsymbol{B}| ) | 12 |
| 704 | If ( A ) be a ( 3 times 3 ) matrix and ( I ) be the unit matrix of that order such that ( boldsymbol{A}= ) ( A^{2}+I ) then ( A^{-1} ) is equal to A . ( A ) в. ( A+I ) c. ( I-A ) D. ( A-I ) | 12 |
| 705 | Find the transpose of matrix ( left[begin{array}{ll}2 & 5 \ 1 & 3end{array}right] ) | 12 |
| 706 | Determine the value of ( (x+y) ) if ( left[begin{array}{cc}2 x+y & 4 x \ 5 x-7 & 4 xend{array}right]=left[begin{array}{cc}7 & 7 y-12 \ y & x+6end{array}right] ) | 12 |
| 707 | If ( A ) and ( B ) are square matrices of the same order, explain, why in general? ( (A+B)^{2} neq A^{2}+2 A B+B^{2} ) | 12 |
| 708 | Let ( A ) and ( B ) be two symmetric matrices of order 3 Statement-1: ( boldsymbol{A}(boldsymbol{B} boldsymbol{A}) ) and ( (boldsymbol{A B}) boldsymbol{A} ) are symmetric matrices. Statement-2: ( A B ) is symmetric matrix if matrix multiplication of ( boldsymbol{A} ) and ( boldsymbol{B} ) is commutative. A. Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1. B. Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement- 1 . c. Statement-1 is true, Statement-2 is false D. Statement-1 is false, Statement-2 is true. | 12 |
| 709 | ( operatorname{Given} boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{2} & mathbf{4} & mathbf{1} \ mathbf{2} & mathbf{3} & mathbf{1}end{array}right], boldsymbol{B}=left[begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{3} & mathbf{4}end{array}right] ) Find ( P ) such that ( B P A=left[begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 0end{array}right] ) | 12 |
| 710 | ( mathrm{IF} boldsymbol{B}=left[begin{array}{cc}mathbf{1} & mathbf{3} \ -mathbf{2} & mathbf{5}end{array}right] ) and ( boldsymbol{C}=left[begin{array}{cc}-mathbf{2} & mathbf{5} \ mathbf{3} & mathbf{4}end{array}right] ) then find the value of ( B-4 C ) | 12 |
| 711 | The element in the second row and third column of the matrix ( left[begin{array}{ccc}mathbf{4} & mathbf{5} & mathbf{- 6} \ mathbf{3} & mathbf{- 4} & mathbf{3} \ mathbf{2} & mathbf{1} & mathbf{0}end{array}right] ) is ( A cdot 3 ) B. ( c cdot 2 ) D. – | 12 |
| 712 | The equation, ( left[begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{y}end{array}right]left[begin{array}{ccc}mathbf{1} & boldsymbol{3} & mathbf{1} \ mathbf{0} & boldsymbol{2} & -mathbf{1} \ mathbf{0} & mathbf{0} & mathbf{1}end{array}right]left[begin{array}{l}mathbf{1} \ boldsymbol{x} \ boldsymbol{y}end{array}right]=[mathbf{0}] ) has for (i) ( y=0 ) (p) rational roots (ii) ( y=-1 ) (q) irrational roots (r) integral roots A ( . ) (i) ( (p) ) (ii) ( (r) ) B. (i) (q) (ii) (p) ( c cdot(i)(p)(text { ii) }(q) ) D. (i) (r) (ii) (p) | 12 |
| 713 | If the product of the matrices [ left[begin{array}{ll} 1 & 1 \ 0 & 1 end{array}right]left[begin{array}{ll} 1 & 2 \ 0 & 1 end{array}right]left[begin{array}{ll} 1 & 3 \ 0 & 1 end{array}right] ] ( left[begin{array}{cc}mathbf{1} & boldsymbol{n} \ mathbf{0} & mathbf{1}end{array}right]=left[begin{array}{cc}mathbf{1} & mathbf{3 7 8} \ mathbf{0} & mathbf{1}end{array}right], ) then ( boldsymbol{n} ) is equal to ( A cdot 27 ) B . 26 ( c .37 ) D. 37 | 12 |
| 714 | If ( A ) is a skew-symmetric matrix and ( n ) is a positive integer, then ( boldsymbol{A}^{boldsymbol{n}} ) is A. a symmetric matrix B. skew-symmetric matrix c. diagonal matrix D. none of these | 12 |
| 715 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{0} & mathbf{0} \ mathbf{4} & mathbf{0}end{array}right], ) then find ( boldsymbol{A}^{16} ) | 12 |
| 716 | Find matrix ( boldsymbol{A} ) such that ( left[begin{array}{cc}2 & -1 \ 1 & 0 \ -3 & 4end{array}right] A=left[begin{array}{cc}-1 & -8 \ 1 & -2 \ 9 & 22end{array}right] ) | 12 |
| 717 | ( mathbf{f} mathbf{A}=left[begin{array}{ccc}mathbf{0} & mathbf{1} & mathbf{4} \ mathbf{- 1} & mathbf{0} & mathbf{7} \ mathbf{- 4} & mathbf{- 7} & mathbf{0}end{array}right] ) then ( mathbf{A}^{mathbf{T}}= ) ( A ) в. – А ( c ) D. ( A^{2} ) | 12 |
| 718 | If ( A ) is a matrix of order ( 3 times 4 ) and ( B ) is a matrix of order ( 4 times 3, ) find the order of the matrix of ( boldsymbol{A B} ) | 12 |
| 719 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{3} & mathbf{7} \ mathbf{2} & mathbf{5}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{cc}-mathbf{3} & mathbf{2} \ mathbf{4} & -mathbf{1}end{array}right] ) find the matrix ( C ) if ( 2 C=A+B ) | 12 |
| 720 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose elements ( a_{i j} ) are given by: ( boldsymbol{a}_{boldsymbol{i} j}=left|frac{boldsymbol{3} boldsymbol{i}-boldsymbol{j}}{boldsymbol{2}}right| ) | 12 |
| 721 | Find the values of ( x, y ) and ( z ) from the matrix equation. ( left(begin{array}{cc}5 x+2 & y-4 \ 0 & 4 z+6end{array}right)=left(begin{array}{cc}12 & -8 \ 0 & 2end{array}right) ) | 12 |
| 722 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1}^{2} & mathbf{2}^{mathbf{2}} & mathbf{3}^{2} \ mathbf{2}^{mathbf{2}} & mathbf{3}^{mathbf{2}} & mathbf{4}^{2} \ mathbf{3}^{mathbf{2}} & mathbf{4}^{mathbf{2}} & mathbf{5}^{mathbf{2}}end{array}right] ) then ( |boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A}|= ) ( A ) B . 16 ( c cdot 64 ) D. 128 | 12 |
| 723 | Using elementary tranormtion, find the [ text { inverse of }left[begin{array}{ccc} mathbf{1} & mathbf{3} & -mathbf{2} \ -mathbf{3} & mathbf{0} & -mathbf{5} \ mathbf{2} & mathbf{5} & mathbf{0} end{array}right] ] | 12 |
| 724 | Find the inverse of the following matrix by using elementary row transformation ( left[begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3end{array}right] ) | 12 |
| 725 | If ( A=left[begin{array}{l}45 \ 21end{array}right], ) then show that ( A^{-1}= ) ( frac{1}{6}(A-5 I) ) | 12 |
| 726 | [1 2 32. If A = 2 1 a 2 2 -2 is a matrix satisfying the equation 6 AAT=91, where I is 3 x 3 identity matrix, then the ordered pair (a, b) is equal to: (JEE M 2015] (a) (2,1) (b) (-2,-1) (c) (2, -1) (d) (-2,1) | 12 |
| 727 | If ( boldsymbol{A}=left(begin{array}{ccc}mathbf{8} & mathbf{5} & mathbf{2} \ mathbf{1} & mathbf{- 3} & mathbf{4}end{array}right), ) then find ( boldsymbol{A}^{boldsymbol{T}} ) and ( left(A^{T}right)^{T} ) | 12 |
| 728 | ( left[begin{array}{ccc}1 & 0 & 2 \ -1 & 1 & -2 \ 0 & 2 & 1end{array}right]+left[begin{array}{ccc}5 & 1 & -2 \ 1 & 1 & 0 \ -2 & -2 & 1end{array}right] ) What will be the sum of the diagonal elements of the resultant matrix. A . 10 B. 6 ( c ) ( D ) | 12 |
| 729 | If ( A ) and ( B ) are two square matrix of order ( n ) then prove that ( :(A B)^{-1}= ) ( boldsymbol{B}^{-1} boldsymbol{A}^{-1} ) | 12 |
| 730 | If ( A ) is a square matrix of order ( n ), then ( |boldsymbol{k} boldsymbol{A}|= ) A ( . k|A| ) в. ( k^{n}|A| ) c. ( k^{-n}|A| ) D. | ( A mid ) | 12 |
| 731 | If ( boldsymbol{A}=left[begin{array}{ccc}-1 & 1 & -1 \ 3 & -3 & 3 \ 5 & 5 & 5end{array}right] ) and ( B= ) ( left[begin{array}{ccc}0 & 4 & 3 \ 1 & -3 & -3 \ -1 & 4 & 4end{array}right], ) then find ( A^{2}-B^{2} ) | 12 |
| 732 | ( mathbf{f}left[begin{array}{cc}boldsymbol{x}+mathbf{3} & mathbf{4} \ boldsymbol{y}-mathbf{4} & boldsymbol{x}+boldsymbol{y}end{array}right]=left[begin{array}{ll}mathbf{5} & mathbf{4} \ mathbf{3} & mathbf{9}end{array}right], ) find ( boldsymbol{x} ) and ( boldsymbol{y} ) | 12 |
| 733 | Show that the elements on the main diagonal of a skew-symmetric matrix are all zero. | 12 |
| 734 | ( left.begin{array}{l}text { Find } boldsymbol{a}, boldsymbol{b}, boldsymbol{c}, boldsymbol{d} text { if }left[begin{array}{cc}boldsymbol{d}+mathbf{1} & mathbf{1 0}+boldsymbol{a} \ mathbf{3} boldsymbol{b}-mathbf{2} & boldsymbol{a}-mathbf{4}end{array}right]= \ mathbf{2} quad mathbf{2} boldsymbol{a}+mathbf{1} \ boldsymbol{b}-mathbf{5} quad boldsymbol{4} boldsymbol{c}end{array}right] ) | 12 |
| 735 | If ( mathbf{0} leq[boldsymbol{x}]<mathbf{2},-mathbf{1} leq[boldsymbol{y}]<mathbf{1} ) and ( mathbf{1} leq ) ( [z]<3([.] ) denotes the greatest integer function) then the maximum value of determinant ( boldsymbol{Delta}=left|begin{array}{ccc}{[boldsymbol{x}]+mathbf{1}} & {[boldsymbol{y}]} & {[boldsymbol{z}]} \ {[boldsymbol{x}]} & {[boldsymbol{y}]+mathbf{1}} & {[boldsymbol{z}]} \ {[boldsymbol{x}]} & {[boldsymbol{y}]} & {[boldsymbol{z}]+mathbf{1}}end{array}right| ) ( A ) B. 2 ( c cdot 3 ) ( D ) | 12 |
| 736 | Find Order of matrix : ( left[begin{array}{lll}mathbf{1} & mathbf{2} & mathbf{6} \ mathbf{2} & mathbf{4} & mathbf{3}end{array}right] ) | 12 |
| 737 | If ( A ) is a square matrix, then ( a d j A^{T}- ) ( (a d j A)^{T} ) is equal to A ( cdot 2|A| ) B . ( 2|A| I ) c. null matrix D. unit matrix | 12 |
| 738 | ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 3} & mathbf{- 4} \ mathbf{- 1} & mathbf{3} & mathbf{4} \ mathbf{1} & mathbf{- 3} & mathbf{- 4}end{array}right] ) and ( mathbf{A}^{2}=boldsymbol{lambda} boldsymbol{I} ) then ( boldsymbol{lambda}= ) ( A cdot 0 ) B. ( c cdot frac{1}{2} ) D. – | 12 |
| 739 | Let three matrices ( A=left[begin{array}{ll}2 & 1 \ 4 & 1end{array}right] ; B= ) ( left[begin{array}{ll}mathbf{3} & mathbf{4} \ mathbf{2} & mathbf{3}end{array}right] ) and ( C=left[begin{array}{cc}mathbf{3} & -mathbf{4} \ -mathbf{2} & mathbf{3}end{array}right] ) then ( boldsymbol{t}_{boldsymbol{r}}(boldsymbol{A})+boldsymbol{t}_{boldsymbol{r}}left(frac{A B C}{2}right)+boldsymbol{t}_{boldsymbol{r}}left(frac{boldsymbol{A}(boldsymbol{B C})^{2}}{4}right)+ ) ( boldsymbol{t}_{boldsymbol{r}}left(frac{boldsymbol{A}(boldsymbol{B} C)^{3}}{boldsymbol{8}}right)+ldots+infty ) ( A cdot 6 ) B. c. 12 D. none of these | 12 |
| 740 | If ( boldsymbol{A}=operatorname{diag}[mathbf{2},-mathbf{3},-mathbf{5}], boldsymbol{B}= ) ( operatorname{diag}[4,-6,-3] ) and ( C=operatorname{diag}[-3,4,1] ) then find ( mathbf{2 A}+boldsymbol{B}-mathbf{5} boldsymbol{C} ) | 12 |
| 741 | ff ( A=left[begin{array}{ccc}-1 & 0 & 0 \ 0 & x & 0 \ 0 & 0 & mend{array}right] ) is a scalar matrix then ( boldsymbol{x}+boldsymbol{m}= ) ( mathbf{A} cdot mathbf{0} ) B. – ( c cdot-2 ) D. -3 | 12 |
| 742 | If ( A_{2 times 3}, B_{4 times 3} ) and ( C_{2 times 4} ) are three matrices then which of the following is/are defined? A ( cdot A C^{T} B ) в. ( B^{T} C^{T} A ) c. ( A B^{T} C ) D. All of these | 12 |
| 743 | Find the vector equation of the plane passing through points ( 4 i-3 j- ) ( k, 3 i+7 j-10 k ) and ( 2 i+5 j-7 k ) and show that the point ( i+2 j-3 k ) lies in the plane. | 12 |
| 744 | If ( A, B ) are square matrices of order ( 3, A ) is non-singular and ( A B=O ), then ( B ) is ( a ) A. Null matrix B. singular matrix c. Unit matrix D. Non-singular matrix | 12 |
| 745 | If ( l ) is an identity matrix and ( A ) is a square matrix such that ( A^{2}=A ), then find the value of ( (boldsymbol{l}+boldsymbol{A})^{2}-mathbf{3} boldsymbol{A} ) | 12 |
| 746 | If ( D_{1} ) and ( D_{2} ) are two ( 3 times 3 ) diagonal matrices, then A. ( D_{1} D_{2} ) is a diagonal matrix B. ( D_{1}+D_{2} ) is a diagonal matrix c. ( D_{1}^{2}+D_{2}^{2} ) is a diagonal matrix D. 1,2,3 are correct | 12 |
| 747 | Construct a ( 2 times 2 ) matrix ( A=left[a_{i j}right] ) whose element ( a_{i j} ) is ( frac{(i+j)^{2}}{2} ) | 12 |
| 748 | ( boldsymbol{A}=left[begin{array}{cc}cos boldsymbol{theta} & -sin boldsymbol{theta} \ sin boldsymbol{theta} & cos boldsymbol{theta}end{array}right] ) and ( boldsymbol{A B}=boldsymbol{B A}= ) ( I, ) then ( B ) is equal to ( mathbf{A} cdotleft[begin{array}{cc}-cos theta & sin theta \ sin theta & cos thetaend{array}right] ) ( mathbf{B} cdotleft[begin{array}{cc}cos theta & sin theta \ -sin theta & cos thetaend{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-sin theta & cos theta \ cos theta & sin thetaend{array}right] ) ( mathbf{D} cdotleft[begin{array}{cc}sin theta & -cos theta \ -cos theta & sin thetaend{array}right] ) | 12 |
| 749 | ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & mathbf{3} \ mathbf{5} & mathbf{7}end{array}right] boldsymbol{B}=left[begin{array}{cc}mathbf{0} & mathbf{4} \ -mathbf{1} & mathbf{7}end{array}right] boldsymbol{C}= ) ( left[begin{array}{cc}mathbf{1} & mathbf{0} \ mathbf{- 1} & mathbf{4}end{array}right] ) find ( boldsymbol{A} boldsymbol{C}+boldsymbol{B}^{2}-mathbf{1 0 C} ) | 12 |
| 750 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right]_{3 times 3} ) is a square matrix so that ( a_{i j}=i^{2}-j^{2}, ) then ( A ) is a A. unit matrix B. symmetric marix c. skew symmetric matrix D. orthogonal matrix | 12 |
| 751 | ( fleft(begin{array}{cc}4 & 8 \ -2 & -4end{array}right] ) find ( A^{2} ) | 12 |
| 752 | ( mathbf{f} mathbf{A}=left{begin{array}{ll}mathbf{0} & mathbf{2} \ mathbf{3} & -mathbf{4}end{array}right}, mathbf{k} mathbf{A}=left{begin{array}{ll}mathbf{0} & mathbf{3} mathbf{a} \ mathbf{2} mathbf{b} & mathbf{2 4}end{array}right} ) then arrange the values of ( k, a, b, ) in ascending order ( mathbf{A} cdot k, a, b ) в. ( b, a, k ) ( mathbf{c} cdot a, k, b ) ( mathbf{D} cdot b, k, a ) | 12 |
| 753 | Given a square matrix ( boldsymbol{A}=left[boldsymbol{a}_{boldsymbol{g}}right], ) where ( a_{g}=hat{i}^{2}-hat{j}^{2} . ) Then matrix ( A ) is a unit matrix or null matrix or a symmetric matrix a skew symmetric matrix. Select with a reason. | 12 |
| 754 | The matrix ( boldsymbol{B} ) is A. Symmetric B. Scalar c. Skew hermitian D. Skew- symmetric | 12 |
| 755 | Find the value of ( mathbf{x}, mathbf{y}, mathbf{z}left[begin{array}{cc}boldsymbol{x}+mathbf{2} & mathbf{6} \ mathbf{3} & mathbf{5} zend{array}right]= ) ( left[begin{array}{cc}mathbf{3} & boldsymbol{y}^{2}+mathbf{4} \ mathbf{3} & mathbf{2 0}end{array}right] ) | 12 |
| 756 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right], boldsymbol{i}, boldsymbol{j}=1,2 ) where ( boldsymbol{a}_{i j} ) is defined as ( a_{i j}=i^{2}+j^{2} ) then write the sum of the elements of the matrix ( boldsymbol{A} ) | 12 |
| 757 | If ( boldsymbol{A}=left[boldsymbol{a}_{boldsymbol{i j}}right] ) is a square matrix of even order such that ( left[a_{i j}right]=i^{2}-j^{2}, ) then A ( cdot A ) is a skew-symmetric matrix and ( |A|=0 ) B. ( A ) is symmetric matrix and |A| is a square C. ( A ) is symmetric matrix and ( |A|=0 ) D. none of these | 12 |
| 758 | ( mathbf{f} boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{0} & mathbf{5} & mathbf{7} \ mathbf{6} & mathbf{8} & mathbf{9}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{lll}mathbf{2} & mathbf{0} & mathbf{3} \ mathbf{3} & mathbf{0} & mathbf{5} \ mathbf{5} & mathbf{7} & mathbf{0}end{array}right] ) then find the value of ( 3 A-2 B ) | 12 |
| 759 | If the matrix ( left[begin{array}{ccc}1 & 3 & lambda+2 \ 2 & 4 & 8 \ 3 & 5 & 10end{array}right] ) is singular ( operatorname{then} lambda= ) A . -2 B. 4 ( c cdot 2 ) D. – | 12 |
| 760 | Find teh adjoint of the matrix ( boldsymbol{A}= ) ( left(begin{array}{ccc}1 & 0 & -1 \ 3 & 4 & 5 \ 0 & -6 & -7end{array}right) ) and hence find the matrix ( boldsymbol{A}^{-1} ) | 12 |
| 761 | If ( A ) and ( B ) are symmetric matrices and ( A B=B A, ) then ( A^{-1} B ) is a A. Symmetric matrix B. Skew-symmetric matrix c. Identity matrix D. None of these | 12 |
| 762 | If a ( M ) matrix ( A ) is such that ( A A^{T}= ) ( boldsymbol{I}=boldsymbol{A}^{boldsymbol{T}} boldsymbol{A}, ) find ( |boldsymbol{A}|=? ) | 12 |
| 763 | If order of matrix ( A ) is ( 4 times 3 ) and order of matrix ( B ) is ( 3 times 5 ) then order of matrix ( boldsymbol{B}^{prime} boldsymbol{A}^{prime} ) is: ( mathbf{A} cdot 5 times 2 ) B. ( 4 times 5 ) ( mathbf{c} cdot 5 times 4 ) D. ( 3 times 2 ) | 12 |
| 764 | ( boldsymbol{A}=left[begin{array}{cc}boldsymbol{a} & boldsymbol{b} \ mathbf{0} & boldsymbol{c}end{array}right] ) then ( boldsymbol{A}^{-1}+(boldsymbol{A}-boldsymbol{a} boldsymbol{I})(boldsymbol{A}-1) ) ( boldsymbol{c} boldsymbol{I})= ) A ( cdot frac{1}{a c}left[begin{array}{cc}a & b \ 0 & -cend{array}right] ) в. ( frac{1}{a c}left[begin{array}{cc}-a & b \ 0 & cend{array}right] ) c. ( frac{1}{a c}left[begin{array}{cc}c & -b \ 0 & aend{array}right] ) D ( cdot frac{1}{a c}left[begin{array}{cc}c & b \ 0 & aend{array}right] ) | 12 |
| 765 | If ( boldsymbol{A}=left(begin{array}{l}832 \ 591end{array}right) ) and ( B=left(begin{array}{c}1-1 \ 0end{array}right) . ) Find ( A+B ) if it exists. | 12 |
| 766 | If ( A ) and ( B ) are square matrices of order n’ such that ( A^{2}-B^{2}=(A-B)(A+ ) ( B ) ), then which of the following will be true? A. Either of A or B is zero matrix в. ( A=B ) c. ( A B=B A ) D. Either of A or B is an identity matrix | 12 |
| 767 | If ( A^{-1}=left[begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2end{array}right] ) and ( B= ) ( left[begin{array}{ccc}1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1end{array}right] ), find ( (A B)^{-1} ) | 12 |
| 768 | Using elementary transformation, find the inverse of the matrix ( boldsymbol{A}= ) ( left[begin{array}{cc}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & left(frac{1+boldsymbol{b} c}{boldsymbol{a}}right)end{array}right] ) ( ^{mathbf{A}} rightarrow A^{-1}=left[begin{array}{cc}frac{1+b c}{a} & b \ -c & aend{array}right] ) ( stackrel{mathbf{B}}{rightarrow} Rightarrow A^{-1}=left[begin{array}{cc}frac{1+b c}{a} & -b \ c & aend{array}right] ) ( stackrel{mathrm{c}}{*} quad, A^{-1}=left[begin{array}{cc}frac{1+b c}{a} & b \ c & aend{array}right] ) D. None of these. | 12 |
| 769 | Is this possible ( a neq ) ( mathbf{0} cdotleft|begin{array}{ccc}boldsymbol{x}+mathbf{1} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{a} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{a}^{2}end{array}right|=mathbf{0} ) represents a straight line parallel to the y-axis A. True B. False | 12 |
| 770 | If ( boldsymbol{A}^{T} boldsymbol{B}^{T}=boldsymbol{C}^{T} ) then ( mathbf{C}= ) ( A cdot A B ) B. BA ( c . ) вс D. ABC | 12 |
| 771 | [ operatorname{ftg}left[begin{array}{ccc} 3 & 2 & -1 \ 2 & -2 & 0 \ 1 & 3 & 1 end{array}right], Bleft[begin{array}{ccc} -3 & -1 & 0 \ 2 & 1 & 3 \ 4 & -1 & 2 end{array}right] ] and ( X=A+B ) then find ( X ) | 12 |
| 772 | The sum and product of matrices A and B exist. Which of the following implications are necessarily true? 1. A and B are square matrices of same order. 2. A and B are non-singular matrices. Select the correct answer using the code given below: A . 1 only B. 2 only c. Both 1 and 2 D. Neither 1 nor 2 | 12 |
| 773 | Find the inverse of the following matrix by using elementary row transformation: ( left[begin{array}{ll}5 & 2 \ 2 & 1end{array}right] ) | 12 |
| 774 | For two ( 3 times 3 ) matrices ( A ) and ( B ), let ( A+B=2 B^{prime} ) and ( 3 A+2 B=I_{3}, ) where ( B^{prime} ) is the transpose of ( B ) and ( I_{3} ) is ( 3 times 3 ) identify matrix. Then : A. ( 5 A+10 B=2 I_{3} ) В . ( 3 A+6 B=2 I_{3} ) c. ( 10 A+5 B=3 I_{3} ) D. ( B+2 A=I_{3} ) | 12 |
| 775 | If ( boldsymbol{A}=left[begin{array}{cc}cos boldsymbol{x} & sin boldsymbol{x} \ -sin boldsymbol{x} & cos boldsymbol{x}end{array}right], ) show that ( A^{2}= ) ( left[begin{array}{cc}cos 2 x & sin 2 x \ -sin 2 x & cos 2 xend{array}right] ) and ( A^{prime} A=I ) | 12 |
| 776 | If ( boldsymbol{A}=left[begin{array}{ccc}0 & 1 & 2 \ 1 & 2 & 3 \ 2 & 3 & 4end{array}right] ) and ( B=left[begin{array}{cc}1 & -2 \ -1 & 0 \ 2 & -1end{array}right] ) Check ( A B=B A ? ) | 12 |
| 777 | If ( boldsymbol{A}=left[begin{array}{lll}mathbf{0} & mathbf{0} & mathbf{1} \ mathbf{0} & mathbf{1} & mathbf{0} \ mathbf{1} & mathbf{0} & mathbf{0}end{array}right], ) then ( boldsymbol{A}^{-1} ) is A . ( -A ) в. ( c cdot 1 ) D. None of these | 12 |
| 778 | The order of ( [x, y, z]left[begin{array}{lll}a & h & g \ h & b & f \ g & f & cend{array}right]left[begin{array}{l}x \ y \ zend{array}right] ) is ( A cdot 3 times 1 ) B. ( 1 times 1 ) c. ( 1 times 3 ) D. ( 3 times 3 ) | 12 |
| 779 | If ( A ) is a scalar matrix ( k I ) with scalar ( k neq 0 ) of order ( 3, ) the ( A^{-1} ) is: A ( cdot frac{1}{k^{2}} I ) B. ( frac{1}{k^{3}} ) c. ( frac{1}{k} I ) D. ( k I ) | 12 |
| 780 | Find the inverse of ( left[begin{array}{ccc}3 & -1 & -2 \ 2 & 0 & -1 \ 3 & -5 & 0end{array}right] ) using elementary row transformations. | 12 |
| 781 | If ( operatorname{Tr}(mathbf{A})=mathbf{2}+mathbf{i}, ) Then ( operatorname{Tr}[(mathbf{2}-mathbf{i}) mathbf{A}]= ) A ( .2+i ) в. ( 2-i ) ( c .3 ) D. 5 | 12 |
| 782 | The numbers of ( 3 times 3 ) matrices A whose entries are either 0 or 1 and for which the system ( boldsymbol{A}left[begin{array}{l}boldsymbol{x} \ boldsymbol{y} \ boldsymbol{z}end{array}right]=left[begin{array}{l}mathbf{1} \ mathbf{0} \ mathbf{0}end{array}right] ) has exactly two distinct solutions is? A . B . ( 2^{9}-1 ) ( c .168 ) D. | 12 |
| 783 | If ( A_{2 times 3}, B_{4 times 3}, C_{2 times 4} ) are three matrices, then which of the following is/are defined ? A ( . A C^{T} B ) B . ( B^{T} C^{T} A ) ( mathbf{c} cdot A B^{T} C ) D. All of these | 12 |
| 784 | If ( boldsymbol{A}=left[begin{array}{c}-1 \ 2 \ mathbf{3}end{array}right], boldsymbol{B}=[-mathbf{2},-mathbf{1},-mathbf{4}] ) verify that ( (boldsymbol{A B})^{T}=boldsymbol{B}^{boldsymbol{T}} boldsymbol{A}^{boldsymbol{T}} ) | 12 |
| 785 | Let ( boldsymbol{A}=left[begin{array}{ll}mathbf{2} & mathbf{4} \ mathbf{3} & mathbf{2}end{array}right], boldsymbol{B}=left[begin{array}{cc}mathbf{1} & mathbf{3} \ -mathbf{2} & mathbf{5}end{array}right] ) and ( boldsymbol{C}=left[begin{array}{cc}mathbf{- 2} & mathbf{5} \ mathbf{3} & mathbf{4}end{array}right] . ) Find: ( boldsymbol{B}-mathbf{4} boldsymbol{C} ) | 12 |
| 786 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{0}end{array}right], ) then ( boldsymbol{A}^{4}= ) A. ( left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) в. ( left[begin{array}{ll}1 & 1 \ 0 & 0end{array}right] ) c. ( left[begin{array}{ll}0 & 0 \ 1 & 1end{array}right] ) D. ( left[begin{array}{ll}0 & 1 \ 1 & 0end{array}right] ) | 12 |
| 787 | If two square matrices ( A ) and ( B ) are of same order and, ( operatorname{Tr}(boldsymbol{A})=mathbf{3}, boldsymbol{T} boldsymbol{r}(boldsymbol{B})=mathbf{5} ) ( operatorname{then} operatorname{Tr}(boldsymbol{A}+boldsymbol{B})= ) A . 15 B. 8 ( c cdot 3 / 5 ) D. cannot be determined | 12 |
| 788 | If the traces of the matrices ( A ) and ( B ) are 20 and ( 8, ) then trace of ( mathbf{A}+mathbf{B}= ) A . 28 B. 20 ( c cdot-8 ) D. 12 | 12 |
| 789 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{3} & mathbf{1} \ mathbf{7} & mathbf{5}end{array}right], ) find the values of ( boldsymbol{x} ) and ( y ) such that ( A^{2}+x I_{2}=y A ) | 12 |
| 790 | If ( P ) is a ( 3 times 3 ) matrix such that ( P^{T}= ) ( 2 P+I ) where ( P^{T} ) is the transpose of ( P ) and lis the ( 3 times 3 ) identify matrix, then there exists a column matrix ( boldsymbol{X}= ) ( left[begin{array}{l}x \ y \ zend{array}right] neqleft[begin{array}{l}0 \ 0 \ 0end{array}right] ) such that ( ^{mathbf{A}} cdot operatorname{IX}=left[begin{array}{l}0 \ 0 \ 0end{array}right] ) B. PX = X ( c cdot ) Рх ( =2 x ) D. PX = – x | 12 |
| 791 | If ( A=left|begin{array}{cc}2 & -3 \ 3 & 2end{array}right| ) and ( B=left|begin{array}{cc}3 & -2 \ 2 & 3end{array}right| ) then ( 2 A-B= ) A. 11 ( begin{array}{ll}1 & 4 \ 4 & 1end{array} ) в. ( mid begin{array}{ll}1 & 4 \ 1 & 4end{array} ) c. ( left|begin{array}{cc}1 & -4 \ 4 & 1end{array}right| ) D. ( mid begin{array}{ll}4 & 1 \ 1 & 4end{array} ) | 12 |
| 792 | Solve for ( x ) and ( y ) if ( left(begin{array}{l}2 x+y \ x-3 yend{array}right)=left(begin{array}{c}5 \ 13end{array}right) ) | 12 |
| 793 | The number of nonzero diagonal matrices of order 3 satisfying ( A^{2}=A ) is | 12 |
| 794 | Using elementary row transformations, find the inverse of the matrix ( boldsymbol{A}= ) ( left[begin{array}{ccc}1 & 2 & 3 \ 2 & 5 & 7 \ -2 & -4 & -5end{array}right] ) | 12 |
| 795 | If ( A ) and ( B ) are matrices given below: [ begin{array}{l} A=left[begin{array}{ccc} 0 & c & -b \ -c & o & a \ b & -a & 0 end{array}right] text { and } B= \ {left[begin{array}{ccc} a^{2} & a b & a c \ a b & b^{2} & b c \ a c & b c & c^{2} end{array}right]} end{array} ] then ( A B ) is a unit matrix. Is this statement true? | 12 |
| 796 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right]_{boldsymbol{m} times boldsymbol{n}^{prime}} boldsymbol{B}=left[boldsymbol{b}_{boldsymbol{i} j}right]_{boldsymbol{m} times boldsymbol{n}^{prime}} ) then the element ( C_{23} ) of the matrix ( C=A+B ) is: A ( . C_{23} ) В. ( a_{23}+b_{32} ) ( mathbf{c} cdot a_{23}+b_{23} ) D ( cdot a_{32}+b_{23} ) | 12 |
| 797 | The transpose of a row matrixis A. zero matrix B. diagonal matrix C. column matrix D. row matrix | 12 |
| 798 | Write the following as a single matrix ( left[begin{array}{cc}-1 & 2 \ 1 & -2 \ 3 & -1end{array}right]+left[begin{array}{cc}0 & 1 \ -1 & 0 \ -2 & 1end{array}right] ) | 12 |
| 799 | Find the inverse of the following matrix by using elementary row transformation: ( left[begin{array}{cc}mathbf{1} & mathbf{2} \ mathbf{2} & -mathbf{1}end{array}right] ) | 12 |
| 800 | If ( boldsymbol{A}=left[begin{array}{cc}cos boldsymbol{alpha} & sin boldsymbol{alpha} \ -sin boldsymbol{alpha} & cos boldsymbol{alpha}end{array}right], ) then verify that ( boldsymbol{A}^{boldsymbol{T}} boldsymbol{A}=boldsymbol{I}_{2} ) | 12 |
| 801 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & -mathbf{2} \ mathbf{3} & mathbf{0}end{array}right], boldsymbol{B}=left[begin{array}{cc}mathbf{- 1} & mathbf{4} \ mathbf{2} & mathbf{3}end{array}right], boldsymbol{C}= ) ( left[begin{array}{cc}mathbf{0} & mathbf{1} \ mathbf{- 1} & mathbf{0}end{array}right], ) then ( mathbf{5} boldsymbol{A}-mathbf{3} boldsymbol{B}+mathbf{2} boldsymbol{C}= ) ( mathbf{A} cdotleft[begin{array}{cc}8 & 20 \ 7 & 9end{array}right] ) ( mathbf{B} cdotleft[begin{array}{cc}8 & -20 \ 7 & -9end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-8 & 20 \ -7 & 9end{array}right] ) ( mathbf{D} cdotleft[begin{array}{cc}8 & 7 \ -20 & -9end{array}right] ) | 12 |
| 802 | Find ( boldsymbol{X} ) and ( boldsymbol{Y}, ) if ( boldsymbol{2}left[begin{array}{ll}mathbf{1} & boldsymbol{3} \ boldsymbol{0} & boldsymbol{x}end{array}right]+left[begin{array}{ll}boldsymbol{y} & boldsymbol{0} \ boldsymbol{1} & boldsymbol{2}end{array}right]= ) ( left[begin{array}{ll}5 & 6 \ 1 & 8end{array}right] ) | 12 |
| 803 | For 3 x 3 matrices M and N, which of the following statement(s) is (are) NOT correct? (JEE Adv. 2013) (a) N’MN is symmetric or skew symmetric, according as Mis symmetric or skew symmetric (b) MN-NM is skew symmetric for all symmetric matrices M and N (c) MN is symmetric for all symmetric matrices Mand N (d) (adj M) (adj N)= adj (MN) for all invertible matrices M and N | 12 |
| 804 | atrices such 26. Let A = 2 1 0. Ifu, and tl, are column matrices 3 2 1 0) that Auy = 0 and Au2 = 1 , then u + uz is equal to : [2012 (a) | 12 |
| 805 | If ( M ) is a ( 3 times 3 ) matrix, where ( M^{T} M= ) ( boldsymbol{I} ) and ( operatorname{det}(boldsymbol{M})=mathbf{1} ) then prove that ( operatorname{det}(M-I)=0 ) | 12 |
| 806 | Find the inverse of the following matrix using transformation method. ( left[begin{array}{cc}mathbf{1} & mathbf{2} \ mathbf{2} & -mathbf{1}end{array}right] ) | 12 |
| 807 | ( operatorname{Given} boldsymbol{F}(boldsymbol{x})=left[begin{array}{ccc}cos boldsymbol{x} & -sin boldsymbol{x} & mathbf{0} \ sin boldsymbol{x} & cos boldsymbol{x} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{1}end{array}right] ) ( boldsymbol{x} in boldsymbol{R} ) Then prove ( boldsymbol{y}, boldsymbol{F}(boldsymbol{x}+boldsymbol{y})= ) ( boldsymbol{F}(boldsymbol{x}) boldsymbol{F}(boldsymbol{y}) ) | 12 |
| 808 | Assertion If ( A ) is a square matrix of order ( n ) then ( operatorname{det}(k A)=k^{n}|A| ) Reason If matrix ( mathrm{B} ) is obtained from ( mathrm{A} ) by multiplying any row (or column) by a non zero scalar ( k ) then ( operatorname{det}(B)= ) ( boldsymbol{k} operatorname{det}(boldsymbol{A}) ) A. Both (A) & (R) are individually true & (R) is correct explanation of ( (A) ) B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A) c. (A)is true but (R) is false, D. (A)is false but (R) is true. | 12 |
| 809 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{3} & mathbf{1} \ -mathbf{1} & mathbf{2}end{array}right], ) show that ( boldsymbol{A}^{2}-mathbf{5} boldsymbol{A}+ ) ( mathbf{7} boldsymbol{I}=boldsymbol{O} . ) Hence find ( boldsymbol{A}^{-1} ) | 12 |
| 810 | Let ( A ) and ( B ) be square matrices of the other ( 3 times 3 . ) Is ( (A B)^{2}=A^{2} B^{2} ? ) Give reasons. | 12 |
| 811 | ff ( left[begin{array}{cc}mathbf{4} & -mathbf{3} \ mathbf{2} & mathbf{1 6}end{array}right]=left[begin{array}{cc}mathbf{4} & -mathbf{3} \ mathbf{2} & mathbf{2}^{t}end{array}right], ) then ( mathbf{t}= ) ( A cdot 2 ) B. 3 ( c cdot 4 ) ( D ) | 12 |
| 812 | If ( A ) is a skew symmetric matrix of order 3, then the value of ( |boldsymbol{A}| ) is A . 3 B. 0 ( c .9 ) D. 27 | 12 |
| 813 | ( fleft(begin{array}{ccc}2 & 3 & 4 \ 5 & -3 & 8 \ 9 & 2 & 16end{array}right], ) then trace of ( A ) is A . 17 B. 25 ( c cdot 8 ) D. 15 | 12 |
| 814 | Inverse of ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & mathbf{3} \ mathbf{2} & -mathbf{2}end{array}right] ) is equal to? ( mathbf{A} ) ( mathbf{A} cdot-frac{1}{8}left[begin{array}{cc}3 & 1 \ -2 & 2end{array}right] ) В. ( -frac{1}{8}left[begin{array}{rr}-2 & -3 \ -2 & 1end{array}right] ) ( ^{mathbf{c}} cdot frac{1}{8}left[begin{array}{cc}-1 & -3 \ -2 & 2end{array}right] ) D. None of these | 12 |
| 815 | The inverse of ( left[begin{array}{lll}1 & a & b \ 0 & x & 0 \ 0 & 0 & 1end{array}right] ) is ( left[begin{array}{ccc}1 & -a & -b \ 0 & 1 & 0 \ 0 & 0 & 1end{array}right] ) then ( x= ) ( A ) в. ( c .0 ) ( D ) | 12 |
| 816 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{3} & -mathbf{2} \ mathbf{4} & -mathbf{2}end{array}right], ) find ( boldsymbol{K} ) such that ( boldsymbol{A}^{mathbf{2}}= ) ( boldsymbol{K} boldsymbol{A}-boldsymbol{2} boldsymbol{I}, ) where ( boldsymbol{I} ) is the identity element. | 12 |
| 817 | ( left[begin{array}{ll}1 & -tan theta \ tan theta & 1end{array}right]left[begin{array}{ll}1 & tan theta \ -tan theta & 1end{array}right] ) ( =left[begin{array}{ll}a & -b \ -b & aend{array}right] ) then ( A cdot a=1 ) B ( cdot a=sec ^{2} theta, b=0 ) c. ( a=0, b=sin ^{2} theta ) ( mathbf{D} cdot mathbf{a}=sin 2 theta, mathbf{b}=cos 2 theta ) | 12 |
| 818 | Let ( A=left(begin{array}{l}3^{2} \ 5end{array}right) ) and ( B=left(begin{array}{c}8-1 \ 3end{array}right) . ) Find the matrix ( mathrm{C} ) if ( boldsymbol{C}=mathbf{2} boldsymbol{A}+boldsymbol{B} ) | 12 |
| 819 | Find the inverse of the following matrix by using elementary row transformation ( left[begin{array}{cc}mathbf{3} & mathbf{1 0} \ mathbf{2} & mathbf{7}end{array}right] ) | 12 |
| 820 | IIf ( mathbf{A}=left[begin{array}{ll}boldsymbol{a} & mathbf{0} \ boldsymbol{a} & mathbf{0}end{array}right], mathbf{B}=left[begin{array}{ll}mathbf{0} & mathbf{0} \ boldsymbol{b} & boldsymbol{b}end{array}right], ) then ( mathbf{A B}= ) ( mathbf{A} cdot mathbf{0} ) в. ВА c. АВ D. ABAB | 12 |
| 821 | ( mathrm{If} mathrm{A}+mathrm{B}=left[begin{array}{lll}1 & 0 & 2 \ 2 & 2 & 2 \ 1 & 1 & 1end{array}right] ) and ( A-B= ) ( left[begin{array}{ccc}1 & 4 & 4 \ 4 & 2 & 0 \ -1 & -1 & 2end{array}right] ) then prove that ( A=left[begin{array}{lll}1 & 2 & 3 \ 3 & 2 & 1 \ 0 & 0 & 2end{array}right] ) and ( B= ) ( left[begin{array}{ccc}0 & -2 & -1 \ -1 & 0 & 1 \ 1 & 1 & 0end{array}right] ) | 12 |
| 822 | ( operatorname{Let} boldsymbol{A}=left(begin{array}{cc}mathbf{3} & mathbf{2} \ mathbf{5} & mathbf{1}end{array}right) ) and ( boldsymbol{B}=left(begin{array}{cc}mathbf{8} & -mathbf{1} \ mathbf{4} & mathbf{3}end{array}right) ) Find the matrix ( C, ) if ( C=2 A+B ) | 12 |
| 823 | ( (boldsymbol{A}+boldsymbol{B})^{boldsymbol{T}}= ) ( A cdot A+B ) В. ( A^{T}+B^{T} ) c. Does not exist D. (a) or (b) | 12 |
| 824 | Solve the equation for ( x, y, z ) and ( t ) if ( mathbf{2}left[begin{array}{ll}boldsymbol{x} & boldsymbol{z} \ boldsymbol{y} & boldsymbol{t}end{array}right]+mathbf{3}left[begin{array}{cc}mathbf{1} & -mathbf{1} \ mathbf{0} & mathbf{2}end{array}right]=mathbf{3}left[begin{array}{ll}mathbf{3} & mathbf{5} \ mathbf{4} & mathbf{6}end{array}right] ) | 12 |
| 825 | Show that square matrix ( A ) and its transpose ( A^{T} ) have the same eigen values. | 12 |
| 826 | If ( A ) is a skew-symmetric matrix and ( n ) is odd positive integer, then ( A^{n} ) is A. a skew-symmetric matrix B. a symmetric matrix c. a diagonal matrix D. none of these | 12 |
| 827 | Find the inverse of the following matrices by the adjoining method [ left[begin{array}{lll} 1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5 end{array}right] ] | 12 |
| 828 | Assertion If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & boldsymbol{pi} \ mathbf{0} & mathbf{1}end{array}right], ) then ( boldsymbol{A}^{100}=left[begin{array}{cc}mathbf{1} & mathbf{1 0 0} boldsymbol{pi} \ mathbf{0} & mathbf{1}end{array}right] ) Reason If ( B ) is a ( 2 times 2 ) matrix such that ( B^{2}=0 ) then ( (I+B)^{n}=I+n B ) for each ( n in ) ( N ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 829 | Write the number of all possible matrices of order ( 2 times 2 ) with each entry 1,2 or 3 | 12 |
| 830 | ( boldsymbol{A}=left(begin{array}{cc}mathbf{5} & mathbf{2} \ mathbf{7} & mathbf{3}end{array}right) ) and ( boldsymbol{B}=left(begin{array}{cc}mathbf{2} & mathbf{- 1} \ mathbf{- 1} & mathbf{1}end{array}right) ) verify that ( (boldsymbol{A B})^{boldsymbol{T}}=boldsymbol{B}^{boldsymbol{T}} boldsymbol{A}^{boldsymbol{T}} ) | 12 |
| 831 | If ( boldsymbol{A}=left[begin{array}{rr}mathbf{2} & -mathbf{3} \ -mathbf{4} & mathbf{1}end{array}right], ) then ad ( left(3 A^{2}+12 Aright) ) is equal to: A. ( left[begin{array}{ll}51 & 63 \ 84 & 72end{array}right] ) B. ( left[begin{array}{ll}51 & 84 \ 63 & 72end{array}right] ) c. ( left[begin{array}{cc}72 & -63 \ -84 & 51end{array}right] ) D. ( left[begin{array}{cc}72 & -84 \ -63 & 51end{array}right] ) | 12 |
| 832 | Two matrices ( A ) and ( B ) are added if A. both are rectangular B. both have same order C. no of columns of ( A ) is equal to columns of ( B ) D. no of rows of A is equal to no of columns of B | 12 |
| 833 | ( mathbf{f}left[begin{array}{rr}boldsymbol{a}+mathbf{4} & mathbf{3} boldsymbol{b} \ mathbf{8} & -mathbf{6}end{array}right]=left[begin{array}{cc}mathbf{2} boldsymbol{a}+mathbf{2} & boldsymbol{b}+mathbf{2} \ boldsymbol{8} & boldsymbol{a}-mathbf{8} boldsymbol{b}end{array}right] ) then write the value of ( a-2 b ) | 12 |
| 834 | Find ( boldsymbol{x} ) and ( boldsymbol{y}, ) when ( boldsymbol{x}+boldsymbol{y}=left[begin{array}{cc}mathbf{7} & mathbf{0} \ mathbf{2} & mathbf{5}end{array}right] ) and ( boldsymbol{x}-boldsymbol{y}=left[begin{array}{cc}mathbf{3} & mathbf{0} \ mathbf{0} & mathbf{3}end{array}right] ) | 12 |
| 835 | Find the inverse of the following matrix by using elementary row transformation ( left[begin{array}{lll}0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1end{array}right] ) | 12 |
| 836 | ( mathbf{f}left[begin{array}{ccc}mathbf{9} & -mathbf{1} & mathbf{4} \ -mathbf{2} & mathbf{1} & mathbf{3}end{array}right]=boldsymbol{A}+left[begin{array}{ccc}mathbf{1} & mathbf{2} & -mathbf{1} \ mathbf{0} & mathbf{4} & mathbf{9}end{array}right] ) then find the matrix ( A ) | 12 |
| 837 | ff ( left[begin{array}{cc}2 & -1 \ 2 & 0end{array}right]+2 A=left[begin{array}{cc}-3 & 5 \ 4 & 3end{array}right], ) then the matrix A equals A. ( left[begin{array}{ll}-5 & 6 \ 2 & 3end{array}right] ) B. ( left[begin{array}{cc}-frac{5}{2} & 3 \ 1 & frac{3}{2}end{array}right] ) ( mathbf{c} cdotleft[begin{array}{cc}-frac{5}{2} & 6 \ 2 & 3end{array}right] ) D. ( left[begin{array}{cc}-5 & 8 \ 1 & 3end{array}right] ) | 12 |
| 838 | If ( A ) is ( 3 times 4 ) matrix and ( B ) is matrix such that ( A^{prime} B ) and ( B A^{prime} ) are both defined, then ( B ) is of the type. ( A cdot 3 times 4 ) B. ( 3 times 3 ) ( mathbf{c} cdot 4 times 4 ) D. ( 4 times 3 ) | 12 |
| 839 | If ( A ) is a real skew-symmetric matrix such that ( A^{2}+I=O, ) then ( mathbf{A} cdot A ) is a square matrix of even order with ( |A|=pm 1 ) B. ( A ) is a square matrix of odd order with ( |A|=pm 1 ) C ( cdot A ) can be a square matrix of any order with ( |A|=pm 1 ) D. ( A ) is a skew-symmetric matrix of even order with ( |A|=1 ) | 12 |
| 840 | If ( boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{3} & mathbf{2} & mathbf{1}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{lll}mathbf{3} & mathbf{2} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3}end{array}right] ) find ( mathbf{3} boldsymbol{B}-mathbf{2} boldsymbol{A} ) | 12 |
| 841 | Let ( A ) be a ( 2 times 2 ) matrix and ( B=A+ ) ( A^{T} . ) Then show that ( B ) is a symmetric matrix. | 12 |
| 842 | If ( A ) is a non zero square matrix of order ( boldsymbol{n} ) with ( operatorname{det}(boldsymbol{I}+boldsymbol{A}) neq mathbf{0}, ) and ( boldsymbol{A}^{3}=mathbf{0} ) where ( I, O ) are unit and null matrices of order ( n times n ) respectively, then ( (boldsymbol{I}+boldsymbol{A})^{-1}= ) A ( cdot I-A+A^{2} ) B . ( I+A+A^{2} ) c. ( I+A^{2} ) ( mathbf{D} cdot I+A ) | 12 |
| 843 | Let ( A ) being a square matrix, then prove that ( boldsymbol{A}+boldsymbol{A}^{T} ) is symmetric. | 12 |
| 844 | If ( boldsymbol{A}=left(begin{array}{cc}1 & -2 \ -mathbf{3} & 4end{array}right) ) and ( boldsymbol{A}+boldsymbol{B}=boldsymbol{O}, ) then ( mathbf{B} ) is A ( cdotleft(begin{array}{cc}1 & -2 \ -3 & 4end{array}right) ) B. ( left(begin{array}{c}-1 \ 3end{array}right) ) c. ( left(begin{array}{cc}-1 & -2 \ -3-4end{array}right) ) D. ( left(begin{array}{c}1 \ 0 \ 0end{array}right) ) | 12 |
| 845 | Determine whether the product of the matrices is defined in each case. If ( s 0 ) state the order of the product. MN, where ( boldsymbol{M}=left[boldsymbol{m}_{i j}right]_{3 times 1}, boldsymbol{N}=left[boldsymbol{n}_{i j}right]_{1 times 5} ) | 12 |
| 846 | Find the value of ( x, y, z ) if ( left[begin{array}{ll}mathbf{4} & mathbf{3} \ boldsymbol{x} & mathbf{5}end{array}right]=left[begin{array}{ll}boldsymbol{y} & boldsymbol{z} \ mathbf{1} & mathbf{5}end{array}right] ) | 12 |
| 847 | If ( boldsymbol{A}=left[begin{array}{cc}boldsymbol{4} & boldsymbol{3} \ boldsymbol{1} & boldsymbol{2}end{array}right] boldsymbol{B}=left[begin{array}{ll}boldsymbol{2} & boldsymbol{1} \ boldsymbol{1} & boldsymbol{2}end{array}right] ) Verify ( (boldsymbol{A} boldsymbol{B})^{prime}=boldsymbol{B}^{prime} boldsymbol{A}^{prime} ) | 12 |
| 848 | If ( A=left[begin{array}{ccc}2 & 3 & 1 \ 0 & -1 & 5end{array}right], B=left[begin{array}{ccc}1 & 2 & -1 \ 0 & -1 & 3end{array}right] ) | 12 |
| 849 | If ( [A] neq 0 ) then which of the following is not true? A ( cdotleft(A^{2}right)^{-1}=left(A^{-1}right)^{2} ) B. ( left(A^{prime}right)^{-1}=left(A^{-1}right)^{prime} ) ( mathbf{c} cdot A^{-1}=|A|^{-1} ) D. None of these | 12 |
| 850 | Find ( boldsymbol{A}^{boldsymbol{T}}: ) ( boldsymbol{A}=left[begin{array}{lll}mathbf{4} & mathbf{3} & mathbf{- 1} \ mathbf{6} & mathbf{8} & mathbf{- 3} \ mathbf{4} & mathbf{1} & mathbf{3}end{array}right] ) | 12 |
| 851 | If ( A ) is ( 4 times 5 ) matrix, if ( A^{T} B ) and ( B A^{T} ) are defined then ( B= ) ( mathbf{A} cdot 5 times 4 ) B. ( 4 times 4 ) ( mathbf{c} .5 times 5 ) D. ( 4 times 5 ) | 12 |
| 852 | ( mathbf{f} mathbf{A}=left[begin{array}{lll}mathbf{1} & mathbf{- 3} & mathbf{- 4} \ mathbf{- 1} & mathbf{3} & mathbf{4} \ mathbf{1} & mathbf{- 3} & mathbf{- 4}end{array}right], ) then ( mathbf{A}^{2}= ) ( A cdot A ) в. -4 c. Null matrix D. ( 2 A ) | 12 |
| 853 | If ( boldsymbol{m}left[begin{array}{ll}-mathbf{3} & mathbf{4}end{array}right]+boldsymbol{n}left[begin{array}{ll}mathbf{4} & -mathbf{3}end{array}right]=left[begin{array}{ll}mathbf{1 0} & -mathbf{1 1}end{array}right] ) then ( 3 m+7 n= ) ( A cdot 3 ) B. 5 c. 10 ( D ) | 12 |
| 854 | ( mathbf{f} boldsymbol{A}=left[begin{array}{lll}mathbf{3} & mathbf{2} & mathbf{0} \ mathbf{1} & mathbf{4} & mathbf{0} \ mathbf{0} & mathbf{0} & mathbf{5}end{array}right] . ) Show that ( boldsymbol{A}^{mathbf{2}} ) [ mathbf{7 A}+mathbf{1 0}=mathbf{0} ] | 12 |
| 855 | Is it possible to define the matrix ( A+B ) when A has 3 rows and ( mathrm{B} ) has 2 columns | 12 |
| 856 | ( mathbf{f}left[begin{array}{ccc}boldsymbol{x}-boldsymbol{y} & mathbf{1} & boldsymbol{z} \ mathbf{2} boldsymbol{x}-boldsymbol{y} & boldsymbol{0} & boldsymbol{w}end{array}right]=left[begin{array}{ccc}-mathbf{1} & mathbf{1} & mathbf{4} \ mathbf{0} & mathbf{0} & mathbf{5}end{array}right] ) find ( boldsymbol{x}, boldsymbol{y}, boldsymbol{z}, boldsymbol{w} ) | 12 |
| 857 | Solve the following system of linear equations using matrix method: ( 3 x+ ) ( boldsymbol{y}+boldsymbol{z}=mathbf{1}, mathbf{2} boldsymbol{x}+mathbf{2} boldsymbol{z}=mathbf{0}, mathbf{5} boldsymbol{x}+mathbf{5} boldsymbol{y}+ ) ( mathbf{2} z=mathbf{2} ) | 12 |
| 858 | If ( boldsymbol{A}=left(boldsymbol{a}_{boldsymbol{i} j}right)_{mathbf{2} times mathbf{2}}, ) where ( boldsymbol{a}_{boldsymbol{i} j}=boldsymbol{i}+boldsymbol{j}, ) then ( A ) is equal to: A. ( left[begin{array}{ll}1 & 2 \ 2 & 3end{array}right] ) B. ( left[begin{array}{ll}0 & 1 \ 1 & 0end{array}right] ) ( mathbf{c} cdotleft[begin{array}{ll}2 & 3 \ 3 & 4end{array}right] ) D. ( left[begin{array}{ll}1 & 2 \ 3 & 4end{array}right] ) | 12 |
| 859 | A square, non-singular matrix ( boldsymbol{A} ) satifies ( boldsymbol{A}^{2}-boldsymbol{A}+mathbf{2} boldsymbol{I}=mathbf{0}, ) then ( boldsymbol{A}^{-1}= ) ( mathbf{A} cdot I-A ) в. ( frac{(I-A)}{2} ) c. ( I+A ) D. ( frac{(I+A)}{2} ) | 12 |
| 860 | If ( boldsymbol{A}+boldsymbol{B}=left[begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5}end{array}right] ) and ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{0} & mathbf{3}end{array}right] ) then matrix ( B ) is ( mathbf{A} cdotleft[begin{array}{ll}1 & 1 \ 4 & 2end{array}right] ) в. ( left[begin{array}{ll}1 & 4 \ 1 & 2end{array}right] ) c. ( left[begin{array}{ll}2 & 4 \ 1 & 1end{array}right] ) D. ( left[begin{array}{ll}4 & 2 \ 1 & 1end{array}right] ) | 12 |
| 861 | If ( mathbf{A}=left[begin{array}{ll}mathbf{3} & -mathbf{4} \ mathbf{1} & -mathbf{1}end{array}right] ) then ( boldsymbol{A}^{k}= ) ( left[begin{array}{cc}mathbf{1}+mathbf{2} boldsymbol{k} & -mathbf{4} boldsymbol{k} \ boldsymbol{k} & mathbf{1}-mathbf{2} boldsymbol{k}end{array}right] ) where ( k ) is any ( + ) ve integer | 12 |
| 862 | If ( A ) is a square matrix of order 3 then ( left|mathbf{A} mathbf{d} mathbf{j}left(A d j A^{2}right)right|= ) ( mathbf{A} cdot|A|^{2} ) B . ( |A|^{4} ) c. ( left.|A|^{8}right|^{8} mid ) D. ( |A|^{16} ) | 12 |
| 863 | If ( boldsymbol{A}=left[begin{array}{cc}cos 2 boldsymbol{theta} & -sin 2 boldsymbol{theta} \ sin 2 boldsymbol{theta} & cos 2 boldsymbol{theta}end{array}right] ) and ( boldsymbol{A}+ ) ( A^{T}=I, ) where ( I ) is the unit of matrix of ( 2 times 2 ) and ( A^{T} ) is the transpose of ( A, ) then the value of ( theta ) is equal to A ( cdot frac{pi}{6} ) в. c. ( pi ) D. ( frac{3 pi}{2} ) | 12 |
| 864 | If ( A ) is a skew-symmetric matrix, then trace of ( A ) is ( mathbf{A} cdot mathbf{1} ) B. – – ( c cdot 0 ) D. none of these | 12 |
| 865 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2}end{array}right], boldsymbol{B}=left[begin{array}{ll}mathbf{3} & mathbf{4}end{array}right] ) then ( boldsymbol{A}- ) ( boldsymbol{B}= ) ( mathbf{A} cdotleft[begin{array}{ll}-2 & -2end{array}right] ) B. ( left[begin{array}{ll}2 & 2end{array}right] ) ( mathbf{c} cdotleft[begin{array}{ll}-3 & -1end{array}right] ) D. None of these | 12 |
| 866 | Let ( A ) be a ( 2 times 2 ) matrix with non-zero entries and let ( mathbf{A}^{2}=boldsymbol{I}, ) where I is ( mathbf{2} times mathbf{2} ) identity matrix. Define ( operatorname{Tr}(mathbf{A})=operatorname{sum} ) of diagonal elements of ( A ) and ( |mathbf{A}|= ) determinant of matrix A. Statement-1 ( operatorname{Tr}(mathrm{A})=0 ) Statement-2: | 12 |
| 867 | f ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{0}end{array}right], boldsymbol{A}^{2}=boldsymbol{I} ) then find ‘ ( boldsymbol{x}^{prime} ) | 12 |
| 868 | Construct a ( 3 times 2 ) matrix whose elements are given by ( a_{i j}=2 i-j ) | 12 |
| 869 | If ( A timesleft(begin{array}{l}1 \ 0end{array}right)=(12), ) then the order of ( A ) is A ( .2 times 1 ) B. ( 2 times 2 ) c. ( 1 times 2 ) D. 3 ( times 2 ) | 12 |
| 870 | if ( A=left[begin{array}{cc}2 & 3 \ 5 & -7end{array}right] ) then ( quadleft(A^{1}right)^{2}= ) A. ( left[begin{array}{ccc}5 & -7 & 12 \ 1 & 4 & 22end{array}right] ) B ( cdotleft[begin{array}{cc}1 & 17 \ 1 & -4 \ 0 & 2end{array}right] ) c. ( left[begin{array}{cc}-19 & -25 \ -15 & 64end{array}right] ) D. ( left[begin{array}{cc}19 & -25 \ -15 & 64end{array}right] ) | 12 |
| 871 | If ( A, B ) are symmetric matrices of the same order then ( mathbf{A B}-mathbf{B A} ) is A. symmetric matrix B. skew symmetric matrix c. Diagonal matrix D. identity matrix | 12 |
| 872 | ( mathbf{f} boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{4} & mathbf{0} \ mathbf{2} & mathbf{5} & mathbf{0} \ mathbf{3} & mathbf{6} & mathbf{0}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{lll}mathbf{3} & mathbf{2} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5} & mathbf{6}end{array}right] ) and ( C=left[begin{array}{lll}3 & 2 & 1 \ 1 & 2 & 3 \ 7 & 8 & 9end{array}right], ) Then evaluate ( operatorname{matrix} boldsymbol{A B}-boldsymbol{B C} ) | 12 |
| 873 | If ( A ) is a ( 3 times 3 ) skew-symmetric matrix, then trace of ( A ) is equal to ( A cdot 1 ) в. ( |A| ) ( c cdot-1 ) D. none of these | 12 |
| 874 | ( mathbf{A}=left[begin{array}{cc}cos alpha & sin alpha \ -sin alpha & cos alphaend{array}right] ) then ( mathbf{A} . mathbf{A}^{mathbf{T}} ) A. Null matrix в. А ( c cdot I ) D. A | 12 |
| 875 | If ( A ) and ( B ) are symmetric matrices, then write the condition for which ( A B ) is also symmetric | 12 |
| 876 | Define a scalar matrix. | 12 |
| 877 | If square matrices ( A ) and ( B ) are such that ( boldsymbol{A} boldsymbol{A}^{prime}=boldsymbol{A}^{prime} boldsymbol{A}, boldsymbol{B} boldsymbol{B}^{prime}=boldsymbol{B}^{prime} boldsymbol{B}, boldsymbol{A} boldsymbol{B}^{prime}= ) ( boldsymbol{B}^{prime} boldsymbol{A} ) then is the statement ( A B(A B)^{prime}= ) ( (A B)^{prime} A B ) is where ( A^{prime} ) is transpose of ( A ) If true enter 1 else enter 0 | 12 |
| 878 | [ text { If } boldsymbol{f}(boldsymbol{x}, boldsymbol{y})=boldsymbol{x}^{2}+boldsymbol{y}^{2}-boldsymbol{2} boldsymbol{x} boldsymbol{y},(boldsymbol{x}, boldsymbol{y} in boldsymbol{R}) ] and ( boldsymbol{A}= ) [ left[begin{array}{lll} fleft(x_{1}, y_{1}right) & fleft(x_{1}, y_{2}right) & fleft(x_{1}, y_{3}right) \ fleft(x_{2}, y_{1}right) & fleft(x_{2}, y_{2}right) & fleft(x_{2}, y_{3}right) \ fleft(x_{3}, y_{1}right) & fleft(x_{3}, y_{2}right) & fleft(x_{3}, y_{3}right) end{array}right] ] such that trace ( (boldsymbol{A})=mathbf{0}, ) then which of the following is true (only one option) A. ( operatorname{det}(A) geq 0 ) в. ( operatorname{det}(A)=0 ) ( mathbf{c} cdot operatorname{det}(A) leq 0 ) D. ( operatorname{det}(A)>0 ) | 12 |
| 879 | Two ( n times n ) square matrices ( A ) and ( B ) are said to be similar if there exists a non- singular matrix ( boldsymbol{P} ) such that ( boldsymbol{P}^{-1} boldsymbol{A} boldsymbol{P}=boldsymbol{B} ) If ( A ) and ( B ) are two non-singular matrices, then ( mathbf{A} cdot A ) is similar to ( B ) B. ( A B ) is similar to ( B A ) C. ( A B ) is similarto ( A^{-1} B ) D. none of these | 12 |
| 880 | If ( boldsymbol{A}-boldsymbol{A}^{prime}=mathbf{0}, ) then ( boldsymbol{A}^{prime} ) is A. orthogonal matrix B. symmetric matrix c. skew-symmetric matrix D. triangular matrix | 12 |
| 881 | If ( boldsymbol{A}=left[begin{array}{cc}cos boldsymbol{alpha} & -sin boldsymbol{alpha} \ sin boldsymbol{alpha} & cos boldsymbol{alpha}end{array}right], ) then ( boldsymbol{A}+ ) ( A^{prime}=I, ) if the value of ( alpha ) is A ( cdot frac{pi}{6} ) в. ( c ) D. ( frac{3 pi}{2} ) | 12 |
| 882 | If ( boldsymbol{A}=left[begin{array}{cc}1 & -1 \ -1 & 1end{array}right], ) satisfies the matrix equation ( A^{2}=k A, ) write the value of ( k ) | 12 |
Hope you will like above questions on matrices and follow us on social network to get more knowledge with us. If you have any question or answer on above matrices questions, comments us in comment box.
Stay in touch. Ask Questions.
Lean on us for help, strategies and expertise.
