Determinants Questions

We provide determinants practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. We focus on determinants skills mastery so, below you will get all questions that are also asking in the competition exam beside that classroom.

Determinants Questions

List of determinants Questions

Question NoQuestionsClass
1Solve:
( left|begin{array}{ccc}mathbf{0} & boldsymbol{a} & -boldsymbol{b} \ -boldsymbol{a} & boldsymbol{0} & -boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{0}end{array}right|=0 )
12
2The value of ( (operatorname{adj} A) ) is equal to
A . ( 2 A )
в. ( 4 A )
( c .8 A )
D. ( 16 A )
12
3Find the values of ( x, ) if ( left|begin{array}{ll}mathbf{3} & boldsymbol{x} \ boldsymbol{x} & mathbf{1}end{array}right|=left|begin{array}{ll}mathbf{3} & mathbf{2} \ mathbf{4} & mathbf{1}end{array}right| )12
4( fleft(begin{array}{ccc}2 & -1 & 3 \ -5 & 3 & 1 \ -3 & 2 & 3end{array}right], ) then
( A .(A d j A)= )
A ( cdotleft(A d j cdot A^{T}right) )
B. ( (text {Adj.A}) . A )
c. ( |A| . A )
D. None of these
12
5set of all values of 2 for which the system of linear
31. The set of all value
equations :
2×1 – 2×2 + x3 = 2×1
2x – 3×2 + 2×3 = 2×2
-x + 2×2=hxz
has a non-trivial solution
[JEEM 2015]
(a) contains two elements
(b) contains more than two elements
(C) is an empty set
(d) is a singleton
12
61
2
3 and (adj M) =
-5
where
3
-1
To 1 al
-1 1 -17
15. Let M= 1
3 and (adi M = 8 -6 2
| 3 6 1
a and b are real numbers. Which of the following options
is/are correct ?
(JEE Adv. 2019)
(a) a+b=3
(6) det (adj M2)=81
© (adjM)-1 + adjM =-M
N
(d)
IfM
, then a-B+y = 3
12
7( f(x)=left|begin{array}{ccc}cos x & x & 1 \ 2 sin x & x^{2} & 2 x \ tan x & x & 1end{array}right| . ) The value of
( lim _{x rightarrow 0} frac{f(x)}{x} ) is equal to
A . 1
B. –
( c cdot 0 )
D. None of these
12
8Using properties of determinants, prove
that:
[
left|begin{array}{ccc}
1+a & 1 & 1 \
1 & 1+b & 1 \
1 & 1 & 1+c
end{array}right|=a b c+b c+
]
( c a+a b )
12
9Find the value of the determinant:
[
left|begin{array}{cc}
mathbf{5} & -mathbf{2} \
-mathbf{3} & mathbf{1}
end{array}right|
]
12
10A. square matrix ( boldsymbol{A} ) of order ( mathbf{3}, ) has ( |boldsymbol{A}|= )
( 5, ) find ( mid A ) adjal
12
11valuate: ( left|begin{array}{cccc}1 & a & a^{2} & a^{3}+b c d \ 1 & b & b^{2} & b^{3}+c d a \ 1 & c & c^{2} & c^{3}+a b d \ 1 & d & d^{2} & d^{3}+a b cend{array}right| )12
12( left|begin{array}{cc}cos 15^{circ} & sin 15^{circ} \ sin 15^{circ} & cos 15^{circ}end{array}right|=? )
A . 1
B.
c. ( frac{sqrt{3}}{2} )
D. none of these
12
13Show that the points
( boldsymbol{A}(mathbf{1}, mathbf{2}, mathbf{7}), boldsymbol{B}(mathbf{2}, mathbf{6}, mathbf{3}) ) and ( boldsymbol{C}(mathbf{3}, mathbf{1 0},-mathbf{1}) )
are collinear.
12
14( boldsymbol{a} neq boldsymbol{p}, boldsymbol{b} neq boldsymbol{q}, boldsymbol{c} neq boldsymbol{r} ) and ( left|begin{array}{lll}boldsymbol{p} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a} & boldsymbol{q} & boldsymbol{c} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{r}end{array}right|=0 )
then the value of ( frac{boldsymbol{p}}{boldsymbol{p}-boldsymbol{a}}+frac{boldsymbol{q}}{boldsymbol{q}-boldsymbol{b}}+ )
( frac{r}{r-c} ) is equal to
( A )
B. 2
( c cdot 3 )
D.
12
15Using properties of determinant solve 🙁 left|begin{array}{ccc}1 & a & a^{2}-b c \ 1 & b & b^{2}-a c \ 1 & c & c^{2}-a bend{array}right|= )12
16Evaluate.
( mid begin{array}{ccc}frac{1}{z} & frac{1}{z} & -frac{(x+y)}{z^{2}} \ -frac{(y+z)}{x^{2}} & frac{1}{x} & frac{1}{x} \ -frac{y(y+z)}{x^{2} z} & frac{x+2 y+z}{x z} & -frac{y(x+y)}{x z^{2}}end{array} )
12
17Expand:
( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a}end{array}right| )
12
18( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{omega}^{2} & mathbf{1} \ boldsymbol{omega}^{2} & mathbf{1} & boldsymbol{omega}end{array}right] ) Find Det ( mathbf{A} )12
19If ( boldsymbol{A}=left[begin{array}{ccc}2 & 52 & 152 \ 4 & 106 & 358 \ 6 & 162 & 620end{array}right], ) then the
determinant of the matrix ( a d j(2 A) ) is equal to:
12
20( fleft|begin{array}{ccc}mathbf{6} i & -mathbf{3} i & mathbf{1} \ mathbf{4} & mathbf{3} i & -mathbf{1} \ mathbf{2 0} & mathbf{3} & boldsymbol{i}end{array}right|=boldsymbol{x}+boldsymbol{i} boldsymbol{y}, ) then
A ( . x=3, y=1 )
B. ( x=1, y=3 )
c. ( x=0, y=3 )
D. ( x=0, y=0 )
12
21Let ( boldsymbol{alpha}, boldsymbol{beta}, boldsymbol{gamma}, boldsymbol{a}, boldsymbol{b}, boldsymbol{c}, boldsymbol{x} boldsymbol{epsilon} boldsymbol{R} ) and let
( Delta= )
( left|begin{array}{ccc}sin (x+alpha) & cos (x+alpha) & a+x sin alpha \ sin (x+beta) & cos (x+beta) & b+x sin beta \ sin (x+gamma) & cos (x+gamma) & c+x sin gammaend{array}right| )
then ( Delta ) is
This question has multiple correct options
A. independent of ( x )
B. dependent of ( alpha, beta, ),
c. dependent on ( a, b, c, c )
D. ( a ) constant
12
22If
[] denotes the greatest integer less than or equal to the real number under
consideration, and ( -1 leq x<0, quad 0 leq )
( y<1,1 leq z<2, ) the value of the determinant ( 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| )
is
( A cdot[x] )
в. ( [y] )
( c cdot[z )
D. none of these
12
23( left|begin{array}{ccc}1+a & 1 & 1 \ 1 & 1+b & 1 \ 1 & 1 & 1+cend{array}right|= )
( boldsymbol{X} boldsymbol{a b c}left(1+frac{1}{boldsymbol{a}}+frac{1}{boldsymbol{b}}+frac{1}{c}right) . ) Find the value
of ( boldsymbol{X} )
12
24if ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & mathbf{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{0} & boldsymbol{a}end{array}right|=mathbf{0} ) then
A. ( a ) is a cube root of 1
B. ( b ) is a cube root of 1
c. ( frac{a}{b} ) is a cube root of 1
D. ( frac{a}{b} ) is a cube root of -1
12
25Find the values of ( K ) if Area of the
triangle is 4 sq. units and vertices are ( (k 0)(40)(02) ) using determinants.
12
26If ( boldsymbol{A}=left[begin{array}{cc}-mathbf{4} & -mathbf{1} \ mathbf{3} & mathbf{1}end{array}right], ) then the determinant
of the matrix ( left(A^{2016}-2 A^{2015}-A^{2014}right) )
is:
A. -175
в. 2014
( c .2016 )
D. -25
12
27If the value of ( left|begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{4} \ -mathbf{1} & mathbf{3} & mathbf{0} \ mathbf{4} & mathbf{1} & mathbf{0}end{array}right| ) is ( k, ) then find
( frac{-boldsymbol{k}}{mathbf{1 3}} )
12
28f ( l, m, n ) are ( p^{t h}, q^{t h}, r^{t h} ) terms of ( G . P . ) al positive, then ( left|begin{array}{lll}log l & p & 1 \ log m & q & 1 \ log n & r & 1end{array}right| ) equals
A . –
B. 2
( c )
( D )
12
29If ( boldsymbol{x}=boldsymbol{c} boldsymbol{y}+boldsymbol{b} boldsymbol{z}, boldsymbol{y}=boldsymbol{a} boldsymbol{z}+boldsymbol{c} boldsymbol{x}, boldsymbol{z}=boldsymbol{b} boldsymbol{x}+ )
( a y, ) where ( x, y, z ) are not all zero, then
the value of ( a^{2}+b^{2}+c^{2}+2 a b c )
( mathbf{A} cdot mathbf{0} )
B.
c. -1
D. None of these
12
30Find the integral value of ( x, ) if ( left|begin{array}{ccc}boldsymbol{x}^{2} & boldsymbol{x} & mathbf{1} \ mathbf{0} & boldsymbol{2} & mathbf{1} \ boldsymbol{3} & boldsymbol{1} & boldsymbol{4}end{array}right|=mathbf{2} mathbf{8} )12
31Let ( a, b, c in R ) be such that ( a+b+ )
( boldsymbol{c}>mathbf{0} ) and ( boldsymbol{a b c}=mathbf{2 .} ) Let ( boldsymbol{A}=left[begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right] )
If ( A^{2}=I, ) then value of ( a^{3}+b^{3}+c^{3} ) is
( A cdot 7 )
B.
( c cdot 0 )
( D )
12
32If the lines ( boldsymbol{a} boldsymbol{x}+boldsymbol{y}+mathbf{1}=mathbf{0}, boldsymbol{x}+boldsymbol{b} boldsymbol{y}+ )
( mathbf{1}=mathbf{0} & boldsymbol{x}+boldsymbol{y}+boldsymbol{c}=mathbf{0} ) where ( mathbf{a}, mathbf{b} & mathbf{c} )
are distinct real numbers different from
are concurrent, then the value of ( frac{1}{1-a}+frac{1}{1-b}+frac{1}{1-c}= )
( A cdot 4 )
B. 3
( c cdot 2 )
( D )
12
33If ( a+b+c neq 0 ) and ( left|begin{array}{lll}a & b & c \ b & c & a \ c & a & bend{array}right|=0 )
then using properties of determinants, prove that ( boldsymbol{a}=boldsymbol{b}=boldsymbol{c} )
12
34The system of linear equations:
( boldsymbol{lambda} boldsymbol{x}+boldsymbol{2} boldsymbol{y}+boldsymbol{2} boldsymbol{z}=mathbf{5} )
( 2 lambda x+3 y+5 z= )
( 4 x+lambda y+6 z=10 ) has
A. no solution when ( lambda=2 )
B. a unique solution when ( lambda=-8 )
c. infinitely many solutions when ( lambda=2 )
D. no solution when ( lambda=8 )
12
35Find the value of determinant.
(i) ( left|begin{array}{cc}cos theta & -sin theta \ sin theta & cos thetaend{array}right| )
(ii) ( left|begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1end{array}right| )
12
36If ( alpha=beta+frac{2 pi}{3} . ) then ( A_{theta} ) is maximum
when ( gamma ) equals
A. ( alpha+pi / 3 )
в. ( alpha-pi / 3 )
c. ( alpha+2 pi / 3 )
D. none of these
12
37begin{tabular}{c|ccc}
( mathbf{1} ) & ( mathbf{s i n Q} ) & ( mathbf{1} ) \
( -mathbf{s i n Q} ) & ( mathbf{1} ) & ( mathbf{s i n Q} ) \
( mathbf{- 1} ) & ( -mathbf{s i n Q} ) & ( mathbf{1} )
end{tabular}
12
38( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{y} boldsymbol{z} & boldsymbol{z} boldsymbol{x} & boldsymbol{x} boldsymbol{y}end{array}right|=boldsymbol{b}(boldsymbol{x}-boldsymbol{y})(boldsymbol{y}- )
( z)(z-x) . ) Find
12
39The points which are not collinear are:
A. (0,1),(8,3) and (6,7)
B. (4,3),(5,1) and (1,9)
C. (2,5),(-1,2) and (4,7)
D. (-3,2)(1,-2) and (9,-10)
12
40( fleft(a_{1}, a_{2}, a_{3} dots, a_{n} cdot ) are in G.P, then the right.
determinant ( Delta= ) ( left|begin{array}{ccc}log a_{n} & log a_{n+1} & log a_{n+2} \ log a_{n+3} & log a_{n+4} & log a_{n+5} \ log a_{n+6} & log a_{n+7} & log a_{n+8}end{array}right| ) is equal
to
A . 0
B. 1
( c cdot 2 )
D.
12
41Consider the system of linear equations in x, y, z:
(sin 30) x-y+z=0
(cos 20) x + 4y + 3z=0
2x+ 7y+ 7z=0
Find the values of for which this system has nontrivial
solutions.
(1986 – 5 Marks)
12
42( fleft(alpha, beta text { are the roots of } x^{2}+x+1=0right. )
( operatorname{then}left|begin{array}{ccc}boldsymbol{y}+mathbf{1} & boldsymbol{beta} & boldsymbol{alpha} \ boldsymbol{beta} & boldsymbol{y}+boldsymbol{alpha} & mathbf{1} \ boldsymbol{alpha} & mathbf{1} & boldsymbol{y}+boldsymbol{beta}end{array}right|=? )
A ( cdot y^{2}-1 )
В . ( yleft(y^{2}-1right) )
c. ( y^{2}-y )
D. ( y^{3} )
12
43What are the values of ( x ) that satisfy the
equation ( left|begin{array}{ccc}boldsymbol{x} & mathbf{0} & mathbf{2} \ mathbf{2} boldsymbol{x} & mathbf{2} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right|+left|begin{array}{ccc}mathbf{3} boldsymbol{x} & mathbf{0} & mathbf{2} \ boldsymbol{x}^{mathbf{2}} & boldsymbol{2} & mathbf{1} \ mathbf{0} & boldsymbol{1} & mathbf{1}end{array}right|=mathbf{0} ? )
B. ( -1 pm sqrt{3} )
c. ( -1 pm sqrt{6} )
( mathbf{D} cdot-2 pm sqrt{6} )
12
44f ( x+y+z=pi ) and
( Delta=left|begin{array}{ccc}sin 3 x & sin 3 y & sin 3 z \ sin x & sin y & sin z \ cos x & cos y & cos zend{array}right| )
then ( Delta ) equals
( A )
в.
( c . )
( D )
12
45f ( omega ) is a cube root of unity, then ( left|begin{array}{ccc}mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{omega}^{2} & boldsymbol{1} \ boldsymbol{omega}^{2} & boldsymbol{1} & boldsymbol{omega}end{array}right| ) is equal to
A .
B. ( omega )
( c cdot omega^{2} )
D.
12
46An equilateral triangle has each of its
sides of length ( 6 mathrm{cm} . ) If ( left(x_{1}, y_{1}right) ;left(x_{2}, y_{2}right) )
( &left(x_{3}, y_{3}right) ) are its vertices then the value
of determinant (in nearest integer value) ( ,left|begin{array}{lll}boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & mathbf{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & mathbf{1}end{array}right| ) is equal to a then
find ( frac{boldsymbol{a}}{mathbf{3 1}} )
12
47$
If A2 – A+I=0, then the inverse of A is
(2) A+1 (b) A (c) A-I
[2005]
(2) I-A
12
48If ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{alpha} & boldsymbol{2} \ boldsymbol{2} & boldsymbol{alpha}end{array}right] ) and ( |boldsymbol{A}|^{3}=125 ) then the
value of ( boldsymbol{alpha} ) is
A. ±1
B. ±2
( c .pm 3 )
D. ±5
12
4939.
If the system of linear equations
x+ky+3z=0
3x + ky-2z=0
2x + 4y-3z=0
(JEE M 2018]
has a non-zero solution (x, y, z), then
XZ
2 is equal to :
(2) 10
(6) – 30
(c) 30
(2) – 10
12
50Show that the following set of points are
collinear.
(2,5),(4,6) and (8,8)
12
51The value of the determinant
( left|begin{array}{ccc}1-alpha & alpha-alpha^{2} & alpha^{2} \ 1-beta & beta-beta^{2} & beta^{2} \ 1-gamma & gamma-gamma^{2} & gamma^{2}end{array}right| ) is equal to
A ( cdot(alpha-beta)(beta-gamma)(alpha-gamma) )
В . ( (alpha-beta)(beta-gamma)(gamma-alpha) )
c. ( (alpha-beta)(beta-gamma)(alpha-gamma)(alpha+beta+gamma) )
D.
12
52√3
1
15.
If P=
and A=
and Q = PAPT and
√3
(2005)
x=PTQ2005 P then x is equal to
fi 2005
(a) 0 1
[4+200573 6015 1
(b) 2005 4-2005/3]
12+√3 1
(c) A 1-1 2-13]
w 1 2005 2-√3]
(d) 4 2+13 2005
12
53( operatorname{Det}left{begin{array}{lll}2 & 45 & 55 \ 1 & 29 & 32 \ 3 & 68 & 87end{array}right}=dots dots )
A . 45
B. 64
( c cdot 54 )
D. 32
12
54The least value of the product xyz for which the determinant ( left|begin{array}{lll}boldsymbol{x} & mathbf{1} & mathbf{1} \ mathbf{1} & boldsymbol{y} & mathbf{1} \ mathbf{1} & mathbf{1} & boldsymbol{z}end{array}right| ) is
non-negative, is :
begin{tabular}{l}
A ( -16 sqrt{2} ) \
hline
end{tabular}
В. ( -2 sqrt{2} )
( c cdot-1 )
D. – –
12
55Solve for ( boldsymbol{lambda} ) if
( left|begin{array}{ccc}boldsymbol{a}^{2}+boldsymbol{lambda} & boldsymbol{a} boldsymbol{b} & boldsymbol{a c} \ boldsymbol{a b} & boldsymbol{b}^{2}+boldsymbol{lambda} & boldsymbol{b c} \ boldsymbol{a c} & boldsymbol{b c} & boldsymbol{c}^{2}+boldsymbol{lambda}end{array}right|=mathbf{0} )
12
56( left|begin{array}{ccc}boldsymbol{a}^{2}+mathbf{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b} & boldsymbol{b}^{2}+mathbf{1} & boldsymbol{b} boldsymbol{c} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c} boldsymbol{b} & boldsymbol{c}^{2}+mathbf{1}end{array}right|=mathbf{1}+boldsymbol{a}^{2}+ )12
57The number of distinct real values of ( alpha )
for which the vectors ( boldsymbol{alpha}^{2} hat{boldsymbol{i}}-hat{boldsymbol{j}}-hat{boldsymbol{k}},-hat{boldsymbol{i}}- )
( boldsymbol{alpha}^{2} hat{boldsymbol{j}}-hat{boldsymbol{k}},-hat{boldsymbol{i}}-hat{boldsymbol{j}}-boldsymbol{alpha}^{2} hat{boldsymbol{k}} ) will lie in the
same place is
( A cdot 1 )
B . 2
( c .3 )
D.
12
58( fleft|begin{array}{ccc}mathbf{6} i & -mathbf{3} i & mathbf{1} \ mathbf{4} & mathbf{3} i & -mathbf{1} \ mathbf{2 0} & mathbf{3} & boldsymbol{i}end{array}right|=boldsymbol{x}+boldsymbol{i} boldsymbol{y}, ) then
A ( . x=3, y=1 )
B. ( x=1, y=3 )
c. ( x=0, y=3 )
D. ( x=0, y=0 )
12
59Prove that ( mid begin{array}{ccc}b c-a^{2} & c a-b^{2} & a b \ -b c+c a+a b & b c-c a+a b & b c+c \ (a+b)(a+c) & (b+c)(b+a) & (c+aend{array} )
( mathbf{3} cdot(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}-boldsymbol{a})(boldsymbol{a}-boldsymbol{b})(boldsymbol{a}+boldsymbol{b}+ )
( c(a b+b c+c a) )
12
60The line ( A x+B y+C=0 ) cuts the
circle ( boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{a} boldsymbol{x}+boldsymbol{b} boldsymbol{y}+boldsymbol{c}=boldsymbol{0} ) in ( boldsymbol{P} )
and ( Q )
The line ( boldsymbol{A}^{prime} boldsymbol{x}+boldsymbol{B}^{prime} boldsymbol{y}+boldsymbol{C}^{prime}=mathbf{0} ) cuts the
circle ( boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{a}^{prime} boldsymbol{x}+boldsymbol{b}^{prime} boldsymbol{y}+boldsymbol{c}^{prime}=mathbf{0} ) in ( boldsymbol{R} )
and ( S ).
If ( P, Q, R, S ) are concyclic, then show
that
( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{a}^{prime} & boldsymbol{b}-boldsymbol{b}^{prime} & boldsymbol{c}-boldsymbol{c}^{prime} \ boldsymbol{A} & boldsymbol{B} & boldsymbol{C} \ boldsymbol{A}^{prime} & boldsymbol{B}^{prime} & boldsymbol{C}^{prime}end{array}right|=mathbf{0} )
12
61( operatorname{Let} boldsymbol{A}(mathbf{1}, mathbf{3}), boldsymbol{B}(mathbf{0}, mathbf{0}) ) and ( boldsymbol{C}(boldsymbol{k}, boldsymbol{0}) ) be
vertices of a triangle ( A B C ) such that
area of ( triangle A B C ) is ( 3 . ) Find the value of ( k )
( A cdot pm 2 )
B. ±3
( c .pm 4 )
D. ±1
12
62( f(a, b, c)=left[begin{array}{ccc}a & 0 & 0 \ {[0.3 e m] 0} & b & 0 \ {[0.3 e m] 0} & 0 & cend{array}right] )
such that ( a b c neq 0 )
( operatorname{then} A^{-1}=operatorname{diag}left(frac{1}{a}, frac{1}{b}, frac{1}{c}right)= )
( left[begin{array}{ccc}frac{1}{a} & 0 & 0 \ {[0.3 e m] 0} & frac{1}{b} & 0 \ {[0.3 e m] 0} & 0 & frac{1}{c}end{array}right] )
12
63If ( D_{x}=25, D=5 ) are the values of the
determinants for certain simultaneous
equations in ( x ) and ( y, ) find ( x )
12
64The value of ( mid begin{array}{cc}mathbf{2} & boldsymbol{a}+boldsymbol{b}+boldsymbol{c}+boldsymbol{d} \ boldsymbol{a}+boldsymbol{b}+boldsymbol{c}+boldsymbol{d} & boldsymbol{2}(boldsymbol{a}+boldsymbol{b})(boldsymbol{c}+boldsymbol{d}) \ boldsymbol{a} boldsymbol{b}+boldsymbol{c} boldsymbol{d} & boldsymbol{a} boldsymbol{b}(boldsymbol{c}+boldsymbol{d})+boldsymbol{c} boldsymbol{d}(boldsymbol{a}+boldsymbol{b})end{array} )
( mathbf{A} cdot mathbf{0} )
B.
( c .-1 )
D. None of these
12
65( left|begin{array}{ccc}2 & -3 & 3 \ 2 & 2 & 3 \ 3 & -2 & 2end{array}right| )12
66Using the properties of determinant and
without expanding, prove that:
( left|begin{array}{lll}2 & 7 & 65 \ 3 & 8 & 75 \ 5 & 9 & 86end{array}right|=0 )
12
67( (1,6),(3 .-2) ) and ( (-2, K) ) are collinear
points. What is ( boldsymbol{K} ) ?
A . -6
B. 2
c. 8
D. 10
E . 18
12
68If 07,a2, az ……., an….. are in G.P., then the value of the
determinant
[2004]
logan log an+1 log an+2|
log An+3 log an+4 log an+5 . is
log an+6 log an+7 log an+8|
(a) -2 (b) 1 (c) 2
(d) 0
.
12
69Find determinant ( boldsymbol{D} boldsymbol{c}=left|begin{array}{ccc}mathbf{1} & mathbf{1} & -mathbf{2} \ mathbf{1} & -mathbf{2} & mathbf{3} \ mathbf{2} & -mathbf{1} & -mathbf{1}end{array}right| )12
70f ( (k, 2-2 k),(-k+1,2 k),(-4- )
( k, 6-2 k) ) are collinear, then ( k= )
( A cdot+1 )
B. –
( c cdot-2 )
( D .2 )
12
71Prove that:
( left|begin{array}{ccc}mathbf{0} & boldsymbol{a} & -boldsymbol{b} \ -boldsymbol{a} & boldsymbol{0} & -boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{0}end{array}right|=mathbf{0} )
12
7213.
If
and|A = 125 then the value of a is
(2004S)
(a) #1
(b) +2
(c) +3
(d) 5
12
73If the points ( boldsymbol{A}(boldsymbol{x}, mathbf{2}), boldsymbol{B}(-mathbf{3},-mathbf{4}) ) and
( C(7,-5) ) are collinear, then the value of ( x ) is :
A . -63
B. 63
( c .60 )
D. – 60
12
74( left|begin{array}{ccc}a^{2}+1 & a b & a c \ a b & b^{2}+1 & b c \ a c & b c & c^{2}+1end{array}right|= )
A . abc
B. atb+c
c. ( 1+a^{2}+b^{2}+c^{2} )
( D cdot a b c(1+a+b+c) )
12
75f ( x, y, z ) are all different and if ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{1}+boldsymbol{x}^{3} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{1}+boldsymbol{y}^{3} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{1}+boldsymbol{z}^{3}end{array}right|=boldsymbol{0} operatorname{then} 1+boldsymbol{x} boldsymbol{y} boldsymbol{z}= )
A . -1
B.
( c cdot 1 )
D. 2
12
76Find the adjoint of matrix ( A= ) ( left[begin{array}{lll}1 & 1 & 2 \ 2 & 3 & 5 \ 2 & 0 & 1end{array}right] )12
77In a triangle ( A B C, ) with usual notations, if ( left|begin{array}{ccc}1 & a & b \ 1 & c & a \ 1 & b & cend{array}right|=0, ) then ( 4 sin ^{2} A+ )
( 24 sin ^{2} B+36 sin ^{2} C ) is equal to
A . 48
B. 50
c. 44
D. 34
12
78Let ( triangle boldsymbol{a}=left|begin{array}{ccc}boldsymbol{a}-mathbf{1} & boldsymbol{n} & boldsymbol{6} \ (boldsymbol{a}-mathbf{1})^{2} & boldsymbol{2} boldsymbol{n}^{2} & boldsymbol{4} boldsymbol{n}-boldsymbol{2} \ (boldsymbol{a}-mathbf{1})^{3} & boldsymbol{3} boldsymbol{n}^{3} & boldsymbol{3} boldsymbol{n}^{2}-boldsymbol{3} boldsymbol{n}end{array}right| )
Then ( sum_{a-1}^{n} triangle a ) is equal to
( mathbf{A} cdot mathbf{0} )
В ( cdot(a-1) sum n^{2} )
c. ( (a-1) n sum n )
D. None of these
12
79If ( boldsymbol{A}_{mathbf{3} times mathbf{3}} ) and ( |boldsymbol{A}| neq mathbf{0} Rightarrow boldsymbol{A} boldsymbol{d} boldsymbol{j}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A})= )
( mathbf{A} cdot|A|^{2} A )
B ( cdot|A| A )
c. ( frac{A}{|A|} )
D. ( frac{A}{|A|^{2}} )
12
80Find the value of the following
determinant:
( left|begin{array}{cc}1.2 & 0.03 \ 0.57 & -0.23end{array}right| )
A. -0.266
B. -0.2471
c. -0.2381
D. -0.2931
12
81( left|begin{array}{ccc}1+sin ^{2} theta & sin ^{2} theta & sin ^{2} theta \ cos ^{2} theta & 1+cos ^{2} theta & cos ^{2} theta \ 4 sin 4 theta & 4 sin 4 theta & 1+4 sin 4 thetaend{array}right|= )
( 0, ) then ( sin 4 theta ) equals to
A. ( 1 / 2 )
B.
( c cdot-1 / 2 )
D. -1
12
82The roots of the equation ( left|begin{array}{ccc}boldsymbol{x}-mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & boldsymbol{x}-mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1} & boldsymbol{x}-mathbf{1}end{array}right|=mathbf{0} operatorname{are} )
A. 1,2
в. -1,2
c. -1,-2
D. 1,-2
12
83Find the value of ( left|begin{array}{ccc}53 & 106 & 159 \ 52 & 65 & 91 \ 102 & 153 & 221end{array}right| )12
84The value of the determinant
( left|begin{array}{lll}k a & k^{2}+a^{2} & 1 \ k b & k^{2}+b^{2} & 1 \ k c & k^{2}+c^{2} & 1end{array}right| ) is
A. ( k(a+b)(b+c)(c+a) )
B. ( k a b cleft(a^{2}+b^{2}+c^{2}right) )
c. ( k(a-b)(b-c)(c-a) )
D. ( k(a+b-c)(b+c-a)(c+a-b) )
12
85If ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{y}end{array}right] ) for ( boldsymbol{x} neq )
( mathbf{0}, boldsymbol{y} neq mathbf{0}, ) then ( boldsymbol{D} ) is:
A. divisible by neither ( x ) nor ( y )
B. divisible by both ( x ) nor ( y )
c. divisible by ( x ) but not ( y )
D. divisible by ( y ) but not ( x )
12
86( mathbf{f} mathbf{Delta}=left|begin{array}{lll}boldsymbol{b}^{2}-boldsymbol{a} boldsymbol{b} & boldsymbol{b}-boldsymbol{c} & boldsymbol{b} boldsymbol{c}-boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b}-boldsymbol{a}^{2} & boldsymbol{a}-boldsymbol{b} & boldsymbol{b}^{2}-boldsymbol{a} boldsymbol{b} \ boldsymbol{b} boldsymbol{c}-boldsymbol{a} boldsymbol{c} & boldsymbol{c}-boldsymbol{a} & boldsymbol{a} boldsymbol{b}-boldsymbol{a}^{2}end{array}right| ) then
( Delta ) equals
A ( cdot(b-c)(c-a)(a-b) )
B. ( a b c(b-c)(c-a)(a-b) )
c. ( (a+b+c)(b-c)(c-a)(a-b) )
D.
12
87( mathbf{f}_{mathbf{0}}=left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r}end{array}right| ) and ( boldsymbol{Delta}_{2}=left|begin{array}{lll}boldsymbol{y} & boldsymbol{b} & boldsymbol{q} \ boldsymbol{x} & boldsymbol{a} & boldsymbol{p} \ boldsymbol{z} & boldsymbol{c} & boldsymbol{r}end{array}right| )
then ( Delta_{1} ) is equal to
( A cdot 2 Delta_{2} )
в. ( Delta_{2} )
( c cdot-Delta_{2} )
D. none of these
12
88The value of the determinant
( left|begin{array}{ccc}cos alpha & -sin alpha & 1 \ sin alpha & cos alpha & 1 \ cos (alpha+beta) & -sin (alpha+beta) & 1end{array}right| )
A. Independent of ( alpha )
B. Independent of ( beta )
c. Independent of ( alpha ) and ( beta )
D. None of the above
12
89f ( a neq b neq c, ) prove that the points
( left(a, a^{2}right),left(b, b^{2}right),left(c, c^{2}right) ) can never be
collinear.
12
90( mathbf{f}left(mathbf{1}+boldsymbol{a} boldsymbol{x}+boldsymbol{b} boldsymbol{x}^{2}right)^{4}=boldsymbol{a}_{mathbf{0}}+boldsymbol{a}_{1} boldsymbol{x}+ )
( a_{2} x^{2}+ldots ldots . .+a_{8} x^{8}, ) where
( a, b, a_{0}, a_{1}, dots dots a_{8} in R ) such that ( a_{0}+ )
( a_{1}+a_{2} neq 0 ) and ( left|begin{array}{lll}a_{0} & a_{1} & a_{2} \ a_{1} & a_{2} & a_{0} \ a_{2} & a_{0} & a_{1}end{array}right|=0, ) then
the value of ( frac{mathbf{5} a}{b} ) is
12
91If ( k ) is a scalar and ( A ) is an ( n times n ) square
matrix, then ( |boldsymbol{k} boldsymbol{A}|= )
A ( cdot k|A|^{n} )
в. ( k|A| )
c ( cdot k^{n}left|A^{n}right| )
D ( cdot k^{n}|A| )
12
92The value of the determinant
( left|begin{array}{c}b^{2}-a b b-c b c-a c \ a b-a^{2} a-b b^{2}-a b \ b c-a c c-a a b-a^{2}end{array}right|= )
A ( . a b c )
B. ( a+b+c )
c. 0
( mathbf{D} cdot a b+b c+c a )
12
93If ( A ) is a ( 3- ) rowed square matrix and
( |boldsymbol{A}|=mathbf{4} ) then ( boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A})=? )
A ( .4 A )
в. ( 16 A )
( c cdot 64 A )
D. None of these
12
94Prove that ( left|begin{array}{lll}boldsymbol{a}+boldsymbol{b} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a} & boldsymbol{b}end{array}right|=mathbf{3} boldsymbol{a} boldsymbol{b} boldsymbol{c}-boldsymbol{1} )
( boldsymbol{a}^{3}-boldsymbol{b}^{3}-boldsymbol{c}^{3} )
12
95If ( A=left|begin{array}{ll}2 & 3 \ 6 & 9end{array}right| ) then ( |A|= )
A .
B. 1
( c cdot 2 )
D.
12
96( fleft|begin{array}{lll}boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & mathbf{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & 1end{array}right|=left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{b}_{1} & mathbf{1} \ boldsymbol{a}_{2} & boldsymbol{b}_{2} & mathbf{1} \ boldsymbol{a}_{3} & boldsymbol{b}_{3} & 1end{array}right| ), then the
two triangles with vertices ( left(x_{1}, y_{1}right),left(x_{2}, y_{2}right),left(x_{3}, y_{3}right) ) and ( left(a_{1}, b_{1}right) )
( left(a_{2}, b_{2}right),left(a_{3}, b_{3}right) ) must be congruent
A. True
B. False
12
97The value of ( left|begin{array}{ccc}(boldsymbol{a}+boldsymbol{d})(boldsymbol{a}+mathbf{2} boldsymbol{d}) & boldsymbol{a}+mathbf{2} boldsymbol{d} & boldsymbol{a} \ mathbf{2} boldsymbol{d}(boldsymbol{a}+mathbf{2} boldsymbol{d}) & boldsymbol{d} & boldsymbol{d} \ mathbf{2} boldsymbol{d}(boldsymbol{a}+mathbf{3} boldsymbol{d}) & boldsymbol{d} & boldsymbol{d}end{array}right| )
A . ( 4 d )
B ( .4 d^{2} )
( c cdot 4 d^{3} )
D. ( 4 d^{4} )
12
98f ( a, b, c ) are non-zero and different from
1, then the value of
( left|begin{array}{ccc}log _{a} 1 & log _{a} b & log _{a} c \ log _{a}left(frac{1}{b}right) & log _{b} 1 & log _{a}left(frac{1}{c}right) \ log _{a}left(frac{1}{c}right) & log _{a} c & log _{c} 1end{array}right| )
A . 0
B. ( 1+log _{a}(a+b+c) )
( mathbf{c} cdot log _{a}(a b+b c+c a) )
D. 1
( E cdot log _{a}(a+b+c) )
12
999.
Which of the following values of a satisfy the equation
| (1+a)2 (1+2a)2 (1+3a)2
(2+ a)2 (2+2a)2 (2+3a)2 = -648a ?
|(3+a)? (3+2a)2 (3+3a)2 |
(a)
4
(6)
9
(c)
9
(JEE Adv. 2015)
(d) 4
12
100Solve:
( left|begin{array}{ccc}mathbf{0} & -mathbf{3} & boldsymbol{x} \ boldsymbol{x}+mathbf{1} & mathbf{3} & mathbf{1} \ mathbf{4} & mathbf{1} & mathbf{5}end{array}right|=mathbf{0} )
12
101solve: ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{2} & -mathbf{1} \ boldsymbol{3} & -boldsymbol{1} & boldsymbol{4} \ boldsymbol{x} & boldsymbol{2} & -boldsymbol{5}end{array}right|=mathbf{0} )12
102Assertion
Let ( A ) be a ( 2 times 2 ) matrix with real
entries. Let ( I ) be the ( 2 times 2 ) identity
matrix. Denote by ( t r(A), ) the sum of
diagonal entries of ( A ). Assume that
( boldsymbol{A}^{2}=boldsymbol{I} )
If ( boldsymbol{A} neq boldsymbol{I} ) and ( boldsymbol{A} neq-boldsymbol{I}, ) then ( operatorname{det}(boldsymbol{A})=-mathbf{1} )
Reason
If ( boldsymbol{A} neq boldsymbol{I} ) and ( boldsymbol{A} neq-boldsymbol{I}, ) then ( boldsymbol{t r}(boldsymbol{A}) neq mathbf{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
103Evaluate: ( left|begin{array}{ccc}mathbf{1} & boldsymbol{b} boldsymbol{c} & boldsymbol{a}(boldsymbol{b}+boldsymbol{c}) \ mathbf{1} & boldsymbol{c} boldsymbol{a} & boldsymbol{b}(boldsymbol{c}+boldsymbol{a}) \ boldsymbol{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{c}(boldsymbol{a}+boldsymbol{b})end{array}right| )
( mathbf{A} cdot mathbf{0} )
B.
( c . a b c )
D. ( a^{2}+b^{2}+c^{2} )
12
104x + 4y + z = 0;
If the system of linear equations
x + 2ay + az = 0 ; x + 3y + bz = 0 ; x + 4y + CZ
has a non – zero solution, then a, b, c.
(a) satisfy a + 2b + 3c = 0 (b) are in A.P
(c) are in G.P
(d) are in H.P.
12
105Find the value of following determinant. ( left|begin{array}{ll}frac{7}{3} & frac{5}{3} \ frac{3}{2} & frac{1}{2}end{array}right| )12
106If the points ( boldsymbol{A}(-2,1), B(a, b) ) and ( C(4,-1) ) are collinear and ( a-b=1 )
find the values of ( a ) and ( b )
A ( . a=1, b=5 )
В. ( a=1, b=0 )
c. ( a=2, b=0 )
D. None of these
12
107Using properties of determinants, prove
the following
[
mid begin{array}{ccc}
1+a^{2}-b^{2} & 2 a b & -2 b \
2 a b & 1-a^{2}+b^{2} & 2 a \
2 b & -2 a & 1-a^{2}-b^{2}
end{array}
]
( left(1+a^{2}+b^{2}right)^{3} )
12
108( left|begin{array}{ccc}mathbf{1}+boldsymbol{x} & mathbf{2} & mathbf{3} \ mathbf{1} & mathbf{2}+boldsymbol{x} & mathbf{3} \ mathbf{1} & mathbf{2} & mathbf{3}+boldsymbol{x}end{array}right|=mathbf{0} operatorname{then} boldsymbol{x}= )
( A )
B. -1
( c .-6 )
( D )
12
109Let ( n ) be a positive integer and ( Delta_{r}= ) ( left|begin{array}{ccc}2 r-1 & n & C_{r} & 1 \ n^{2}-1 & 2^{n} & n+1 \ cos ^{2}left(n^{2}right) & cos ^{2} n & cos ^{2}(n+1)end{array}right| ) then
( sum_{r=0}^{n} Delta_{r}=dots )
( A cdot 0 )
B.
( c cdot 2 )
( D )
12
110(i) Solve the equation ( left|begin{array}{ccc}boldsymbol{x}-mathbf{1} & boldsymbol{2} & boldsymbol{3} \ mathbf{0} & boldsymbol{x}-boldsymbol{2} & boldsymbol{4} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{x}-boldsymbol{3}end{array}right| )
(ii) Show that ( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{b} boldsymbol{c} \ mathbf{1} & boldsymbol{b} & boldsymbol{c a} \ mathbf{1} & boldsymbol{c} & boldsymbol{a b}end{array}right|=left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2} \ mathbf{1} & boldsymbol{b} & boldsymbol{b}^{2} \ mathbf{1} & boldsymbol{c} & boldsymbol{c}^{2}end{array}right| )
12
111Prove the following:
( left|begin{array}{lll}boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c}end{array}right|=mathbf{2}left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a}end{array}right| )
12
112( mathbf{a}=left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a}end{array}right|, ) then the value of
( left|begin{array}{ccc}a^{2}-b c & b^{2}-c a & c^{2}-a b \ c^{2}-a b & a^{2}-b c & b^{2}-c a \ b^{2}-c a & c^{2}-a b & a^{2}-b cend{array}right| )
( A cdot Delta^{2} )
B ( .2 Delta^{2} )
( c cdot Delta^{3} )
D. none of these
12
113Solve: Value of ( boldsymbol{D}=left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2}end{array}right| ) is12
114If ( a neq 6, b, c ) satisfy ( left|begin{array}{ccc}a & 2 b & 2 c \ 3 & b & c \ 4 & a & bend{array}right|=0 )
then ( a b c= )
( mathbf{A} cdot a+b+c )
B.
( c cdot b^{3} )
( mathbf{D} cdot a b+b-c )
12
115Using properties of determinants, show that triangle ( A B C ) is isosceles, if :
( mid begin{array}{ccc}mathbf{1} & mathbf{1} \ mathbf{1}+cos boldsymbol{A} & mathbf{1}+cos boldsymbol{B} & mathbf{1} \ cos ^{2} boldsymbol{A}+cos boldsymbol{A} & cos ^{2} boldsymbol{B}+cos boldsymbol{B} & mathbf{c o s}^{2}end{array} )
0
12
116Let ( S ) be the sample space of all ( 3 times 3 )
matrices with entries from the set
( {0,1} . ) Let the events ( E_{1}= )
( {A in S: operatorname{det} A=0} ) and ( E_{2}= )
( {A in S: S u m text { of entries of } A text { is } 7} )
If a matrix is chosen at random from ( boldsymbol{S} )
then the conditional probability
( boldsymbol{P}left(boldsymbol{E}_{1} mid boldsymbol{E}_{2}right) ) equals.
12
117( left|begin{array}{lll}mathbf{a}+mathbf{b} & mathbf{a} & mathbf{b} \ mathbf{a} & mathbf{a}+mathbf{c} & mathbf{c} \ mathbf{b} & mathbf{c} & mathbf{b}+mathbf{c}end{array}right|= )
( A cdot 4 ) abc
B. abç
( c cdot 2 a^{2} b^{2} c^{2} )
D. ( 4 a^{2} b^{2} c^{2} )
12
118( boldsymbol{D}=left|begin{array}{ccc}mathbf{1 8} & mathbf{4 0} & mathbf{8 9} \ mathbf{4 0} & mathbf{8 9} & mathbf{1 9 8} \ mathbf{8 9} & mathbf{1 9 8} & mathbf{4 4 0}end{array}right|= )
( A )
B. –
c. zero
( D )
12
119Let
( boldsymbol{f}(boldsymbol{x})=left|begin{array}{ccc}2 cot boldsymbol{x} & -mathbf{1} & mathbf{0} \ mathbf{1} & cot boldsymbol{x} & -mathbf{1} \ mathbf{0} & mathbf{1} & mathbf{2} cot boldsymbol{x}end{array}right| )
then
12
120( left|begin{array}{lll}a^{2}+lambda^{2} & a b+c lambda & c a-b lambda \ a b-c lambda & b^{2}+lambda^{2} & b c+a lambda \ c a+b lambda & b c-a lambda & c^{2}+lambda^{2}end{array}right| mid begin{array}{cc}lambda & c \ -c & lambda \ b & -aend{array} )
( left(1+a^{2}+b^{2}+c^{2}right)^{3}, ) then the value of ( lambda )
is
( A cdot 8 )
в. 2
( c )
( D )
12
121Using properties of deteminants, prove
[
operatorname{that}left|begin{array}{ccc}
frac{(boldsymbol{a}+boldsymbol{b})^{2}}{boldsymbol{c}} & boldsymbol{c} & boldsymbol{c} \
boldsymbol{a} & frac{(boldsymbol{b}+boldsymbol{c})^{2}}{boldsymbol{a}} & boldsymbol{a} \
boldsymbol{b} & boldsymbol{b} & frac{(boldsymbol{c}+boldsymbol{a})^{2}}{boldsymbol{b}}
end{array}right|
]
( 2(a+b+c)^{3} )
12
122Show that ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c}end{array}right|= )
( boldsymbol{a}^{3}+boldsymbol{b}^{3}+boldsymbol{c}^{3}-boldsymbol{3} boldsymbol{a} boldsymbol{b} boldsymbol{c} )
12
123Assertion If ( A=left(begin{array}{ll}cos alpha & sin alpha \ cos alpha & sin alphaend{array}right) ) and ( B= )
( left(begin{array}{cc}cos alpha & cos alpha \ sin alpha & sin alphaend{array}right) ) then ( A B neq I )
Reason
The product of two matrices can never be equal to an identity matrix
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
124Solve: ( left|begin{array}{ccc}-1 & x & 2 \ 3 & 4 & -2 \ 4 & x & -3end{array}right|=0 )12
125( fleft|begin{array}{ccc}6 i & -3 i & 1 \ 4 & 3 i & -1 \ 20 & 3 & iend{array}right|=x+i y ) then
A ( . x=3, y=1 )
B. ( x=1, y=3 )
c. ( x=0, y=3 )
D. ( x=0, y=0 )
12
126The value of ( left|begin{array}{ll}cos 15^{circ} & sin 15^{circ} \ sin 75^{circ} & cos 75^{circ}end{array}right| )12
127If ( A=left[begin{array}{cc}-8 & 5 \ 2 & 4end{array}right] ) satisfies the equation ( x^{2}+4 x-p=0, ) then ( p= )
A . 64
B. 42
( c . ) 36
D. 24
12
128( mathrm{ft} mathrm{A}=left[begin{array}{ccc}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4end{array}right] ) then show that
( |3 A|=27|A| )
12
129If the points ( (-3,6),(-9, a) ) and (0,15) are collinear, then find a12
130f ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{4} & boldsymbol{7} \ boldsymbol{6} & boldsymbol{5}end{array}right], ) find ( |boldsymbol{3} boldsymbol{A}| )12
131If ( A A^{T}=I ) and ( operatorname{det}(A)=1, ) then
A. Every element of ( A ) is equal to it’s co-factor
B. Every element of A and it’s co-factor are additive inverse of each other
C. Every element of A and it’s co-factor are multiplicative inverse of each other
D. None of these
12
132Find the non-zero roots of the equation ( boldsymbol{Delta}=left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{a} boldsymbol{x}+boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b} boldsymbol{x}+boldsymbol{c} \ boldsymbol{a} boldsymbol{x}+boldsymbol{b} & boldsymbol{b} boldsymbol{x}+boldsymbol{c} & boldsymbol{c}end{array}right|=0 )12
133Find the values of ( x, ) if ( left|begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5}end{array}right|=left|begin{array}{ll}boldsymbol{x} & mathbf{3} \ mathbf{2} boldsymbol{x} & mathbf{5}end{array}right| )12
1343.
If 1,0,02 are the cube roots of unity, then
1
on
w2n|
A=0″
@21
1
is equal to
[2003]
(a) 0²
(b) 0
(0)
1
(d) w
12
135( boldsymbol{A}=left[begin{array}{ccc}mathbf{5} & mathbf{5} boldsymbol{a} & boldsymbol{a} \ mathbf{0} & boldsymbol{a} & mathbf{5} boldsymbol{a} \ mathbf{0} & mathbf{0} & mathbf{5}end{array}right] ) If ( left|boldsymbol{A}^{2}right|=mathbf{2 5} ) then
( |boldsymbol{a}|= )
( A cdot 5 )
В. ( 5^{2} )
( c )
( D )
12
136f ( boldsymbol{y}=sin boldsymbol{p} boldsymbol{x} ) prove that
( boldsymbol{Delta}=left|begin{array}{lll}boldsymbol{y} & boldsymbol{y}_{1} & boldsymbol{y}_{2} \ boldsymbol{y}_{3} & boldsymbol{y}_{4} & boldsymbol{y}_{5} \ boldsymbol{y}_{6} & boldsymbol{y}_{7} & boldsymbol{y}_{8}end{array}right|=0 )
where ( y_{r}, ) means ( r-t h ) differential
coefficient of ( boldsymbol{y} )
12
137( f(x)=left|begin{array}{ccc}cos x & 1 & 0 \ 1 & cos x & 1 \ 0 & 1 & cos xend{array}right| ) the
( f^{prime}left(frac{pi}{3}right) ) equals
A ( frac{11 sqrt{3}}{8} )
( B cdot frac{5 sqrt{3}}{8} )
( c cdot-frac{5 sqrt{3}}{8} )
D. none of these
12
138Find the relation between ( a ) and ( b ). If the
points ( boldsymbol{P}(1,2), Q(0,0) ) and ( P(a, b) ) are collinear
12
139Show that ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{y} boldsymbol{z} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{z} boldsymbol{x} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{x} boldsymbol{y}end{array}right|=(boldsymbol{x}-boldsymbol{y})(boldsymbol{y}- )
( z(z-x)(x y+y z+z x) )
12
140If the points (0,4),(4,0) and ( (5, p) ) are collinear, then value of ( p ) is
A . -1
B. 7
( c cdot 6 )
D.
12
141If ( boldsymbol{u}_{n}= )
then ( boldsymbol{u}_{n}=boldsymbol{a}_{n} boldsymbol{u}_{n-1}+boldsymbol{u}_{n-2} )
f true enter 1 else enter
12
142Find the largest value of a third-order
determinant whose element are 0 or 1
12
143Using properties of determinant, prove
[
text { that }left|begin{array}{lll}
boldsymbol{b}+boldsymbol{c} & boldsymbol{a}-boldsymbol{b} & boldsymbol{a} \
boldsymbol{c}+boldsymbol{a} & boldsymbol{b}-boldsymbol{c} & boldsymbol{b} \
boldsymbol{a}+boldsymbol{b} & boldsymbol{c}-boldsymbol{a} & boldsymbol{c}
end{array}right|=mathbf{3} boldsymbol{a} boldsymbol{b} boldsymbol{c}-boldsymbol{a}^{3}-
]
( b^{3}-c^{3} )
12
144If ( boldsymbol{A}=[boldsymbol{a} boldsymbol{i} boldsymbol{j}] ) is a matrix of order ( 2 x boldsymbol{2} )
such that ( |A|=15 ) and cij represents
the co factor of aij then find ( a_{21} c_{21}+ )
( boldsymbol{a}_{22} boldsymbol{c}_{22} )
12
145The value of ( left|begin{array}{ccc}boldsymbol{a}+boldsymbol{p} boldsymbol{d} & boldsymbol{a}+boldsymbol{q} boldsymbol{d} boldsymbol{a}+boldsymbol{r} boldsymbol{d} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r} \ boldsymbol{d} & boldsymbol{f} & boldsymbol{d}end{array}right| )
( mathbf{A} cdot mathbf{0} )
B. –
( c cdot 1 )
D. ( p+q+r )
12
146The value of determinant ( left|begin{array}{ccc}mathbf{1 9} & mathbf{6} & mathbf{7} \ mathbf{2 1} & mathbf{3} & mathbf{1 5} \ mathbf{2 8} & mathbf{1 1} & mathbf{6}end{array}right| )
is :
A. 150
B. -110
( c cdot 0 )
D. None of these
12
147Find the Adjoint matrix of the matrix ( left|begin{array}{lll}1 & 2 & 3 \ 2 & 3 & 2 \ 3 & 3 & 4end{array}right| )12
148Find the value of the determinant with
out expanding:
( left|begin{array}{lll}5 & 2 & 3 \ 7 & 3 & 4 \ 9 & 4 & 5end{array}right| )
12
149Let ( a, b, c ) be such that ( b(a+c) neq 0 ) ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{a}+mathbf{1} & boldsymbol{a}-mathbf{1} \ -boldsymbol{b} & boldsymbol{b}+mathbf{1} & boldsymbol{b}-mathbf{1} \ boldsymbol{c} & boldsymbol{c}-mathbf{1} & boldsymbol{c}+mathbf{1}end{array}right|+ )
( left|begin{array}{ccc}boldsymbol{a}+mathbf{1} & boldsymbol{b}+mathbf{1} & boldsymbol{c}-mathbf{1} \ boldsymbol{a}-mathbf{1} & boldsymbol{b}-mathbf{1} & boldsymbol{c}+mathbf{1} \ (-mathbf{1})^{n+mathbf{2}} boldsymbol{a} & (-mathbf{1})^{n+mathbf{1}} boldsymbol{b} & (-mathbf{1})^{n} boldsymbol{c}end{array}right|=mathbf{0} )
then the value of ( n ) is
A. Any integer
B. zero
c. Any even integer
D. Any odd integer
12
1505
15. Let A = 10
Το
5α α
α 5α
O 5
then la equals
(a) 1/5
(6) 5
(0) 52
[2007]
(d) 1
12
151f ( x, y z ) are all different and if ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{boldsymbol{2}} & boldsymbol{1}+boldsymbol{x}^{boldsymbol{3}} \ boldsymbol{y} & boldsymbol{y}^{boldsymbol{2}} & boldsymbol{1}+boldsymbol{y}^{boldsymbol{3}} \ boldsymbol{z} & boldsymbol{z}^{boldsymbol{2}} & boldsymbol{1}+boldsymbol{z}^{boldsymbol{3}}end{array}right|=0 ) then ( mathbf{x y z}= )
A . -1
B.
( c . )
D. ±1
12
152Evaluate the following determinant : ( = ) ( begin{array}{|lll|}mathbf{6 7} & mathbf{1 9} & mathbf{2 1} \ mathbf{3 9} & mathbf{1 3} & mathbf{1 4} \ mathbf{8 1} & mathbf{2 4} & mathbf{2 6}end{array} )12
153( begin{array}{|ccc|}text { If } boldsymbol{x}+boldsymbol{y}+boldsymbol{z}= & mathbf{0} text { ,find } \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{x}^{mathbf{2}} & boldsymbol{y}^{mathbf{2}} & boldsymbol{z}^{mathbf{2}} \ boldsymbol{y}+boldsymbol{z} & boldsymbol{z}+boldsymbol{x} & boldsymbol{x}+boldsymbol{y}end{array} )12
154Using the properties of determinant and without expanding, prove that:
( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{a} & boldsymbol{x}+boldsymbol{a} \ boldsymbol{y} & boldsymbol{b} & boldsymbol{y}+boldsymbol{b} \ boldsymbol{z} & boldsymbol{c} & boldsymbol{z}+boldsymbol{c}end{array}right|=mathbf{0} )
12
155Find ( ^{prime} x^{prime} ) if
( left|begin{array}{ccc}mathbf{4} & boldsymbol{x} & mathbf{6} \ mathbf{2} & mathbf{3} & mathbf{4} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right|=mathbf{1 0} )
12
156If ( a>0, b>0, c>0 ) are respectively
the ( p^{t h}, q^{t h}, r^{t h} ) terms of a G.P., then the value of the deteminant ( left|begin{array}{lll}log boldsymbol{a} & boldsymbol{p} & 1 \ log boldsymbol{b} & boldsymbol{q} & 1 \ log boldsymbol{c} & boldsymbol{r} & 1end{array}right| ) is
A . 1
B. 0
( c cdot-1 )
D. None of these
12
157Let ( 0<theta<pi / 2 ) and
( boldsymbol{Delta}(boldsymbol{x}, boldsymbol{theta})=left|begin{array}{ccc}boldsymbol{x} & tan boldsymbol{theta} & cot boldsymbol{theta} \ -tan boldsymbol{theta} & -boldsymbol{x} & mathbf{1} \ cot boldsymbol{theta} & boldsymbol{1} & boldsymbol{x}end{array}right| )
then
This question has multiple correct options
A ( cdot Delta(0, theta)=0 )
B. ( Deltaleft(x, frac{pi}{4}right)=x-x^{3} )
c. ( operatorname{Min}_{0<theta<pi / 2} Delta(1, theta)=0 )
D. ( Delta(x, theta) ) is independent of ( x )
12
158The determinant ( left|begin{array}{ccc}sin alpha & cos alpha & 1 \ sin beta & cos beta & 1 \ sin gamma & cos gamma & 1end{array}right| ) is
equal to
This question has multiple correct options
( ^{mathrm{A}} cdot_{-4 sin } frac{alpha-beta}{2} sin frac{alpha-gamma}{2} sin frac{gamma-alpha}{2} )
( mathbf{B} cdot sin alpha+sin beta+sin gamma )
c. ( sin (alpha-beta)+sin (beta-gamma)+sin (gamma-alpha) )
D. none of these
12
159The vertices of the triangle ( A B C ) are
( (2,1,1),(3,1,2),(-4,0,1) . ) The area of
triangle is
A ( cdot frac{3 sqrt{38}}{2} )
B. ( sqrt{38} )
c. ( frac{sqrt{38}}{2} )
D. 4
12
160( fleft(a_{1}, a_{2}, a_{3}, dots ) from a geometric right.
progression and ( a_{i}>0 ) for all ( i geq 1 )
[
operatorname{then}left|begin{array}{ccc}
log boldsymbol{a}_{m} & log boldsymbol{a}_{m+1} & log boldsymbol{a}_{m+2} \
log boldsymbol{a}_{m+3} & log boldsymbol{a}_{m+4} & log boldsymbol{a}_{m+5} \
log boldsymbol{a}_{m+boldsymbol{6}} & log boldsymbol{a}_{m+boldsymbol{7}} & log boldsymbol{a}_{m+8}
end{array}right| text { is }
]
equal to
( mathbf{A} cdot log a_{m+8}-log a_{m} )
B. ( log a_{m} )
( mathbf{c} cdot 2 log a_{m+1} )
D.
12
161( A_{3 * 3} ) is a non – singular matrix ( Rightarrow )
( boldsymbol{A}^{2}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A})= )
( mathbf{A} cdot|A| A )
B. ( I )
c. ( |A| )
D・ ( |A|^{2} I )
12
162( left|begin{array}{ccc}mathbf{0} & boldsymbol{p}-boldsymbol{q} & boldsymbol{p}-boldsymbol{r} \ boldsymbol{q}-boldsymbol{p} & boldsymbol{0} & boldsymbol{q}-boldsymbol{r} \ boldsymbol{r}-boldsymbol{p} & boldsymbol{r}-boldsymbol{q} & boldsymbol{0}end{array}right| ) is equal to
A. ( p+q+r )
B.
c. ( p-q-r )
( mathbf{D} cdot-p+q+r )
12
163If ( t_{1}, t_{2} ) and ( t_{3} ) distinct. and the points
( left(t_{1} cdot 2 a t_{1}+a t_{1}^{3}right) cdotleft(t_{2} cdot 2 a t_{2}+a t_{2}^{3}right),left(t_{3} cdot 2 aright. )
are collinear, then ( t_{1}+t_{2}+t_{3}= )
A ( cdot t_{1} t_{2} t_{3}=-1 )
B ( cdot t_{1}+t_{2}+t_{3}=t_{1} t_{2} t )
( mathbf{c} cdot t_{1}+t_{2}+t_{3}=0 )
D. ( t_{1}+t_{2}+t_{3}=-1 )
12
164Prove that ( :=(a-b)(b-c)(c-a)(a+ )
( b+c) )
[
begin{array}{ll}1 & a \ 1 & b \ 1 & c & a^{3}end{array}
]
12
165Iet a b c be the real numbers. Then following system of
equations in x, y and z
(1995)
IN
NI
r2
y2 + 3 = 1 has
22
(a) no solution
(b) unique solution
(c) infinitely many solutions(d) finitely many solutions
Q2 x 22
62. x
12
166f ( x, y, z ) are different and ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{1}+boldsymbol{x}^{2} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{1}+boldsymbol{y}^{2} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{1}+boldsymbol{z}^{2}end{array}right|=mathbf{0} ) then prove that
( mathbf{1}+boldsymbol{x} boldsymbol{y} boldsymbol{z}=mathbf{0} )
12
167Prove that
[
left|begin{array}{ccc}
mathbf{1} & mathbf{1} & mathbf{1}+mathbf{3} boldsymbol{x} \
mathbf{1}+mathbf{3} boldsymbol{y} & mathbf{1} & mathbf{1} \
mathbf{1} & mathbf{1}+mathbf{3} boldsymbol{z} & mathbf{1}
end{array}right|=mathbf{9}(mathbf{3} boldsymbol{x} boldsymbol{y} boldsymbol{z}+
]
( boldsymbol{x} boldsymbol{y}+boldsymbol{y} boldsymbol{z}+boldsymbol{z} boldsymbol{x} )
12
168For what value of ( x ) the matrix ( A ) is
singular? ( boldsymbol{A}=left[begin{array}{ll}1+x & 7 \ 3-x & 8end{array}right] )
12
169f ( boldsymbol{A}=left|begin{array}{cc}2 & mathbf{3} \ -mathbf{1} & mathbf{2}end{array}right| ) Find ( boldsymbol{A} )12
170Consider the points ( boldsymbol{P}=(-sin (boldsymbol{beta}- )
( boldsymbol{alpha}),-cos beta), boldsymbol{Q}=(cos (beta-boldsymbol{alpha}), sin beta) )
and ( boldsymbol{R}=(cos (boldsymbol{beta}-boldsymbol{alpha}+boldsymbol{theta}), sin (boldsymbol{beta}-boldsymbol{theta})) )
where ( 0<alpha, beta<frac{pi}{4} ) then
A. ( P ) lies on the line segment ( R Q )
B. ( Q ) lies on the line segment ( P R )
c. ( R ) lies on the line segment ( Q P )
D. ( P, Q, R ) are non-collinear
12
171( mathbf{A}=left[begin{array}{lll}b^{2} c^{2} & b c & b+c \ c^{2} a^{2} & c a & c+a \ a^{2} b^{2} & a b & a+bend{array}right] ) then ( |A|=? )
A ( cdot a b c )
B . ( a b c-1 )
( c cdot a b c+1 )
D.
12
172Evaluate:
( left|begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1end{array}right| )
12
173If ( A ) is a square matrix of order ( 3 times 3 )
such that ( |boldsymbol{A}|=mathbf{5}, ) then find ( |boldsymbol{4} boldsymbol{A}| )
12
174( A_{3 times 3} ) is a matrix such that ( |A|= ) ( boldsymbol{a}, boldsymbol{B}=(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}) ) such that ( |boldsymbol{B}|=boldsymbol{b} . ) Find
the value of ( frac{left(a b^{2}+a^{2} b+1right) S}{25} ) where ( frac{1}{2} S=frac{a}{b}+frac{a^{2}}{b^{3}}+frac{a^{3}}{b^{5}}+dots dots dots u p t o infty )
and ( boldsymbol{a}=mathbf{3} )
12
175The 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
176Consider the following statements in respect of the determinant ( left|begin{array}{cc}cos ^{2} frac{alpha}{2} & sin ^{2} frac{alpha}{2} \ sin ^{2} frac{beta}{2} & cos ^{2} frac{beta}{2}end{array}right| ) where ( alpha, beta ) are
complementary angles
1. The value of the determinant is ( frac{1}{sqrt{2}} cos left(frac{alpha-beta}{2}right) )
2. The maximum value of the determinant is ( frac{mathbf{1}}{sqrt{mathbf{2}}} )
Which of the above statements is/are
correct?
A. 1 only
B. 2 only
c. Both 1 and 2
D. Neither 1 nor 2
12
177( mathbf{a}=left|begin{array}{ccc}cos theta / 2 & 1 & 1 \ 1 & cos theta / 2 & -cos theta / 2 \ -cos theta / 2 & 1 & -1end{array}right| )
If the minimun of ( Delta ) is ( m_{1} ) and
maximum of ( Delta ) is ( m_{2}, ) then ( left[m_{1}, m_{2}right] ) are
related
A. [-4,-2]
B. [2, 4]
( c cdot[-4,0] )
D. [0, 2]
12
178( left|begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5end{array}right| )12
179Prove that ( left|begin{array}{lll}boldsymbol{b} boldsymbol{c} & boldsymbol{b} boldsymbol{c}^{prime}+boldsymbol{b}^{prime} boldsymbol{c} & boldsymbol{b}^{prime} boldsymbol{c}^{prime} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c} boldsymbol{a}^{prime}+boldsymbol{c}^{prime} boldsymbol{a} & boldsymbol{c}^{prime} boldsymbol{a}^{prime} \ boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{b}^{prime}+boldsymbol{a}^{prime} boldsymbol{b} & boldsymbol{a}^{prime} boldsymbol{b}^{prime}end{array}right|= )
( left(boldsymbol{a} boldsymbol{b}^{prime}-boldsymbol{a}^{prime} boldsymbol{b}right)left(boldsymbol{b} boldsymbol{c}-boldsymbol{b}^{prime} boldsymbol{c}right)left(boldsymbol{c} boldsymbol{a}^{prime}-boldsymbol{c}^{prime} boldsymbol{a}right) )
12
180Evaluate the determinant to the closest nteger: ( boldsymbol{A}=left[begin{array}{cc}log _{3} mathbf{5 1 2} & log _{4} mathbf{3} \ log _{3} mathbf{8} & log _{4} mathbf{9}end{array}right] )12
181( mathbf{f} mathbf{Delta}=left|begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{2} & mathbf{0} & mathbf{1} \ mathbf{5} & mathbf{3} & mathbf{8}end{array}right|, ) write the minor of the
elements ( a_{22} )
12
182( fleft|begin{array}{lll}boldsymbol{x}+mathbf{1} & boldsymbol{x}+boldsymbol{2} & boldsymbol{x}+boldsymbol{a} \ boldsymbol{x}+mathbf{2} & boldsymbol{x}+boldsymbol{3} & boldsymbol{x}+boldsymbol{b} \ boldsymbol{x}+mathbf{3} & boldsymbol{x}+boldsymbol{4} & boldsymbol{x}+boldsymbol{c}end{array}right|=mathbf{0} ) then show
that ( a, b, c ) are in ( A . P )
12
183( mathrm{f}left|begin{array}{lll}boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b}end{array}right|=boldsymbol{t} times operatorname{det} ) of
circulant matrix whose elements of first
column are ( a, b, c ) then ( t ) equals
A. 5
B. 6
( c cdot-2 )
( D )
12
184(a) JL
How many 3 x 3 matrices M with entries from {0, 1, 2} are
there, for which the sum of the diagonal entries of M Mis
5?
(JEE Adv. 2017)
(a) 126 (b) 198 (c) 162 (d) 135
12
185If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) find ( |mathbf{2} boldsymbol{A}| )12
186Evaluate the following deteminants:
i) ( left|begin{array}{cc}boldsymbol{x} & -mathbf{7} \ boldsymbol{x} & mathbf{5} boldsymbol{x}+mathbf{1}end{array}right| )
ii) ( mid begin{array}{cc}cos 15^{circ} & sin 15^{circ} \ sin 75^{circ} & cos 75^{circ}end{array} )
12
18720. Let a, b, c be such that b(a + c)+ 0 if
[2009
a a+1 a 1 a+1 6+1 c-1
1-6 6+1 6-1 – a-1 6-1 ct1=0,
c c-1 c+1 |(–1)n+2a (+1) +1b (-1)” cl
then the value of n is :
(a) any even integer (b) any odd integer
(c) any integer
(d) zero
12
18815.
(2003 – 2 Marks)
If Mis a 3 x 3 matrix, where det M=1 and MM=I, where I
is an identity matrix, prove that det (M-1)=0.
(2004 2 Maula
12
189Let ( boldsymbol{d} in boldsymbol{R}, ) and ( boldsymbol{A}= )
( left[begin{array}{ccc}-2 & 4+d & (sin theta)-2 \ 1 & (sin theta)+2 & d \ 5 & (2 sin theta)-d & (-sin theta)+2+2 dend{array}right. )
( boldsymbol{theta} in[mathbf{0}, mathbf{2} boldsymbol{pi}] . ) If the minimum value of
( operatorname{det}(A) ) is ( 8, ) then a value of d is?
A . -7
B. ( 2(sqrt{2}+2) )
( c .-5 )
D. ( 2(sqrt{2}+1) )
12
190( mathbf{a}_{r}=left|begin{array}{ccc}mathbf{2}^{r}-mathbf{1} & mathbf{2} mathbf{.} mathbf{3}^{r}-mathbf{1} & mathbf{4} . mathbf{5}^{r}-mathbf{1} \ boldsymbol{alpha} & boldsymbol{beta} & boldsymbol{gamma} \ mathbf{2}^{n}-mathbf{1} & mathbf{3}^{n}-mathbf{1} & mathbf{5}^{n}-mathbf{1}end{array}right| )
then find the value of ( sum_{r=1}^{n} Delta_{r} )
( A )
в. ( alpha beta gamma )
( mathbf{c} cdot-alpha beta gamma )
( D )
12
191If ( alpha, beta ) and ( gamma ) are the roots of the
equation ( x^{3}+p x+q=0 ) then the value of the determinant ( left|begin{array}{lll}boldsymbol{alpha} & boldsymbol{beta} & gamma \ boldsymbol{beta} & gamma & boldsymbol{alpha} \ boldsymbol{gamma} & boldsymbol{alpha} & boldsymbol{beta}end{array}right| ) is
2
( mathbf{A} cdot underline{p} )
B. ( q )
c. ( p^{2}-2 q )
( D )
12
192One factor of ( Delta= )
( left|begin{array}{ccc}a^{2}+lambda & a b & a c \ a b & b^{2}+lambda & c b \ c a & c b & c^{2}+lambdaend{array}right| )
( A cdot lambda^{2} )
B. ( 1 / lambda )
c. ( left(a^{2}+lambdaright)left(b^{2}+lambdaright)left(c^{2}+lambdaright) )
D. none
12
193Evaluate the following determinant:
( begin{array}{|ccc|}15 & 11 & 7 \ 11 & 17 & 14 \ 10 & 16 & 13end{array} )
12
194Find the equation of line passing through the points (3,2) and (-1,3) by using determinants.12
195If ( A ) is a square matrix of order ( n times n )
and ( k ) is a scalar, then ( a d j(k A) ) is equal
to
A ( cdot k^{n-1} a d j A )
в. ( k^{n} ) adj ( A )
c. ( k^{n+1} a d j A )
D. kadj A
12
196If ( f(x)=left|begin{array}{ccc}a & -1 & 0 \ a x & a & -1 \ a x^{2} & a x & aend{array}right|, ) using
properties of determinets find the value
of ( f(2 x)-f(x) ? )
12
197The repeated factor of the determinant ( left|begin{array}{lll}boldsymbol{y}+boldsymbol{z} & boldsymbol{x} & boldsymbol{y} \ boldsymbol{z}+boldsymbol{x} & boldsymbol{z} & boldsymbol{x} \ boldsymbol{x}+boldsymbol{y} & boldsymbol{y} & boldsymbol{z}end{array}right| )
A. ( z-x )
в. ( x-y )
( mathbf{c} cdot y-z )
D. none of these
12
198Evaluate ( left|begin{array}{ll}cos 65^{circ} & sin 65^{circ} \ sin 25^{circ} & cos 25^{circ}end{array}right| )12
199( operatorname{lf} Delta=left|begin{array}{lll}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}end{array}right| ) and ( A_{2}, B_{2}, C_{2} ) are
respectively cofactors of ( a_{2}, b_{2}, c_{2} ) then
( a_{1} A_{2}+b_{1} B_{2}+c_{1} C_{2} ) is equal to
( A cdot-Delta )
B.
( c cdot Delta )
D. none of these
12
200In the diagram on a lunar eilpse, if the positions od sun,Earth and moon are shown by ( (-4,6),(k,-2) ) and (5,-6)
respectively, then find the value of ( mathrm{k} )
12
201If ( A=left[begin{array}{ll}alpha & 2 \ 2 & alphaend{array}right] ) and ( left|A^{3}right|=125 ) then ( alpha ) is
( mathbf{A} cdot pm 1 )
B. =2
( c .pm 3 )
D. ±5
12
202Find the values of the following determinants
( left|begin{array}{cc}mathbf{1}+mathbf{3} i & boldsymbol{i}-mathbf{2} \ -boldsymbol{i}-mathbf{2} & mathbf{1}-mathbf{3} boldsymbol{i}end{array}right| )
where ( i=sqrt{-1} )
12
203[2005]
10. If a? +62 +62 =-2 and
| 1+a’x (1+62)x (1+c%)x
f(x) = (1 + a²)x 1 +6²x (1+ c²)x.
(1 + a²)x (1+6²)x 1+ c²x
then f(x) is a polynomial of degree
(a) 1 (b) 0 (c) 3
(d) 2
12
204( left|begin{array}{ccc}left(a^{x}+a^{-x}right)^{2} & left(a^{x}-a^{-x}right)^{2} & 1 \ left(b^{x}+b^{-x}right)^{2} & left(b^{x}-b^{-x}right)^{2} & 1 \ left(c^{x}+c^{-x}right)^{2} & left(c^{x}-c^{-x}right)^{2} & 1end{array}right| ) is equa
to
( A )
B. ( 2 a b c )
( mathbf{c} cdot a^{2} b^{2} c^{2} )
D. None of these
12
205( mathbf{A}=left|begin{array}{lll}mathbf{5} & mathbf{3} & mathbf{8} \ mathbf{2} & mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3}end{array}right|, ) write the cofactor of
the element ( a_{32} )
12
206( |f(x)|=left[begin{array}{ccc}sin x & operatorname{cosec} x & tan x \ sec x & x sin x & x tan x \ x^{2}-1 & cos x & x^{2}+1end{array}right] )
then ( ldots int_{-a}^{a}|boldsymbol{f}(boldsymbol{x})| boldsymbol{d} ) equals
( A )
в.
( c cdot 2 a )
( D )
12
207Let ( Delta= )
( left|begin{array}{ccc}sin theta cos phi & sin theta sin phi & cos theta \ cos theta cos phi & cos theta sin phi & -sin theta \ -sin theta sin phi & sin theta cos phi & 0end{array}right|, ) ther
( A cdot Delta ) is independent of ( theta )
B. ( Delta ) is independent of ( phi )
( c cdot Delta ) is a constant
D. ( Delta ) is dependent of ( phi )
12
208Find the values of ( x, ) if ( left|begin{array}{cc}mathbf{2 x} & mathbf{5} \ mathbf{8} & boldsymbol{x}end{array}right|=left|begin{array}{ll}mathbf{6} & mathbf{5} \ mathbf{8} & mathbf{3}end{array}right| )12
209( left|begin{array}{ccc}mathbf{0} & boldsymbol{a} boldsymbol{b}^{2} & boldsymbol{a} boldsymbol{c}^{2} \ boldsymbol{a}^{2} boldsymbol{b} & boldsymbol{0} & boldsymbol{b} boldsymbol{c}^{2} \ boldsymbol{a} boldsymbol{c}^{2} & boldsymbol{b}^{2} boldsymbol{c} & boldsymbol{0}end{array}right|= )
( mathbf{A} cdot a^{3} b^{3} c^{3}+a^{2} b^{3} c^{4} )
B ( cdot a^{3} b^{3} c^{3} )
( mathbf{c} cdot 2 a^{3} b^{3} c^{3} )
D・ ( (2 a b c)^{3} )
12
210ff ( boldsymbol{D}_{boldsymbol{r}}=left|begin{array}{ccc}boldsymbol{r}-mathbf{1} & boldsymbol{n} & boldsymbol{6} \ (boldsymbol{r}-mathbf{1})^{2} & boldsymbol{2} boldsymbol{n}^{2} & boldsymbol{4} boldsymbol{n}-boldsymbol{2} \ (boldsymbol{r}-mathbf{1})^{3} & boldsymbol{3} boldsymbol{n}^{3} & boldsymbol{3} boldsymbol{n}^{2}-boldsymbol{3} boldsymbol{n}end{array}right| )
then ( sum_{r=1}^{n} D_{r}= )
( A )
B.
c. ( frac{n(n-1)}{2}-r^{2} )
D. ( 2 n-n^{2} )
12
211If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{2} & mathbf{3}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{0} & mathbf{0}end{array}right] )
then what is determinant of AB?
( mathbf{A} cdot mathbf{0} )
B.
c. 10
D. 20
12
21212.
Given 2x – y + 2z=2, x – 2y+z=-4, x+y+nza
en the value of 2 such that the given system of equation
has NO solution, is
(a) 3 (6) 1 (c) 0 (d) -3
(2004S)
12
213Consider the matrix ( boldsymbol{A}=left(begin{array}{ll}mathbf{3} & -mathbf{2} \ mathbf{4} & -mathbf{1}end{array}right) )
Then all possible values of ( lambda ) such that
the determinant of ( B=A-lambda I ) is 0
where ( boldsymbol{I}=left(begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right) ) and ( i=sqrt{-mathbf{1}} )
( mathbf{A} cdot 1 pm 2 i )
B . ( 2 pm 3 i )
c. ( 3 pm 4 i )
( mathbf{D} cdot 5 pm 6 i )
12
214Find the area of ( triangle P Q R ) whose vertices ( operatorname{are} boldsymbol{P}(mathbf{2}, mathbf{1}), boldsymbol{Q}(mathbf{3}, mathbf{4}) ) and ( boldsymbol{R}(mathbf{5}, mathbf{2}) )12
215Evaluate ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ -mathbf{1} & mathbf{1} & -mathbf{1} \ mathbf{1} & mathbf{- 1} & mathbf{1}end{array}right| )12
216The values of lying between O=0 and O=TJ2 and satisfying
the equation
(1988 – 2 Marks)
1+ sine cose
sin 1+ cos2 e
sin cos
4sin 40
4sin 40
1+4 sin 40
= 0 are
12
217( begin{array}{ccc}(boldsymbol{b}+boldsymbol{c})^{2} & boldsymbol{a}^{2} & boldsymbol{a}^{2} \ boldsymbol{b}^{2} & (boldsymbol{c}+boldsymbol{a})^{2} & boldsymbol{b}^{2} \ boldsymbol{c}^{2} & boldsymbol{c}^{2} & (boldsymbol{a}+boldsymbol{b})^{2}end{array} mid= )12
218Find a value of ( boldsymbol{x} ) if ( left|begin{array}{cc}boldsymbol{x} & boldsymbol{2} \ boldsymbol{1} boldsymbol{8} & boldsymbol{x}end{array}right|=left|begin{array}{cc}boldsymbol{6} & boldsymbol{2} \ boldsymbol{1} boldsymbol{8} & boldsymbol{6}end{array}right| )12
219Find the values of ( k, ) if the points ( boldsymbol{A}(boldsymbol{k}+ ) 1, ( 2 k ) ), ( B(3 k, 2 k+3) ) and ( C(5 k+1,5 k) )
are collinear.
This question has multiple correct options
A .
B. ( frac{1}{2} )
( c cdot 2 )
D. 2.
12
220Prove that the points ( (a, 0),(0, b) ) and
(1,1) are collinear if ( left(frac{1}{a}+frac{1}{b}=1right) )
12
221Let ( A ) be the matrix of order ( 3 times 3 ) such
that ( |boldsymbol{A}|=mathbf{1}, boldsymbol{B}=mathbf{2} boldsymbol{A}^{-1} ) and ( boldsymbol{C}=frac{(boldsymbol{a} d boldsymbol{j} boldsymbol{A})}{sqrt[3]{2}} )
then the value of ( left|A B^{2} . C^{3}right| ) is [Note : ( |A| ) represent determinant value of matrix A.]
12
222The cofactors of elements in second row
of the determinant ( left|begin{array}{ccc}mathbf{2} & mathbf{- 1} & mathbf{4} \ mathbf{4} & mathbf{2} & mathbf{- 3} \ mathbf{1} & mathbf{1} & mathbf{2}end{array}right| ) are
( mathbf{A} cdot 5,6,4 )
В. 6,0,-3
c. 5,1,8
( mathbf{D} cdot 6,0,3 )
12
223The number of positive integral
solutions of the equation ( left|begin{array}{ccc}boldsymbol{x}^{boldsymbol{3}}+mathbf{1} & boldsymbol{x}^{boldsymbol{2}} boldsymbol{y} & boldsymbol{x}^{boldsymbol{2}} boldsymbol{z} \ boldsymbol{x} boldsymbol{y}^{boldsymbol{2}} & boldsymbol{y}^{boldsymbol{3}}+boldsymbol{1} & boldsymbol{y}^{boldsymbol{2}} boldsymbol{z} \ boldsymbol{x} boldsymbol{z}^{boldsymbol{2}} & boldsymbol{y} boldsymbol{z}^{boldsymbol{2}} & boldsymbol{z}^{boldsymbol{3}}+mathbf{1}end{array}right|=mathbf{1 1} ) is
A.
B. 3
( c cdot 6 )
D. 12
12
224Evaluate the following determinant:
( left|begin{array}{ccc}1 & 3 & 5 \ 2 & 6 & 10 \ 31 & 11 & 38end{array}right| )
12
225ff ( Delta=left|begin{array}{ccc}mathbf{3} & mathbf{5} & mathbf{7} \ mathbf{2} & -mathbf{3} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{2}end{array}right|, ) find it’s value12
226Let
( boldsymbol{f}(boldsymbol{x}, boldsymbol{y})=left|begin{array}{lll}boldsymbol{y}-boldsymbol{x}^{2} & boldsymbol{x}-boldsymbol{y}^{2} & boldsymbol{x} boldsymbol{y}-mathbf{1} \ boldsymbol{x}-boldsymbol{y}^{2} & boldsymbol{x} boldsymbol{y}-mathbf{1} & boldsymbol{y}-boldsymbol{x}^{2} \ boldsymbol{x} boldsymbol{y}-mathbf{1} & boldsymbol{y}-boldsymbol{x}^{2} & boldsymbol{x}-boldsymbol{y}^{2}end{array}right| )
find ( boldsymbol{f}(boldsymbol{2}, boldsymbol{2}) )
12
227( left|begin{array}{cc}boldsymbol{x}+mathbf{2} & boldsymbol{x} \ mathbf{4} & mathbf{3}end{array}right|=mathbf{0} ) find the value of ( mathbf{x} )12
228If the points ( (3,-2),(x, 2) ) and (8,8) are collinear find ( 10 x ) using determinant12
229If ( (3,2),left(x, frac{22}{5}right),(8,8) ) lie on a line, then ( x ) is equal to
A . -5
B. 2
( c cdot 4 )
D. 5
12
230Prove that:
( left|begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \ 2 a+1 & a+2 & 1 \ 3 & 3 & 1end{array}right|=(a-1)^{2} )
12
231Prove that ( :left|begin{array}{ccc}boldsymbol{a} & boldsymbol{c} & boldsymbol{a}+boldsymbol{c} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}end{array}right|= )
( 4 a b c )
12
232Without expanding, show that the value of each of the following determinants is
zero:
( left|begin{array}{ccc}a & b & c \ a+2 x & b+2 y & c+2 z \ x & y & zend{array}right| )
12
233Let ( omega neq 1 ) be a cube root of unity and ( S )
be the set of all non-singular matrices of the form ( left[begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{omega} & boldsymbol{1} & boldsymbol{c} \ boldsymbol{omega}^{2} & boldsymbol{omega} & boldsymbol{1}end{array}right] ) Where each of
( a, b ) and ( c ) is either ( omega ) or ( omega^{2} ). Then the
number of distinct matrices in the set
( boldsymbol{S} ) is
( A cdot 2 )
B. 6
( c cdot 4 )
D. 8
12
234How do I find
( boldsymbol{A}=left[begin{array}{ccc}1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1end{array}right]=|A|=? )
12
235The number of solutions of equations ( left|begin{array}{ccc}sin 3 theta & -1 & 1 \ cos 2 theta & 4 & 3 \ 2 & 7 & 7end{array}right|=0 ) in ( [0,2 pi] ) is
A .2
B. 3
( c cdot 4 )
D. 5
12
236If ( boldsymbol{u}=boldsymbol{a} boldsymbol{x}^{2}+boldsymbol{2 b} boldsymbol{x} boldsymbol{y}+boldsymbol{c} boldsymbol{y}^{2}, boldsymbol{u}^{prime}=boldsymbol{a}^{prime} boldsymbol{x}^{2}+ )
( 2 b^{prime} x y+c^{prime} y^{2}, ) then prove that
( left|begin{array}{ccc}boldsymbol{y}^{2} & -boldsymbol{x} boldsymbol{y} & boldsymbol{x}^{2} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{prime} & boldsymbol{b}^{prime} & boldsymbol{c}^{prime}end{array}right|= )
( left|begin{array}{cc}boldsymbol{a x}+boldsymbol{b y} & boldsymbol{b x}+boldsymbol{c y} \ boldsymbol{a}^{prime} boldsymbol{x}+boldsymbol{b}^{prime} boldsymbol{y} & boldsymbol{b}^{prime} boldsymbol{x}+boldsymbol{c}^{prime} boldsymbol{y}end{array}right|= )
( -frac{1}{y}left|begin{array}{cc}u & u^{prime} \ a x+b y & a^{prime} x+b^{prime} yend{array}right| )
12
237zoaluate ( left|begin{array}{ccc}log _{x} x y z & log _{x} y & log _{x} z \ log _{y} x y z & 1 & log _{y} z \ log _{z} x y z & log _{z} y & 1end{array}right| )12
238If ( 1, omega, omega^{2} ) are two cube roots of unity
then ( Delta=left|begin{array}{ccc}1 & omega^{n} & omega^{2 n} \ omega^{2 n} & 1 & omega^{n} \ omega^{n} omega^{2 n} & 1end{array}right| ) has the value
( mathbf{A} cdot mathbf{0} )
B.
( c cdot omega^{2} )
D.
12
239If ( left.left.right|_{2} ^{4} 1right|^{2}=left|begin{array}{cc}3 & 2 \ 1 & xend{array}right|-left|begin{array}{cc}x & 3 \ -2 & 1end{array}right|, ) then ( x= )
( A cdot 6 )
B. 7
( c cdot 8 )
D. 16
12
240( 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{3} boldsymbol{x} & mathbf{6}end{array}right|, ) then ( boldsymbol{x} ) is equal to
( A cdot 6 )
B. ±6
( c .-6 )
( D )
12
241ff ( x_{1}, x_{2}, x_{3} ) as well as ( y_{1}, y_{2}, y_{3} ) are in
G.P. with same common ratio, then the
points ( boldsymbol{P}left(boldsymbol{x}_{1}, boldsymbol{y}_{1}right), boldsymbol{Q}left(boldsymbol{x}_{2}, boldsymbol{y}_{2}right) ) and
( boldsymbol{R}left(boldsymbol{x}_{3}, boldsymbol{y}_{3}right) )
A. lies on a straight line
B. lie on an ellipse
c. lie on a circle
D. are vertices of a triangle
12
242Show that:
( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right|^{2}= )
[
left|begin{array}{ccc}
2 b c-a^{2} & c^{2} & b^{2} \
c^{2} & 2 a c-b^{2} & a^{2} \
b^{2} & a^{2} & 2 a b-c^{2}
end{array}right|=
]
( left(a^{3}+b^{3}+c^{3}-3 a b cright)^{2} )
12
243Find the value of: ( left|begin{array}{ll}-3 & -5 \ -2 & -1end{array}right| )12
244<Tt,
1. The number of all possible values of 0, where 0 <
for which the system of equations
(y + 2) cos 30 = (xyz) sin 30
2 cos 30 2 sin 30
x sin 30= =
Y Z
(xyz) sin 30= (y + 2z) cos 30+ y sin30
has a solution (xo, Yo, zo) with yożo 70 is
– +-
12
245( operatorname{Let} boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{2} & mathbf{1} & mathbf{0} \ mathbf{3} & mathbf{2} & mathbf{1}end{array}right] ) and ( boldsymbol{U}_{1}, boldsymbol{U}_{2}, boldsymbol{U}_{3} ) be
column
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] . ) If ( U ) is
( 3 times 3 ) matrix whose columns are
( boldsymbol{U}_{1}, boldsymbol{U}_{2}, boldsymbol{U}_{3}, ) then ( |boldsymbol{U}|= )
A . 3
B. -3
( c cdot frac{3}{2} )
( D )
12
246Without expanding at any stage,
evaluate the value of determinant
12
247( left|begin{array}{ccc}boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & mathbf{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & 1end{array}right|=left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{b}_{1} & boldsymbol{b}_{2} & boldsymbol{b}_{3} \ boldsymbol{a}_{1} & boldsymbol{a}_{2} & boldsymbol{a}_{3}end{array}right| ) then the
two triangles whose vertices are
( left(x_{1}, y_{1}right),left(x_{2}, y_{2}right),left(left(x_{3}, y_{3}right) ) and right.
( left(a_{1}, b_{1}right),left(a_{2}, b_{2}right),left(a_{13}, b_{3}right), ) are
A . congruent
B. similar
c. equal in area
D. none of these
12
248If each row of a determinant of third
order of value ( Delta ) is multipled by ( 3, ) then the value of new determinant is
A. ( Delta )
B. ( 27 Delta )
c. ( 21 Delta )
D. ( 54 Delta )
12
249Adj ( left[begin{array}{ccc}1 & 0 & 2 \ -1 & 5 & -2 \ 0 & 2 & 1end{array}right]= )
( left[begin{array}{ccc}mathbf{9} & boldsymbol{a} & mathbf{- 2} \ mathbf{- 1} & mathbf{1} & mathbf{0} \ mathbf{- 2} & mathbf{2} & boldsymbol{b}end{array}right] Rightarrowleft[begin{array}{ll}boldsymbol{a} & boldsymbol{b}end{array}right]= )
( left.begin{array}{ll}text { A. }[-4 & 5end{array}right] )
B ( cdotleft[begin{array}{ll}-4 & -1end{array}right] )
( mathbf{c} cdotleft[begin{array}{ll}4 & 1end{array}right] )
D・[4 -1
12
250Find the value of ( x ) for which the points
( (x,-1),(2,1) ) and (4,5) are collinear.
12
251If ( mathbf{A} ) is an unitary matrix then ( |boldsymbol{A}| ) is
equal to:
( mathbf{A} cdot mathbf{1} )
B. –
( c .pm 1 )
D. 2
12
252begin{tabular}{|ccc}
If ( boldsymbol{f}(boldsymbol{x})= ) & \
( mathbf{1} ) & ( boldsymbol{x} ) & ( boldsymbol{x} ) \
( boldsymbol{2} boldsymbol{x} ) & ( boldsymbol{x}(boldsymbol{x}-mathbf{1}) ) & ( (boldsymbol{x}-1) ) \
( boldsymbol{3} boldsymbol{x}(boldsymbol{x}-mathbf{1}) ) & ( boldsymbol{x}(boldsymbol{x}-mathbf{1})(boldsymbol{x}-mathbf{2}) ) & ( (boldsymbol{x}+mathbf{1}) )
end{tabular}
then ( f(100) ) is equal to
A.
B.
( c cdot 100 )
D. -100
12
253Let ( A ) be a ( 3 times 3 ) matrix and ( B ) be its
adjoint matrix. If ( |boldsymbol{B}|=mathbf{6 4}, ) then ( |boldsymbol{A}|= )
( A cdot pm 2 )
B. ±4
( c .pm 8 )
D. ±12
12
254The value of determinant
( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{2} boldsymbol{y} \ boldsymbol{x}+boldsymbol{2} boldsymbol{y} & boldsymbol{x} & boldsymbol{x}+boldsymbol{y} \ boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{2} boldsymbol{y} & boldsymbol{x}end{array}right| ) is:
A ( cdot 9 x^{2}(x+y) )
В ( cdot 9 y^{2}(x+y) )
c. ( 3 y^{2}(x+y) )
D. ( 7 x^{2}(x+y) )
12
255If ( Delta_{1}=left|begin{array}{ll}mathbf{1} & mathbf{0} \ boldsymbol{a} & boldsymbol{b}end{array}right| ) and ( boldsymbol{Delta}_{2}=left|begin{array}{ll}mathbf{1} & mathbf{0} \ boldsymbol{c} & boldsymbol{d}end{array}right| ) then
( Delta_{2} Delta_{1} ) is equal to
( mathbf{A} cdot a c )
B. ( b d )
c. ( (b-a)(d-c) )
D. none of these
12
256Evaluate ( sum_{n=1}^{N} U_{n} ) if
[
boldsymbol{U}_{boldsymbol{n}}=mid begin{array}{ccc}boldsymbol{n} & mathbf{1} & mathbf{5} \ boldsymbol{n}^{2} & boldsymbol{2} boldsymbol{N}+mathbf{1} & boldsymbol{2} boldsymbol{N}+mathbf{1} \ boldsymbol{n}^{boldsymbol{3}} & boldsymbol{3} boldsymbol{N}^{2} & boldsymbol{3} boldsymbol{N}end{array}
]
12
257f ( x, y, z ) are complex numbers, then ( Delta= ) ( left|begin{array}{ccc}mathbf{0} & -boldsymbol{y} & -boldsymbol{z} \ overline{boldsymbol{y}} & mathbf{0} & -boldsymbol{x} \ overline{boldsymbol{z}} & overline{boldsymbol{x}} & boldsymbol{0}end{array}right| ) is equal
A. Purely real
B. Purely imaginary
c. complex
D.
12
258( operatorname{Let} A=left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z}end{array}right| ) and suppose that
det. ( (A)=2 ) then the det.(B) equals, where ( boldsymbol{B}=left|begin{array}{ccc}mathbf{4} boldsymbol{x} & mathbf{2} boldsymbol{a} & -boldsymbol{p} \ mathbf{4} boldsymbol{y} & mathbf{2} boldsymbol{b} & -boldsymbol{q} \ boldsymbol{4} boldsymbol{z} & boldsymbol{2} boldsymbol{c} & -boldsymbol{t}end{array}right| )
A. ( operatorname{det}(B)=-2 )
B. ( operatorname{det}(B)=-8 )
c. ( operatorname{det}(B)=-16 )
D. ( operatorname{det}(B)=8 )
12
259If ( a, b, c ) are real, then ( f(x)= ) ( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{a}^{2} & boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b} & boldsymbol{x}+boldsymbol{b}^{2} & boldsymbol{b} boldsymbol{c} \ boldsymbol{a} boldsymbol{c} & boldsymbol{b} boldsymbol{c} & boldsymbol{x}+boldsymbol{c}^{2}end{array}right| ) is decreasing
in
A ( cdotleft(-frac{2}{3}left(a^{2}+b^{2}+c^{2}right), 0right) )
B. ( left(0, frac{2}{3}left(a^{2}+b^{2}+c^{2}right)right) )
( left(frac{a^{2}+b^{2}+c^{2}}{3}, 0right) )
D. None of these
12
260Using determinants show that points ( boldsymbol{A}(boldsymbol{a}, boldsymbol{b}+boldsymbol{c}), boldsymbol{B}(boldsymbol{b}, boldsymbol{c}+boldsymbol{a}) ) and ( boldsymbol{C}(boldsymbol{c}, boldsymbol{a}+boldsymbol{b}) )
are col-linear.
12
261Evaluate the following determinant:
( left|begin{array}{lll}a & h & g \ h & b & f \ g & f & cend{array}right| )
12
262Using properties of determinant, prove
[
text { that }left|begin{array}{lll}
boldsymbol{b}+boldsymbol{c} & boldsymbol{a}-boldsymbol{b} & boldsymbol{a} \
boldsymbol{c}+boldsymbol{a} & boldsymbol{b}-boldsymbol{c} & boldsymbol{b} \
boldsymbol{a}+boldsymbol{b} & boldsymbol{c}-boldsymbol{a} & boldsymbol{c}
end{array}right|=mathbf{3} boldsymbol{a} boldsymbol{b} boldsymbol{c}-boldsymbol{a}^{3}-
]
( b^{3}-c^{3} )
12
263( mathrm{ff}=left[begin{array}{ccc}boldsymbol{a} & mathbf{0} & mathbf{0} \ {[mathbf{0 . 3 e m}] mathbf{0}} & boldsymbol{a} & mathbf{0} \ {[mathbf{0 . 3 e m}] mathbf{0}} & mathbf{0} & boldsymbol{a}end{array}right], ) then the value
of |A| |Adj. A|
A ( cdot a^{3} )
в. ( a^{6} )
( c cdot a^{9} )
( mathbf{D} cdot a^{2} )
12
264Solve ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{y}end{array}right| )
( A )
B.
( c )
D. ( x y )
12
265Using the method of slope, show that the following points are collinear:
( (i) A(4,8), B(5,12), C(9,28) )
( (i i) A(16,-18), B(3,-6), C(-10,6) )
12
266If ( A ) is a singular matrix, then ( A(operatorname{adj} A) )
is a
A. scalar matrix
B. zero matrix
c. identity matrix
D. orthogonal matrix
12
267Find the value of ( mathbf{x} ) for which ( left|begin{array}{ll}3 & x \ x & 1end{array}right|= )
( left|begin{array}{ll}3 & 2 \ 8 & 1end{array}right| )
12
268The line ( A x+B y+C=0 ) cuts the
circle ( boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{a} boldsymbol{x}+boldsymbol{b} boldsymbol{y}+boldsymbol{c}=boldsymbol{0} ) in ( boldsymbol{P} )
and ( Q )
The line ( boldsymbol{A}^{prime} boldsymbol{x}+boldsymbol{B}^{prime} boldsymbol{y}+boldsymbol{C}^{prime}=mathbf{0} ) cuts the
circle ( boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{a}^{prime} boldsymbol{x}+boldsymbol{b}^{prime} boldsymbol{y}+boldsymbol{c}^{prime}=mathbf{0} ) in ( boldsymbol{R} )
and ( S ).
If ( P, Q, R, S ) are concyclic, then show
that
( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{a}^{prime} & boldsymbol{b}-boldsymbol{b}^{prime} & boldsymbol{c}-boldsymbol{c}^{prime} \ boldsymbol{A} & boldsymbol{B} & boldsymbol{C} \ boldsymbol{A}^{prime} & boldsymbol{B}^{prime} & boldsymbol{C}^{prime}end{array}right|=mathbf{0} )
12
269Find the value of the following
determinant:
( left|begin{array}{cc}mathbf{3} sqrt{mathbf{6}} & -mathbf{4} sqrt{mathbf{2}} \ mathbf{5} sqrt{mathbf{3}} & mathbf{2}end{array}right| )
A. ( 20 sqrt{6} )
B. ( 16 sqrt{6} )
c. ( 26 sqrt{6} )
D. ( 10 sqrt{6} )
12
270Evaluate ( Delta=mid begin{array}{cc}cos 2 alpha & sin 2 alpha \ -sin 3 alpha & cos 3 alphaend{array} )12
271ff ( f(x), g(x), h(x) ) are polynomials in ( x ) of ( operatorname{degree} 2 ) and ( F(x)=left|begin{array}{ccc}boldsymbol{f} & boldsymbol{g} & boldsymbol{h} \ boldsymbol{f}^{prime} & boldsymbol{g}^{prime} & boldsymbol{h}^{prime} \ boldsymbol{f}^{prime prime} & boldsymbol{g}^{prime prime} & boldsymbol{h}^{prime prime}end{array}right|, ) then
( F(x) ) is equal to
( A cdot 1 )
B.
( c cdot-1 )
D. ( f(x) . g(x) . h(x) )
12
272Let ( t ) be a positive integer and ( Delta_{t}= ) ( left|begin{array}{lll}mathbf{2} t-1 & boldsymbol{m}^{2}-mathbf{1} & cos ^{2}left(boldsymbol{m}^{2}right) \ boldsymbol{m}_{C_{t}} & mathbf{2}^{m} & cos ^{2}(boldsymbol{m}) \ mathbf{1} & boldsymbol{m}+mathbf{1} & cos left(boldsymbol{m}^{2}right)end{array}right| ) then the
value of ( sum_{t=0}^{m} Delta_{t} ) is equal to:
( A cdot 2^{m} )
B.
( mathbf{c} cdot 2^{m} cos ^{2}left(2^{m}right) )
D. ( m^{2} )
12
273The point ( (-a,-b),(0,0),(a, b) ) and
( left(a^{2}, a bright) ) are
A. collinear
B. concyclic
c. vertices of a rectangle
D. vertices of a parallelogram
12
274ff ( f(x)= )
( f(x-a)(x-b)(x-c)(x-d) ) then
prove that
( Delta=left|begin{array}{cccc}boldsymbol{a} & boldsymbol{x} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{b} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{c} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x} & boldsymbol{d}end{array}right|=boldsymbol{f}(boldsymbol{x})-boldsymbol{x} boldsymbol{f}^{prime}(boldsymbol{x}) )
12
275Let ( m ) be a positive integer and ( Delta_{r}= ) ( left|begin{array}{ccc}mathbf{2 r}-mathbf{1} & boldsymbol{m} boldsymbol{C}_{boldsymbol{r}} & mathbf{1} \ boldsymbol{m}^{mathbf{2}}-mathbf{1} & mathbf{2}^{boldsymbol{m}} & boldsymbol{m}+mathbf{1} \ sin ^{2}left(boldsymbol{m}^{2}right) & sin ^{2}(boldsymbol{m}) & sin ^{2}(boldsymbol{m}+mathbf{1})end{array}right| )
( boldsymbol{r} leq boldsymbol{m}) . ) Then the value of ( sum_{r=0}^{m} boldsymbol{Delta}_{boldsymbol{r}} )
( A . )
B. ( m^{2}-1 )
( c cdot 2^{m} )
D ( cdot 2^{m} sin ^{2}left(2^{m}right) )
12
276Using the property of determinants and with out expanding prove that ( left|begin{array}{lll}boldsymbol{a}-boldsymbol{b} & boldsymbol{b}-boldsymbol{c} & boldsymbol{c}-boldsymbol{a} \ boldsymbol{b}-boldsymbol{c} & boldsymbol{c}-boldsymbol{a} & boldsymbol{a}-boldsymbol{b} \ boldsymbol{c}-boldsymbol{a} & boldsymbol{a}-boldsymbol{b} & boldsymbol{b}-boldsymbol{c}end{array}right|=mathbf{0} )12
277a b c]
14. If matrix A= b c a where a, b, c are real positive
Lc a b
numbers, abc = 1 and ATA = 1, then find the value of
a3 + b3 + c3
(2003 – 2 Marks)
12
278( f(x)=left|begin{array}{ccc}2 x & x^{2} & x^{3} \ x^{2}+2 x & 1 & 3 x+1 \ 2 x & 1-3 x^{2} & 5 xend{array}right| )
then find ( boldsymbol{f}^{prime}(mathbf{1}) )
12
279( left|begin{array}{ccc}mathbf{1} & boldsymbol{w} & boldsymbol{w}^{2} \ boldsymbol{w} & boldsymbol{w}^{2} & mathbf{1} \ boldsymbol{w}^{2} & boldsymbol{w} & mathbf{1}end{array}right| ) Where ( mathbf{w} ) is a complex
cube root of unity
12
280Prove
[
left|begin{array}{ccc}
x^{2} & y^{2} & z^{2} \
(x+1)^{2} & (y+1)^{2} & (z+1)^{2} \
(x-1)^{2} & (y-1)^{2} & (z-1)^{2}
end{array}right|=
]
( -4(x-y)(y-z)(x-z) )
12
281If ( f(x)= ) ( left|begin{array}{ccc}(mathbf{1}+mathbf{3} boldsymbol{x})^{2 boldsymbol{a}} & (mathbf{1}+mathbf{2} boldsymbol{x})^{boldsymbol{3} boldsymbol{b}} & mathbf{1} \ mathbf{1} & (mathbf{1}+mathbf{3} boldsymbol{x})^{2 boldsymbol{a}} & (mathbf{1}+mathbf{2} boldsymbol{x})^{boldsymbol{3} boldsymbol{b}} \ (mathbf{1}+mathbf{2} boldsymbol{x})^{boldsymbol{3} boldsymbol{b}} & mathbf{1} & (mathbf{1}+mathbf{3} boldsymbol{x})^{mathbf{2} boldsymbol{a}}end{array}right| )
then
A. ( f(x) ) has constant term 1
B. constant term is ( 2 a-3 b )
c. coefficient of ( x ) in ( f(x) ) is zero
D. constant term is ( 2 a+3 b )
12
282f ( a, b, c ) are real numbers such that ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a}end{array}right|=0, ) then show
that either ( boldsymbol{a}+boldsymbol{b}+boldsymbol{c}=boldsymbol{0} ) or, ( boldsymbol{a}=boldsymbol{b}=boldsymbol{c} )
12
2831.
If a > 0 and discriminant of ax2+2bx+c is-ve,
[2001
a
b
ax+b
b
c
bx+c
ax + bl.
bx+c is equal to
o
(b) (ac-b2)(ax2 +2bx+c)
(d) 0
(a) +ve
(C) -ve
12
284Assertion
If ( A ) is skew symmetric matrix of order 3 then its determinant should be zero
Reason

If ( A ) is square matrix, then ( d e t A= )
( operatorname{det} A^{prime}=operatorname{det}left(-A^{prime}right) )
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
285( begin{array}{lllll}mathbf{0} & & mathbf{c o s} boldsymbol{alpha} & mathbf{c o s} & boldsymbol{beta} \ mathbf{c o s} & boldsymbol{alpha} & mathbf{0} & mathbf{c o s} & gamma \ mathbf{c o s} & boldsymbol{beta} & mathbf{c o s} boldsymbol{gamma} & mathbf{0} & end{array} mid= )
A ( cdot cos alpha+cos beta+cos gamma )
( mathbf{B} cdot cos alpha cos beta cos gamma )
c. ( 2 cos alpha cos beta cos gamma )
D. ( 2 sum cos alpha cos beta )
12
286Solve for ( boldsymbol{x} )
( left|begin{array}{ccc}1-x & 2 & 3 \ 0 & x & 0 \ 0 & 0 & xend{array}right|=0 )
( mathbf{A} cdot 1,0,0 )
B. 1,1,0
c. 1,1,1
D. 0,0,0
12
287Points (1,5),(2,3) and (-2,-11) are
A. Non-collinear
B. Collinear
c. vertices of equilateral triangle
D. Vertices of right angle triangle
12
288Verify whether the points (1,5),(2,3) ,and (-2,-1) are collinear or not.12
289If ( left.A=left[begin{array}{ll}a & b \ c & dend{array}right] text { (where } b neq cright) ) and satisfies the equation ( A^{2}+k I=0, ) then
( mathbf{A} cdot a+d=0 )
B . ( k=-|A| )
c. ( k=|A| )
D. None of the above
12
290Evaluate the following:
( left|begin{array}{ccc}1 & a & b c \ 1 & b & c a \ 1 & c & a bend{array}right| )
12
291Find the equation of the line joining ( A(1,3) ) and ( B(0,0) ) using determinants.12
292( left|begin{array}{ccc}boldsymbol{x}+mathbf{2} & mathbf{2} boldsymbol{x}+mathbf{3} & mathbf{3} boldsymbol{x}+mathbf{4} \ mathbf{2} boldsymbol{x}+mathbf{3} & mathbf{3} boldsymbol{x}+mathbf{4} & mathbf{4} boldsymbol{x}+mathbf{5} \ mathbf{3} boldsymbol{x}+mathbf{5} & mathbf{5} boldsymbol{x}+mathbf{8} & mathbf{1 0} boldsymbol{x}+mathbf{1 7}end{array}right|=mathbf{0} ) then
( x ) is equal to
A ( .-1,-2 )
в. 1,2
( c cdot 1,-2 )
D. -1,2
12
293If ( A ) is a matrix of order ( 3 times 3 ) then find
( |boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}| ) where ( |boldsymbol{A}|=boldsymbol{2} )
12
294Show that
( begin{array}{l}sin 10^{circ} quad-cos 10^{circ} \ sin 80^{circ} quad cos 80^{circ}end{array} mid=1 )
12
295Solve
( left|begin{array}{ccc}1 & 1 & -1 \ 6 & 4 & -5 \ -4 & -2 & 3end{array}right| )
12
296( begin{array}{ccc}(boldsymbol{b}+boldsymbol{c})^{2} & boldsymbol{a}^{2} & boldsymbol{a}^{2} \ boldsymbol{b}^{2} & (boldsymbol{c}+boldsymbol{a})^{2} & boldsymbol{b}^{2} \ boldsymbol{c}^{2} & boldsymbol{c}^{2} & (boldsymbol{a}+boldsymbol{b})^{2}end{array} mid= )12
297Using properties of determinants, find
the following:
( left|begin{array}{ccc}boldsymbol{alpha} & boldsymbol{beta} & boldsymbol{gamma} \ boldsymbol{alpha}^{2} & boldsymbol{beta}^{2} & boldsymbol{gamma}^{2} \ boldsymbol{beta}+boldsymbol{gamma} & boldsymbol{gamma}+boldsymbol{alpha} & boldsymbol{alpha}+boldsymbol{beta}end{array}right| )
A ( cdot(alpha+beta)(beta+gamma)(gamma-alpha)(alpha+beta+gamma) )
B . ( (alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma) )
c.
D. None of these
12
298( left|begin{array}{ccc}1 & sin theta & 1 \ -sin theta & 1 & sin theta \ -1 & -sin theta & 1end{array}right| ) then
( mathbf{A} cdot Delta=0 )
B. ( Delta in(0, infty) )
c. ( Delta in[-1,2] )
D. ( Delta in[2,4] )
12
299The number of ordered triplets of positive integral solutions of ( left|begin{array}{ccc}boldsymbol{y}^{3}+mathbf{1} & boldsymbol{y}^{2} boldsymbol{z} & boldsymbol{y}^{2} boldsymbol{x} \ boldsymbol{y} boldsymbol{z}^{2} & boldsymbol{z}^{3}+mathbf{1} & boldsymbol{z}^{2} boldsymbol{x} \ boldsymbol{y} boldsymbol{x}^{2} & boldsymbol{x}^{2} boldsymbol{z} & boldsymbol{x}^{3}+mathbf{1}end{array}right|=mathbf{1 1} )12
300If ( omega ) is a non-real cube root of unity and
n is not a multiple of ( 3, ) then ( Delta= ) ( left|begin{array}{ccc}mathbf{1} & boldsymbol{omega}^{n} & boldsymbol{omega}^{2 n} \ boldsymbol{omega}^{2 boldsymbol{n}} & boldsymbol{1} & boldsymbol{omega}^{n} \ boldsymbol{omega}^{boldsymbol{n}} & boldsymbol{omega}^{2 boldsymbol{n}} & boldsymbol{1}end{array}right| ) is equal to
( mathbf{A} cdot mathbf{0} )
B. ( omega )
( c cdot omega^{2} )
D.
12
301( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{- 1} \ -mathbf{1} & mathbf{1} & mathbf{2} \ mathbf{2} & mathbf{- 1} & mathbf{1}end{array}right], ) then
( operatorname{det}(operatorname{adj}(operatorname{adj} A)) ) is equal to
A ( cdot 14^{4} )
B. ( 14^{text {? }} )
( c cdot 14^{2} )
D. 14
12
302Show that ( (x-a) ) is a factor of
( left|begin{array}{lll}boldsymbol{x} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{x} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a} & boldsymbol{x}end{array}right| )
12
303Write minors and cofactors of the
elements of following determinants
(i) ( left|begin{array}{cc}2 & -4 \ 0 & 3end{array}right| )
(ii) ( left|begin{array}{ll}boldsymbol{a} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{d}end{array}right| )
12
304( 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
305If ( A=left[begin{array}{cc}2 & -1 \ -1 & 2end{array}right], ) then the general
solution of ( sin theta=left|A^{2}-4 A+3 Iright| ) is
A ( . n pi )
B ( cdot 2 n+1 frac{pi}{2} )
c. ( n pi+(-1)^{n} frac{pi}{2} )
D. ( 2 n pi, n epsilon z )
12
306( operatorname{Det}left[begin{array}{ccc}mathbf{4 3} & mathbf{1} & mathbf{6} \ mathbf{3 5} & mathbf{7} & mathbf{4} \ mathbf{1 7} & mathbf{3} & mathbf{2}end{array}right]=dots )
( A )
B. – –
( c cdot 0 )
( D )
12
307Prove 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| )
divisible by ( (x+y+z) ) and hence find
the quotient
12
308The remainder when the determinant
( left|begin{array}{lll}2014^{2014} & 2015^{2015} & 2016^{2016} \ 2017^{2017} & 2018^{2018} & 2019^{2019} \ 2020^{2020} & 2021^{2021} & 2022^{2022}end{array}right| )
is divided by 5 is.
A . 1
B . 2
( c cdot 4 )
( D )
12
309Prove that: ( left|begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{y}+boldsymbol{z} \ mathbf{1} & boldsymbol{y} & boldsymbol{z}+boldsymbol{x} \ mathbf{1} & boldsymbol{z} & boldsymbol{x}+boldsymbol{y}end{array}right|=mathbf{0} )12
310Let ( boldsymbol{A}=left[boldsymbol{a}_{i j}right]_{n times n} ) be a square matirx and
let ( c_{i j} ) be cofactor of ( a_{i j} ) in A. If ( C=left[c_{i j}right] )
then
B . ( |C|=|A|^{n-1} )
c. ( |C|=|A|^{n-2} )
D. none of these
12
311( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{b} boldsymbol{c} \ mathbf{1} & boldsymbol{b} & boldsymbol{c a} \ mathbf{1} & boldsymbol{c} & boldsymbol{a b}end{array}right|=boldsymbol{lambda}left|begin{array}{ccc}boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{1} & boldsymbol{1} & boldsymbol{1}end{array}right| ) then
( lambda ) is equal to
( A )
B . –
( c )
( D .- )
12
312Calculate the values of the
determinants:
( left|begin{array}{cccc}mathbf{0} & boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ -boldsymbol{x} & boldsymbol{0} & boldsymbol{c} & boldsymbol{b} \ -boldsymbol{y} & -boldsymbol{c} & boldsymbol{0} & boldsymbol{a} \ -boldsymbol{z} & -boldsymbol{b} & -boldsymbol{a} & boldsymbol{0}end{array}right| )
12
313If ( boldsymbol{A}=left[begin{array}{ccc}boldsymbol{x} & mathbf{1} & -boldsymbol{x} \ mathbf{0} & mathbf{1} & -mathbf{1} \ boldsymbol{x} & mathbf{0} & mathbf{7}end{array}right] ) and ( operatorname{det}(boldsymbol{A})= )
( left|begin{array}{ccc}mathbf{3} & mathbf{0} & mathbf{1} \ mathbf{2} & mathbf{- 1} & mathbf{0} \ mathbf{0} & mathbf{6} & mathbf{7}end{array}right| ) then the value of ( boldsymbol{x} ) is
A . -3
B. 3
( c cdot 2 )
D. -8
E. -2
12
314Evaluate
( left|begin{array}{ccc}265 & 240 & 219 \ 240 & 225 & 198 \ 219 & 198 & 181end{array}right| )
12
315Find the values of ( x, ) if
( left|begin{array}{ll}boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{1} \ boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{2}end{array}right|=left|begin{array}{ll}boldsymbol{4} & -mathbf{1} \ mathbf{1} & mathbf{3}end{array}right| )
12
316f ( a, b, c>1, Delta= )
( left|begin{array}{ccc}log _{a}(a b c) & log _{a} b & log _{a} c \ log _{b}(a b c) & 1 & log _{b} c \ log _{c}(a b c) & log _{c} b & 1end{array}right| ) is
( mathbf{A} cdot mathbf{0} )
B ( cdot log _{a} b+log _{b} c+log _{c} a )
( mathbf{c} cdot log _{a b c}(a+b+c) )
D. none of these
12
317The value of the determinant
( left|begin{array}{ccc}left(a^{x}+a^{-x}right)^{2} & left(a^{x}-a^{-x}right)^{2} & 1 \ left(b^{x}+b^{-x}right)^{2} & left(b^{x}-b^{-x}right)^{2} & 1 \ left(c^{x}+c^{-x}right)^{2} & left(c^{x}-c^{-x}right)^{2} & 1end{array}right| ) is
( mathbf{A} cdot mathbf{0} )
B. 2abc
( mathbf{c} cdot a^{2} b^{2} c^{2} )
D. None of these
12
318what is the value of
( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{a}^{2} & boldsymbol{a}^{boldsymbol{3}}-mathbf{1} \ boldsymbol{b} & boldsymbol{b}^{2} & boldsymbol{b}^{3}-mathbf{1} \ boldsymbol{c} & boldsymbol{c}^{2} & boldsymbol{c}^{boldsymbol{3}}-mathbf{1}end{array}right|=? )
12
319If ( A ) is a square matrix of order 3 such
that ( |boldsymbol{a} boldsymbol{d} boldsymbol{j} cdot boldsymbol{A}|=boldsymbol{3} boldsymbol{6}, ) find ( |boldsymbol{A}| )
12
320Without expanding, show that the value
of the following determinant is zero:
( left|begin{array}{ccc}a & b & c \ a+2 x & b+2 y & c+2 z \ x & y & zend{array}right| )
12
321Prove that
[
left|begin{array}{ccc}
boldsymbol{x}+boldsymbol{y}+mathbf{2} boldsymbol{z} & boldsymbol{x} & boldsymbol{y} \
boldsymbol{z} & boldsymbol{y}+boldsymbol{z}+mathbf{2} boldsymbol{x} & boldsymbol{y} \
boldsymbol{z} & boldsymbol{x} & boldsymbol{z}+boldsymbol{x}+boldsymbol{2} boldsymbol{y}
end{array}right|
]
( 2(x+y+z)^{3} )
12
322The adjoint of the matrix ( boldsymbol{A}= )
[
left[begin{array}{ccc}
1 & 1 & 1 \
2 & 1 & -3 \
-1 & 2 & 3
end{array}right] text { is }
]
( A )
[
left.begin{array}{l|lcc}
& 9 & -1 & -4 \
frac{1}{11} & -3 & 4 & 5 \
& 5 & -3 & -1
end{array}right]
]
B.
[
left[begin{array}{ccc}
9 & 1 & -4 \
3 & 4 & -5 \
5 & 3 & -1
end{array}right]
]
( c )
[
left[begin{array}{ccc}
9 & -3 & 5 \
-1 & 4 & -3 \
-4 & 5 & -1
end{array}right]
]
D.
[
left[begin{array}{ccc}
9 & -1 & -4 \
-3 & 4 & 5 \
5 & -3 & -1
end{array}right]
]
12
323If ( boldsymbol{A}=int_{1}^{sin theta} frac{boldsymbol{t}}{1+boldsymbol{t}^{2}} boldsymbol{d} boldsymbol{t} ) and
( B=int_{1}^{operatorname{cosec} theta} frac{1}{tleft(1+t^{2}right)} d t, ) then the value
of determinant ( left|begin{array}{ccc}boldsymbol{A} & boldsymbol{A}^{2} & boldsymbol{B} \ boldsymbol{e}^{boldsymbol{A}+boldsymbol{B}} & boldsymbol{B}^{2} & -mathbf{1} \ mathbf{1} & boldsymbol{A}^{2}+boldsymbol{B}^{2} & -mathbf{1}end{array}right| )
is
( A cdot sin theta )
B. ( operatorname{cosec} theta )
( c )
D.
12
324If ( left(begin{array}{lll}1 & 2 & 4 \ 1 & 3 & 5 \ 1 & 4 & aend{array}right) ) is singular, the value
of ( a ) is
A ( . a=-6 )
B. ( a=5 )
c. ( a=-5 )
( mathbf{D} cdot a=6 )
12
325The value of the determinant ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & mathbf{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{0} & boldsymbol{a}end{array}right| )
is equal to
A ( cdot a^{3}-b^{3} )
В . ( a^{3}+b^{3} )
( c cdot 0 )
D. none of these
12
326For a positive numbers ( x, y ) and ( z ) the numerical value of the determinant ( left[begin{array}{ccc}1 & log _{x} y & log _{x} z \ log _{y} x & 1 & log _{y} z \ log _{z} x & log _{z} y & 1end{array}right] ) is:
A. 0
B. 1
( mathbf{c} cdot log _{e} x y z )
( mathbf{D} cdot-log _{e} x y z )
12
327( mathbf{A}=left|begin{array}{ccc}cos frac{theta}{2} & mathbf{1} & mathbf{1} \ mathbf{1} & cos frac{theta}{2} & -cos frac{theta}{2} \ -cos frac{theta}{2} & mathbf{1} & -mathbf{1}end{array}right| )
the minimum of ( Delta ) is ( m_{1} ) and
maximum of ( Delta ) is ( m_{2} ) then ( left[m_{1}, m_{2}right] ) are
related to
A. [4, 2]
B. [2,4]
( c cdot[4,0] )
D. [0,2]
12
328( left|begin{array}{ccc}boldsymbol{a}+boldsymbol{x} & boldsymbol{a}-boldsymbol{x} & boldsymbol{a}-boldsymbol{x} \ boldsymbol{a}-boldsymbol{x} & boldsymbol{a}+boldsymbol{x} & boldsymbol{a}-boldsymbol{x} \ boldsymbol{a}-boldsymbol{x} & boldsymbol{a}-boldsymbol{x} & boldsymbol{a}+boldsymbol{x}end{array}right|=mathbf{0} ) then
the non-zero value of ( x=dots )
( A )
B. ( 3 a )
( c .2 a )
D. ( 4 a )
12
329Calculate the values of the
determinants:
( 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| )
12
330Let ( triangle_{1}, triangle_{2}, triangle_{3}, dots, triangle_{k} ) be the set of
third-order determinants that can be
made with the distinct nonzero real
numbers ( a_{1}, a_{2}, a_{3}, dots a_{9}, ) and ( Sigma a_{i}=0 )
Then
This question has multiple correct options
( mathbf{A} cdot k=9 ! )
B . ( sum_{i=1}^{k} Delta_{i}=0 )
C . at least one ( Delta_{i}=0 )
D. none of these
12
331( left|begin{array}{cc}mathbf{0} & boldsymbol{a}-boldsymbol{b} \ -boldsymbol{a} mathbf{0}-boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} mathbf{0}end{array}right|=mathbf{0} )12
332The value of the determinant
[
left|begin{array}{ccc}
mathbf{1} & cos (boldsymbol{alpha}-boldsymbol{beta}) & mathbf{c o s} boldsymbol{alpha} \
cos (boldsymbol{alpha}-boldsymbol{beta}) & mathbf{1} & cos beta \
cos boldsymbol{alpha} & cos beta & mathbf{1}
end{array}right|
]
A ( cdot alpha^{2}+beta^{2} )
B ( cdot alpha^{2}-beta^{2} )
( c .1 )
( D )
12
333frue Enter ( ^{prime} 1^{prime} ) else ( ^{prime} 0^{prime} )
( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{y} & boldsymbol{x}+boldsymbol{y} \ boldsymbol{y} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x} \ boldsymbol{x}+boldsymbol{y} & boldsymbol{x} & boldsymbol{y}end{array}right|=-boldsymbol{2}(boldsymbol{x}+ )
( boldsymbol{y})left(boldsymbol{x}^{2}+boldsymbol{y}^{2}-boldsymbol{x} boldsymbol{y}right) )
12
334Let ( Delta(x)= )
( left|begin{array}{ccc}(boldsymbol{x}-mathbf{2}) & (boldsymbol{x}-mathbf{1})^{2} & boldsymbol{x}^{mathbf{3}} \ (boldsymbol{x}-mathbf{1}) & boldsymbol{x}^{2} & (boldsymbol{x}+mathbf{1})^{3} \ boldsymbol{x} & (boldsymbol{x}+mathbf{1})^{2} & (boldsymbol{x}+mathbf{2})^{3}end{array}right| ) Then the
coefficient of ( x ) in ( Delta(x) ) is ( -k . ) Find ( k )
( A cdot-1 )
B. 2
( c cdot-2 )
( D )
12
335( mathbf{f} boldsymbol{u}_{boldsymbol{r}}=boldsymbol{a}_{boldsymbol{r}} boldsymbol{x}+boldsymbol{b}_{boldsymbol{r}} boldsymbol{y}+, boldsymbol{c}_{boldsymbol{r}}=mathbf{0}(r=1,2,3 )
be the three sides of a triangle them the equations of the circumcircle of this triangle is ( mid begin{array}{ccc}frac{mathbf{1}}{boldsymbol{u}_{1}} & frac{mathbf{1}}{boldsymbol{u}_{2}} & frac{mathbf{1}}{boldsymbol{u}_{3}} \ boldsymbol{a}_{2} boldsymbol{a}_{3}-boldsymbol{b}_{2} boldsymbol{b}_{3} & boldsymbol{a}_{3} boldsymbol{a}_{1}-boldsymbol{b}_{3} boldsymbol{b}_{1} & boldsymbol{a}_{1} boldsymbol{a}_{2}-boldsymbol{b}_{4} \ boldsymbol{a}_{2} boldsymbol{b}_{3}+boldsymbol{a}_{3} boldsymbol{b}_{2} & boldsymbol{a}_{3} boldsymbol{b}_{1}+boldsymbol{a}_{1} boldsymbol{b}_{3} & boldsymbol{a}_{1} boldsymbol{b}_{2}+boldsymbol{a}_{2}end{array} )
( =0 )
12
336If ( A ) is a skew-symmetric matrix of
order 3 then find ( |A| )
12
33714.
11
If D= 1
1
1
1+ x
1
1
1
1+y
for x = 0, y 70, then D is
(a) divisible by x but not y
(b) divisible by y but not x
(c) divisible by neither x nor y
(d) divisible by both x and y
12
338( f(x)= )
[
begin{array}{ccc}
cos ^{x} & cos x sin x & -sin x \
cos x sin x & sin ^{x} & cos x \
sin x & -cos x & 0
end{array}
]
Prove that ( f(x)=1 ) is an identity
12
339Prove that ( triangle= )
( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{a} boldsymbol{alpha}+boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b} boldsymbol{alpha}+boldsymbol{c} \ boldsymbol{a} boldsymbol{alpha}+boldsymbol{b} & boldsymbol{b} boldsymbol{alpha}+boldsymbol{c} & boldsymbol{0}end{array}right|=boldsymbol{0} ) if ( mathbf{a}, mathbf{b}, mathbf{c} )
( operatorname{arcin} mathrm{G} . mathrm{P} )
12
340( mathbf{f}left|begin{array}{ccc}boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b}end{array}right|=boldsymbol{t} times operatorname{det} ) of
circulant matrix whose elements of first
column are ( a, b, c )
then ‘t’ equals
A . 5
B. 6
( c cdot-2 )
( D )
12
341Three lines ( boldsymbol{p} boldsymbol{x}+boldsymbol{q} boldsymbol{y}+boldsymbol{r}=mathbf{0}, boldsymbol{q} boldsymbol{x}+boldsymbol{r} boldsymbol{y}+ )
( boldsymbol{p}=mathbf{0} ) and ( boldsymbol{r} boldsymbol{x}+boldsymbol{p} boldsymbol{y}+boldsymbol{q}=mathbf{0} ) are
concurrent if
This question has multiple correct options
A ( . p+q+r=0 )
B . ( p^{2}+q^{2}+r^{2}=p r+q r+p q )
c. ( p^{3}+q^{3}+r^{3}=3 p q r )
D. none of these
12
342( operatorname{Solve}left|begin{array}{ccc}mathbf{0} & boldsymbol{p}-boldsymbol{q} & boldsymbol{p}-boldsymbol{r} \ boldsymbol{q}-boldsymbol{p} & boldsymbol{0} & boldsymbol{q}-boldsymbol{r} \ boldsymbol{r}-boldsymbol{p} & boldsymbol{r}-boldsymbol{q} & boldsymbol{0}end{array}right|= )
( mathbf{A} cdot p+q+r )
в. ( p q+q r+r p )
( c cdot c )
D. ( p^{2}+q^{2}+r^{2} )
12
343Without expanding prove that the determinant
( left|begin{array}{ccc}sin A & operatorname{Cos} A & sin (A+theta) \ sin B & cos B & sin (B+theta) \ sin C & cos C & sin (c+theta)end{array}right|=0 )
12
344If planes ( boldsymbol{x}-boldsymbol{c} boldsymbol{y}-boldsymbol{b} boldsymbol{z}=boldsymbol{0}, boldsymbol{c} boldsymbol{x}-boldsymbol{y}+ )
( boldsymbol{a} boldsymbol{z}=mathbf{0} ) and ( boldsymbol{b} boldsymbol{x}+boldsymbol{a} boldsymbol{y}-boldsymbol{z}=mathbf{0} ) pass
through a straight line then ( a^{2}+b^{2}+ )
( boldsymbol{c}^{2}= )
A. ( 1-a b c )
B . ( a b c-1 )
c. ( 1-2 a b c )
D. ( 2 a b c-1 )
12
3451.
Consider the set A of all determinants of order 3 with entries
0 or 1 only. Let B be the subset of A consisting of all
determinants with value 1. Let C be the subset of A consisting
of all determinants with value-1. Then
(a C is empty
(1981 – 2 Marks)
(b) B has as many elements as C
(c) A= BUC
(d) B has twice as many elements as elements as C
12
346( A=left[begin{array}{lll}3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3end{array}right], ) then ( operatorname{Adj}(A) )
( A cdot 3 A )
B. ( 6 A )
( mathrm{c} cdot 9 A^{T} )
D. ( 2 A^{text {? }} )
12
3479.
(1992 – 4
For a fixed positive integer n, if
n! (n+1)! (n + 2)!|
D= (n+1)! (n+ 2)! (n+3)!
|(n+2)! (n+3)! (n+4)!
then show that

-4
is divisible by n.
n13
12
348Find the values of ( a ) and ( b ) so that the
points ( (boldsymbol{a}, boldsymbol{b}, mathbf{3}),(mathbf{2}, mathbf{0},-mathbf{1}) ) and
(1,-1,-3) are collinear. This question has multiple correct options
A ( . a=4, b=2 )
В. ( a=0, b=2 )
c. ( a=4, b=-2 )
begin{tabular}{l}
D. ( a=-4, b=-2 ) \
hline
end{tabular}
12
3491.
For what value of k do the following system of equations
possess a non trivial (i.e., not all zero) solution over the set
of rationals Q?
x + ky + 3z=0
3x + ky-2z=0
2x+3y – 4z=0
For that value of k, find all the solutions for the system.
12
350Prove that ( left|begin{array}{lll}mathbf{1} & boldsymbol{x} & boldsymbol{x}^{3} \ mathbf{1} & boldsymbol{y} & boldsymbol{y}^{3} \ mathbf{1} & boldsymbol{z} & boldsymbol{z}^{3}end{array}right|=(boldsymbol{x}+boldsymbol{y}+ )
( z(x-y)(y-z)(z-x) )
12
351If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & mathbf{2} \ mathbf{2} & mathbf{1}end{array}right] ) then ( boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{A})=? )
( mathbf{A} cdotleft[begin{array}{cc}1 & -2 \ -2 & 1end{array}right] )
в. ( left[begin{array}{cc}2 & 1 \ 1 & 1end{array}right] )
c. ( left[begin{array}{cc}1 & -2 \ -2 & -1end{array}right] )
D. ( left[begin{array}{cc}-1 & 2 \ 2 & -1end{array}right] )
12
352Solve:det ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{y} boldsymbol{z} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{z} boldsymbol{x} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{x} boldsymbol{y}end{array}right| )12
353If ( |boldsymbol{A}|=mathbf{3} ) and ( boldsymbol{A}^{-1}=left[begin{array}{cc}mathbf{3} & -mathbf{1} \ -mathbf{5} & mathbf{2} \ hline mathbf{3} & mathbf{3}end{array}right] ) then
( boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}=? )
( mathbf{A} cdotleft[begin{array}{cc}9 & 3 \ -5 & -2end{array}right] )
В. ( left[begin{array}{cc}9 & -3 \ -5 & 2end{array}right] )
begin{tabular}{lll}
C. ( left[begin{array}{cc}-9 & 3 \
5 & -2end{array}right] ) \
hline
end{tabular}
D. ( left[begin{array}{cc}9 & -3 \ 5 & -2end{array}right] )
12
354The value of ( Delta=left|begin{array}{llll}1^{2} & 2^{2} & 3^{2} & 4^{2} \ 2^{2} & 3^{2} & 4^{2} & 5^{2} \ 3^{2} & 4^{2} & 5^{2} & 6^{2} \ 4^{2} & 5^{2} & 6^{2} & 7^{2}end{array}right| )
equals to
( mathbf{A} cdot mathbf{0} )
B ( cdot 1^{2}+2^{2}+3^{2}+4^{2}+ldots+7^{2} )
( c cdot 1 )
D. –
12
355The area of the triangle whose vertices ( operatorname{are} A(1,2,3), B(2,-1,1) ) and
( boldsymbol{C}(mathbf{1}, mathbf{2},-mathbf{4}) ) is
A . ( 7 sqrt{10} ) sq units
B. ( frac{1}{2} sqrt{10} ) sq units
c. ( frac{7}{2} sqrt{10} ) sq units
D. None of these
12
356If ( left(x_{1}-x_{2}right)^{2}+left(y_{1}-y_{2}right)^{2}=a^{2} )
( left(x_{2}-x_{3}right)^{2}+left(y_{2}-y_{3}right)^{2}=b^{2} )
( left(x_{3}-x_{1}right)^{2}+left(y_{3}-y_{1}right)^{2}=c^{2} ) and
( kleft|begin{array}{lll}boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & mathbf{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & mathbf{1}end{array}right|=(boldsymbol{a}+boldsymbol{b}+boldsymbol{c})(boldsymbol{b}+boldsymbol{c}- )
( a)(c+a-b) times(a+b-c), ) then the
value of ( k ) is
( A )
в. 2
( c )
D. none of these
12
357( left|begin{array}{ccc}x^{2}+x & x+1 & x-2 \ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \ x^{2}+2 x+3 & 2 x-1 & 2 x-1end{array}right|= )
( A x+B ) then
A. A and B are independent of ( x )
B. A and B are dependent of ( x )
C. A dependent on x but B does not depend on x
D. B depends on x but A does not depend on x
12
358Let ( 0(0,0), P(3,4), Q(6,0) ) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR,PQR, OQR are of equal area. The coordinates of ( mathrm{R} ) are
( ^{mathrm{A}} cdotleft(frac{4}{3}, 3right) )
в. ( left(3, frac{2}{3}right) )
c. ( left(3, frac{4}{3}right) )
D ( cdotleft(frac{4}{3}, frac{2}{3}right) )
12
359If ( omega ) is a complex cube root of unity, the ( left|begin{array}{ccc}mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{omega}^{2} & boldsymbol{1} \ boldsymbol{omega}^{2} & boldsymbol{1} & boldsymbol{omega}end{array}right| ) is equal to
A . -1
B.
( c cdot 0 )
D.
12
360( f(x)=left|begin{array}{ccc}cos x & x & 1 \ 2 sin x & x^{2} & 2 x \ tan x & x & 1end{array}right| ) then
( lim _{x rightarrow 0} f(x)= )
( A cdot O )
B.
( c cdot-2 )
( D )
12
361If ( f(x)=left|begin{array}{cc}x & lambda \ 2 lambda & xend{array}right|, ) then ( f(lambda x)-f(x) ) is
equal to:
A ( cdot xleft(lambda^{2}-1right) )
B ( cdot 2 lambdaleft(x^{2}-1right) )
C ( cdot lambda^{2}left(x^{2}-1right) )
D. ( lambdaleft(x^{2}-1right) )
E ( cdot x^{2}left(lambda^{2}-1right) )
12
362Value of ( left|begin{array}{lll}sin alpha & cos alpha & sin alpha+cos beta \ sin beta & cos alpha & sin beta+cos beta \ sin gamma & cos alpha & sin gamma+cos betaend{array}right| ) is
( A cdot sin alpha+sin beta+sin gamma )
B. ( cos alpha+cos beta+cos gamma )
( mathbf{c} cdot sin alpha-sin (alpha+beta)-cos alpha+cos (gamma+beta) )
( D )
12
363Find the of ( lambda, ) so that the matrix ( left[begin{array}{cc}mathbf{5}-boldsymbol{lambda} & boldsymbol{lambda}+mathbf{1} \ mathbf{2} & mathbf{4}end{array}right] ) may be singular12
36428. If P=
1 a 37
1 3 3 is the adjoint of a 3 x 3 matrix A and
2 4 4
JA) = 4, then a is equal to :
(a) 4 (6) 11
[JEEM 2013
(d) 0
(C) 5
12
365U
TIUTUU
ay = 0, az + y = 0 and
If the system of equations x + ay = 0, az + y =
ax + z = 0 has infinite solutions, then the value of a 15
(2003)
(a) -1
(b) 1
(c) o
(d) no real values
12
366ff ( alpha, beta ) are the roots of ( left|begin{array}{lll}x & 1 & 2 \ 0 & 1 & 1 \ 1 & x & 2end{array}right|=0 )
( operatorname{then} boldsymbol{alpha}^{boldsymbol{n}}+boldsymbol{beta}^{boldsymbol{n}}=? )
A.
в.
( c cdot 2 )
D. 2n
12
367If ( D_{p}=left|begin{array}{ccc}boldsymbol{p} & mathbf{1 5} & mathbf{8} \ boldsymbol{p}^{2} & mathbf{2 5} & mathbf{9} \ boldsymbol{p}^{mathbf{3}} & mathbf{4 5} & mathbf{1 0}end{array}right|, ) then ( operatorname{det}left(boldsymbol{D}_{1}+right. )
( D_{2} ) ) is equal to
( mathbf{A} cdot mathbf{0} )
B . 25
( c .625 )
D. None of these
12
368Find the values of ( x ) for which ( left|begin{array}{ll}3 & x \ x & 1end{array}right|= ) ( left|begin{array}{ll}mathbf{3} & mathbf{2} \ mathbf{4} & mathbf{1}end{array}right| )12
3696.
x +1
If fx) = 2x x (x-1) (x+1)x then
3x(x – 1) x(x – 1) (x – 2) (x+1) x(x – 1)|
f(100) is equal to
(1999 – 2 Marks)
(a) o (6) 1 (c) 100 (d) -100
12
370The number of distinct real roots of the
quation ( left|begin{array}{ccc}cos x & sin x & sin x \ sin x & cos x & sin x \ sin x & sin x & cos xend{array}right|=0 ) in
the interval ( left[-frac{pi}{4}, frac{pi}{4}right] ) is
( A )
B. 4
( c cdot 2 )
( D )
12
371( fleft|begin{array}{ccc}6 i & -3 i & 1 \ 4 & 3 i & -1 \ 20 & 3 & iend{array}right|=x+i y ) then
A ( . x=3, y=1 )
B. ( x=1, y=3 )
c. ( x=0, y=3 )
D. ( x=0, y=0 )
12
372Let ( A, B, C, D, E ) be the interior angles of
convex pentagon and ( boldsymbol{Delta}=left|begin{array}{ccc}cos boldsymbol{A} & sin boldsymbol{A} & sin (boldsymbol{A}+boldsymbol{D}+boldsymbol{E}) \ cos boldsymbol{B} & sin boldsymbol{B} & sin (boldsymbol{B}+boldsymbol{D}+boldsymbol{E}) \ cos boldsymbol{C} & sin boldsymbol{C} & sin (boldsymbol{C}+boldsymbol{D}+boldsymbol{E})end{array}right| )
find ( Delta(pi / 3)+Delta^{prime}(pi / 6) )
12
373Find the area of the triangle formed by
the lines ( x=3, y=2 ) and ( 3 x+4 y=29 )
12
374( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2} \ mathbf{1} & boldsymbol{b} & boldsymbol{b}^{2} \ mathbf{1} & boldsymbol{c} & boldsymbol{c}^{2}end{array}right|=(boldsymbol{a}-boldsymbol{b})(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}-boldsymbol{a}) )12
375( fleft(begin{array}{ccc}-1 & -3 & -3 \ 3 & 1 & -3 \ 3 & -3 & 1end{array}right] ) then adj ( (A) ) is
A.
[
=4left[begin{array}{ccc}-2 & 3 & 3 \ -3 & 2 & -3 \ -3 & 3 & 2end{array}right]
]
B. ( quadleft[begin{array}{ccc}-2 & 3 & 3 \ 3 & 2 & -3 \ -3 & -3 & 2end{array}right] )
C.
[
=4left[begin{array}{ccc}-2 & -3 & 3 \ -3 & 2 & -3 \ -3 & -3 & 2end{array}right]
]
D.
[
=4left[begin{array}{ccc}-2 & 3 & 3 \ -3 & 2 & -3 \ -3 & -3 & 2end{array}right]
]
12
376( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{0}end{array}right|=mathbf{0}, ) then ( boldsymbol{a}, boldsymbol{b}, boldsymbol{c} )
( operatorname{are} ) in
A. A.P
B. G.P.
( c . ) н.
D. None of these
12
377Prove the following:
( left|begin{array}{ccc}boldsymbol{a}^{2}+boldsymbol{b}^{2} & boldsymbol{c} & boldsymbol{c} \ boldsymbol{c} & frac{boldsymbol{b}^{2}+boldsymbol{c}^{2}}{boldsymbol{a}} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{b} & frac{boldsymbol{c}^{2}+boldsymbol{a}^{2}}{boldsymbol{b}}end{array}right|=boldsymbol{4 a b c} )
12
378( operatorname{Matrix} boldsymbol{A}=left[begin{array}{ccc}boldsymbol{x} & boldsymbol{3} & boldsymbol{2} \ boldsymbol{1} & boldsymbol{y} & boldsymbol{4} \ boldsymbol{2} & boldsymbol{2} & boldsymbol{z}end{array}right], ) if ( boldsymbol{x} boldsymbol{y} boldsymbol{z}=boldsymbol{6} mathbf{0} )
and ( 8 x+4 y+3 z=20, ) then ( A(a d j A) )
is equal to
A. ( left|begin{array}{lll}64 & 0 & 0 \ 0 & 64 & 0 \ 0 & 0 & 64end{array}right| )
B. ( mid begin{array}{ccc}68 & 0 & 0 \ 0 & 68 & 0 \ 0 & 0 & 68end{array} )
begin{tabular}{l|lll}
38 & 0 & 0 \
0 & 38 & 0 \
0 & 0 & 38
end{tabular}
D. ( mid begin{array}{ccc}32 & 0 & 0 \ 0 & 32 & 0 \ 0 & 0 & 32end{array} )
12
379Find the values of ( x, ) if ( left|begin{array}{cc}mathbf{3 x} & mathbf{7} \ mathbf{2} & mathbf{4}end{array}right|=mathbf{1 0} )12
380If ( boldsymbol{D}_{boldsymbol{r}}=left|begin{array}{ccc}boldsymbol{r} & boldsymbol{x} & frac{boldsymbol{n}(boldsymbol{n}+mathbf{1})}{mathbf{2}} mid \ boldsymbol{2 r}-boldsymbol{1} & boldsymbol{y} & boldsymbol{n}^{2} \ boldsymbol{3 r}-boldsymbol{2} & boldsymbol{z} & frac{boldsymbol{n}(boldsymbol{3} boldsymbol{n}-mathbf{1})}{mathbf{2}}end{array}right|, ) then
( sum_{r=1}^{n} D_{r} ) is equal to
A ( cdot frac{1}{6} n(n+1)(2 n+1) )
B. ( frac{1}{4} n^{2}(n+1)^{2} )
( c cdot 0 )
D. none of these
12
381The co-ordinates of the vertices ( A, B, C )
of a triangle are ( (mathbf{6}, mathbf{3}),(-mathbf{3}, mathbf{5}),(mathbf{4},-mathbf{2}) )
respectively and ( P ) is any point ( (x, y) )
then the ratio of areas of triangles PBC and ABC is
A ( cdot|x-y-2|: 7 )
B cdot | ( x+y+2 mid: 7 )
c. ( |x+y-2|: 7 )
D. None of these
12
382For what value of ( x ) the matrix ( A ) is
singular? ( boldsymbol{A}=left[begin{array}{ll}mathbf{1}+boldsymbol{x} & mathbf{7} \ mathbf{3}-boldsymbol{x} & mathbf{8}end{array}right] )
A ( cdot frac{12}{15} )
в. ( frac{13}{15} )
c. ( frac{14}{15} )
D. none of these
12
383The value of the determinant
( Delta=left|begin{array}{lll}log x & log y & log z \ log 2 x & log 2 y & log 2 z \ log 3 x & log 3 y & log 3 zend{array}right| )
( A cdot 0 )
( mathbf{B} cdot log (x y z) )
( mathbf{C} cdot log (6 x y z) )
( mathbf{D} cdot 6 log (x y z) )
12
384If ( f(x)= )
( left|begin{array}{ccc}cos ^{2} x & cos x cdot sin x & -sin x \ cos x cdot sin x & sin ^{2} x & cos x \ sin x & -cos x & 0end{array}right|, ) then
for all ( x epsilon R, ) the value of ( f(x)= )
( mathbf{A} cdot mathbf{0} )
B.
( c )
D. None of the above
12
385For what values of ( m ) will the expression ( y^{2}+2 x y+2 x+m y-3 ) be capable of
resolution into two rational factors?
12
386The maximum value of ( left|begin{array}{ccc}1+sin ^{2} x & cos ^{2} x & 4 cos 2 x \ sin ^{2} x & 1+cos ^{2} x & 4 sin 2 x \ sin ^{2} x & cos ^{2} x & 1+4 sin 2 xend{array}right| )
( A )
B.
( c .5 )
( D )
12
387If ( D=left|begin{array}{cc}3 sqrt{5} & 6 \ 5 & mend{array}right|=0 )
Find the value of ( boldsymbol{m} )
A . ( sqrt{5} )
B. ( 4 sqrt{5} )
( c cdot sqrt{3} )
D. ( 2 sqrt{5} )
12
38813. Prove that for all values of o,
sine
cose
sin20
| sin ( 1 + 2 ) cos( 1+ 2+ sin(20 + 4) = 0
sin(0-20) cos(0 – 21) sin( 20 – 49)
COS
12
389Find the value of ( x, ) if ( left|begin{array}{cc}boldsymbol{x}+mathbf{2} & boldsymbol{x} \ boldsymbol{x}-boldsymbol{4} & boldsymbol{x}+mathbf{3}end{array}right|=left|begin{array}{cc}boldsymbol{4} & boldsymbol{2} \ -boldsymbol{2} & boldsymbol{5}end{array}right| )12
390The value of ( left|begin{array}{lllll}mathbf{1} & mathbf{0} & mathbf{0} & mathbf{0} & mathbf{0} \ mathbf{2} & mathbf{2} & mathbf{0} & mathbf{0} & mathbf{0} \ mathbf{4} & mathbf{4} & mathbf{3} & mathbf{0} & mathbf{0} \ mathbf{5} & mathbf{5} & mathbf{5} & mathbf{4} & mathbf{0} \ mathbf{6} & mathbf{6} & mathbf{6} & mathbf{6} & mathbf{5}end{array}right| )
( A cdot 6 )
B. 5
c. ( 1.2^{2} .3 .4^{3} .5^{4} .6^{4} )
D. None of these
12
391Find the value of following determinant. ( left|begin{array}{cc}-1 & 7 \ 2 & 4end{array}right| )12
392Find the equation of line joining (1,2) and (3,6) using determinants.12
393If ( operatorname{in} ) a ( Delta A B C ; frac{cos A}{7}=frac{cos B}{19}= )
( frac{cos C}{25}=k, ) then
( left|begin{array}{ccc}-1 / k & 25 & 19 \ 25 & -1 / k & 7 \ 19 & 7 & -1 / kend{array}right|= )
( A )
B.
( c cdot 2 )
( D )
12
394The number of distinct real roots of ( left|begin{array}{lll}sin x & cos x & cos x \ cos x & sin x & cos x \ cos x & cos x & sin xend{array}right|=0 ) in the
nterval ( -frac{pi}{4}2 )
12
395Find values of ( k ) if area of triangle is 4
sq. units and vertices are
(i) ( (k, 0),(4,0),(0,2) )
(ii) ( (2,0),(0,4),(0, k) )
12
396A determinant of second order is made
with the elements 0 and ( 1 . ) Find the
number of determinants with non-
negative values.
12
397( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{2}-boldsymbol{b} boldsymbol{c} & boldsymbol{b}^{2}-boldsymbol{c} boldsymbol{a} & boldsymbol{c}^{2}-boldsymbol{a} boldsymbol{b}end{array}right|= )
( A . )
( B )
( c . a b c )
D. ( (a-b),(b-c),(c-a) )
12
398Consider three points ( boldsymbol{P}=(-sin (boldsymbol{beta}- )
( boldsymbol{alpha}),-cos beta), boldsymbol{Q}=(cos (beta-boldsymbol{alpha}), sin beta) )
and ( boldsymbol{R}=(cos (boldsymbol{beta}-boldsymbol{alpha}+boldsymbol{theta}), sin (boldsymbol{beta}-boldsymbol{theta})) )
where ( 0<alpha, beta, theta<frac{pi}{4} . ) Then
A. ( P ) lies on the line segment ( R Q )
B. ( Q ) lies on the line segment ( P R )
c. ( R ) lies on the line segment ( Q P )
D. ( P, Q, R ) are non-collinear
12
399The value of ( left|begin{array}{ccc}boldsymbol{y} boldsymbol{z} & boldsymbol{z} boldsymbol{x} & boldsymbol{x} boldsymbol{y} \ boldsymbol{p} & boldsymbol{2} boldsymbol{q} & boldsymbol{3} boldsymbol{r} \ boldsymbol{1} & boldsymbol{1} & boldsymbol{1}end{array}right| ) where ( boldsymbol{x}, boldsymbol{y}, boldsymbol{z} )
are respectively, ( p t h,(2 q) t h, a n d(3 r) t h )
terms of an H.P., is
A . -1
B.
c. 1
D. none of these
12
400Calculate the value of the following
determinant:
( left|begin{array}{cccc}mathbf{1}+boldsymbol{a} & mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{b} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{c} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{d}end{array}right| )
12
401If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & log _{boldsymbol{b}} boldsymbol{a} \ log _{boldsymbol{a}} boldsymbol{b} & mathbf{1}end{array}right] ) then ( |boldsymbol{A}| ) is equal
to
A. 0
B ( cdot log _{a} b )
( c cdot-1 )
( mathbf{D} cdot log _{b} a )
12
402Prove that ( left|begin{array}{lll}1 ! & 2 ! & 3 ! \ 2 ! & 3 ! & 4 ! \ 3 ! & 4 ! & 5 !end{array}right|=4 ! )12
403Let ( A ) be a square matrix of order 3 write the value of ( |2 A|, ) where ( |A|=4 )12
404Find determinant of ( left|begin{array}{lll}mathbf{1} & mathbf{0} & mathbf{2} \ mathbf{2} & mathbf{1} & mathbf{0} \ mathbf{3} & mathbf{2} & mathbf{1}end{array}right| )12
405( mathbf{f} mathbf{Delta}=left|begin{array}{ccc}-boldsymbol{a} & mathbf{2} boldsymbol{b} & mathbf{0} \ mathbf{0} & -boldsymbol{a} & mathbf{2} boldsymbol{b} \ mathbf{2} boldsymbol{b} & mathbf{0} & -boldsymbol{a}end{array}right|=mathbf{0}, ) then
A ( cdot frac{1}{b} ) is a cube root of unity
B. ( a ) is one of the cube roots of unity
( mathrm{c} . b ) is one of the cube roots of 8
D. ( frac{a}{b} ) is a cube root of 8
12
406Prove the following:
[
left|begin{array}{ccc}
boldsymbol{a}^{2} & boldsymbol{b} boldsymbol{c} & boldsymbol{a} boldsymbol{c}+boldsymbol{c}^{2} \
boldsymbol{a}^{2}+boldsymbol{a} boldsymbol{b} & boldsymbol{b}^{2} & boldsymbol{a} boldsymbol{c} \
boldsymbol{a b} & boldsymbol{b}^{2}+boldsymbol{b} boldsymbol{c} & boldsymbol{c}^{2}
end{array}right|=boldsymbol{4} boldsymbol{a}^{2} boldsymbol{b}^{2} boldsymbol{c}^{2}
]
12
407( left|begin{array}{lll}boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c}end{array}right|=boldsymbol{K}left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right| )
( operatorname{then} K= )
( A )
B. 2
( c )
( D )
12
408f ( alpha, beta, gamma ) are the roots of the equation
( x^{3}+p x+q=0 ) then the value of the
determinant ( left|begin{array}{lll}boldsymbol{alpha} & boldsymbol{beta} & gamma \ boldsymbol{beta} & gamma & boldsymbol{alpha} \ boldsymbol{gamma} & boldsymbol{alpha} & boldsymbol{beta}end{array}right| ) is
( mathbf{A} cdot mathbf{q} )
B.
c. ( p )
D. ( p^{2}-2 q )
12
409( fleft(a_{1}, a_{2}, a_{3}, dots . . ) are in G.P. then the value right.
of determinant ( left|begin{array}{ccc}log a_{n} & log a_{n+1} & log a_{n+2} \ log a_{n+3} & log a_{n+4} & log a_{n+5} \ log a_{n+6} & log a_{n+7} & log a_{n+8}end{array}right| ) equal
( mathbf{A} cdot mathbf{0} )
B. 1
( c cdot 2 )
D. 4
12
410What is the area of the triangle formed by the points ( (a, c+a),(a, c) ) and
( (-a, c-a) ? )
A. ( -a^{2} )
B. ( frac{1}{a^{2}} )
c. ( a^{2}+a )
D. zero
12
411f the determinant ( D= )
[
left|begin{array}{ccc}
mathbf{1} & mathbf{1} & mathbf{1} \
boldsymbol{alpha}+boldsymbol{beta} & boldsymbol{alpha}^{2}+boldsymbol{beta}^{2} & mathbf{2} boldsymbol{alpha} boldsymbol{beta} \
boldsymbol{alpha}+boldsymbol{beta} & mathbf{2} boldsymbol{alpha} boldsymbol{beta} & boldsymbol{alpha}^{2}+boldsymbol{beta}^{2}
end{array}right| text { and }
]
( boldsymbol{D}_{1}=left|begin{array}{ccc}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{0} & boldsymbol{alpha} & boldsymbol{beta} \ mathbf{0} & boldsymbol{beta} & boldsymbol{alpha}end{array}right|, ) then find the
determinant of ( D_{2} ) such that ( D_{2}=frac{D}{D_{1}} )
12
412f ( boldsymbol{A}+boldsymbol{B}+boldsymbol{C}=boldsymbol{pi}, ) then
( left|begin{array}{ccc}tan (boldsymbol{A}+boldsymbol{B}+boldsymbol{C}) & tan boldsymbol{B} & tan C \ tan (boldsymbol{A}+boldsymbol{C}) & boldsymbol{0} & tan boldsymbol{A} \ tan (boldsymbol{A}+boldsymbol{B}) & -tan boldsymbol{A} & boldsymbol{0}end{array}right| )
equal to
( mathbf{A} cdot mathbf{0} )
B.
( c . ) tan AtanBtan
D. -2
12
413With out expanding show that ( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2} \ mathbf{1} & boldsymbol{b} & boldsymbol{b}^{2} \ mathbf{1} & boldsymbol{c} & boldsymbol{c}^{2}end{array}right|=(boldsymbol{a}-boldsymbol{b})(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}-boldsymbol{a}) )12
414If ( a, b, c ) are positive and are the ( p ) th ( , q t h )
and ( r t h ) terms, respectively, of a G.P. ( operatorname{then} Delta=left|begin{array}{lll}log a & p & 1 \ log b & q & 1 \ log c & r & 1end{array}right| ) is
( A )
B. ( log (a b c) )
c. ( -(p+q+r) )
D. none of these
12
415Find area of triangle with vertices at the point given in each of the following
( (mathrm{i})(1,0),(6,0),(4,3) )
( (i i)(2,7),(1,1),(10,8) )
( (text { iii) }(-2,-3),(3,2),(-1,-8) )
12
41623. Consider the system of linear equations ;
x₂ + 2x₂ + x₃ = 3
2x + 3x₂+xz=3
3x, + 5×2 + 2×3 = 1
The system has
(a) exactly 3 solutions
(b) a unique solution
(C) no solution
(d) infinite number of solutions
12
417begin{tabular}{|ccc}
if ( boldsymbol{f}(boldsymbol{x})= ) & \
( mathbf{1} ) & ( boldsymbol{x} ) & ( boldsymbol{x}+ ) \
( boldsymbol{2} boldsymbol{x} ) & ( boldsymbol{x}(boldsymbol{x}-mathbf{1}) ) & ( (boldsymbol{x}+ ) \
( boldsymbol{3} boldsymbol{x}(boldsymbol{x}-mathbf{1}) ) & ( boldsymbol{x}(boldsymbol{x}-mathbf{1})(boldsymbol{x}-mathbf{2}) ) & ( (boldsymbol{x}+mathbf{1}) boldsymbol{x} )
end{tabular}
then ( f(100) ) is equal to
( mathbf{A} cdot mathbf{0} )
B. 1
( c .100 )
D. –
12
418Using the properties of determinants & without expanding ( left[begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{q}+boldsymbol{r} & boldsymbol{y}+boldsymbol{z} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{r}+boldsymbol{p} & boldsymbol{z}+boldsymbol{x} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{p}+boldsymbol{q} & boldsymbol{x}+boldsymbol{y}end{array}right]=boldsymbol{2}left[begin{array}{ccc}boldsymbol{a} & boldsymbol{p} & boldsymbol{x} \ boldsymbol{b} & boldsymbol{q} & boldsymbol{y} \ boldsymbol{c} & boldsymbol{r} & boldsymbol{z}end{array}right] )12
419If each element in a row of a
determinant is multiplied by the same
factor ( r, ) then the value of the
determinant:
A . Is multiplied by ( r^{3} )
B. Is increased by ( 3 r )
c. Remains unchanged
D. Is multiplied by ( r )
12
42016. Let A be a 2 x 2 matrix with real entries. Let I be the 2 x 2
identity matrix. Denote by tr(A), the sum of diagonal entries
of a. Assume that A2 = 1.
[2008]
Statement-1: IfA #I and A#-I, then det(A)=-1
Statement-2: If A #I and A#-I, then tr (A) +0.
(a) Statement-1 is false, Statement-2 is true
(b) Statement-1 is true, Statement-2 is true; Statement -2 is
a correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2
is not a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
12
421STATEMENT 1: In a ( Delta A B C, a, b, c ) denotes lengths of the sides and ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right|=mathbf{0} ) then the triangle is
equilateral triangle.
STATEMENT 2: Sum of three non-
negative numbers ( =0 Rightarrow ) each number
is zero.
A. Statement-1 is true, Statement-2 is true, Statement-2 is correct explanation of Statement-
B. Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation for Statement-
c. Statement-1 is true, Statement-2 is false
D. Statement-1 is false, Statement-2 is true
12
422ff ( x_{i}=a_{i} b_{i} c_{i}, i=1,2,3 ) are three-digit
positive integers such that each ( x_{i} ) is a
multiple of ( 19, ) then for some integer ( n ) prove that ( left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{a}_{2} & boldsymbol{a}_{3} \ boldsymbol{b}_{1} & boldsymbol{b}_{2} & boldsymbol{b}_{3} \ boldsymbol{c}_{1} & boldsymbol{c}_{2} & boldsymbol{c}_{3}end{array}right| ) is divisible
by 19
12
423Evaluate the determinant: ( left|begin{array}{ccc}mathbf{8} & mathbf{2} & mathbf{7} \ mathbf{1 2} & mathbf{3} & mathbf{5} \ mathbf{1 6} & mathbf{4} & mathbf{3}end{array}right| )12
4242. Let a>0, d>0. Find the value of the determinant
(1996 – 5 Marks)
a(a +d)
(a+d)(a +2d)
(a + d)
(a +d)(a +2d)
(a +2d)(a +3d)
(a + 2d)
(a +2d)(a +3d)
(a +3d)(a +40)
12
425Evaluate :
[
boldsymbol{Delta}=left|begin{array}{ccc}
cos alpha cos beta & cos alpha sin beta & -sin alpha \
-sin beta & cos beta & 0 \
sin alpha cos beta & sin alpha sin beta & cos alpha
end{array}right|
]
12
426( left|begin{array}{ccc}-mathbf{2} boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{a}+boldsymbol{c} \ boldsymbol{b}+boldsymbol{a} & -boldsymbol{2 b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{c}+boldsymbol{b} & -boldsymbol{2 c}end{array}right|=boldsymbol{4}(boldsymbol{b}+boldsymbol{c})(boldsymbol{c}+ )
( boldsymbol{a})(boldsymbol{a}+boldsymbol{b}) )
12
427( left|begin{array}{ccc}mathbf{8} & mathbf{- 5} & mathbf{1} \ mathbf{5} & boldsymbol{x} & mathbf{3} \ mathbf{6} & mathbf{3} & mathbf{1}end{array}right|=mathbf{2} ) then what is the
value of ( x ) ?
A . 44
B. 55
c. 61
D. 84
12
428( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{1} \ mathbf{2} & mathbf{1} & -mathbf{3} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right] ) and ( boldsymbol{B}= )
( left[begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & alpha \ 1 & -2 & 3end{array}right] ) If ( B ) is the adjoint of ( A )
then ( alpha ) equals
A .2
в.
( c cdot-2 )
D.
12
429If ( A ) is ( 3 x 3 ) matrix and ( operatorname{det} operatorname{adj}(A)=k )
then
( operatorname{det}(operatorname{adj} 2 A)= )
( A cdot 2 k )
в. 8k
( c cdot 16 k )
D. ( 64 k^{2} )
12
430First row of the matrix ( boldsymbol{A} ) is ( left[begin{array}{lll}mathbf{1} & mathbf{3} & mathbf{2}end{array}right] . )
[
begin{array}{l}
boldsymbol{a d j}(boldsymbol{A})= \
{left[begin{array}{ccc}
-mathbf{2} & mathbf{4} & boldsymbol{a} \
-mathbf{1} & mathbf{2} & mathbf{1} \
mathbf{3} boldsymbol{a} & mathbf{- 5} & mathbf{- 2}
end{array}right]}
end{array}
]
then a possible value of ( operatorname{det}(boldsymbol{A}) ) is
A . 1
B . 2
( c cdot-1 )
D. -2
12
43133. The system of linear equations
x+2y-z=0
ax-y-z=0
xty-dz=0
has a non-trivial solution for:
(a) exactly two values of 2.
(b) exactly three values of .
(c) infinitely many values of 2.
(d) exactly one value of 2.
12
432f ( 2 s=a+b+c, ) prove that ( left|begin{array}{ccc}a^{2} & (s-a)^{2} & (s-a)^{2} \ (s-b)^{2} & b^{2} & (s-b)^{2} \ (s-c)^{2} & (s-c)^{2} & c^{2}end{array}right| )
( =2 s^{3}(s-a)(s-b)(s-c) )
12
433If ( boldsymbol{A}=left[begin{array}{ccc}5 & 5 x & x \ 0 & x & 5 x \ 0 & 0 & 5end{array}right] ) and ( left|A^{2}right|=25 )
then ( |x| ) is equal to?
A ( cdot frac{1}{5} )
B. 5
( c cdot 5^{2} )
D.
12
434Find the ratio in which the line segment
joining the points ( P(x, 2) ) divides the line segment joining the points ( boldsymbol{A}(mathbf{1} mathbf{2}, mathbf{5}) ) and ( B(4,-3) . ) Also find the value of ( x )
12
435The value of the determinant
( left|begin{array}{ccc}5 & 5 & 14 \ ^{5} C_{1} & ^{5} C_{4} & 1 \ ^{5} C_{2} & ^{5} C_{5} & 1end{array}right| ) is
( mathbf{A} cdot mathbf{0} )
B. -576
c. 80
D. none of these
12
436Find the values of the following determinants where ( i=sqrt{-1} )
( (i)left|begin{array}{cc}2 i & -3 i \ i^{3} & -2 i^{5}end{array}right| )
( (i i)left|begin{array}{cc}mathbf{1}+mathbf{3} i & boldsymbol{i}-mathbf{2} \ boldsymbol{i}+mathbf{2} & mathbf{1}-mathbf{3} iend{array}right| )
12
437f ( a, b, c epsilon R, ) find the number of real
roots of the equation given by ( Delta=0 )
where ( boldsymbol{Delta}=left|begin{array}{ccc}boldsymbol{x} & boldsymbol{c} & boldsymbol{- b} \ -boldsymbol{c} & boldsymbol{x} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{-} boldsymbol{a} & boldsymbol{x}end{array}right| )
A . 0
B.
( c cdot 2 )
D. 3
12
438( left|begin{array}{cc}boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{1} \ boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{2}end{array}right|=left|begin{array}{cc}mathbf{4} & -mathbf{1} \ mathbf{1} & mathbf{3}end{array}right|, ) then write
the value of ( x )
12
439If the points ( (k, 2 k),(3 k, 3 k) ) and (3,1) are collinear then the value of ( k ) is
A ( cdot frac{7}{9} )
B. ( frac{2}{3} )
c. ( frac{-2}{3} )
D. ( frac{-1}{3} )
12
440Let
( boldsymbol{Delta}_{1}=left|begin{array}{ccc}mathbf{1} & cos boldsymbol{alpha} & cos beta \ cos boldsymbol{alpha} & boldsymbol{1} & cos gamma \ cos beta & cos gamma & 1end{array}right| )
and ( Delta_{2}=left|begin{array}{ccc}0 & cos alpha & cos beta \ cos alpha & 0 & cos gamma \ cos beta & cos gamma & 0end{array}right| )
If ( Delta_{1}=Delta_{2}, ) find ( sin ^{2} alpha+sin ^{2} beta+sin ^{2} gamma )
12
441Solve this ( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{b}-boldsymbol{c} & boldsymbol{2} boldsymbol{a} & boldsymbol{2} boldsymbol{a} \ boldsymbol{2} boldsymbol{b} & boldsymbol{b}-boldsymbol{c}-boldsymbol{a} & boldsymbol{2} boldsymbol{b} \ boldsymbol{2} boldsymbol{c} & boldsymbol{2} boldsymbol{c} & boldsymbol{c}-boldsymbol{a}-boldsymbol{b}end{array}right| )12
442Assertion
The area of the triangle formed by the
points ( boldsymbol{A}(mathbf{2 0 0 7}, mathbf{2 0 0 9}), boldsymbol{B}(mathbf{2 0 0 8}, mathbf{2 0 1 1}) )
( C(2009,2010) ) will be same as the
area formed by the points ( boldsymbol{P}(mathbf{0}, mathbf{0}) )
( Q(1,2), R(2,1) )
Reason
The area of the triangle remains same
w.r.t to transition of co-ordinate axes.
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
443( operatorname{det}left{begin{array}{lll}18 & 40 & 89 \ 40 & 89 & 198 \ 89 & 198 & 440end{array}right}= )
A . -8
B. –
( c .-1 )
( D )
12
444Solve:
( mid begin{array}{ccc}x^{2}-1 & x^{2}+2 x+1 & 2 x^{2}+3 x \ 2 x^{2}+x-1 & 2 x^{2}+5 x-3 & 4 x^{2}+4 x \ 6 x^{2}-x-2 & 6 x^{2}-7 x+2 & 12 x^{2}-5end{array} )
0. Let sum of all values of ( x ) be ( k ). Find
( -2 k )
12
445Find the minor of ( left[begin{array}{ccc}2 & 7 & 3 \ -4 & 3 & -1 \ 0 & -3 & 7end{array}right] )12
446If ( (8,1),(k,-4),(2,-5) ) are collinaer,
then ( k= )
( mathbf{A} cdot mathbf{1} )
B . 2
( c .3 )
D. 4
12
447The value of
( left|begin{array}{ccc}a_{1} x+b_{1} y & a_{2} x+b_{2} y & a_{3} x+b_{3} y \ b_{1} x+a_{1} y & b_{2} x+a_{2} y & b_{3} x+a_{3} y \ b_{1} x+a_{1} & b_{2} x+a_{2} & b_{3} x+a_{3}end{array}right| ) is
equal to
A ( cdot x^{2}+y^{2} )
B.
( mathbf{c} cdot a_{1} a_{2} a_{3} x^{2}+b_{1} b_{2} b_{3} y^{2} )
D. none of these
12
448Solve
( left|begin{array}{ccc}boldsymbol{a}^{2}+mathbf{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b} & boldsymbol{b}^{2}+mathbf{1} & boldsymbol{b} boldsymbol{c} \ boldsymbol{a} boldsymbol{c} & boldsymbol{b} boldsymbol{c} & boldsymbol{c}^{2}+1end{array}right|=boldsymbol{a}^{2}+boldsymbol{b}^{2}+ )
( c^{2}+1 )
12
449If ( boldsymbol{A}=left(begin{array}{ccc}1 & 2 & 1 \ -1 & 0 & 3 \ 2 & -1 & 1end{array}right) ) then
characteristic equation is given by
A. ( -lambda^{3}+2 lambda^{2}-4 lambda+18=0 )
0
B . ( lambda^{3}+2 lambda^{2}+4 lambda+18=0 )
c. ( 2 lambda^{3}-lambda^{2}+6 lambda-2=0 )
D. None of these
12
450Write the value of the determinant ( left|begin{array}{cc}mathbf{3} & mathbf{- 1} \ mathbf{2} & mathbf{1}end{array}right| )12
451If ( f(x)=left|begin{array}{ccc}a & -1 & 0 \ a x & a & -1 \ a x^{2} & a x & aend{array}right|, ) by using
properties of determinants, find the
value ( boldsymbol{f}(mathbf{2 x})-boldsymbol{f}(boldsymbol{x}) )
12
452If the determinant
( left|begin{array}{ccc}boldsymbol{b}-boldsymbol{c} & boldsymbol{c}-boldsymbol{a} & boldsymbol{a}-boldsymbol{b} \ boldsymbol{b}^{prime}-boldsymbol{c}^{prime} & boldsymbol{c}^{prime}-boldsymbol{a}^{prime} & boldsymbol{a}^{prime}-boldsymbol{b}^{prime} \ boldsymbol{b}^{prime prime}-boldsymbol{c}^{prime prime} & boldsymbol{c}^{prime prime}-boldsymbol{a}^{prime prime} & boldsymbol{a}^{prime prime}-boldsymbol{b}^{prime prime}end{array}right| )
is expressible as ( mleft|begin{array}{lll}a & b & c \ a^{prime} & b^{prime} & c^{prime} \ a^{prime prime} & b^{prime prime} & c^{prime prime}end{array}right|, ) then
the value of ( mathrm{m} ) is
A . –
B.
( c )
( D )
12
453Find ( left|begin{array}{lll}log e & log e^{2} & log e^{3} \ log e^{2} & log e^{3} & log e^{4} \ log e^{3} & log e^{4} & log e^{5}end{array}right| )
( A )
B.
( c .4 log e )
( mathbf{D} cdot 5 log e )
12
454If ( boldsymbol{A}=left[begin{array}{lll}boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{a}end{array}right], ) then the value of
( |boldsymbol{A}||boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}| ) is ( boldsymbol{a}^{boldsymbol{k}} )
What is k?
12
455(
19, 1U MOHON
17.
The number of 3 x 3 matrices A whose entries are either 0 or
1 and for which the system
A y = 0
has exactly two
(2010)
distinct solutions, is
(a) 0 (b) 22–1
(c) 168
(d) 2
10
Lot 1
1
12
456( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{x}^{2} & boldsymbol{y}^{2} & boldsymbol{z}^{2} \ boldsymbol{x}^{boldsymbol{3}} & boldsymbol{y}^{boldsymbol{3}} & boldsymbol{z}^{boldsymbol{3}}end{array}right|=boldsymbol{x} boldsymbol{y} boldsymbol{z}(boldsymbol{x}-boldsymbol{y})(boldsymbol{y}-boldsymbol{z})(boldsymbol{z} )
( boldsymbol{x}) )
A. True
B. Falss
12
457ff ( a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} in R ) and are such
( operatorname{that} a_{i} b_{j} neq 1 ) for ( 1 leq i, j leq 3, ) then
( begin{array}{|lll|}frac{1-a_{1}^{3} b_{1}^{3}}{1-a_{1} b_{1}} & frac{1-a_{1}^{3} b_{2}^{3}}{1-a_{1} b_{2}} & frac{1-a_{1}^{3} b_{3}^{3}}{1-a_{1} b_{3}} \ frac{1-a_{2}^{3} b_{1}^{3}}{1-a_{2} b_{1}} & frac{1-a_{2}^{3} b_{2}^{3}}{1-a_{2} b_{2}} & frac{1-a_{2}^{3} b_{3}^{3}}{1-a_{2} b_{3}} \ frac{1-a_{3}^{3} b_{1}^{3}}{1-a_{3} b_{1}} & frac{1-a_{3}^{3} b_{2}^{3}}{1-a_{3} b_{2}} & frac{1-a_{3}^{3} b_{3}^{3}}{1-a_{3} b_{3}}end{array} )
either ( a_{1}<a_{2}<a_{3} ) and ( b_{1}<b_{2}a_{2} a_{3} ) and
( boldsymbol{b}_{1}>boldsymbol{b}_{2}>boldsymbol{b}_{3} )
then show ( left(a_{1}-a_{2}right)left(a_{2}-a_{3}right)left(a_{3}-right. )
( left.boldsymbol{a}_{1}right)left(boldsymbol{b}_{1}-boldsymbol{b}_{2}right)left(boldsymbol{b}_{2}-boldsymbol{b}_{3}right)left(boldsymbol{b}_{3}-boldsymbol{b}_{1}right)<mathbf{0} )
12
458Find the solution set of
( left|begin{array}{ccc}2+x & 2-x & 2-x \ 2-x & 2+x & 2-x \ 2-x & 2-x & 2+xend{array}right|=0 )
12
459( mathrm{ff}=left[begin{array}{ccc}boldsymbol{a} & mathbf{0} & mathbf{0} \ {[mathbf{0 . 3 e m}] mathbf{0}} & boldsymbol{a} & mathbf{0} \ {[mathbf{0 . 3 e m}] mathbf{0}} & mathbf{0} & boldsymbol{a}end{array}right], ) then the value
of IAdj. Al is equal to
A ( cdot a^{3} )
в. ( a^{6} )
( c cdot a^{9} )
( mathbf{D} cdot a^{2} )
12
460( A ) and ( B ) are two points and ( C ) is any
point collinear with ( A ) and ( B . ) IF ( A B= )
( mathbf{1 0}, boldsymbol{B C}=mathbf{5}, ) then ( boldsymbol{A} boldsymbol{C} ) is equal to:
A. either 15 or 5
B. necessarily 5
c. necessarily 16
D. none of these
12
461If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) then ( boldsymbol{A}^{-mathbf{1}}= )
A ( cdot frac{-1}{2}left[begin{array}{cc}4 & -2 \ -3 & 1end{array}right] )
в. ( frac{1}{2}left[begin{array}{cc}4 & -2 \ -3 & 1end{array}right] )
c. ( left[begin{array}{cc}-2 & 4 \ 1 & 3end{array}right] )
D. ( left[begin{array}{ll}2 & 4 \ 1 & 3end{array}right] )
12
462If ( f(x) ) and ( g(x) ) are functions such that ( boldsymbol{f}(boldsymbol{x}+)=boldsymbol{f}(boldsymbol{x}) boldsymbol{g}(boldsymbol{y})+boldsymbol{g}(boldsymbol{x}) boldsymbol{f}(boldsymbol{y}), ) then the
value of ( left|begin{array}{lll}boldsymbol{f}(boldsymbol{alpha}) & boldsymbol{g}(boldsymbol{alpha}) & boldsymbol{f}(boldsymbol{alpha}+boldsymbol{theta}) \ boldsymbol{f}(boldsymbol{beta}) & boldsymbol{g}(boldsymbol{beta}) & boldsymbol{f}(boldsymbol{beta}+boldsymbol{theta}) \ boldsymbol{f}(boldsymbol{gamma}) & boldsymbol{g}(boldsymbol{gamma}) & boldsymbol{f}(boldsymbol{gamma}+boldsymbol{theta})end{array}right| )
A ( cdot f(alpha) cdot f(beta) cdot f(gamma) )
B.
( mathbf{c} cdot g(alpha) cdot g(beta) cdot g(gamma) )
D. None of these
12
463( operatorname{Let} A=left|begin{array}{ll}mathbf{5} & mathbf{0} \ mathbf{1} & mathbf{0}end{array}right| ) and ( boldsymbol{B}=left|begin{array}{ll}mathbf{2 0} & mathbf{5} mid \ -mathbf{1} & mathbf{0}end{array}right| )
( mathbf{4} boldsymbol{A}+mathbf{5} boldsymbol{B}-boldsymbol{C}=mathbf{0}, ) then ( boldsymbol{C} ) is
( mathbf{A} cdotleft|begin{array}{cc}5 & 25 \ -1 & 0end{array}right| )
( mathbf{B} cdotleft|begin{array}{cc}120 & 25 \ -1 & 0end{array}right| )
( mathbf{c} cdotleft|begin{array}{cc}5 & -1 \ 0 & 25end{array}right| )
( mathbf{D} cdotleft|begin{array}{cc}5 & 25 \ -1 & 5end{array}right| )
12
464( operatorname{Let} boldsymbol{A}=left(begin{array}{cc}boldsymbol{x}+mathbf{2} & mathbf{3} boldsymbol{x} \ boldsymbol{3} & boldsymbol{x}+mathbf{2}end{array}right), boldsymbol{B}= )
( left(begin{array}{cc}boldsymbol{x} & mathbf{0} \ mathbf{5} & boldsymbol{x}+mathbf{2}end{array}right) . ) Then all solutions of the
equation ( operatorname{det}(A B)=0 ) is
A. 1,-1,0,2
B. 1,4,0,-2
c. 1,-1,4,3
D. -1,4,0,3
12
465The value of the determinant ( mid begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \ 2 a b & 1-a^{2}+b^{2} & 2 a \ 2 b & -2 a & 1+a^{2}-b^{2}end{array} )
is equal to
( A )
B ( cdotleft(1+a^{2}+b^{2}right) )
c. ( left(1+a^{2}+b^{2}right)^{2} )
D. ( left(1+a^{2}+b^{2}right)^{3} )
12
466( mathbf{f} A=left[begin{array}{cc}frac{-mathbf{1}+i sqrt{mathbf{3}}}{mathbf{2} i} & frac{-mathbf{1}-i sqrt{mathbf{3}}}{mathbf{2} i} \ frac{mathbf{1}+i sqrt{mathbf{3}}}{mathbf{2} i} & frac{mathbf{1}-i sqrt{mathbf{3}}}{mathbf{2} i}end{array}right], i= )
( sqrt{-i} ) and ( f(x)=x^{2}+2, ) then ( f(A) ) is
equal to
( ^{A} cdotleft(frac{5-i sqrt{3}}{2}right)left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] )
( ^{text {В }} cdotleft(frac{3-i sqrt{3}}{2}right)left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] )
c. ( left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] )
D. ( (2+i sqrt{3})left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] )
12
467( left|begin{array}{lll}a^{2}+2 a & 2 a+1 & 1 \ 2 a+1 & a+2 & 1 \ 3 & 3 & 1end{array}right|= )
( A cdot(1-a)^{3} )
В – ( (a-1)^{2} )
c. ( (a-1)^{3} )
D・ ( (a+1)^{2} )
12
468Let ( A ) be a square matrix of order ( 3 times 3 )
then ( |k A| ) is equal to
( mathbf{A} cdot k mid A )
в. ( k^{2}|A| )
c. ( k^{3}|A| )
D. ( 3 k|A| )
12
469Evaluate the following ( begin{array}{|ccc|}15 & 11 & 7 \ 11 & 17 & 14 \ 10 & 16 & 13end{array} )12
470Evaluate the following determinant:
( left|begin{array}{ccc}1 & 4 & 9 \ 4 & 9 & 16 \ 9 & 16 & 25end{array}right| )
12
471If ( omega ) is an imaginary cube root of
unity,then the value of ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} boldsymbol{omega}^{2} & boldsymbol{a} boldsymbol{omega} \ boldsymbol{b} boldsymbol{c} & boldsymbol{c} & boldsymbol{b} boldsymbol{omega}^{2} \ boldsymbol{c} boldsymbol{omega}^{2} & boldsymbol{a} boldsymbol{omega} & boldsymbol{c}end{array}right|, ) is
( mathbf{A} cdot a^{3}+b^{3}+c^{3} )
B . ( a^{2} b-b^{2} c )
( c cdot 0 )
D. ( a^{3} b+b^{3}+3 a b c )
12
472( left|begin{array}{ccc}text { If } & \ cos (boldsymbol{A}+boldsymbol{B}) & -sin (boldsymbol{A}+boldsymbol{B}) & cos 2 boldsymbol{B} \ sin boldsymbol{operatorname { s o s }} boldsymbol{operatorname { s i n }} boldsymbol{A} & boldsymbol{operatorname { s i n } boldsymbol { operatorname { m o s } }}end{array}right| )
( =0 ) then ( B= )
( A cdot(2 n+1) frac{pi}{2} )
B. ( n pi )
( c cdot(2 n+1) pi )
D. 2 nn
12
473If the lines ( boldsymbol{p}_{1} boldsymbol{x}+boldsymbol{q}_{1} boldsymbol{y}=mathbf{1}, boldsymbol{p}_{2} boldsymbol{x}+boldsymbol{q}_{2} boldsymbol{y}= )
1 and ( p_{3} x+q_{3} y=1 ) be concurrent
show that the points ( left(p_{1}, q_{1}right),left(p_{2}, q_{2}right) )
and ( left(p_{3}, q_{3}right) ) are collinear
A. vertices of right angle triangle
B. vertices of an equilateral triangle
c. vertices of an isosceles triangle
D. Collinear
12
474If points ( (a, 0),(0, b) ) and ( (x, y) ) are
collinear, prove that ( frac{x}{a}+frac{y}{b}=1 )
12
475If ( left|begin{array}{ccc}2 a & x_{1} & y_{1} \ 2 b & x_{2} & y_{2} \ 2 c & x_{3} & y_{3}end{array}right|=frac{a b c}{2} neq 0, ) then the
area of the triangle whose vertices are ( left(frac{x_{1}}{a}, frac{y_{1}}{a}right),left(frac{x_{2}}{b}, frac{y_{2}}{b}right) ) and ( left(frac{x_{3}}{c}, frac{y_{3}}{c}right) ) is
A ( cdot frac{1}{4} a b c )
B. ( frac{1}{8} a b c )
( c cdot frac{1}{4} )
D. ( frac{1}{8} )
E ( cdot frac{1}{12} )
12
476If ( omega ) is cube root of unity, then ( boldsymbol{Delta}=left|begin{array}{cccc}boldsymbol{x}+mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{x}+boldsymbol{omega}^{2} & mathbf{1} \ boldsymbol{omega}^{2} & boldsymbol{1} & boldsymbol{x}+boldsymbol{omega}end{array}right|= )
( mathbf{A} cdot x^{3}+1 )
B. ( x^{3}+omega )
c. ( x^{3}+omega^{2} )
D. ( x^{3} )
12
477If ( boldsymbol{A}=left|begin{array}{ll}mathbf{0} & mathbf{0} \ mathbf{1} & mathbf{1}end{array}right| ) then the value of ( boldsymbol{A}+ )
( boldsymbol{A}^{2}+boldsymbol{A}^{3}+ldots+boldsymbol{A}^{n}=? )
( A cdot A )
B. nA
c. ( (n+1) A )
( D )
12
478Find the values of ( x, ) if ( left|begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5}end{array}right|=left|begin{array}{ll}boldsymbol{x} & mathbf{3} \ mathbf{2} boldsymbol{x} & mathbf{5}end{array}right| )12
47976. Consider the system of linear equations in x, y, and z:
(sin 30) x – y + z = 0
(cos 20) x + 4y + 3z = 0
2 x + 7y + z = 0
Which of the following can be the values of O for which
the system has a non-trivial solution?
a. nit+ (-1)” īc/6, Vnez
b. nn + (-1)” 7/3, ne z
c. nit+ (-1)” Tt/9, ne z
d. none of these
12
480Consider the system of linear equations in x, y, z:
(sin 30) x-y+z=0
(cos 20) x+4y + 3z=0
2x+7y+z=0
12
481f maximum and minimum values of ( boldsymbol{D}=left|begin{array}{ccc}mathbf{1} & -cos boldsymbol{theta} & mathbf{1} \ cos boldsymbol{theta} & mathbf{1} & -cos boldsymbol{theta} \ mathbf{1} & cos boldsymbol{theta} & mathbf{1}end{array}right| ) are ( p ) and
respectively, then the value of ( 2 p+3 q )
is
A . 16
B. 6
( c .14 )
( D )
12
482( A ) is a ( 3 times 3 ) matrix and ( B ) is its adjoint
matrix. If the determinant of ( B ) is 64
then the det ( A ) is
( A cdot 4 )
B. ±4
( c .pm 8 )
D. 8
12
483( mathrm{f}left|begin{array}{ccc}boldsymbol{x}^{2}+boldsymbol{x} & boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{2} \ mathbf{2} boldsymbol{x}^{2}+mathbf{3} boldsymbol{x}-mathbf{1} & mathbf{3} boldsymbol{x} & mathbf{3} boldsymbol{x}-mathbf{3} \ boldsymbol{x}^{mathbf{2}}+mathbf{2} boldsymbol{x}+mathbf{3} & mathbf{2} boldsymbol{x}-mathbf{1} & mathbf{2} boldsymbol{x}-mathbf{1}end{array}right|= )
( A x-12, ) then the value of ( A ) is
( A cdot 12 )
в. 24
( c .-12 )
( D .-24 )
12
484Let ( boldsymbol{P}=left[boldsymbol{a}_{i j}right] ) be a ( boldsymbol{3} times boldsymbol{3} ) matrix and let
( Q=left(b_{i j}right) ) where ( b_{i j}=2^{i+j} a_{i j} ) for ( 1 leq )
( i, j leq 3 . ) If the determinant of ( P ) is 2 then
the determinant of the matrix ( Q ) is?
A ( cdot 2^{10} )
B . ( 2^{11} )
( c cdot 2^{12} )
D. ( 2^{13} )
12
485A point ( boldsymbol{P}(mathbf{2},-mathbf{1}) ) is equidistant from
points ( (a, 7) ) and ( (-3, a) ). Find ( a )
12
486The value of ( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{b}-boldsymbol{c} & boldsymbol{2} boldsymbol{a} & boldsymbol{2} boldsymbol{a} \ boldsymbol{2} boldsymbol{b} & boldsymbol{b}-boldsymbol{c}-boldsymbol{a} & boldsymbol{2} boldsymbol{b} \ boldsymbol{2} boldsymbol{c} & boldsymbol{2} boldsymbol{c} & boldsymbol{c}-boldsymbol{a}-boldsymbol{b}end{array}right| ) will
n ( mathbf{2 C} )
A ( cdot(a+b+c)^{2} )
B ( cdot(a+b+c)^{3} )
c. ( (a-b-c)^{2} )
D ( cdot(a+b-c)^{2} )
12
487( mathbf{1 f} mathbf{A}=left[begin{array}{ccc}mathbf{1} & mathbf{5} & mathbf{- 6} \ mathbf{- 8} & mathbf{0} & mathbf{4} \ mathbf{3} & mathbf{- 7} & mathbf{2}end{array}right] ) then the
cofactors of the elements 3,-7,2 are
p,q,r respectively their ascending order
is
( mathbf{A} cdot mathbf{p}, mathbf{r}, mathbf{q} )
B. ( mathrm{q}, mathrm{r}, mathrm{p} )
( c cdot p, q, r )
( D cdot r, p, q )
12
488Find the value of ( k ) if
( boldsymbol{A}(mathbf{4}, mathbf{1 1}), boldsymbol{B}(mathbf{2}, mathbf{5}), boldsymbol{C}(boldsymbol{6}, boldsymbol{k}) ) are collinear
points.
12
489f ( x ) is a positive integer, then ( left|begin{array}{ccc}boldsymbol{x} ! & (boldsymbol{x}+mathbf{1}) ! & (boldsymbol{x}+mathbf{2}) ! \ (boldsymbol{x}+mathbf{1}) ! & (boldsymbol{x}+mathbf{2}) ! & (boldsymbol{x}+mathbf{3}) ! \ (boldsymbol{x}+mathbf{2}) ! & (boldsymbol{x}+mathbf{3}) ! & (boldsymbol{x}+mathbf{4}) !end{array}right| ) is equa
to
A ( cdot 2 x !(x+1) ! )
в. ( 2 x !(x+1) !(x+2) )
c. ( 2 x !(x+3) ! )
D. ( 2(x+1) !(x+2) !(x+3) ! )
12
490( mathbf{f} mathbf{Delta}=left|begin{array}{lll}boldsymbol{a} & mathbf{0} & mathbf{0} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right| ) then ( left|begin{array}{lll}boldsymbol{p}^{2} boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{p} boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{p c} & boldsymbol{a} & boldsymbol{b}end{array}right| ) is
equal to
A ( . p Delta )
B . ( p^{2} Delta )
c. ( p^{3} Delta )
D. ( 2 p Delta )
12
491Prove the following:
( left|begin{array}{ccc}boldsymbol{b} boldsymbol{c} & boldsymbol{b} boldsymbol{c}^{prime}+boldsymbol{b}^{prime} boldsymbol{c} & boldsymbol{b}^{prime} boldsymbol{c}^{prime} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c} boldsymbol{a}^{prime}+boldsymbol{c}^{prime} boldsymbol{a} & boldsymbol{c}^{prime} boldsymbol{a}^{prime} \ boldsymbol{a b} & boldsymbol{a b}^{prime}+boldsymbol{d}^{prime} boldsymbol{b} & boldsymbol{d}^{prime} boldsymbol{b}^{prime}end{array}right|= )
[
left(b c^{prime}-b^{prime} cright)left(c a^{prime}-c^{prime} aright)left(a b^{prime}-d bright)
]
12
492( boldsymbol{B}_{1}+boldsymbol{B}_{2}+ldots ldots+boldsymbol{B}_{49} ) is equal to
A. ( B_{0} )
В. ( 7 B_{0} )
( mathbf{c} cdot 49 B_{0} )
D. 49
12
493( left|begin{array}{ccc}mathbf{2}^{3} & mathbf{3}^{3} & mathbf{3 . 2}^{mathbf{2}}+mathbf{3 . 2}+mathbf{1} \ mathbf{3}^{mathbf{3}} & mathbf{4}^{mathbf{3}} & mathbf{3 . 3}^{mathbf{2}}+mathbf{3 . 3}+mathbf{1} \ mathbf{4}^{mathbf{3}} & mathbf{5}^{mathbf{3}} & mathbf{3 . 4}^{mathbf{2}}+mathbf{3 . 4}+mathbf{1}end{array}right| ) is equal to
( A )
B.
( c cdot 2 )
D. 3
12
494Suppose ( x, y, z ) are positive integers ( (neq )
1) if ( Delta=operatorname{det} ) of ( left[begin{array}{ccc}1 & log _{x} y & log _{x} z \ log _{y} x & 1 & log _{y} z \ sin (x+y) & -cos (x+y) & sin ^{2} zend{array}right] )
then ( Delta ) is 1 ) Independent of ( x ) II) Independent of ( y ) IIII Independent of ( z ) The which of the above statement is
are correct
A. only land II
D. all the three I,II,III
12
495( operatorname{et} Delta_{1}=left|begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & yend{array}right| )
( boldsymbol{Delta}_{2}=left|begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{y} \ mathbf{1} & boldsymbol{x}+boldsymbol{y} & boldsymbol{y} \ mathbf{1} & boldsymbol{x} & boldsymbol{x}+boldsymbol{y}end{array}right| )
( boldsymbol{Delta}_{3}=left|begin{array}{ccc}boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{2} boldsymbol{y} \ -boldsymbol{x} & boldsymbol{x} & boldsymbol{0} \ boldsymbol{0} & -boldsymbol{x} & boldsymbol{x}end{array}right| )
( boldsymbol{Delta}_{4}=left|begin{array}{lll}1 & boldsymbol{x} & boldsymbol{x}^{2} \ mathbf{1} & boldsymbol{y} & boldsymbol{y}^{2} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right| )
then
12
496Find the maximum value of ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+sin boldsymbol{theta} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+mathbf{c} mathbf{o} mathbf{s} boldsymbol{theta}end{array}right| )12
497The minors and cofactors of -4 and 9 in
determinant ( left|begin{array}{ccc}-1 & -2 & 3 \ -4 & -5 & -6 \ -7 & 8 & 9end{array}right| ) are
respectively
A. 42,( 42 ; 3,3 )
В. -42,( 42 ;-3,-3 )
c. 42,( -42 ; 3,-3 )
D. 42,3: 42,3
12
498The straight lines ( mathbf{x}+mathbf{2 y}-mathbf{9}=mathbf{0}, mathbf{3 x}+ )
( 5 y-5=0 ) and ( a x+b y-1=0 ) are
concurrent if the straight line ( 22 x- )
( 35 y-1=0 ) passes through the point
( A cdot(a, b) )
в. ( (b, a) )
c. ( (-a, b) )
D. (-a,-b)
12
499The value of the determinant ( Delta= )
A ( a_{1} a_{2}+b_{1} b_{2} )
a
В ( cdotleft(a_{1} a_{2} a_{3}right)+left(b_{1} b_{2} b_{3}right) )
c. ( a_{1} a_{2} b_{1} b_{2}+a_{2} a_{3} b_{2} b_{3}+a_{3} a_{1} b_{3} b_{1} )
D. none of these
12
500If ( boldsymbol{A}=left[begin{array}{cc}boldsymbol{a} & boldsymbol{p} \ boldsymbol{b} & boldsymbol{q} \ boldsymbol{c} & boldsymbol{r}end{array}right]_{mathbf{3} times mathbf{2}} ) then determinant
( left(A A^{T}right) ) is equal to
( mathbf{A} cdot mathbf{0} )
B. ( a^{2}+b^{2}+c^{2} )
c. ( p^{2}+q^{2}+r^{2} )
D. ( p^{2}+q^{2} )
12
501Evaluate
( left|begin{array}{ccc}cos alpha cos beta & cos alpha sin beta & -sin alpha \ -sin beta & cos beta & 0 \ sin alpha cos beta & sin alpha sin beta & cos alphaend{array}right| )
12
502Prove the following identities
( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2} \ boldsymbol{a}^{2} & mathbf{1} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a}^{2} & mathbf{1}end{array}right|=left(boldsymbol{a}^{3}-mathbf{1}right)^{2} )
12
503If ( boldsymbol{a}=sin boldsymbol{theta}, boldsymbol{b}=sin (boldsymbol{theta}+boldsymbol{2} boldsymbol{pi} / mathbf{3}), boldsymbol{c}= )
( sin (theta+4 pi / 3), x=cos theta, y= )
( cos (theta+2 pi / 3), z=cos (theta+4 pi / 3) )
then value of
( boldsymbol{Delta}=left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{b} boldsymbol{c} & boldsymbol{c} boldsymbol{a} & boldsymbol{a} boldsymbol{b}end{array}right| )
is
A . 1 ( overline{8} )
B. ( frac{3 sqrt{3}}{4} )
( c cdot frac{3 sqrt{3}}{8} )
D. ( frac{sqrt{3}}{4} )
12
504The value of the determinant
( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ m_{mathbf{C}} & m+mathbf{C}_{1} & m+mathbf{2} \ m & mathbf{C}_{1} \ m & boldsymbol{m}+mathbf{1} & boldsymbol{C}_{2}end{array}right| mathbf{~ i s ~ e q u a l ~ t o ~} )
( mathbf{A} cdot mathbf{1} )
B. –
( c cdot 0 )
D. None of these
12
505( left|begin{array}{ccc}x^{2}+x & x+1 & x+2 \ x^{2}+3 x-1 & 3 x & 3 x-3 \ x^{2}+2 x+3 & 2 x-1 & 2 x-1end{array}right|= )
( A x+B ) where ( A ) and ( B ) are
determinants of order ( 3 . ) Then ( A+2 B ) is
equal to
12
506Find the values of ( x, ) if ( left|begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5}end{array}right|=left|begin{array}{ll}boldsymbol{x} & mathbf{3} \ mathbf{2} boldsymbol{x} & mathbf{5}end{array}right| )12
507If ( A ) is a ( 3 times 3 ) matrix and ( operatorname{det}(3 A)= )
( boldsymbol{k}{boldsymbol{d e t}(boldsymbol{A})}, boldsymbol{k}= )
( mathbf{A} cdot mathbf{9} )
B. 6
c. 1
D. 27
12
508Which of the following are correct in
respect of the system of equations ( boldsymbol{x}+ )
( boldsymbol{y}+boldsymbol{z}=mathbf{8}, boldsymbol{x}-boldsymbol{y}+mathbf{2} boldsymbol{z}=boldsymbol{6} ) and ( boldsymbol{3} boldsymbol{x}- )
( boldsymbol{y}+mathbf{5} boldsymbol{z}=boldsymbol{k} ? )
1. They have no solution, if ( k=15 )
2. They have infinitely many solutions, if
( k=20 )
3. They have unique solution, if ( k=25 )
Select the correct answer using the code given below
( A cdot 1 ) and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1,2 and 3
12
509The graph of ( f(x) ) is shown above in the
( x y ) -plane. The points ( (0,3),(5 b, b) ) and
( (10 b,-b) ) are on the line described by
( f(x) . ) If ( b ) is a positive constant, find the
coordinates of point ( C )
( mathbf{A} cdot(5,1) )
B. (10,-1)
c. (15,-0.5)
(15, ( -0.5) )
D. (20,-2)
12
510Applying ( boldsymbol{C}_{1} rightarrow boldsymbol{C}_{1}-^{8} boldsymbol{C}_{3}, ) we
getWithout expanding the determinant, prove that ( left|begin{array}{ccc}mathbf{4 1} & mathbf{1} & mathbf{5} \ mathbf{7 9} & mathbf{7} & mathbf{9} \ mathbf{2 9} & mathbf{5} & mathbf{3}end{array}right|=mathbf{0} )
12
511If ( A ) is a singular matrix, then adj ( A ) is
A. non- singular
B. singular
c. symmetric
D. not defined
12
512( left|begin{array}{cc}mathbf{2} & -mathbf{4} \ mathbf{9} & boldsymbol{d}-mathbf{3}end{array}right|=mathbf{4} ) then ( boldsymbol{d}= )
A . 1
B. -11
c. 12
D. -13
12
513s
I a-1 n 6
Let Aa=/(a 1)2 2n² 4n-2
(a – 1)3 3n² 3n² – 3n
Show that Aa = c , a constant.
a = 1
12
514. If ( A, B, C ) are angles of angle and ( mid begin{array}{ccc}mathbf{1} & mathbf{1} \ mathbf{1}+boldsymbol{s} mathbf{i} boldsymbol{n} boldsymbol{A} & mathbf{1}+boldsymbol{operatorname { s i n }} boldsymbol{B} & mathbf{1}+ \ boldsymbol{s i n} boldsymbol{A}+boldsymbol{s i n}^{2} boldsymbol{A} & boldsymbol{s i n} boldsymbol{B}+boldsymbol{s i n}^{2} boldsymbol{B} & boldsymbol{s i n} boldsymbol{C}end{array} )
( =0 ) then triangle is isosceles
II. If ( boldsymbol{a}=mathbf{1}+mathbf{2}+mathbf{4}+— ) upto ( mathbf{n} )
terms ( boldsymbol{b}=mathbf{1}+mathbf{3}+mathbf{9}+— ) up to ( mathbf{n} )
terms ( c=1+5+25+—- ) up to ( n ) terms then ( Deltaleft|begin{array}{ccc}a & 2 b & 4 c \ 2 & 2 & 2 \ 2^{n} & 3^{n} & 5^{n}end{array}right|=0 )
A. I, II both are true
B. only lis true
c. only I I is true
D. neither of them are true
12
515f ( P(x, y) ) is such that ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{y} & mathbf{1} \ boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & boldsymbol{1}end{array}right|+left|begin{array}{ccc}boldsymbol{x} & boldsymbol{y} & mathbf{1} \ boldsymbol{x}_{1} & boldsymbol{y}_{1} & boldsymbol{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & 1end{array}right|=mathbf{0} ) then the
line through ( A ) and ( P ) is
A. median of ( Delta A B C )
B. bisector of ( angle A )
c. altitude through vertex ( A )
D. perpendicular bisector of the side ( B C )
12
516Solve for ( boldsymbol{x} )
( left|begin{array}{ccc}boldsymbol{x}-mathbf{2} & mathbf{2} boldsymbol{x}-mathbf{3} & mathbf{3} boldsymbol{x}-mathbf{4} \ boldsymbol{x}-mathbf{4} & mathbf{2} boldsymbol{x}-mathbf{9} & mathbf{3} boldsymbol{x}-mathbf{1} mathbf{6} \ boldsymbol{x}-mathbf{8} & mathbf{2} boldsymbol{x}-mathbf{2 7} & mathbf{3} boldsymbol{x}-mathbf{6 4}end{array}right|=mathbf{0} )
( A cdot frac{7}{4} )
B. ( frac{28}{3} )
( c cdot frac{28}{13} )
D. ( frac{14}{0} )
12
517Prove that:
( 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|=4 a^{2} b^{2} c^{2} )
12
51811.
For all values of A, B, C and P, Q, R show that
(1994 – 4 Marks)
cos(A-P) cos(A-2) cos(A – R)
cos(B-P) cos(B-Q) cos(B – R) = 0
cos(C-P) cos(C-0) cos(C-R)
12
519If the points ( (k, 2-2 k)(1-k, 2 k) ) and ( (-k ) ( -4,6-2 x) ) be collinear the possible values of k are
A ( cdot frac{1}{2} )
B. ( frac{1}{2} )
( c )
D. – –
12
520Find the value of ( x ) if
( left|begin{array}{ccc}boldsymbol{x}-mathbf{2} & mathbf{2} boldsymbol{x}-mathbf{3} & mathbf{3} boldsymbol{x}-mathbf{4} \ boldsymbol{x}-mathbf{4} & mathbf{2} boldsymbol{x}-mathbf{9} & mathbf{3} boldsymbol{x}-mathbf{1 6} \ boldsymbol{x}-mathbf{8} & mathbf{2} boldsymbol{x}-mathbf{2 7} & mathbf{3} boldsymbol{x}-mathbf{6 4}end{array}right|=mathbf{0} ? )
12
521Show that ( triangle A B C ) is an isosceles
triangle, if the determinant ( mid begin{array}{ccc}mathbf{1} & mathbf{1} \ mathbf{1}+cos boldsymbol{A} & mathbf{1}+cos boldsymbol{B} & mathbf{1} \ cos ^{2} boldsymbol{A}+cos boldsymbol{A} & cos ^{2} boldsymbol{B}+cos boldsymbol{B} & mathbf{c o s}^{2}end{array} )
( mathbf{D} )
12
522Let A=(12)
– ( a
and B=1
0
,
(3
4
10 b) 0,6 EN. Then
[2006]
(a) there cannot exist any B such that AB = BA
(b) there exist more then one but finite number of B’s such
that AB=BA
(c) there exists exactly one B such that AB=BA
(d) there exist infinitely many B’s such that AB=BA
12
523( mid begin{array}{cccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} & boldsymbol{d} \ boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{a}+boldsymbol{b}+boldsymbol{c} & boldsymbol{a}+boldsymbol{b}+boldsymbol{c} \ boldsymbol{a} & boldsymbol{2} boldsymbol{a}+boldsymbol{b} & boldsymbol{3} boldsymbol{a}+boldsymbol{2} boldsymbol{b}+boldsymbol{c} & boldsymbol{4} boldsymbol{a}+boldsymbol{3} boldsymbol{b}+boldsymbol{2} \ boldsymbol{a} & boldsymbol{3} boldsymbol{a}+boldsymbol{b} & boldsymbol{6} boldsymbol{a}+boldsymbol{3} boldsymbol{b}+boldsymbol{c} & boldsymbol{1} boldsymbol{0} boldsymbol{a}+boldsymbol{6} boldsymbol{b}+boldsymbol{3}end{array} )12
52417. Let x e R and let
(
11 17 [2 x x
P=10 2 2 8 = 0 4 ol
and R= PQP-1
10 0 3 X x 6]
Then which of the following options is/are correct?
(JEE Adv. 2019)
[2 x x
(a) det R=det 0 4 0 +8, for all x ER
[x x 5
(b) For x = 1, there exists a unit vector ai+Bj+ył for
To
There exists a real number x such that PQ=QP
(C)
(d)
For x = 0, if
= a
[b]
a
. then a+b=5
[b]
12
525Find the values of ( x, ) if ( left|begin{array}{cc}mathbf{2 x} & mathbf{5} \ mathbf{8} & boldsymbol{x}end{array}right|=left|begin{array}{ll}mathbf{6} & mathbf{5} \ mathbf{8} & mathbf{3}end{array}right| )12
526If in the determinant ( Delta= )
( left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{b}_{1} & boldsymbol{c}_{1} \ boldsymbol{a}_{2} & boldsymbol{b}_{2} & boldsymbol{c}_{2} \ boldsymbol{a}_{3} & boldsymbol{b}_{3} & boldsymbol{c}_{3}end{array}right|, boldsymbol{A}_{i}, boldsymbol{B}_{i}, boldsymbol{C}_{i} ) etc. be the co-
factors of ( a_{i}, b_{i}, c_{i} ) etc., then which of the
following relations is incorrect?
A ( cdot a_{1} A_{1}+b_{1} B_{1}+c_{1} C_{1}=Delta )
B . ( a_{2} A_{2}+b_{2} B_{2}+c_{2} C_{2}=Delta )
c. ( a_{3} A_{3}+b_{3} B_{3}+c_{3} C_{3}=Delta )
D. ( a_{1} A_{2}+b_{1} B_{2}+c_{1} C_{2}=Delta )
12
527f ( a neq b neq c ) such that
( left|begin{array}{ccc}boldsymbol{a}^{3}-mathbf{1} & boldsymbol{b}^{mathbf{3}}-mathbf{1} & boldsymbol{c}^{mathbf{3}}-mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{mathbf{2}} & boldsymbol{b}^{mathbf{2}} & boldsymbol{c}^{mathbf{2}}end{array}right|=mathbf{0} ) ther
A ( . a b+b c+c a=0 )
B . ( a+b+c=0 )
( c cdot a b c=1 )
D. ( a+b+c=1 )
12
528Find the values of ( x, ) if ( left|begin{array}{ll}boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{1} \ boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{2}end{array}right|=left|begin{array}{cc}mathbf{4} & mathbf{- 1} \ mathbf{1} & mathbf{3}end{array}right| )12
529( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r}+mathbf{1}end{array}right| ) is equal to
A ( cdot q-p )
B. ( q+p )
( c cdot q )
( D )
12
530Find the values of ( x ), if ( left|begin{array}{cc}boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{1} \ boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{2}end{array}right|=left|begin{array}{cc}mathbf{4} & mathbf{- 1} \ mathbf{1} & mathbf{3}end{array}right| )12
531If ( boldsymbol{A}=left[begin{array}{lll}boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{a}end{array}right], ) then the value of
( |boldsymbol{A}||mathbf{A} mathbf{d} mathbf{j} boldsymbol{A}| )
A ( cdot a^{3} )
в. ( a^{6} )
( c cdot a^{9} )
( D cdot a^{2} )
12
532Find the values of the following determinant, where ( boldsymbol{i}=sqrt{-1} ) ( left|begin{array}{cc}mathbf{2} boldsymbol{i} & -mathbf{3} i \ boldsymbol{i}^{mathbf{3}} & -mathbf{2} boldsymbol{i}^{5}end{array}right| )12
53323. Let @ =-
23. Let o

+
, then the value of the det
.
, then the value of the det.
1
-1-0-
(2002 – 2 Marks
(a) 30
© 36?
(b) 30(0-1)
(d) 30(1-w)
12
534The points ( (-a,-b),(0,0),(a, b) ) and ( left(a^{2},right. )
ab) are
A. collinear
B. vertices of a rectangle
c. vertices of a parallelogram
D. None of these
12
535The value of
( begin{array}{|ccc|}1 & cos alpha-sin alpha & cos alpha+sin alpha \ 1 & cos beta-sin beta & cos beta+sin beta \ 1 & cos gamma-sin gamma & cos gamma+sin gammaend{array} mid )
A. ( left|begin{array}{ccc}1 & cos alpha & sin alpha \ 1 & cos beta & sin beta \ 1 & cos gamma & sin gammaend{array}right| )
B. ( left|begin{array}{ccc}1 & cos alpha & sin alpha \ 1 & cos beta & sin beta \ 1 & cos gamma & sin gammaend{array}right| )
( mathbf{c} cdotleft|begin{array}{lll}cos alpha & cos beta & 1 \ cos beta & cos gamma & 1 \ cos gamma & cos alpha & 1end{array}right| )
D. ( left|begin{array}{ccc}1 & cos alpha & sin alpha \ 1 & cos beta & sin beta \ 1 & cos gamma & sin gammaend{array}right| )
12
536Match the entries of List – ( A ) and List ( -B )12
537be a square matrix all of whose entries are integers.
[2008]
then A- exists but all its entries are not
18. Let A be a square matrix all
Then which one of the following is true?
(a) If det A=+1, then A-1 exists but all its entri
necessarily integers
(b) If det A++1, then A-1 exists and all its entries are non
integers
(©) If det A = + 1, then A-1 exists but all its entries are
integers
(d) If det A=+1, then A-1 need not exists
12
538( f f(x)=left|begin{array}{ccc}sin x & 1 & 0 \ 1 & 2 sin x & 1 \ 0 & 1 & 2 sin xend{array}right| ) then
( int_{-pi / 2}^{pi / 2} f(x) d x ) equals to
( A )
B.
( c )
D. ( frac{3 pi}{2} )
12
539Find the value of the determinant:
[
left|begin{array}{ccc}
cos (theta+phi) & -sin (theta+phi) & cos 2 phi \
sin theta & cos theta & sin phi \
-cos theta & sin theta & cos phi
end{array}right|
]
12
540If ( A ) is a square matrix of order 3 with
( |A|=4, ) then write the value of ( |-2 A| )
12
541If the points ( (a, 0),(0, b) ) and (1,1) are collinear, then ( frac{1}{a}+frac{1}{b} ) equal to –
A . 1
B. 2
( c cdot 3 )
D. 4
12
542Assertion
Points ( boldsymbol{P}(-sin (boldsymbol{beta}- )
( boldsymbol{alpha}),-cos beta), boldsymbol{Q}(cos (boldsymbol{beta}-boldsymbol{alpha}), sin beta) ) and
( boldsymbol{R}(cos (boldsymbol{beta}-boldsymbol{alpha}+boldsymbol{theta}), sin (boldsymbol{beta}-boldsymbol{theta})), ) where
( beta=frac{pi}{4}+frac{alpha}{2} ) are non-collinear.
Reason
Three given points are non-collinear if they form a triangle of non-zero area.
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 the Reason is correct
12
543Let the three din
the three digit numbers A 28, 309, and 62 C, where A, B,
Care integers between 0 and 9, be divisible by a fixed
integer k Show that the determinant 8
12
9 C is divisible
B 2
(1990) – 4 Marks)
by ki
12
544If ( D=left|begin{array}{ccc}a^{2}+1 & a b & a c \ b a & b^{2}+1 & b c \ c a & c b & c^{2}+1end{array}right| ) then
( D= )
12
545( boldsymbol{A}=left[begin{array}{lll}mathbf{4} & -mathbf{2} & mathbf{5}end{array}right], boldsymbol{B}=left[begin{array}{l}mathbf{2} \ mathbf{0} \ mathbf{3}end{array}right], ) then
( boldsymbol{A} boldsymbol{d} boldsymbol{j}(boldsymbol{B} boldsymbol{A})= )
( A cdotleft{begin{array}{lll}0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{array}right} )
( mathbf{B} cdotleft{begin{array}{lll}8 & -4 & 10 \ 0 & 0 & 0 \ 12 & -6 & 15end{array}right} )
( mathbf{c} cdotleft{begin{array}{ccc}8 & 0 & 12 \ -4 & 0 & -6 \ 10 & 0 & 5end{array}right} )
D. None of the above
12
546If ( A ) and ( B ) are square matrices of order
3 such that ( |boldsymbol{A}|=-mathbf{1},|boldsymbol{B}|=mathbf{3}, ) then
( |mathbf{3} boldsymbol{A} boldsymbol{B}|= )
A . -9
B. -81
c. -27
D. 81
12
547( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{a}-boldsymbol{c} & boldsymbol{a}-boldsymbol{b} \ boldsymbol{b}-boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{b}-boldsymbol{a} \ boldsymbol{c}-boldsymbol{b} & boldsymbol{c}-boldsymbol{a} & boldsymbol{a}+boldsymbol{b}end{array}right|= )
A. ( 4 a b c )
B. ( 6 a b c )
( c .8 a b c )
D. ( 2 a b c )
12
548If ( Delta=left|begin{array}{lll}mathbf{5} & mathbf{3} & mathbf{8} \ mathbf{2} & mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3}end{array}right|, ) white the cofactor of
the element ( a_{32} )
12
549If ( boldsymbol{x}^{boldsymbol{a}} boldsymbol{y}^{boldsymbol{b}}=boldsymbol{e}^{boldsymbol{m}}, boldsymbol{x}^{boldsymbol{c}} boldsymbol{y}^{boldsymbol{d}}=boldsymbol{e}^{boldsymbol{n}}, Delta_{mathbf{1}}= )
( left|begin{array}{ll}boldsymbol{m} & boldsymbol{b} \ boldsymbol{n} & boldsymbol{d}end{array}right|, triangle_{2}=left|begin{array}{ll}boldsymbol{a} & boldsymbol{m} \ boldsymbol{c} & boldsymbol{n}end{array}right| ) and ( Delta_{boldsymbol{3}}=left|begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right| )
the value of ( x ) and ( y ) are respectively
A ( cdot frac{Delta_{1}}{Delta_{3}} ) and ( frac{Delta_{2}}{Delta_{3}} )
B. ( frac{Delta_{2}}{Delta_{1}} ) and ( frac{Delta_{3}}{Delta_{1}} )
( ^{mathbf{c}} cdot log left(frac{Delta_{1}}{Delta_{3}}right) a n d log left(frac{Delta_{2}}{Delta_{3}}right) )
( mathbf{D} cdot e^{Delta_{1} / Delta_{3}} ) and ( e^{Delta_{2} / Delta_{3}} )
12
550If ( A ) is a square matrix of order 3 and ( |boldsymbol{A}|=4 . ) Find the value of ( |mathbf{2 A}| )12
551|6i -3i 1
7. If 4 3i -1 = x+iy, then (1998 – 2 Marks)
en 20 3 il
(a) x=3, y=2
(b)x=1, y=3
(C) x=0, y=3
(d) x=0, y=0
12
552( fleft|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{b} boldsymbol{c} \ mathbf{1} & boldsymbol{b} & boldsymbol{c a} \ mathbf{1} & boldsymbol{c} & boldsymbol{a b}end{array}right|=boldsymbol{lambda}left|begin{array}{ccc}boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ mathbf{1} & boldsymbol{1} & boldsymbol{1}end{array}right|, ) then ( lambda ) is
equal to
( A cdot 1 )
B. -1
( c cdot 2 )
( D .-3 )
12
553Show that the area of the triangle on the Argand diagram
formed by the complex numbers z, iz and z+iz is — Iz.
12
554( mathbf{f} mathbf{Delta}=left|begin{array}{lll}boldsymbol{a}_{11} & boldsymbol{a}_{12} & boldsymbol{a}_{13} \ boldsymbol{a}_{21} & boldsymbol{a}_{22} & boldsymbol{a}_{23} \ boldsymbol{a}_{31} & boldsymbol{a}_{32} & boldsymbol{a}_{33}end{array}right| ) and ( boldsymbol{c}_{i j}= )
( (-1)^{i+j} ) (determinant obtained by
deleting ith row and jth column), then ( left|begin{array}{lll}c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \ c_{31} & c_{32} & c_{33}end{array}right|=Delta^{2} )
( left|begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{x}^{2} \ boldsymbol{x} & boldsymbol{x}^{2} & mathbf{1} \ boldsymbol{x}^{2} & boldsymbol{1} & boldsymbol{x}end{array}right|=mathbf{7} ) and ( boldsymbol{Delta}= )
( left|begin{array}{ccc}x^{3}-1 & 0 & x-x^{4} \ 0 & x-x^{4} & x^{3}-1 \ x-x^{4} & x^{3}-1 & 0end{array}right|, ) then
A ( . Delta=7 )
B. ( Delta=343 )
c. ( Delta=-49 )
D. ( Delta=49 )
12
555Prove that ( left[begin{array}{ccc}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}end{array}right]= )
( (a-b)(b-c)(c-a) )
12
556Let a, b, c be positive and not all equal. Show that the value
а ь с
of the determinant
b ca is negative
с а ь
12
557( fleft|begin{array}{lll}boldsymbol{x} & boldsymbol{2} & boldsymbol{8} \ boldsymbol{2} & boldsymbol{8} & boldsymbol{x} \ boldsymbol{8} & boldsymbol{x} & boldsymbol{2}end{array}right|=left|begin{array}{lll}boldsymbol{3} & boldsymbol{x} & boldsymbol{7} \ boldsymbol{x} & boldsymbol{7} & boldsymbol{3} \ boldsymbol{7} & boldsymbol{3} & boldsymbol{x}end{array}right|= )
( left|begin{array}{ccc}mathbf{5} & mathbf{5} & boldsymbol{x} \ mathbf{5} & boldsymbol{x} & mathbf{5} \ boldsymbol{x} & mathbf{5} & mathbf{5}end{array}right|=mathbf{0} ) then ( boldsymbol{x} ) is equal to
( mathbf{A} cdot mathbf{0} )
B. -10
( c .3 )
D. None of these
12
558Match the entries in column I with
column II
12
559if ( a, b, c ) are unequal what is the condition that the value of following determinat is zero ( delta=left|begin{array}{lll}a & a^{2} & a^{3}+1 \ b & b^{2} & b^{3}+1 \ c & c^{2} & c^{3}+1end{array}right| )
A ( .1+a b c=0 )
B. ( a+b+c+1=0 )
c. ( (a-b)(b-c)(c-a)=0 )
D. None of these
12
560Find the value of the determinant
without expansion ( left|begin{array}{ccc}b^{2}-a b & b-c & b c-a c \ a b-a^{2} & a-b & b^{2}-a b \ b c-a c & c-a & a b-a^{2}end{array}right| )
12
561Show that the points ( boldsymbol{A}(-mathbf{3}, mathbf{3}), boldsymbol{B}(mathbf{0}, mathbf{0}) )
( C(3,-3) ) are collinear
12
562( left|begin{array}{lll}boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c}end{array}right|=boldsymbol{K}left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right| )
( operatorname{then} K= )
A . 1
B. 2
( c )
( D )
12
563If ( a^{2}+b^{2}+c^{2}=-2 ) and ( f(x)= )
( left|begin{array}{ccc}mathbf{1}+boldsymbol{a}^{2} boldsymbol{x} & left(mathbf{1}+boldsymbol{b}^{2}right) boldsymbol{x} & left(mathbf{1}+boldsymbol{c}^{2}right) boldsymbol{x} \ left(mathbf{1}+boldsymbol{a}^{2}right) boldsymbol{x} & mathbf{1}+boldsymbol{b}^{2} boldsymbol{x} & left(mathbf{1}+boldsymbol{c}^{2}right) boldsymbol{x} \ left(mathbf{1}+boldsymbol{a}^{2}right) boldsymbol{x} & left(mathbf{1}+boldsymbol{b}^{2}right) boldsymbol{x} & mathbf{1}+boldsymbol{c}^{mathbf{2}} boldsymbol{x}end{array}right| ) then
( f(x) ) is a polynomial of degree
( A cdot )
B.
( c .3 )
( D )
12
564If ( boldsymbol{A}=left[begin{array}{cc}mathbf{0} & mathbf{1} \ -mathbf{1} & mathbf{0}end{array}right] ) then determinant of ( [boldsymbol{A}] )
is
( mathbf{A} cdot mathbf{1} )
B. –
c. 0
D.
12
565If the value of the determinant
( left|begin{array}{lll}boldsymbol{a} & boldsymbol{1} & boldsymbol{1} \ boldsymbol{1} & boldsymbol{b} & boldsymbol{1} \ boldsymbol{1} & boldsymbol{1} & boldsymbol{c}end{array}right| ) is positive, then
A ( . a b c>1 )
B . ( a b c>-8 )
c. ( a b c-2 )
12
566f ( A+B+C=pi ), then
( left|begin{array}{ccc}sin (boldsymbol{A}+boldsymbol{B}+boldsymbol{C}) & sin boldsymbol{B} & cos boldsymbol{C} \ -sin boldsymbol{B} & boldsymbol{0} & boldsymbol{t a n} boldsymbol{A} \ cos (boldsymbol{A}+boldsymbol{B}) & -boldsymbol{t a n} boldsymbol{A} & boldsymbol{0}end{array}right| )
equals
A .
B. 2 ( sin B tan A cos C )
( c )
D. none of these
12
567Find the value of ( K ) if the point ( A(2,3), B ) ( (4, K) ) and ( C(6,-3) ) are collinear?12
568For how many real values of ( ^{prime} m^{prime} ) the points ( boldsymbol{A}(boldsymbol{m}+mathbf{1}, mathbf{1}), boldsymbol{B}(boldsymbol{2 m}+mathbf{1}, boldsymbol{3}) ) and
( C(2 m+2,2 m) ) are collinear.
12
569Let ( mathbf{P} ) and ( mathbf{Q} ) be ( mathbf{3} times mathbf{3} ) matrices with
( boldsymbol{P} neq boldsymbol{Q} . ) If ( boldsymbol{P}^{3}=boldsymbol{Q}^{3} ) and ( boldsymbol{P}^{2} boldsymbol{Q}=boldsymbol{Q}^{2} boldsymbol{P} )
then determinant of ( left(P^{2}+Q^{2}right) ) is equal
to:
A . -2
B.
( c cdot 0 )
D. –
12
570( left|begin{array}{ccc}boldsymbol{x}^{boldsymbol{n}} & boldsymbol{x}^{boldsymbol{n}+mathbf{2}} & boldsymbol{x}^{boldsymbol{n}+mathbf{3}} \ boldsymbol{y}^{boldsymbol{n}} & boldsymbol{y}^{boldsymbol{n}+mathbf{2}} & boldsymbol{y}^{boldsymbol{n}+boldsymbol{3}} \ boldsymbol{z}^{boldsymbol{n}} & boldsymbol{z}^{boldsymbol{n}+mathbf{2}} & boldsymbol{z}^{boldsymbol{n}+mathbf{3}}end{array}right|=(boldsymbol{x}-boldsymbol{y})(boldsymbol{y}- )
( (z-x)left(frac{1}{x}+frac{1}{y}+frac{1}{z}right), ) then the value
of ( n ) is
( A )
B. –
( c cdot 1 )
( D )
12
571Find the integral value of ( x, ) if ( left|begin{array}{ccc}boldsymbol{x}^{2} & boldsymbol{x} & mathbf{1} \ mathbf{0} & boldsymbol{2} & mathbf{1} \ boldsymbol{3} & boldsymbol{1} & boldsymbol{4}end{array}right|=mathbf{2} mathbf{8} )12
572The value of ( left|begin{array}{llll}boldsymbol{p} & boldsymbol{0} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{a} & boldsymbol{q} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{r} & boldsymbol{0} \ boldsymbol{d} & boldsymbol{e} & boldsymbol{f} & boldsymbol{s}end{array}right| ) is
( A cdot p+q+r+s )
B.
( c cdot a b+c d+e f )
D. pqrs
12
573Prove that:
[
left|begin{array}{ccc}
boldsymbol{a}-boldsymbol{b}-boldsymbol{c} & boldsymbol{2} boldsymbol{a} & boldsymbol{2} boldsymbol{a} \
boldsymbol{2} boldsymbol{b} & boldsymbol{b}-boldsymbol{c}-boldsymbol{a} & boldsymbol{2} boldsymbol{b} \
boldsymbol{2} boldsymbol{c} & boldsymbol{2} boldsymbol{c} & boldsymbol{c}-boldsymbol{a}-boldsymbol{b}
end{array}right|=
]
( (a+b+c)^{3} )
12
574The coefficient of ( x^{2} ) in the expansion of
the determinant
( left|begin{array}{ccc}x^{2} & x^{3}+1 & x^{5}+2 \ x^{3}+3 & x^{2}+x & x^{3}+x^{4} \ x+4 & x^{3}+x^{4} & 2^{3}end{array}right| ) is
A . -10
B. -8
( c .-2 )
D. – 6
E.
12
575The characteristic equation of a matrix ( A ) is ( lambda^{3}-5 lambda^{2}-3 lambda+2 I=0 ) then
( |boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}|= )
( mathbf{A} cdot mathbf{4} )
B . 25
c. 9
D. 30
12
576If ( P=left[begin{array}{lll}1 & c & 3 \ 1 & 3 & 3 \ 2 & 4 & 4end{array}right] ) is the adjoint of a
( 3 times 3 ) matrix ( Q ) and ( operatorname{det} .(Q)=4, ) then ( c ) is
equal to.
( mathbf{A} cdot mathbf{0} )
B. 4
c. 5
D. 11
12
577( left|begin{array}{cc}sin ^{2} theta & cos ^{2} theta \ -cos ^{2} theta & sin ^{2} thetaend{array}right|= )
( mathbf{A} cdot cos 2 theta )
в. ( frac{1}{2}left(1+cos ^{2} 2 thetaright) )
c. ( frac{1}{2}left(1-sin ^{2} 2 thetaright) )
D. ( frac{1}{2} sin ^{2} 2 theta )
12
578( mathbf{f} mathbf{f}_{mathbf{3}}=left[begin{array}{ccc}mathbf{0} & mathbf{1} & mathbf{- 1} \ mathbf{2} & mathbf{1} & mathbf{3} \ mathbf{3} & mathbf{2} & mathbf{1}end{array}right], ) then
( left[boldsymbol{A}(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}) boldsymbol{A}^{-1}right] boldsymbol{A}= )
( mathbf{A} cdotleft[begin{array}{lll}6 & 0 & 0 \ 0 & 6 & 0 \ 0 & 0 & 6end{array}right] )
( mathbf{B} cdotleft[begin{array}{lll}4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4end{array}right] )
( mathbf{C} cdotleft[begin{array}{lll}2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2end{array}right] )
D. ( I )
12
579Find the value of ( lambda ) if the following equations are consistent ( boldsymbol{x}+boldsymbol{y}-mathbf{3}=mathbf{0} )
( (1+lambda) x+(2+lambda) y-8=0 )
( boldsymbol{x}-(mathbf{1}+boldsymbol{lambda}) boldsymbol{y}+(boldsymbol{2}+boldsymbol{lambda})=mathbf{0} )
12
58018.
sinx cos x cos x
sin x cos xl
The number of distinct real roots of COS X
cos x cos x sin x
VI
s
is
= 0 in the interval
(a) 0 (6) 2
(20015)
(d) 3
(
1
12
581The value of ( 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| ) is
A . a perfect cube
B. zero
c. a perfect square
D. negative
12
582Evaluate ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{x}^{2} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{y}^{2} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{z}^{3}end{array}right| )12
583f ( p lambda^{4}+p lambda^{3}+p lambda^{2}+s lambda+t= )
( left|begin{array}{ccc}lambda^{2}+3 lambda & lambda+1 & lambda+3 \ lambda+1 & 2-lambda & lambda-4 \ lambda-3 & lambda+4 & 3 lambdaend{array}right|, ) then value
of t is
A . 16
B . 18
( c .17 )
D. 19
12
584The value of determinant ( left|begin{array}{lll}1 / a & b c & a^{3} \ 1 / b & c a & b^{3} \ 1 / c & a b & c^{3}end{array}right| )
A ( cdot a^{2} b^{2} c^{2}(a-b)(b-c)(c-a) )
B. 0
c ( cdot(a-b)(b-c)(c-a) )
D. None of the above
12
585Solve ( left.boldsymbol{D}=mid begin{array}{ccc}mathbf{1} & -mathbf{2} & mathbf{1} \ mathbf{2} & mathbf{1} & -mathbf{1} \ mathbf{1} & mathbf{3} & mathbf{1}end{array}right] )12
586f ( triangle_{1}= )
( mid begin{array}{ccc}a_{1}^{2}+b_{1}+c_{1} & a_{1} a_{2}+b_{2}+c_{2} & a_{1} a_{3}+c \ b_{1} b_{2}+c_{1} & b_{2}^{2}+c_{2} & b_{2} b_{3} \ c_{3} c_{1} & c_{3} c_{2} & c_{3}^{2}end{array} )
and ( triangle_{2}=left|begin{array}{lll}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}end{array}right|, ) then ( frac{triangle_{1}}{triangle_{2}} ) is
equal to
A ( cdot a_{1} b_{2} c_{3} )
B. ( a_{1} a_{2} a_{3} )
( c cdot a_{3} b_{2} c )
D. ( a_{1} b_{1} c_{1}+a_{2} b_{2} c_{2}+a_{3} b_{3} c_{3} )
12
587In each of the following find the value of
( k, ) for which the points are collinear.
(ii) ( (8,1),(k,-4),(2,-5) )
12
588( (k, k),(2,3) ) and (4,-1) are collimear
So find the value of ( k )
12
589( mathbf{f}_{mathbf{1}}=left|begin{array}{ccc}mathbf{7} & boldsymbol{x} & mathbf{2} \ -mathbf{5} & boldsymbol{x}+mathbf{1} & mathbf{3} \ mathbf{4} & boldsymbol{x} & mathbf{7}end{array}right|, boldsymbol{Delta}_{2}= )
( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{2} & boldsymbol{7} \ boldsymbol{x}+boldsymbol{1} & boldsymbol{3} & -boldsymbol{5} \ boldsymbol{x} & boldsymbol{7} & boldsymbol{4}end{array}right| ) then ( boldsymbol{Delta}_{1}-boldsymbol{Delta}_{2}=boldsymbol{0} ) for
( mathbf{A} cdot x=2 )
B. all real ( x )
c. ( x=0 )
D. none of these
12
590Find ( left|begin{array}{ll}cos alpha & -sin alpha \ sin alpha & cos alphaend{array}right| )12
591Ta 1 o7 [a 1 17 [f]
16. If A= 1 b d ,B= 0 d c ,U = 8 V = 0 ,x = y
1 b c f g h [h lol
and AX = U has infinitely many solutions, prove that
BX=V has no unique solution. Also show that if afd + 0,
then BX= V has no solution.
(2004 – 4 Marks)
12
592( f(x)=left|begin{array}{ccc}x & cos x & e^{x^{2}} \ sin x & x^{2} & sec x \ tan x & 1 & 2end{array}right| . ) Find
( boldsymbol{f}(mathbf{0}) ? )
( mathbf{A} cdot mathbf{0} )
( B )
( c .-r )
D. ( 2 pi )
12
593Using properties of determinant, prove
the following:
[
mid begin{array}{ccc}
1+a^{2}-b^{2} & 2 a b & -2 b \
2 a b & 1-a^{2}+b^{2} & 2 a \
2 b & -2 a & 1-a^{2}-b^{2}
end{array}
]
( left(1+a^{2}+b^{2}right)^{3} )
12
594Without expanding a determinant at any stage, show that
x²+x x+1 x-2
2×2 + 3x -1 3x 3x – 3) = xA+B , where A and B are
x2 + 2x + 3 2x-1 2x – 1
12
5956.
Let A=
(1
2
(1
-1
1
1
1)
( 4
-3. and 10 B = -5
1
(1
2
0

2)
a . If B is
3)
1
the inverseof matrix A, then a is
(a) 5 (6) 1 (c) 2
[2004]
(d) 2
12
596If ( a_{k}, b_{k}, c_{k} epsilon R ) for ( k=1,2,3 ) and
( left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{b}_{1} & boldsymbol{c}_{1} \ boldsymbol{a}_{2} & boldsymbol{b}_{2} & boldsymbol{c}_{2} \ boldsymbol{a}_{3} & boldsymbol{b}_{3} & boldsymbol{c}_{3}end{array}right| )
then sum of the roots of the equation ( left|begin{array}{lll}a_{1}+i b_{1} x & i a_{1} x+b_{1} & c_{1} \ a_{2}+i b_{2} x & i a_{2} x+b_{2} & c_{2} \ a_{3}+i b_{3} x & i a_{3} x+b_{3} & c_{3}end{array}right| )
is
( mathbf{A} cdot a_{1}+a_{2}+a_{3} )
B. ( b_{1}+b_{2}+b_{3} )
( mathbf{c} cdot a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3} )
D. none of these
12
597Solve the determinant ( left|begin{array}{ll}boldsymbol{x} & boldsymbol{y} \ -boldsymbol{y} & boldsymbol{x}end{array}right| )12
598Write the minors and cofactors of each
element of the first column of the
following matrices ( boldsymbol{A}=left[begin{array}{lll}1 & a & b c \ 1 & b & c a \ 1 & c & a bend{array}right] )
12
599( mid begin{array}{ccc}mathbf{2} boldsymbol{a}_{1} boldsymbol{b}_{1} & boldsymbol{a}_{1} boldsymbol{b}_{1}+boldsymbol{a}_{2} boldsymbol{b}_{1} & boldsymbol{a}_{1} boldsymbol{b}_{3}+boldsymbol{a}_{3} boldsymbol{b}_{1} \ boldsymbol{a}_{1} boldsymbol{b}_{2}+boldsymbol{a}_{2} boldsymbol{b}_{1} & boldsymbol{2} boldsymbol{a}_{2} boldsymbol{b}_{2} & boldsymbol{a}_{2} boldsymbol{b}_{3}+boldsymbol{a}_{3} boldsymbol{b}_{2} \ boldsymbol{a}_{1} boldsymbol{b}_{3}+boldsymbol{a}_{3} boldsymbol{b}_{1} & boldsymbol{a}_{3} boldsymbol{b}_{2}+boldsymbol{b}_{3} boldsymbol{a}_{2} & boldsymbol{2} boldsymbol{a}_{3} boldsymbol{b}_{3}end{array} )
( A )
B.
( mathbf{c} cdot a_{1} b_{1} a_{2} b )
D. ( a_{1} b_{1}+a_{2} b_{2} )
12
600If ( boldsymbol{A}=left|begin{array}{ll}mathbf{1 0} & mathbf{2} \ mathbf{3 0} & mathbf{6}end{array}right| )
then ( |boldsymbol{A}|= )
( A cdot O )
B. 10
( c cdot 12 )
D. 60
12
601( 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
602If every element of third order determinant of ( Delta ) is multiplied by 5 then value of new determinant equals
to,
A ( . Delta )
B. ( 5 Delta )
c. ( 25 Delta )
D. ( 125 Delta )
12
60327. Let P and Q be 3 x 3 matrices P+Q. If P3= Q and P20 =
Q2P then determinant of (P2 +Q) is equal to: [2012]
(a) -2 (6) 1 (c) o (d) -1.
12
604Expand: ( left|begin{array}{ccc}1 & -7 & 3 \ 5 & -6 & 0 \ 1 & 2 & -3end{array}right| )12
605Evaluate :
( Delta=left|begin{array}{ccc}0 & sin alpha & -cos alpha \ -sin alpha & 0 & sin beta \ cos alpha & -sin beta & 0end{array}right| )
12
606( left|begin{array}{ccc}boldsymbol{a}+boldsymbol{b} & boldsymbol{a}+mathbf{2} boldsymbol{b} & boldsymbol{a}+boldsymbol{3} boldsymbol{b} \ boldsymbol{a}+mathbf{2} boldsymbol{b} & boldsymbol{a}+boldsymbol{3} boldsymbol{b} & boldsymbol{a}+boldsymbol{4} boldsymbol{b} \ boldsymbol{a}+boldsymbol{4} boldsymbol{b} & boldsymbol{a}+boldsymbol{5} boldsymbol{b} & boldsymbol{a}+boldsymbol{6} boldsymbol{b}end{array}right|= )
A ( cdot a^{2}+b^{2}+c^{2}-3 a b )
B. ( 3 a b c )
( c cdot 3 a+5 b )
( D )
12
607( mathrm{f}left|begin{array}{ccc}boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{y}+boldsymbol{z} \ mathbf{2} boldsymbol{x} & mathbf{3} boldsymbol{x}+mathbf{2} boldsymbol{y} & boldsymbol{4} boldsymbol{x}+boldsymbol{3} boldsymbol{y}+boldsymbol{2} boldsymbol{z} \ boldsymbol{3} boldsymbol{x} & boldsymbol{6} boldsymbol{x}+boldsymbol{3} boldsymbol{y} & boldsymbol{1} mathbf{0} boldsymbol{x}+boldsymbol{6} boldsymbol{y}+boldsymbol{3} boldsymbol{z}end{array}right|=mathbf{6} boldsymbol{4} )
then find ( x )
12
608If ( [mathbf{x}] ) stands greatest integer ( leq mathbf{x} ) then
the value of ( left|begin{array}{ccc}{[boldsymbol{e}]} & {[boldsymbol{pi}]} & {left[boldsymbol{pi}^{2}-boldsymbol{6}right]} \ {[boldsymbol{pi}]} & boldsymbol{pi}^{2}-boldsymbol{6} & {[boldsymbol{e}]} \ {left[boldsymbol{pi}^{2}-boldsymbol{6}right]} & {[boldsymbol{e}]} & {[boldsymbol{pi}]}end{array}right| ) equals
( A cdot-8 )
B. 8
( c cdot-1 )
( D )
12
609Calculate the values of the
determinants:
( left|begin{array}{cccc}mathbf{0} & mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & boldsymbol{b}+boldsymbol{c} & boldsymbol{a} & boldsymbol{a} \ mathbf{1} & boldsymbol{b} & boldsymbol{c}+boldsymbol{a} & boldsymbol{b} \ mathbf{1} & boldsymbol{c} & boldsymbol{c} & boldsymbol{a}+boldsymbol{b}end{array}right| )
12
610Let ( Delta= )
( left|begin{array}{ccc}sin theta cos phi & sin theta sin phi & cos theta \ cos theta cos phi & cos theta sin phi & -sin theta \ -sin theta sin phi & sin theta cos phi & 0end{array}right|, ) then
A. ( Delta ) is independent of ( theta )
B. ( Delta ) is independent of ( phi )
( mathrm{c} cdot Delta ) is a constant
D. none of these
12
61137. 14-
37. IfA=
), then adj(x2 + 124) is equal to
, then adj (3A2 + 12A) is equal to :
(JEE M 2017]
[ 72 -847
L-63 51
a)
-84
51
a [51 84]
() [84 72
12
612If ( boldsymbol{A}=left[begin{array}{cc}-mathbf{5} & mathbf{2} \ mathbf{1} & -mathbf{3}end{array}right], ) then adj ( mathbf{A} ) is equal to
A. ( left[begin{array}{ll}-3 & -2 \ -1 & -5end{array}right] )
(年) 0
В. ( left[begin{array}{cc}3 & -2 \ -1 & 5end{array}right] )
c. ( left[begin{array}{ll}5 & 1 \ 2 & 3end{array}right] )
D. ( left[begin{array}{ll}3 & 2 \ 1 & 5end{array}right] )
12
613Let ( boldsymbol{A}=left[begin{array}{cc}mathbf{5} & mathbf{8} \ mathbf{8} & mathbf{1 3}end{array}right] ) then show that ( boldsymbol{A} )
satisfies the equation ( x^{2}-18 x+1= )
( mathbf{0} )
12
614( operatorname{Let} boldsymbol{D}_{1}=left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c} & boldsymbol{d} & boldsymbol{c}+boldsymbol{d} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{a}-boldsymbol{b}end{array}right|, quad boldsymbol{D}_{2}= )
( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{c} & boldsymbol{a}+boldsymbol{c} \ boldsymbol{b} & boldsymbol{d} & boldsymbol{b}+boldsymbol{d} \ boldsymbol{a} & boldsymbol{c} & boldsymbol{a}+boldsymbol{b}+boldsymbol{c}end{array}right|, ) then the value of
( left|frac{boldsymbol{D}_{1}}{boldsymbol{D}_{2}}right|, ) where ( boldsymbol{b} neq mathbf{0} ) and ( boldsymbol{a} boldsymbol{d} neq boldsymbol{b} boldsymbol{c}, ) is
12
615f ( a, b, c ) are ( p ) th ( , q ) thand ( r ) th terms of a
( mathrm{GP}, ) then ( left|begin{array}{lll}log boldsymbol{a} & boldsymbol{p} & mathbf{1} \ log boldsymbol{b} & boldsymbol{q} & mathbf{1} \ log boldsymbol{c} & boldsymbol{r} & mathbf{1}end{array}right| ) is equal to
( mathbf{A} cdot mathbf{0} )
B.
c. ( log a b c )
D. none of these
12
616( f(a, b, c text { are in } A P ) then Prove that ( left|begin{array}{ccc}boldsymbol{x}+mathbf{2} & boldsymbol{x}+mathbf{3} & boldsymbol{x}+mathbf{2} boldsymbol{a} \ boldsymbol{x}+mathbf{3} & boldsymbol{x}+mathbf{4} & boldsymbol{x}+mathbf{2} boldsymbol{b} \ boldsymbol{x}+mathbf{4} & boldsymbol{x}+mathbf{5} & boldsymbol{x}+mathbf{2} boldsymbol{c}end{array}right|=mathbf{0} )12
617If ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{alpha} & boldsymbol{2} \ boldsymbol{2} & boldsymbol{alpha}end{array}right] ) and ( left|boldsymbol{A}^{3}right|=mathbf{1 2 5}, ) then the
value of ( alpha ) is
( A cdot pm 1 )
( B cdot pm 2 )
( c cdot pm 3 )
( D .pm 5 )
12
618Find ( (5 sqrt{3}+3 sqrt{2}) Delta ) where
( Delta=left|begin{array}{ccc}sqrt{13}+sqrt{3} & sqrt{5} & 2 sqrt{5} \ sqrt{15}+sqrt{26} & sqrt{10} & 5 \ 3+sqrt{65} & 5 & sqrt{15}end{array}right| )
12
619( f )
( a, b, c ) are all different and if ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{a}^{2} & mathbf{1}+boldsymbol{a}^{3} \ boldsymbol{b} & boldsymbol{b}^{2} & mathbf{1}+boldsymbol{b}^{3} \ boldsymbol{c} & boldsymbol{c}^{2} & boldsymbol{1}+boldsymbol{c}^{3}end{array}right|=mathbf{0} ) then ( -boldsymbol{a} boldsymbol{b} boldsymbol{c}= )
12
620If the entries in a ( 3 times 3 ) determinant are
either 0 or 1 , then the greatest value of their determinats is:
A
B. 2
( c cdot 3 )
D.
12
621Evaluate the following determinants
without expansion as far as possible. ( left|begin{array}{ccc}1 & b c & b c(b+c) \ 1 & c a & c a(c+a) \ 1 & a b & a b(a+b)end{array}right| )
12
622If ( left|begin{array}{lll}a & a^{3} & a^{4}-1 \ b & b^{3} & b^{4}-1 \ c & c^{3} & c^{4}-1end{array}right|=0 ) and ( a, b, c ) are all
distinct then ( a b c(a b+b c+c a) ) is equal
to
A ( . a+b+c )
в. ( a b c )
c. 0
D. none of these
12
623If ( A ) is a square matrix of order ( n ), then
( |mathbf{A} mathbf{d} mathbf{j} A| ) is
( mathbf{A} cdot|A|^{2} )
B cdot ( |A|^{n} )
C ( cdot|A|^{n-1} )
D・ |A|
12
624The value of determinant ( left|begin{array}{lll}mathbf{1} & boldsymbol{x} & boldsymbol{y}+boldsymbol{z} \ mathbf{1} & boldsymbol{y} & boldsymbol{z}+boldsymbol{x} \ mathbf{1} & boldsymbol{z} & boldsymbol{x}+boldsymbol{y}end{array}right| )
( mathbf{A} cdot mathbf{0} )
B. ( x+y+z )
c. ( 1+x+y+z )
D. ( (x-y)(y-z)(z-x) )
12
625Match the statements in Column I with
statements in column II
12
626Find the value of the determinant:
[
left|begin{array}{ccc}
cos (theta+phi) & -sin (theta+phi) & cos 2 phi \
sin theta & cos theta & sin phi \
-cos theta & sin theta & cos phi
end{array}right|
]
12
627If ( m ) and ( p ) are positive ( (m geq p) ) and
( boldsymbol{Delta}(boldsymbol{m}, boldsymbol{p})= )
( left|begin{array}{ccc}m & C_{p} & m \ m+1 & C_{p} \ m+2 & m+1 & C_{p+1} & m+1 \ m_{p} & m+2 & m_{p+2} \ & & m+2end{array}right| ) and
( m_{boldsymbol{p}}=mathbf{0} ) if ( mathbf{m}<mathbf{p}, ) then
This question has multiple correct options
A ( . Delta(2,1) / Delta(1,0)=4 )
B. ( Delta(4,3) / Delta(3,2)=2 )
( mathbf{c} cdot Delta(4,3) / Delta(2,1)=5 )
D. ( Delta(4,3) / Delta(1,0)=10 )
12
628Evaluate the following:
( left|begin{array}{ccc}1 & a & b c \ 1 & b & c a \ 1 & c & a bend{array}right| )
12
629Find the values of ( x, ) if
(i) ( left|begin{array}{ll}2 & 4 \ 5 & 1end{array}right|=left|begin{array}{ll}2 x & 4 \ 6 & xend{array}right| )
(ii) ( left|begin{array}{ll}2 & 3 \ 4 & 5end{array}right|= )
( left|begin{array}{ll}boldsymbol{x} & boldsymbol{3} \ boldsymbol{2} boldsymbol{x} & boldsymbol{5}end{array}right| )
12
630The value of
( A cdot O )
B. ( 30^{text {th }} )
( c cdot 30^{-x} )
D. None of these
12
631Assertion
Let ( p<0 ) and ( alpha_{1}, alpha_{2}, dots, alpha_{9} ) be the nine
roots of ( boldsymbol{x}^{mathbf{9}}=boldsymbol{p}, ) then
( boldsymbol{Delta}=left|begin{array}{lll}boldsymbol{alpha}_{1} & boldsymbol{alpha}_{2} & boldsymbol{alpha}_{3} \ boldsymbol{alpha}_{4} & boldsymbol{alpha}_{5} & boldsymbol{alpha}_{6} \ boldsymbol{alpha}_{4} & boldsymbol{alpha}_{8} & boldsymbol{alpha}_{9}end{array}right|=0 )
Reason
If two rows of a determinant are
identical, then determinant equals zero
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
632ut
7
Let a, b, c be any real numbers. Suppose that there are real
numbers x, y, z not all zero such that x=cy + bz, y=az + cx,
and z=bx + ay. Then a2 + b2 + c2 + 2abc is equal to
[2008]
(a) 2 (b) 1 (c) 0 (d) 1
12
633x – 4 2x 2x
38. If 2X X-4 2x
28 2X X-4
=(A+Bx)(x- A)2 , then the ordered
pair (A, B) is equal to :
(a) (-4, 3) (b) (-4,5)
[JEE M 2018]
(d) (-4,-5)
(c) (4,5)
12
634Let ( Delta_{mathrm{o}}=left[begin{array}{lll}a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33}end{array}right] ) and let ( Delta_{1} )
denote the determinant formed by the
cofactors of elements of ( Delta_{0} ) and ( Delta_{2} )
denote the determinant formed by the
cofactor of ( Delta_{1}, ) similarly ( Delta_{n} ) denotes the
determinant formed by the cofactors of
( Delta_{n-1} ) then the determinant value of ( Delta_{n} )
is
( mathbf{A} cdot Delta_{0}^{2 n} )
B . ( Delta_{0}^{2^{2}} )
( mathbf{C} cdot Delta_{0}^{n^{2}} )
D. ( Delta^{2}_{0} )
12
635Find the value of determinant, ( left|begin{array}{ccc}1 & a & a^{2}-b c \ 1 & b & b^{2}-c a \ 1 & c & c^{2}-a bend{array}right| )12
63614.
[1 o o [10 of
A= 0 1 1 and 1 = 0 1 0 and
To 2 4. Lo o 1
dI), then the value of c and d are
(20055)
(a) (6,-11) (b) (6,11) (C) (-6,11) (d) (6,-11)
12
637( mathbf{y}=sin mathbf{x}, boldsymbol{y}_{n}=frac{boldsymbol{d}^{n}(sin boldsymbol{x})}{boldsymbol{d} boldsymbol{x}^{n}} )
( operatorname{then}left|begin{array}{lll}boldsymbol{y} & boldsymbol{y}_{1} & boldsymbol{y}_{2} \ boldsymbol{y}_{3} & boldsymbol{y}_{4} & boldsymbol{y}_{5} \ boldsymbol{y}_{6} & boldsymbol{y}_{7} & boldsymbol{y}_{8}end{array}right|=? )
( A cdot-sin x )
B.
( c cdot sin x )
D. ( cos )
12
638If ( triangle(r)=left|begin{array}{cc}r & r^{3} \ 1 & n(n+1)end{array}right|, ) then ( sum_{r=1}^{n} triangle(r) )
is equal to
( ^{A} cdot sum_{i=1}^{n} r )
B. ( sum_{n=1}^{n} r )
( c cdot sum^{n} )
D. ( sum_{n=1}^{n} r^{4} )
12
639( begin{array}{c}text { If } boldsymbol{A}+boldsymbol{B}+boldsymbol{C}=boldsymbol{pi}, text { then } \ left|begin{array}{ccc}sin (boldsymbol{A}+boldsymbol{B}+boldsymbol{C}) & sin boldsymbol{B} & cos C \ -sin boldsymbol{0} & boldsymbol{operatorname { t a n }} boldsymbol{A}end{array}right|= \ cos (boldsymbol{A}+boldsymbol{B}) quad-boldsymbol{operatorname { t a n } boldsymbol { A }} & boldsymbol{0}end{array} mid )12
640If ( left|begin{array}{lll}a & a^{3} & a^{4}-1 \ b & b^{3} & b^{4}-1 \ c & c^{3} & c^{4}-1end{array}right|=0 ) and ( a, b, c ) are all
distinct, then ( a b c(a b+b c+c a) ) is
equal to
A. ( a+b+c )
B. abc
( c cdot 0 )
D. none of these
12
641( mathbf{f} mathbf{Delta}=left|begin{array}{cc}mathbf{f}(boldsymbol{x}) & boldsymbol{f}left(frac{mathbf{1}}{boldsymbol{x}}right)+boldsymbol{f}(boldsymbol{x}) \ mathbf{1} & boldsymbol{f}left(frac{mathbf{1}}{boldsymbol{x}}right)end{array}right|=mathbf{0} )
where;
( boldsymbol{f}(boldsymbol{x})=boldsymbol{a}+boldsymbol{b} boldsymbol{x}^{n} ) and ( boldsymbol{f}(boldsymbol{2})=mathbf{1 7}, ) then
( boldsymbol{f}(mathbf{5}) ) is:
A . 126
в. 326
( c .428 )
D. 626
12
642( fleft|begin{array}{ccc}boldsymbol{x}+mathbf{1} & mathbf{3} & mathbf{5} \ mathbf{2} & boldsymbol{x}+mathbf{2} & mathbf{5} \ mathbf{2} & mathbf{3} & boldsymbol{x}+mathbf{4}end{array}right|=mathbf{0}, ) then ( boldsymbol{x}=? )
A . 1,9
В. -1,9
c. -1,-9
D. 1,-9
12
643If ( A=left[begin{array}{ll}1 & 1 \ 0 & 1end{array}right], ) then ( operatorname{det}left(A+A^{2}+right. )
( left.boldsymbol{A}^{3}+boldsymbol{A}^{4}+boldsymbol{A}^{5}right) ) is
( mathbf{A} cdot mathbf{1} )
B. 32
c. 25
D. 30
12
644If the vectors ( vec{a}, vec{b}, vec{c} ) are coplanar, then the value of ( left|begin{array}{ccc}overrightarrow{boldsymbol{a}} & overrightarrow{boldsymbol{b}} & overrightarrow{boldsymbol{c}} \ overrightarrow{boldsymbol{a}} cdot overrightarrow{boldsymbol{a}} & overrightarrow{boldsymbol{a}} cdot overrightarrow{boldsymbol{b}} & overrightarrow{boldsymbol{a}} cdot overrightarrow{boldsymbol{c}} \ overrightarrow{boldsymbol{b}} cdot overrightarrow{boldsymbol{a}} & overrightarrow{boldsymbol{b}} cdot overrightarrow{boldsymbol{b}} & overrightarrow{boldsymbol{b}} cdot overrightarrow{boldsymbol{c}}end{array}right|= )
( mathbf{A} cdot mathbf{1} )
B.
( c cdot-1 )
D. ( vec{a}+vec{b}+vec{c} )
12
645What is the value of ( y ) if ( (y, 3),(-5,6) )
and (-8,8) are collinear?
A . -1
B . 2
c. ( frac{1}{2} )
D. ( -frac{1}{2} )
12
646( fleft(a_{1}, a_{2}, a_{3}, dots, a_{n}, dots a r e text { in } G P ) then the right.
determinant ( Delta= )
( left|begin{array}{ccc}log a_{n} & log a_{n+1} & log a_{n+2} \ log a_{n+3} & log a_{n+4} & log a_{n+5} \ log a_{n+6} & log a_{n+7} & log a_{n+8}end{array}right| ) is equal
to
A . 0
B. 1
( c cdot 2 )
D.
12
647If ( A ) is an invertible matrix of order ( n )
then the determinant of adj ( A ) is equal
to:
( mathbf{A} cdot|A|^{n} )
В ( cdot|A|^{n+1} )
c. ( |A|^{n-1} )
D. ( |A|^{n+2} )
12
648The number of distinct real roots of the
[
text { guation, }left|begin{array}{ccc}
cos x & sin x & sin x \
sin x & cos x & sin x \
sin x & sin x & cos x
end{array}right|=0
]
in the interval ( left[-frac{pi}{4}, frac{pi}{4}right] ) is/are
( A )
B . 2
( c cdot 1 )
( D )
12
649Suppose ( f(x) ) is a function satisfying the following conditions.
( (a) f(0)=2, f(1)=1 )
(b) ( f(x) ) has a maximum at ( x=5 / 2, ) and
(c) for all ( x in R )
( boldsymbol{f}^{prime}(boldsymbol{x})= )
( mid begin{array}{ccc}2 a x & 2 a x-1 & 2 a x+b+ \ b & b+1 & -1 \ 2(a x+b) & 2 a x+2 b+1 & 2 a x+bend{array} )
where ( a, b ) are constants, then
12
650The value of the determinant
( left|begin{array}{ccc}b^{2}-a b & b-c & b c-a c \ a b-a^{2} & a-b & b^{2}-a b \ b c-a c & c-a & a b-a^{2}end{array}right| )
A ( . a b c )
B. ( a+b+c )
( c )
( mathbf{D} cdot a b+b c+c a )
12
651( operatorname{Det}left{begin{array}{ccc}1^{2} & 2^{2} & 3^{2} \ 2^{2} & 3^{2} & 4^{2} \ 3^{2} & 4^{2} & 5^{2}end{array}right}=dots )
A . -8
B. -7
( c .-6 )
D. ( -21 / 4 )
12
65224.
The number of values of k for which the linear equations
4x + ky + 2z=0, kx + 4y+z=0 and 2x +2y+z= 0 possess
a non-zero solution is
[2011]
(a) 2 (b) 1 (C) zero (d) 3
12
653Find the value of ( x ) if ( left|begin{array}{ccc}3 & 5 & x \ 2 & 4 & 1 \ -1 & 2 & 3end{array}right|=0 )12
654( left|begin{array}{ccc}frac{1}{a} & a^{2} & b c \ frac{1}{b} & b^{2} & c a \ frac{1}{c} & c^{2} & a bend{array}right|= )
A ( cdot ) abc
B. ( a+b+c )
( c cdot 0 )
D. ( 4 a b c )
12
655When the determinant ( left|begin{array}{lll}cos 2 x & sin ^{2} x & cos 4 x \ sin ^{2} x & cos 2 x & cos ^{2} x \ cos 4 x & cos ^{2} x & cos 2 xend{array}right| ) is expanded in
powers of ( sin x, ) then the constant term in that expression is
( A )
B.
( c cdot-1 )
D.
12
656( f e^{i theta}=cos theta+i sin theta, ) find the value of
( -left|begin{array}{ccc}mathbf{1} & boldsymbol{e}^{i boldsymbol{pi} / mathbf{3}} & boldsymbol{e}^{i boldsymbol{pi} / mathbf{4}} \ boldsymbol{e}^{-i boldsymbol{pi} / mathbf{3}} & mathbf{1} & boldsymbol{e}^{boldsymbol{i} mathbf{2} boldsymbol{pi} / mathbf{3}} \ boldsymbol{e}^{-boldsymbol{i} boldsymbol{pi} / mathbf{4}} & boldsymbol{e}^{-boldsymbol{i} mathbf{2} boldsymbol{pi} / mathbf{3}} & mathbf{1}end{array}right|-mathbf{2}^{frac{1}{2}} )
12
657( f(a+b+c=0, ) then one root of ( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{x} & boldsymbol{c} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{b}-boldsymbol{x} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{a} & boldsymbol{c}-boldsymbol{x}end{array}right|=0 ) is
( A cdot a+b )
B.
( c cdot b+c )
D. atca
12
658If ( D_{P}=left|begin{array}{ccc}boldsymbol{P} & mathbf{1 5} & mathbf{8} \ boldsymbol{P}^{2} & mathbf{3 5} & mathbf{9} \ boldsymbol{P}^{3} & mathbf{2 5} & mathbf{1 0}end{array}right|, ) then ( boldsymbol{D}_{1}+ )
( D_{2}+D_{3}+D_{4}+D_{5} ) is equal to
A. -29000
в. -25000
c. 25000
D. none of these
12
65919. Let A be a 2 x 2 matrix
Statement-1: adj (adj A)=A
[2009]
Statement-2: (adj A F|A|
(a) Statement-1 is true, Statement-2 is true.
Statement-2 is 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
660Find the value of the following determinants. ( left|begin{array}{cc}mathbf{5} & mathbf{3} \ -mathbf{7} & mathbf{0}end{array}right| )12
661The value of ( triangle= )
( begin{array}{cc}mathbf{2} & boldsymbol{a}+boldsymbol{r}+mathbf{2} \ boldsymbol{a}+boldsymbol{r}+mathbf{2} & mathbf{2}(boldsymbol{a}+mathbf{1})(boldsymbol{r}+mathbf{1}) \ boldsymbol{a}+boldsymbol{r} & boldsymbol{a}(boldsymbol{r}+mathbf{1})+boldsymbol{r}(boldsymbol{a}+mathbf{1})end{array} )
( A )
в. ( -2 a(r+1) )
c. ( a(a r+r+a) )
D. -1
12
662Find equation of line joining (1,2) and
(3,6) using determinants.
12
663( mathbf{f} boldsymbol{A}=left|begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{1} \ mathbf{0} & mathbf{2} & -mathbf{3} \ mathbf{2} & mathbf{1} & mathbf{0}end{array}right| ) and ( boldsymbol{B}=(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}) )
and ( C=5 A, ) then ( frac{|a d j B|}{|C|} ) is?
( mathbf{A} cdot mathbf{5} )
B . 25
c. -1
D.
12
664, then prove that
Illustration 3.84 If x+y+z=
sin x sin y sin z
cosx cos y cos z = 0.
[cos x cos y cos z
12
6654.
[1 4 4
If the adjoint of a 3 x 3 matrix P is
2 1 7
, then the
[
113]
(2012)
possible value(s) of the determinant of Pis (are)
(a) 2 (6) -1 (c) 1 (d) 2
12
666Find the value of the determinant
( left|begin{array}{cc}mathbf{2} boldsymbol{i} & -mathbf{3} boldsymbol{i} \ boldsymbol{i}^{3} & -mathbf{2} boldsymbol{i}end{array}right| ) where ( boldsymbol{i}=sqrt{-boldsymbol{i}} )
12
667Evaluate ( mid begin{array}{ccc}cos left(x+x^{2}right) & sin left(x+x^{2}right) & -cos (x+1) \ sin left(x-x^{2}right) & cos left(x-x^{2}right) & sin (x-x) \ sin 2 x & 0 & sin 2 x^{2}end{array} )
A ( cdot sin left(2 x+2 x^{2}right) )
B. ( -sin left(2 x+2 x^{2}right) )
( c cdot cos left(2 x+2 x^{2}right) )
D. ( -cos left(2 x+2 x^{2}right) )
12
668If the system of linear equations
( 2 x+2 a y+a z=0 )
( 2 x+3 b y+b z=0 )
( 2 x+4 c y+c z=0 )
where ( a, b, c epsilon R ) are non-zero and
distinct; has a non-zero solution, then:
A ( cdot frac{1}{a}, frac{1}{b}, frac{1}{c} ) are in A.P
B . ( a+b+c=0 )
c. ( a, b, c ) are in A.P
( mathbf{D} cdot a, b, c ) are in G.P
12
669If ( A ) is an ( n times n ) non-singular matrix,
then ( |boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A}| ) is:
( mathbf{A} cdot|A|^{n} )
B ( cdot|A|^{n+1} )
c. ( |A|^{n-1} )
D. ( |A|^{n-2} )
12
670( mathbf{f} mathbf{a}^{2}+boldsymbol{b}^{2}+boldsymbol{c}^{2}=-mathbf{2} ) and ( boldsymbol{f}(boldsymbol{x})= )
( left|begin{array}{ccc}mathbf{1}+boldsymbol{a}^{2} boldsymbol{x} & left(mathbf{1}+boldsymbol{b}^{2}right) boldsymbol{x} & left(mathbf{1}+boldsymbol{c}^{2}right) boldsymbol{x} \ left(mathbf{1}+boldsymbol{a}^{2}right) boldsymbol{x} & mathbf{1}+boldsymbol{b}^{2} boldsymbol{x} & left(mathbf{1}+boldsymbol{c}^{2}right) boldsymbol{x} \ left(mathbf{1}+boldsymbol{a}^{2}right) boldsymbol{x} & left(mathbf{1}+boldsymbol{b}^{2}right) boldsymbol{x} & mathbf{1}+boldsymbol{c}^{2} boldsymbol{x}end{array}right| )
Then, ( f(x) ) is a polynomial of degree
A .2
B. 3
( c .4 )
( D )
12
671If 3 points ( A(1, a, b), B(a, 2, b), C(a, b, 3) ) are
collinear, then find ( a+b=? )
12
672Consider the following statements:
1. Determinant is a square matrix.
2. Determinant is a number associated
with a square matrix. Which of the above statements is/are
correct?
A . 1 only
B. 2 only
c. Both 1 and 2
D. Neither 1 nor 2
12
673Solve:
( left|begin{array}{ccc}mathbf{0} & -mathbf{3} & boldsymbol{x} \ boldsymbol{x}+mathbf{1} & mathbf{3} & mathbf{1} \ mathbf{4} & mathbf{1} & mathbf{5}end{array}right|=mathbf{0} )
12
674If the determinant ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{a} boldsymbol{t}-boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b} boldsymbol{t}-boldsymbol{c} \ boldsymbol{2} & boldsymbol{1} & boldsymbol{0}end{array}right|=0 )
if ( a, b, c ) are in
A. ( A . P )
в. G.Р.
c. ( H . P )
D. ( k=1 / 2 )
12
675( operatorname{Show}left|begin{array}{ccc}boldsymbol{a x} & boldsymbol{b y} & boldsymbol{c z} \ boldsymbol{x}^{2} & boldsymbol{y}^{2} & boldsymbol{z}^{2} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right|=left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{y} boldsymbol{z} & boldsymbol{z} boldsymbol{x} & boldsymbol{x} boldsymbol{y}end{array}right| )12
676( left(begin{array}{lll}mathbf{7} & mathbf{1} & mathbf{5} \ mathbf{8} & mathbf{0} & mathbf{0}end{array}right)left(begin{array}{l}mathbf{2} \ mathbf{3} \ mathbf{1}end{array}right)+mathbf{5}left(begin{array}{l}mathbf{1} \ mathbf{0}end{array}right) ) is equal to
A ( cdotleft(begin{array}{l}16 \ 27end{array}right. )
B. ( left(begin{array}{l}27 \ 16end{array}right. )
c. ( left(begin{array}{l}15 \ 16end{array}right. )
D. ( left(begin{array}{l}16 \ 15end{array}right. )
12
677Calculate the values of the
determinants:
( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{c}+boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{c} & boldsymbol{a}+boldsymbol{b}end{array}right| )
12
678Evaluate the determinant :
( left|begin{array}{ccc}mathbf{1}+boldsymbol{a} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{1}+boldsymbol{a} & mathbf{0} \ mathbf{0} & mathbf{0} & mathbf{1}+boldsymbol{a}end{array}right| )
12
679( left|begin{array}{ccc}(boldsymbol{b}+boldsymbol{c})^{2} & boldsymbol{a}^{2} & boldsymbol{a}^{2} \ boldsymbol{b}^{2} & (boldsymbol{c}+boldsymbol{a})^{2} & boldsymbol{b}^{2} \ boldsymbol{c}^{2} & boldsymbol{c}^{2} & (boldsymbol{a}+boldsymbol{b})^{2}end{array}right| ) is equal
to
( mathbf{A} cdot a b c(a+b+c)^{2} )
B ( cdot 2 a b c(a+b+c)^{2} )
( mathbf{c} cdot 2 a b c(a+b+c)^{3} )
( mathbf{D} cdot 2 a b c(a+b+c) )
12
680Prove the following identities :
[
begin{array}{ccc}
boldsymbol{a}+boldsymbol{b}+mathbf{2 c} & boldsymbol{a} & boldsymbol{b} \
boldsymbol{c} & boldsymbol{b}+boldsymbol{c}+mathbf{2} boldsymbol{a} & boldsymbol{b} \
boldsymbol{c} & boldsymbol{a} & boldsymbol{c}+boldsymbol{a}+boldsymbol{2} boldsymbol{b}
end{array} mid=
]
( mathbf{2}(boldsymbol{a}+boldsymbol{b}+boldsymbol{c})^{3} )
12
681Prove that: ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b} & boldsymbol{c}end{array}right|=(boldsymbol{a}+boldsymbol{b}+ )
( c)(a-c)^{2} )
12
682Suppose ( a, b, c epsilon R ) and ( a b c=1 . ) If ( A= ) ( left[begin{array}{ccc}2 a & b & c \ b & 2 c & a \ c & a & 2 bend{array}right] ) is such that ( |A|left|A^{prime}right|=64 )
and ( |A|>0, ) then find the value of
( left(a^{3}+b^{3}+c^{3}right)^{4} )
( left(A^{prime} text { denotes transpose of a matrix } A .right) )
A . 1
B. 2
( c .3 )
( D )
12
683Find the integral value of ( x, ) if ( left|begin{array}{ccc}boldsymbol{x}^{2} & boldsymbol{x} & mathbf{1} \ mathbf{0} & boldsymbol{2} & mathbf{1} \ boldsymbol{3} & boldsymbol{1} & boldsymbol{4}end{array}right|=mathbf{2} mathbf{8} )12
684( mathbf{f}left|begin{array}{lll}boldsymbol{x}^{boldsymbol{k}} & boldsymbol{x}^{boldsymbol{k}+boldsymbol{2}} & boldsymbol{x}^{boldsymbol{k}+boldsymbol{3}} \ boldsymbol{y}^{boldsymbol{k}} & boldsymbol{y}^{boldsymbol{k}+boldsymbol{2}} & boldsymbol{y}^{boldsymbol{k}+boldsymbol{3}} \ boldsymbol{z}^{boldsymbol{k}} & boldsymbol{z}^{boldsymbol{k}+boldsymbol{2}} & boldsymbol{z}^{boldsymbol{k}+boldsymbol{3}}end{array}right| )
三室 ( (x-y)(y-z)(z-x)left(frac{1}{x}+frac{1}{y}+frac{1}{z}right) )
then
A. ( k=-3 )
B. ( k=-1 )
( mathbf{c} cdot k=1 )
D. ( k=3 )
12
685For positive numbers ( boldsymbol{x}, boldsymbol{y}, boldsymbol{z} ) the
numerical value of the determinant ( left|begin{array}{ccc}mathbf{1} & log _{x} boldsymbol{y} & log _{boldsymbol{x}} boldsymbol{z} \ log _{boldsymbol{y}} boldsymbol{x} & boldsymbol{3} & log _{boldsymbol{y}} boldsymbol{z} \ log _{boldsymbol{z}} boldsymbol{x} & log _{boldsymbol{z}} boldsymbol{y} & boldsymbol{5}end{array}right| ) is
( mathbf{A} cdot mathbf{0} )
B. ( log x log y log z )
( c .1 )
D. 8
12
686If ( I ) is the unit matrix of order ( n, ) where
( k neq 0 ) is a constant then ( operatorname{adj}(k I)= )
A ( cdot k^{n}(operatorname{adj} I) )
B. ( k(operatorname{adj} I) )
c. ( k^{2}(operatorname{adj} I) )
D. ( k^{n-1}(text { adj } I) )
12
687Find the value of: ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{x}^{2} & boldsymbol{y}^{2} & boldsymbol{z}^{2} \ boldsymbol{x}^{3} & boldsymbol{y}^{3} & boldsymbol{z}^{3}end{array}right| )
( mathbf{A} cdot mathbf{0} )
B. ( (x-y)(y-z)(z-x)(x y+y z+z x) )
C. ( (x+y x+z x)(x+y)(y+z)(z+x) )
D. ( (x+y+z) )
12
688If the area of the triangle formed by ( (0,0),(a, 0) ) and ( left(frac{1}{2}, aright) ) is equal to ( frac{1}{2} s q ) unit, then the values of ( a ) are :
( A ldots pm 2 )
B. ±3
( c .pm 1 )
D. ±4
( mathrm{E} cdot pm 5 )
12
689If ( boldsymbol{A}=left[begin{array}{rrr}3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1end{array}right] ) then find
( boldsymbol{A} boldsymbol{d} boldsymbol{j}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A}) )
A. ( left[begin{array}{lll}3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1end{array}right] )
В. ( left[begin{array}{ccc}3 & 3 & 4 \ 2 & -3 & -4 \ 0 & -1 & 1end{array}right] )
( begin{array}{llll}text { C. } & {left[begin{array}{ccc}3 & 3 & 4 \ 2 & -3 & 4 \ 0 & 1 & 1end{array}right]}end{array} )
D. ( left[begin{array}{ccc}3 & -3 & 4 \ 2 & -3 & -4 \ 0 & 1 & 1end{array}right] )
12
690A set of points which do not lie on the same line are called as
A. collinear
B. non-collinear
c. concurrent
D. square
12
691If ( boldsymbol{A}(boldsymbol{x})=left|begin{array}{ccc}boldsymbol{x}+mathbf{1} & mathbf{2} boldsymbol{x}+mathbf{1} & mathbf{3} boldsymbol{x}+mathbf{1} \ mathbf{2} boldsymbol{x}+mathbf{1} & mathbf{3} boldsymbol{x}+mathbf{1} & boldsymbol{x}+mathbf{1} \ boldsymbol{3} boldsymbol{x}+mathbf{1} & boldsymbol{x}+mathbf{1} & boldsymbol{2} boldsymbol{x}+mathbf{1}end{array}right| )
( operatorname{then} int_{0}^{1} boldsymbol{A}(boldsymbol{x}) boldsymbol{d} boldsymbol{x}= )
A . -15
( B cdot frac{-1}{2} )
( c .-30 )
( D )
12
692( fleft|begin{array}{ll}boldsymbol{x} & boldsymbol{y} \ boldsymbol{4} & boldsymbol{2}end{array}right|=mathbf{7} ) and ( left|begin{array}{ll}boldsymbol{2} & boldsymbol{3} \ boldsymbol{y} & boldsymbol{x}end{array}right|=boldsymbol{4} ) then
A ( cdot x=-3, y=-frac{5}{2} )
B. ( x=-frac{5}{2}, y=-3 )
c. ( _{x}=-3, y=frac{5}{2} )
D. ( x=-frac{5}{2}, y=3 )
12
6935.
The parameter, on which the value of the determinant
1
a
cos(p-d)x cos px
sin(p-d)x sin px
cos(p+d)x
sin(p+.d)x
does not depend
upon is
(a) a
(b) p
(c) d
(1997 – 2 Marks)
(d) x
12
694Evaluate the determinant :
( left|begin{array}{cc}cos theta & -sin theta \ sin theta & cos thetaend{array}right| )
12
695Prove the following:
[
left|begin{array}{ccc}
boldsymbol{a}+boldsymbol{b}+mathbf{2} boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \
boldsymbol{c} & boldsymbol{b}+boldsymbol{c}+mathbf{2} boldsymbol{a} & boldsymbol{b} \
boldsymbol{c} & boldsymbol{a} & boldsymbol{c}+boldsymbol{a}+mathbf{2} boldsymbol{b}
end{array}right|=
]
( mathbf{2}(boldsymbol{a}+boldsymbol{b}+boldsymbol{c})^{3} )
12
696f ( a, b, c ) are real numbers such that ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a}end{array}right|=0, ) then show
that either ( boldsymbol{a}+boldsymbol{b}+boldsymbol{c}=boldsymbol{0} ) or, ( boldsymbol{a}=boldsymbol{b}=boldsymbol{c} )
12
697Let ( omega=-frac{1}{2}+i frac{sqrt{3}}{2}, ) then the value of the
determinant
( left|begin{array}{ccc}1 & 1 & 1 \ 1 & -1-omega^{2} & omega^{2} \ 1 & omega^{2} & omega^{4}end{array}right|, ) is
This question has multiple correct options
( A .3 omega )
В. ( 3 omega(omega-1) )
( c cdot 3 omega^{2} )
D. ( 3(-2 omega-1) )
12
698Adj ( left(A d jleft[begin{array}{cc}2 & -3 \ 4 & 6end{array}right]right)= )
A ( cdotleft[begin{array}{cc}2 & -3 \ 4 & 6end{array}right] )
в. ( left[begin{array}{cc}6 & 3 \ -4 & 2end{array}right] )
с. ( left[begin{array}{cc}-6 & 3 \ -4 & -2end{array}right] )
D ( cdotleft[begin{array}{cc}-6 & -3 \ 4 & -2end{array}right] )
12
699If ( boldsymbol{A}=left|begin{array}{ll}mathbf{0} & mathbf{0} \ mathbf{1} & mathbf{1}end{array}right| ) then the value of ( boldsymbol{A}+ )
( boldsymbol{A}^{2}+boldsymbol{A}^{3}+ldots+boldsymbol{A}^{n}=? )
( A cdot A )
B. nA
c. ( (n+1) A )
( D )
12
700Prove the following:
( left|begin{array}{llll}boldsymbol{x} & boldsymbol{a} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{x} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a} & boldsymbol{x} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a} & boldsymbol{a} & boldsymbol{x}end{array}right|=(boldsymbol{x}+boldsymbol{3} boldsymbol{a})(boldsymbol{x}-boldsymbol{a})^{3} )
12
7012. The values of a lying between 0 = 0 and 0 = 7/2 and
satisfying the equation
1+sine cose 4sin 40
sin 1+cos e 4 sin 40 = 0 are
sin’e cosạe 1+4 sin 40
a. 77/24
& b. 57/24 08
c. 117/24
d. Tt/24 (IIT-JEE 1988)
12
702( left|begin{array}{ccc}cos C & tan A & 0 \ sin B & 0 & -tan A \ 0 & sin B & cos Cend{array}right| ) has the value
( mathbf{A} cdot mathbf{0} )
B.
( c cdot sin A sin B cos B )
D. none of these
12
703( mathbf{f} boldsymbol{x}, boldsymbol{y}, boldsymbol{z} in boldsymbol{R} & boldsymbol{Delta}= )
( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{y}+boldsymbol{z} \ mathbf{2} boldsymbol{x} & mathbf{5} boldsymbol{x}+mathbf{2} boldsymbol{y} & mathbf{7} boldsymbol{x}+mathbf{5} boldsymbol{y}+mathbf{2} z \ mathbf{3} boldsymbol{x} & mathbf{7} boldsymbol{x}+mathbf{3} boldsymbol{y} & mathbf{9} boldsymbol{x}+mathbf{7} boldsymbol{y}+mathbf{3} boldsymbol{z}end{array}right|=-mathbf{1} mathbf{6} )
then value of ( x ) is
A . -2
B. -3
( c cdot 2 )
( D )
12
704If ( A ) is a ( 3 * 3 ) singular matrix then
( boldsymbol{A}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A})= )
( A cdot ) Det ( A )
B.
( c cdot c )
D. ±1
12
705If ( A, B, C ) are the angles of a triangle,
then the value of determinant
( left|begin{array}{ccc}-1+cos B & cos C+cos B & cos B \ cos C+cos A & -1+cos A & cos A \ -1+cos B & -1+cos A & -1end{array}right| )
( mathbf{A} cdot mathbf{0} )
B.
( c .-1 )
( D )
12
70620.
Tf Pisa 3 x 3 matrix such that pl = 2P+1, where PT is the
transpose of P and I is the 3 x 3 identity matrix, then there
x] rol
exists a column matrix X = y 20 such that
z] [0]
(2012)
(b) PX=X
(a) PX=0
0
(c) PX= 2X
(d) PX=-X
12
707Minor ( m_{33} ) of the determinant
( left|begin{array}{ccc}2 & 3 & 5 \ 2 & -1 & 8 \ 1 & 2 & 4end{array}right| ) is
12
7082.
l, m, n are the pth, qth and th term of a G. P. all positive,
log 1 p 1
then log m q 1 equals
[2002]
log nr 1
(a) -1 (b) 2 (C) 1 (d) 0
12
709( mathbf{A}=left[begin{array}{ccc}-boldsymbol{q} boldsymbol{r} & boldsymbol{p}(boldsymbol{q}+boldsymbol{r}) & boldsymbol{p} boldsymbol{r}+boldsymbol{p} boldsymbol{q} \ boldsymbol{p} boldsymbol{q}+boldsymbol{q} boldsymbol{r} & -boldsymbol{p} boldsymbol{r} & boldsymbol{p} boldsymbol{q}+boldsymbol{q} boldsymbol{r} \ boldsymbol{q} boldsymbol{r}+boldsymbol{p} boldsymbol{r} & boldsymbol{q} boldsymbol{r}+boldsymbol{p} boldsymbol{r} & boldsymbol{-} boldsymbol{p} boldsymbol{q}end{array}right] )
then ( |boldsymbol{A}| ) equals:
( ^{A} cdotleft(sum p qright)^{2} )
( ^{mathrm{B}}left(sum p^{2} q^{2}right)^{2} )
( c cdotleft[sum(q r)^{3}right. )
( ^{mathrm{D}} cdotleft(sum p qright)^{frac{3}{2}} )
12
710Let ( a, b, c ) be real numbers with ( a^{2}+ )
( b^{2}+c^{2}=1 )
show that the equation ( mid begin{array}{ccc}boldsymbol{a} boldsymbol{x}-boldsymbol{b} boldsymbol{y}-boldsymbol{c} & boldsymbol{b} boldsymbol{x}+boldsymbol{a} boldsymbol{y} & boldsymbol{c} boldsymbol{x}+ \ boldsymbol{b} boldsymbol{x}+boldsymbol{a} boldsymbol{y} & boldsymbol{-} boldsymbol{a} boldsymbol{x}+boldsymbol{b} boldsymbol{y}-boldsymbol{c} & boldsymbol{c} boldsymbol{y}+ \ boldsymbol{c} boldsymbol{x}+boldsymbol{a} & boldsymbol{c} boldsymbol{y}+boldsymbol{b} & boldsymbol{-} boldsymbol{a} boldsymbol{x}-boldsymbol{b}end{array} )
represents a straight line.
12
711If ( boldsymbol{A}=left[begin{array}{ll}mathbf{3} & mathbf{2} \ mathbf{1} & mathbf{4}end{array}right], ) then ( boldsymbol{A}(boldsymbol{A} boldsymbol{d} boldsymbol{j} cdot boldsymbol{A}) ) equals-
A. ( left[begin{array}{cc}10 & 0 \ 0 & 10end{array}right] )
В. ( left[begin{array}{cc}0 & 10 \ 10 & 0end{array}right] )
c. ( left[begin{array}{ll}10 & 1 \ 1 & 10end{array}right] )
D. none of these
12
712( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{alpha} & boldsymbol{beta} & gamma \ boldsymbol{alpha} & boldsymbol{x}+boldsymbol{beta} & gamma \ boldsymbol{alpha} & boldsymbol{beta} & boldsymbol{x}+boldsymbol{gamma}end{array}right|=mathbf{0}, ) then ( boldsymbol{x} ) is
equal to
( mathbf{A} cdot 0,-(alpha+beta+gamma) )
B. ( 0, alpha+beta+gamma )
c. ( 1, alpha+beta+gamma )
D. ( 0, alpha^{2}+beta^{2}+gamma^{2} )
12
713Find ( left|begin{array}{lll}mathbf{4} & mathbf{6} & mathbf{2} \ mathbf{3} & mathbf{0} & mathbf{9} \ mathbf{7} & mathbf{1} & mathbf{5}end{array}right| )12
714( mathbf{f} mathbf{A}=left[begin{array}{lll}mathbf{1} & mathbf{5} & mathbf{- 6} \ mathbf{- 8} & mathbf{0} & mathbf{4} \ mathbf{3} & mathbf{- 7} & mathbf{2}end{array}right], ) then the
cofactor of ( -7= )
A . 44
B. 43
( c cdot 40 )
D. 39
12
715Let
( f(x)=left|begin{array}{ccc}sec ^{2} x & 1 & 1 \ cos ^{2} x & cos ^{2} x & operatorname{cosec}^{2} x \ 1 & cos ^{2} x & cot ^{2} xend{array}right| )
then
12
716Solve the following determinant :
( begin{array}{|ccc|}15 & 11 & 7 \ 11 & 17 & 14 \ 10 & 16 & 13end{array} )
12
717(2000 – TV US)
25. Let a, b, c be real numbers with a2 + b2 + c2= 1. Show that
the equation
Jax – by-c
bx + ay
cxta
b6+ay
ax + by-c
cytb
axta
cy+b 1=0
– ax – bytc.
represents a straight line
(20016 Marts)
12
718A matrix A of order ( 3 times 3 ) has
determinant 6. What is the value of ( |mathbf{3} boldsymbol{A}| )
( ? )
12
719If ( a neq b neq c, ) the value of ( x ) which
satisfies the question ( left|begin{array}{ccc}mathbf{0} & boldsymbol{x}-boldsymbol{a} & boldsymbol{x}-boldsymbol{b} \ boldsymbol{x}+boldsymbol{a} & mathbf{0} & boldsymbol{x}-boldsymbol{c} \ boldsymbol{x}+boldsymbol{b} & boldsymbol{x}+boldsymbol{c} & mathbf{0}end{array}right|=mathbf{0} ) is
( mathbf{A} cdot x=0 )
B. ( x=a )
( mathbf{c} cdot x=b )
D. ( x=c )
12
720Find the value of the following determinant:
( left|begin{array}{cc}-4 & frac{-6}{35} \ 7 & frac{35}{5} \ 5 & 5end{array}right| )
A ( cdot frac{15}{34} )
в. ( frac{32}{45} )
c. ( frac{25}{33} )
D. ( frac{38}{35} )
12
721Eliminating ( a, b, c ) from ( x=frac{a}{b-c}, y= )
( frac{b}{c-a}, z=frac{c}{a-b} ) we get
( mathbf{A} cdotleft|begin{array}{lll}1 & -x & x \ 1 & -y & y \ 1 & -z & zend{array}right|=0 )
B. ( left|begin{array}{ccc}1 & -x & x \ 1 & 1 & -y \ 1 & z & 1end{array}right|=0 )
c. ( left|begin{array}{ccc}1 & -x & x \ y & 1 & -y \ -z & z & 1end{array}right|=0 )
D. none of these
12
722Using the factor theorem it is found
that ( b+c, c+a ) and ( a+b ) are three
factors of the determinant ( left|begin{array}{ccc}-2 a & a+b & a+c \ b+a & -2 b & b+c \ c+a & c+b & -2 cend{array}right| . ) The other factor
in the value of the determinant is
A . 4
B. 2
( mathbf{c} cdot a+b+c )
D. none of these
12
723( left|begin{array}{lll}mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{omega}^{2} & mathbf{1} \ boldsymbol{omega}^{2} & mathbf{1} & boldsymbol{omega}end{array}right|=ldots )
( A )
B.
( c cdot 2 )
D. -1
12
724Numbers of ways in which 75600 can be
resolved as product of two divisors which are relatively prime?
A .44
B. 8
( c .9 )
D. 16
12
725If
( a, b, c ) all are non-zero and unequal and ( left|begin{array}{ccc}mathbf{1}+boldsymbol{a} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{b} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{c}end{array}right|=mathbf{0}, ) then
( 1+frac{1}{a}+frac{1}{b}+frac{1}{c} ) is equal to?
12
726Prove that ( left|begin{array}{ccc}mathbf{1}+boldsymbol{a} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{b} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{c}end{array}right|= )
( boldsymbol{a b c}left(mathbf{1}+frac{mathbf{1}}{boldsymbol{a}}+frac{mathbf{1}}{boldsymbol{b}}+frac{mathbf{1}}{boldsymbol{c}}right) )
12
727f ( 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{B} & boldsymbol{0} & tan boldsymbol{A} \ cos (boldsymbol{A}+boldsymbol{B}) & -tan boldsymbol{A} & 0end{array}right| )
equal to
( mathbf{A} cdot mathbf{0} )
B. ( 2 sin B tan A cos C )
( c .1 )
D. None of these
12
728Evaluate the following determinant:
[
left|begin{array}{ccc}
102 & 18 & 36 \
1 & 3 & 4 \
17 & 3 & 6
end{array}right|
]
12
729Evaluate the following:
( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda}end{array}right| )
12
730Evaluate the following determinant:
( left|begin{array}{ccc}1 & -3 & 2 \ 4 & -1 & 2 \ 3 & 5 & 2end{array}right| )
12
731f the matrix ( boldsymbol{A}=left[begin{array}{ccc}mathbf{6} & boldsymbol{x} & mathbf{2} \ mathbf{2} & mathbf{- 1} & mathbf{2} \ mathbf{- 1 0} & mathbf{5} & mathbf{2}end{array}right] ) is
singular matrix. Find the value of ( x )
12
732How that the points ( boldsymbol{P}(-boldsymbol{2}, boldsymbol{3}, boldsymbol{5}), boldsymbol{Q}(boldsymbol{1}, boldsymbol{2}, boldsymbol{3}) ) and ( boldsymbol{R}(boldsymbol{7}, boldsymbol{0},-boldsymbol{1}) )
are collinear.
12
733Evaluate ( left[begin{array}{cc}sqrt{mathbf{3}} & sqrt{mathbf{5}} \ -sqrt{mathbf{5}} & mathbf{3} sqrt{mathbf{3}}end{array}right] )12
734( boldsymbol{I f} mathbf{A}=left[begin{array}{ll}mathbf{1} & mathbf{3} \ mathbf{2} & mathbf{1}end{array}right], ) then the determinant
( mathbf{A}^{2}-2 mathbf{A}: )
A . 5
B . 25
( c .-5 )
D. -25
12
735Using properties of determinants, prove
the following:
( left|begin{array}{ccc}a^{2} & b c & a c+c^{2} \ a^{2}+a b & b^{2} & a c \ a b & b^{2}+b c & c^{2}end{array}right|=4 a^{2} b^{2} c^{2} )
12
736rove: ( left|begin{array}{ccc}-cos alpha & sin beta & 0 \ 0 & -sin alpha & cos alpha \ sin alpha & 0 & -sin betaend{array}right|=0 )12
737Evaluate the determinants ( left|begin{array}{cc}2 & 4 \ -5 & -1end{array}right| )12
738Evaluate the following:
( left|begin{array}{ccc}boldsymbol{x} & mathbf{1} & mathbf{1} \ mathbf{1} & boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{1} & boldsymbol{x}end{array}right| )
12
739Using the properties of determinants,
find the value of
( left|begin{array}{ccc}mathbf{0} & boldsymbol{a} & -boldsymbol{b} \ -boldsymbol{a} & boldsymbol{0} & -boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{0}end{array}right| )
12
740If a + p, b #9, C # r and
TP
a
a
b cl
q c = 0. Then find the
b rl
P
value of
(1991 – 4 Marks)
p-aq-br-c
12
741If ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & -mathbf{3} \ mathbf{- 4} & mathbf{1}end{array}right], ) then adj ( left(mathbf{3} boldsymbol{A}^{2}+right. )
( 12 A) ) is equal to.
( mathbf{A} cdotleft[begin{array}{cc}72 & -84 \ -63 & 51end{array}right] )
В. ( left[begin{array}{ll}51 & 63 \ 84 & 72end{array}right] )
c. ( left[begin{array}{ll}51 & 84 \ 63 & 72end{array}right] )
D. ( left[begin{array}{cc}72 & -63 \ -84 & 51end{array}right] )
12
742Maximum value of a second order
determinant whose every element is either 0,1 or 2 only is:
A.
B.
( c cdot 2 )
D. 4
12
743Find the determinant:
( left|begin{array}{ccc}1 & x & x^{2} \ 1 & y & y^{2} \ 1 & z & z^{2}end{array}right| )
12
744Write the value of the determinant
[
left[begin{array}{cc}
boldsymbol{p} & boldsymbol{p}+mathbf{1} \
boldsymbol{p}-mathbf{1} & boldsymbol{p}
end{array}right]
]
when ( p=1342 )
12
745( mathbf{a}=left|begin{array}{ccc}boldsymbol{a} & mathbf{5}-boldsymbol{i} & mathbf{7}+boldsymbol{i} \ mathbf{5}+boldsymbol{i} & boldsymbol{b} & mathbf{3}+boldsymbol{i} \ mathbf{7}-boldsymbol{i} & boldsymbol{3}-boldsymbol{i} & boldsymbol{c}end{array}right|, ) then ( boldsymbol{Delta} ) is
always
A . real
B. imaginary
( c cdot 0 )
D. None of these
12
746Evaluate the following determinant:
( left|begin{array}{ccc}1 & 3 & 5 \ 2 & 6 & 10 \ 31 & 11 & 38end{array}right| )
12
747If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) is a ( mathbf{4} times mathbf{4} ) matrix and ( boldsymbol{c}_{i j} ) is
the co-factor of the element ( boldsymbol{a}_{boldsymbol{i} j} ) in ( |boldsymbol{A}| )
then the expression ( a_{11} c_{11}+a_{12} c_{12}+ )
( boldsymbol{a}_{13} boldsymbol{c}_{13}+boldsymbol{a}_{14} boldsymbol{c}_{14} ) equals
( mathbf{A} cdot mathbf{0} )
B. –
c. 1
D. ( |A| )
12
748toppr
Q Type your question
List I
The value of the determinant
A. ( quadleft|begin{array}{lll}x+2 & x+3 & x+5 \ x+4 & x+6 & x+9 \ x+8 & x+11 & x+15end{array}right| ) is
If one of the roots of the
equation
B. ( left|begin{array}{ccc}7 & 6 & x^{2}-13 \ 2 & x^{2}-13 & 2 \ x^{2}-13 & 3 & 7end{array}right|=0 )
is ( x+2 ), then the sum of all other
five roots is
The value of
C. ( quadleft|begin{array}{ccc}sqrt{6} & 2 i & 3+sqrt{6} \ sqrt{12} & sqrt{3}+sqrt{8} i & 3 sqrt{2}+sqrt{6} i \ sqrt{18} & sqrt{2}+sqrt{12} i & sqrt{27}+2 iend{array}right| )
If ( f(theta)= )
D. ( quadleft|begin{array}{ccc}cos ^{2} theta & cos theta sin theta & -sin theta \ cos theta sin theta & sin ^{2} theta & cos theta \ sin theta & -cos theta & 0end{array}right| )
then ( f(pi / 3) )
( mathbf{A} cdot A-i v, B-i i i, C-i i, D-i )
B . ( A-i, B-i i i, C-i i, D-i v )
c. ( A-i i, B-i i i, C-i v, D-i )
D. ( A-i i i, B-i v, C-i i, D-i )
12
749If ( boldsymbol{a} boldsymbol{x}_{1}^{2}+boldsymbol{b} boldsymbol{y}_{1}^{2}+boldsymbol{c} boldsymbol{z}_{1}^{2}=boldsymbol{a} boldsymbol{x}_{2}^{2}+boldsymbol{b} boldsymbol{y}_{2}^{2}+ )
( boldsymbol{c} boldsymbol{z}_{2}^{2}=boldsymbol{a} boldsymbol{x}_{3}^{2}+boldsymbol{b} boldsymbol{y}_{3}^{2}+boldsymbol{c} boldsymbol{z}_{3}^{2}=boldsymbol{d} ) and
( boldsymbol{a} boldsymbol{x}_{2} boldsymbol{x}_{3}+boldsymbol{b} boldsymbol{y}_{2} boldsymbol{y}_{3}+boldsymbol{c} boldsymbol{z}_{2} boldsymbol{z}_{3}=boldsymbol{a} boldsymbol{x}_{3} boldsymbol{x}_{1}+ )
( boldsymbol{b} boldsymbol{y}_{3} boldsymbol{y}_{1}+boldsymbol{c} boldsymbol{z}_{3} boldsymbol{z}_{1}=boldsymbol{a} boldsymbol{x}_{1} boldsymbol{x}_{2}+boldsymbol{b} boldsymbol{y}_{1} boldsymbol{y}_{2}+ )
( boldsymbol{c} boldsymbol{z}_{1} boldsymbol{z}_{2}=boldsymbol{f}, ) then prove that
( left|begin{array}{lll}boldsymbol{x}_{1} & boldsymbol{y}_{1} & boldsymbol{z}_{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & boldsymbol{z}_{2} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & boldsymbol{z}_{3}end{array}right|=(boldsymbol{d}- )
( f)left[frac{d+2 f}{a b c}right]^{1 / 2}(boldsymbol{a}, boldsymbol{b}, boldsymbol{c} neq mathbf{0}) )
12
750Prove that ( left|begin{array}{ccc}boldsymbol{a}^{2}+mathbf{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b} & boldsymbol{b}^{2}+mathbf{1} & boldsymbol{b} boldsymbol{c} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c b} & boldsymbol{c}^{2}+1end{array}right|= )
( mathbf{1}+boldsymbol{a}^{2}+boldsymbol{b}^{2}+boldsymbol{c}^{2} )
12
751Assertion
( begin{array}{ccc}mathbf{Delta}= & & \ & sin pi & cos (boldsymbol{x}+boldsymbol{pi} / mathbf{4}) & tan (boldsymbol{x}- \ sin (boldsymbol{x}-boldsymbol{pi} / mathbf{4}) & -cos (boldsymbol{pi} / mathbf{2}) & log (boldsymbol{x} \ cot (boldsymbol{pi} / mathbf{4}+boldsymbol{x}) & log (boldsymbol{y} / boldsymbol{x}) & tan end{array} )
0
Reason
A skew symmetric determinant of odd
order equals 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
752The system of equations
a x + y +z= a – 1
x + ay+z= a -1
x+y+ az= a – 1
has infinite solutions, if a is
(a) -2
(c) not-2
2
(b) either – 2 or 1
(d) 1
12
753If ( n ) is a positive integer, then the value
of the determinant ( left|begin{array}{ccc}boldsymbol{a}^{n}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+1}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+mathbf{2}}-boldsymbol{x} \ boldsymbol{a}^{boldsymbol{n}+mathbf{3}}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+mathbf{4}}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+mathbf{5}}-boldsymbol{x} \ boldsymbol{a}^{boldsymbol{n}+mathbf{6}}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+mathbf{7}}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+boldsymbol{8}}-boldsymbol{x}end{array}right|= )
( mathbf{A} cdot mathbf{1} )
B.
( c .-1 )
D. None of these
12
754f ( x, y ) and ( z ) are all distinct and ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{1}+boldsymbol{x}^{3} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{1}+boldsymbol{y}^{3} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{1}+boldsymbol{z}^{3}end{array}right|=mathbf{0}, ) then the value of
( x y z ) is.
A . -4
B. –
( c .-2 )
D. – –
12
755Assertion ( operatorname{Let} boldsymbol{A}=left[begin{array}{ccc}mathbf{0} & boldsymbol{c} & -boldsymbol{b} \ -boldsymbol{c} & boldsymbol{0} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{-} boldsymbol{a} & boldsymbol{0}end{array}right] ) and ( boldsymbol{t} in boldsymbol{C} )
( boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{t} boldsymbol{I}-boldsymbol{A})=boldsymbol{t}^{2} boldsymbol{I}+boldsymbol{t} boldsymbol{A}+boldsymbol{A}^{2}+left(boldsymbol{a}^{2}+right. )
( left.b^{2}+c^{2}right) I )
Reason
( boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}=boldsymbol{A}^{2}+left(boldsymbol{a}^{2}+boldsymbol{b}^{2}+boldsymbol{c}^{2}right) boldsymbol{I} )
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
756Evaluate ( left|begin{array}{cc}sin 60^{circ} & cos 60^{circ} \ -sin 30^{circ} & cos 30^{circ}end{array}right| )12
757For what value of ( x, ) will the points ( (-1, x) ) (-3,2) and (-4,4) lie on a line?
( A cdot-3 )
B. 3
( c cdot-2 )
D.
12
758If ( boldsymbol{A} ) is matrix of order ( boldsymbol{3}, ) then ( boldsymbol{d} boldsymbol{e} boldsymbol{t}(boldsymbol{k} boldsymbol{A}) )
is:
A ( cdot k^{3} d e t(A) )
B . ( k^{2} d e t(A) )
c. ( k d e t(A) )
D. ( operatorname{det}(A) )
12
759If ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{a}^{2} & mathbf{1}+boldsymbol{a}^{3} \ boldsymbol{b} & boldsymbol{b}^{2} & mathbf{1}+boldsymbol{b}^{3} \ boldsymbol{c} & boldsymbol{c}^{2} & boldsymbol{1}+boldsymbol{c}^{3}end{array}right|=mathbf{0} ) and vectors
( left(1, a, a^{2}right),left(1, b, b^{2}right) ) and ( left(1, c, c^{2}right) ) are non
coplanar then ( a b c ) equals
A . -1
B. 1
( c cdot 0 )
D.
12
7608.
Let @=-
XV
. Then the value of the determinant
-1-02
(2002)
(a) 30
(b) 3ola -1) (c) 302
(d) 3011–@)
Ti
12
761Prove that ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{b} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c}end{array}right|=boldsymbol{a}(boldsymbol{b}-boldsymbol{c})(boldsymbol{a}-boldsymbol{b}) )
Hence find the value of ( left|begin{array}{lll}mathbf{3} & mathbf{3} & mathbf{3} \ mathbf{3} & mathbf{5} & mathbf{5} \ mathbf{3} & mathbf{5} & mathbf{7}end{array}right| )
12
762Let ( n ) and ( r ) be two positive integers such that ( n geq r+2 . ) Suppose ( Delta(n, r)= )
( left|begin{array}{ccc}n & C_{r} & ^{n} C_{r+1} & ^{n} C_{r+2} \ ^{n+1} C_{r} & ^{n+1} C_{r+1} & ^{n+1} C_{r+2} \ ^{n+2} C_{r} & ^{n+2} C_{r+1} & ^{n+2} C_{r+2}end{array}right| ) Show that
( Delta(n, r)=frac{n+2}{n+2} C_{3} Delta(n-2, r-1) ) Hence
or otherwise,
A ( cdot frac{left(^{n+2} C_{3}right)left(^{n+1} C_{3}right) ldots . .left(^{n-r+3} C_{3}right)}{left.left(r+2 C_{3}right)left(r+1 C_{3}right) ldots . .^{3} C_{3}right)} )
B. ( -frac{left(^{n+2} C_{3}right)left(^{n+1} C_{3}right) ldots . .left(^{n-r+3} C_{3}right)}{left(r+2 C_{3}right)left(^{r+1} C_{3}right) ldotsleft(^{3} C_{3}right)} )
c. ( frac{(n+3}{(r+3)}(sqrt[n+2]{(r+2)}) ldots . .left(^{n-r+3} C_{3}right) )
D. ( -frac{left.left(^{n+3} C_{3}right)(n+2)_{3}right) ldots . .left(^{n-r+3} C_{3}right)}{left(r+3 C_{3}right)left(r+2 C_{3}right) ldotsleft(^{3} C_{3}right)} )
12
763Show that ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2}end{array}right|=(boldsymbol{a}-boldsymbol{b})(boldsymbol{b}- )
( c)(c-a) )
12
764Calculate the values of the
determinants:
( left|begin{array}{cccc}mathbf{7} & mathbf{1 3} & mathbf{1 0} & mathbf{6} \ mathbf{5} & mathbf{9} & mathbf{7} & mathbf{4} \ mathbf{8} & mathbf{1 2} & mathbf{1 1} & mathbf{7} \ mathbf{4} & mathbf{1 0} & mathbf{6} & mathbf{3}end{array}right| )
12
765Consider the matrix ( A, B, C, D ) with
order ( 2 times 3,3 times 4,4 times 4,4 times 2 )
respectively. Let ( boldsymbol{x}=left(boldsymbol{alpha} boldsymbol{A} boldsymbol{B} boldsymbol{gamma} boldsymbol{C}^{2} boldsymbol{D}right)^{3} )
where ( alpha ) and ( gamma ) are scalars. Let ( |x|= )
( kleft|A B C^{2} Dright|^{3} . ) Then the value of ( k )
A ( cdot alpha^{6} gamma^{6} )
в. ( alpha gamma )
c. ( alpha^{3} gamma^{3} )
D. ( alpha^{6} gamma^{12} )
12
766Which of these are correct?
(a) If any two rows or columns of a determinant are identical, then the
value of the determinant is zero.
(b) If the corresponding rows and columns of a determinant are
interchanged, then the value of the determinant does not change.
(c) If any two rows (or column) of a determinant are interchanged, then the value of the determinant changes in
( operatorname{sign} )
A ( cdot(a) ) and ( (b) )
B. ( (b) ) and ( (c) )
c. ( (a) ) and ( (c) )
D. ( (a),(b) ) and ( (c) )
12
767( begin{array}{ccc}sin ^{2} x & cos ^{2} x & 1 \ cos ^{2} x & sin ^{2} x & 1 \ -10 & 12 & 2end{array} mid= )
( mathbf{A} cdot mathbf{0} )
B. ( 12 cos ^{2} x-10 sin ^{2} x )
c. ( 12 sin ^{2} x-10 cos ^{2} x-2 )
D. ( 10 sin 2 x )
12
768Using properties of determinants, prove
that
[
boldsymbol{omega}left|begin{array}{ccc}
boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \
boldsymbol{x}^{2} & boldsymbol{y}^{2} & boldsymbol{z}^{2} \
boldsymbol{y}+boldsymbol{z} & boldsymbol{z}+boldsymbol{x} & boldsymbol{x}+boldsymbol{y}
end{array}right|=(boldsymbol{x}-
]
( boldsymbol{y})(boldsymbol{y}-boldsymbol{z})(boldsymbol{z}-boldsymbol{x})(boldsymbol{x}+boldsymbol{y}+boldsymbol{z}) )
12
769Using properties of determinants, prove
[
operatorname{that}left[begin{array}{ccc}
boldsymbol{b}+boldsymbol{c} & boldsymbol{q}+boldsymbol{r} & boldsymbol{y}+boldsymbol{z} \
boldsymbol{c}+boldsymbol{a} & boldsymbol{r}+boldsymbol{p} & boldsymbol{z}+boldsymbol{x} \
boldsymbol{a}+boldsymbol{b} & boldsymbol{p}+boldsymbol{q} & boldsymbol{x}+boldsymbol{y}
end{array}right]=
]
( mathbf{2}left[begin{array}{lll}boldsymbol{a} & boldsymbol{p} & boldsymbol{x} \ boldsymbol{b} & boldsymbol{q} & boldsymbol{y} \ boldsymbol{c} & boldsymbol{r} & boldsymbol{z}end{array}right] )
12
770If area of a triangle is 35 sq units with
vertices (2,-6),(5,4) and ( (k, 4), ) then find the value of ( k )
12
771( left|begin{array}{ccc}a^{2} & b^{2} & c^{2} \ (a+1)^{2} & (b+1)^{2} & (c+1)^{2} \ (a-1)^{2} & (b-1)^{2} & (c-1)^{2}end{array}right|= )
( k(a-b)(b-c)(c-a), ) then find the
value of ( -boldsymbol{k} )
12
772Evaluate
[
left|begin{array}{ccc}
2 & 7 & 3 \
-4 & 3 & -1 \
0 & -3 & 7
end{array}right|
]
12
773Prove that:
[
begin{array}{ccc}
boldsymbol{a}+boldsymbol{b}+mathbf{2 c} & boldsymbol{a} & boldsymbol{b} \
boldsymbol{c} & boldsymbol{b}+boldsymbol{c}+mathbf{2} boldsymbol{a} & boldsymbol{b} \
boldsymbol{c} & boldsymbol{a} & boldsymbol{c}+boldsymbol{a}+boldsymbol{2} boldsymbol{b}
end{array} mid=
]
( mathbf{2}(boldsymbol{a}+boldsymbol{b}+boldsymbol{c})^{3} )
12
774If ( boldsymbol{A}=left|begin{array}{ll}mathbf{3} & mathbf{4} \ mathbf{1} & mathbf{2}end{array}right|, ) find the value of ( mathbf{3}|boldsymbol{A}| )12
775Let the three digit numbers ( A 28,3 B 9 )
and ( 62 C, ) where ( A, B, C ) are integers
between 0 and 9 , be divisible by a fixed integer ( k, ) Show that the determinant ( left|begin{array}{lll}boldsymbol{A} & boldsymbol{3} & boldsymbol{6} \ boldsymbol{8} & boldsymbol{9} & boldsymbol{C} \ boldsymbol{2} & boldsymbol{B} & boldsymbol{2}end{array}right| ) is also divisible by the
same integer ( boldsymbol{k} )
12
776( P, Q, R ) are three collinear points. The coordinates of ( mathrm{P} ) and ( mathrm{R} ) are (3,4) and (11
10) respectively and PQ is equal to 2.5 units. Coordinates of ( Q ) are-
A ( cdot(5,11 / 2) )
B. (11,5/2)
c. ( (5,-11 / 2) )
D. ( (-5,11 / 2) )
12
777( fleft(begin{array}{cc}x+y & x-y \ 2 x+z & x+zend{array}right)=left(begin{array}{cc}0 & 0 \ 1 & 1end{array}right), ) then
the values of ( x, y ) and ( z ) are respectively
( mathbf{A} cdot 0,0,1 )
в. 1,1,0
( mathrm{c} cdot-1,0,0 )
( mathbf{D} cdot 0,0,0 )
12
778If the system of equations
x-ky – z= 0, kx – y – z=0, x + y – z = 0 has a non-zero
solution, then the possible values of k are (20005)
(a) -1,2 (6) 1,2 (c) 0,1 (d) -1, 1
12
779( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{2} \ mathbf{3} & mathbf{0} & -mathbf{2} \ mathbf{1} & mathbf{0} & mathbf{3}end{array}right] ) verify that
( boldsymbol{A}(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A})=|boldsymbol{A}| boldsymbol{I} )
12
780The value of determinant is
[
left|begin{array}{ccc}
boldsymbol{a}+boldsymbol{b} & boldsymbol{a}+mathbf{2 b} & boldsymbol{a}+mathbf{3} boldsymbol{b} \
boldsymbol{a}+mathbf{2} boldsymbol{b} & boldsymbol{a}+mathbf{3} boldsymbol{b} & boldsymbol{a}+mathbf{4} boldsymbol{b} \
boldsymbol{a}+mathbf{4} boldsymbol{b} & boldsymbol{a}+mathbf{5} boldsymbol{b} & boldsymbol{a}+boldsymbol{6 b}
end{array}right|
]
12
781Using properties of determinants, prove
the following:
( left(begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{x} \ mathbf{2} boldsymbol{x} & boldsymbol{x}(boldsymbol{x}-mathbf{1}) & boldsymbol{x}(boldsymbol{x} boldsymbol{x} \ mathbf{3} boldsymbol{x}(mathbf{1}-boldsymbol{x}) & boldsymbol{x}(boldsymbol{x}-mathbf{1})(boldsymbol{x}-mathbf{2}) & boldsymbol{x}(boldsymbol{x}+mathbf{1})end{array}right. )
( 6 x^{2}left(1-x^{2}right) )
12
782A square matrix ( mathbf{B} ) of order ( mathbf{3}, ) has ( |boldsymbol{B}|= )
( mathbf{7}, ) find ( mid boldsymbol{B} ) adj ( mathbf{B} mid )
12
783Find the maximum value of ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+sin boldsymbol{theta} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+cos boldsymbol{pi}end{array}right| )12
784Assertion
consider the determinant ( Delta= ) ( left|begin{array}{lll}a_{1}+b_{1} x^{2} & a_{1} x^{2}+b_{1} & c_{1} \ a_{2}+b_{2} x^{2} & a_{2} x^{2}+b_{2} & c_{2} \ a_{3}+b_{3} x^{2} & a_{3} x^{2}+b_{3} & c_{3}end{array}right|=0, ) where
( boldsymbol{x}_{i}, boldsymbol{y}_{i}, boldsymbol{z}_{i} in boldsymbol{R} quad(boldsymbol{i}=mathbf{1}, boldsymbol{2}, boldsymbol{3}), quad boldsymbol{x} in boldsymbol{R} )
The values of ( x ) satisfying ( Delta=0 ) are
( boldsymbol{x}=mathbf{1},-mathbf{1} )
Reason
( left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{b}_{1} & boldsymbol{c}_{1} \ boldsymbol{a}_{2} & boldsymbol{b}_{2} & boldsymbol{c}_{2} \ boldsymbol{a}_{3} & boldsymbol{b}_{3} & boldsymbol{c}_{3}end{array}right|=mathbf{0}, ) then ( boldsymbol{Delta}=mathbf{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 not the correct explanation for Assertion
c. Assertion is correct but Reason is incorrect
D. Assertion is incorrect but Reason is correct
12
785If ( boldsymbol{A}=left[begin{array}{ll}2 & 0 \ 0 & 3end{array}right], ) then show that ( |3 A|= )
( mathbf{9}|boldsymbol{A}| )
12
786f ( f(x)= ) ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{2} boldsymbol{x} & boldsymbol{x}-mathbf{1} & boldsymbol{x} \ mathbf{3} boldsymbol{x}(boldsymbol{x}-mathbf{1}) & (boldsymbol{x}-mathbf{1})(boldsymbol{x}-mathbf{2}) & boldsymbol{x}(boldsymbol{x}-mathbf{1})end{array}right| )
then ( boldsymbol{f}(mathbf{5 0})= )
4.0
B. 2
( c_{1} )
( D )
( E )
12
787If ( a neq b neq c, ) the value of ( x ) which
satisfies the question ( left|begin{array}{ccc}mathbf{0} & boldsymbol{x}-boldsymbol{a} & boldsymbol{x}-boldsymbol{b} \ boldsymbol{x}+boldsymbol{a} & mathbf{0} & boldsymbol{x}-boldsymbol{c} \ boldsymbol{x}+boldsymbol{b} & boldsymbol{x}+boldsymbol{c} & mathbf{0}end{array}right|=mathbf{0} ) is
( mathbf{A} cdot x=0 )
B. ( x=a )
( mathbf{c} cdot x=b )
D. ( x=c )
12
788uet (Padj ())=2 (u) uute
12. Let a, 2. uR. Consider the system
ax +2y=a
3x – 2y = u
T the system of linear equations
ment(s) is (are) correct?
(JEE Adv. 2016)
tem has a unique solution for all
1= -3, then the system has infinitely many solutio
for all values of 2 and u.
(b) Ifa -3, then the system has a unique solution
values of 2 andu.
© If 2 +u = 0, then the system has infinitely many
solutions for a = -3.
(d) If a+u = 0, then the system has no solution for
a=-3.
12
789( mathrm{f}left|begin{array}{ccc}boldsymbol{x} & boldsymbol{2} & boldsymbol{x} \ boldsymbol{x}^{2} & boldsymbol{x} & boldsymbol{6} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{6}end{array}right|=boldsymbol{a} boldsymbol{x}^{4}+boldsymbol{b} boldsymbol{x}^{3}+boldsymbol{c} boldsymbol{x}^{2}+ )
( boldsymbol{d} boldsymbol{x}+boldsymbol{e}, ) then ( boldsymbol{5} boldsymbol{a}+boldsymbol{4} boldsymbol{b}+boldsymbol{3} boldsymbol{c}+boldsymbol{2} boldsymbol{d}+boldsymbol{e} ) is
equal to
A. 11
B. -11
c. 12
D. -12
E . 13
12
790If ( boldsymbol{A}=left[begin{array}{ll}mathbf{4} & mathbf{2} \ mathbf{3} & mathbf{3}end{array}right], ) then adj ( (operatorname{adj} boldsymbol{A}) ) is equal
to
( mathbf{A} cdotleft[begin{array}{cc}3 & -2 \ -3 & 4end{array}right] )
в. ( left[begin{array}{ll}4 & 2 \ 3 & 3end{array}right] )
D. None of these
12
791If ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & -mathbf{3} \ mathbf{4} & mathbf{1}end{array}right], ) then adjoint of matrix
( A ) is
A. ( left[begin{array}{cc}1 & 3 \ -4 & 2end{array}right] )
В. ( left[begin{array}{cc}1 & -3 \ -4 & 2end{array}right] )
c. ( left[begin{array}{cc}1 & 3 \ 4 & -2end{array}right] )
D. ( left[begin{array}{cc}-1 & -3 \ -4 & 2end{array}right] )
12
792Show that ( left|begin{array}{lll}boldsymbol{b} boldsymbol{c} & boldsymbol{b}+boldsymbol{c} & mathbf{1} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c}+boldsymbol{a} & mathbf{1} \ boldsymbol{a} boldsymbol{b} & boldsymbol{a}+boldsymbol{b} & mathbf{1}end{array}right|= )
( (boldsymbol{a}-boldsymbol{b})(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}-boldsymbol{a}) )
12
793If ( |boldsymbol{A}|=2, ) where ( boldsymbol{A} ) is a ( mathbf{3} times mathbf{3} ) matrix.
Then find ( |boldsymbol{3} boldsymbol{A}| )
12
794Find cofactors of the elements of the ( operatorname{matrix} boldsymbol{A}=left[begin{array}{ll}-1 & 2 \ -3 & 4end{array}right] )12
795Prove that ( left|begin{array}{ccc}a^{2} & b c & a c+c^{2} \ a^{2}+a b & b^{2} & a c \ a b & b^{2}+b c & c^{2}end{array}right| )
( =4 a^{2} b^{2} c^{2} )
12
796If ( d ) is the determinant of a square
matrix A of order ( n ), then the
determinant of its adjoint is:
A . ( d^{n} )
B. ( d^{n-1} )
( c cdot d^{n-2} )
D. ( d )
12
797In a third order determinant, each
element of the first column consists of
sum of two terms, each element of the second column consists of sum of three
terms and each element of the
third column consists of sum of four
terms. Then it can be decomposed into n determinants, where n has the value
( mathbf{A} cdot mathbf{1} )
B. 9
c. 16
D. 24
12
798Using properties of determinants prove
that :
( left|begin{array}{lll}boldsymbol{x} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{x} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a} & boldsymbol{x}end{array}right|=(boldsymbol{x}+mathbf{2} boldsymbol{a})(boldsymbol{x}-boldsymbol{a})^{2} )
12
799Let ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) and ( boldsymbol{B}=left[boldsymbol{b}_{boldsymbol{i} j}right] ) be two ( boldsymbol{3} times boldsymbol{3} )
real matrices such that ( b_{i j}= )
( (3)^{(i+j-2)} a_{i j}, ) where ( i, j=1,2,3 . ) If the
determinant of ( B ) is 81 , then the
determinant of A is :
A. ( 1 / 3 )
B. 3
c. ( 1 / 81 )
D. ( 1 / 9 )
12
800Evaluate the following determinant:
( 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| )
12
80130. If A is a 3 x 3 non-singular matrix such that AA’ = A’A and
B=A-1 A’, then BB’ equals:
[JEE M 2014
(a) B-1
(b) (B-1)’ (©) I+B
(d) I
12
802( left|begin{array}{ccc}1 & a & a^{2}-b c \ 1 & b & b^{2}-c a \ 1 & c & c^{2}-a bend{array}right| ) is equal to
( A )
B . ( sum a^{2}(b-c) )
c. ( 2 sum a^{2}(b-c) )
D. ( -2 sum a b(a-b) )
12
80343.
Let a and ß be the roots of the equation x2 + x + 1 = 0. Then
for y#0 in R,
y+1 a B1
a y+B 1
is equal to: [JEEM 2019-9 April (M)
IB 1 yta
(a) y(y2 – 1)
(b) y(y – 3)
C) y
(d) y – 1
12
804( f x neq 0 ) and ( left|begin{array}{ccc}1 & x & 2 x \ 1 & 3 x & 5 x \ 1 & 3 & 4end{array}right|=0, ) then ( x= )
( A )
B. –
( c cdot 2 )
( D ldots-2 )
12
805Evaluate the following:
( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda}end{array}right| )
12
806If ( boldsymbol{x}+boldsymbol{a}+boldsymbol{b}+boldsymbol{c}=boldsymbol{0}, ) then what is the
value of ( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a} & boldsymbol{x}+boldsymbol{b} & boldsymbol{c} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{x}+boldsymbol{c}end{array}right| ? )
( mathbf{A} cdot mathbf{0} )
B ( cdot(A+B+C)^{2} )
c. ( A^{2}+B^{2}+C^{2} )
D. ( A+B+C-2 )
12
807Evaluate the following:
( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda}end{array}right| )
12
808Calculate the values of the
determinants:
( left|begin{array}{cccc}mathbf{1} & mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3} & mathbf{4} \ mathbf{1} & mathbf{3} & mathbf{6} & mathbf{1 0} \ mathbf{1} & mathbf{4} & mathbf{1 0} & mathbf{2 0}end{array}right| )
12
809( operatorname{Det}left{begin{array}{ccc}-2 a & a+b & c+a \ b+a & -2 b & b+c \ c+a & c+b & -2 cend{array}right}= )
( A cdot(a+b)(b+c)(c+a) )
В . ( (a-b)(b-c)(c-a) )
( c cdot 4(a+b)(b+c)(c+a) )
D. ( 4(a-b)(b-c)(c-a) )
12
810[
begin{array}{c}
operatorname{Let}left|begin{array}{ccc}
a^{2}+1 & a b & a c \
a b & b^{2}+1 & b c \
a c & b c & c^{2}+1
end{array}right|=k+ \
a^{2}+b^{2}+c^{2}
end{array}
]
then ( 4 k ) is
12
811Evaluate ( left|begin{array}{cc}sqrt{mathbf{6}} & sqrt{mathbf{5}} \ sqrt{mathbf{2 0}} & sqrt{mathbf{2 4}}end{array}right| )12
812Find the value of ( x ) if
( left|begin{array}{ccc}mathbf{3} & mathbf{4} & mathbf{1} \ mathbf{0} & boldsymbol{x} & mathbf{8} \ mathbf{3} & mathbf{- 1} & mathbf{4}end{array}right|=mathbf{0} )
12
813( boldsymbol{A}=left[begin{array}{ccc}mathbf{5} & mathbf{5} boldsymbol{alpha} & boldsymbol{alpha} \ mathbf{0} & boldsymbol{alpha} & mathbf{5} boldsymbol{alpha} \ mathbf{0} & mathbf{0} & mathbf{5}end{array}right] ; ) If ( left|boldsymbol{A}^{2}right|=mathbf{2 5}, ) then
( |boldsymbol{alpha}|= )
A . 5
B ( .5^{2} )
( c cdot 1 )
( D )
( therefore )
12
814Solve for ( mathbf{y}:left|begin{array}{ccc}boldsymbol{x}+boldsymbol{y} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{y}end{array}right|= )
( mathbf{1 6}(boldsymbol{3} boldsymbol{x}+boldsymbol{4}) )
12
815Show that:
( left|begin{array}{ccc}boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right|=-(boldsymbol{a}-boldsymbol{b})(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}- )
( boldsymbol{a}) )
12
816( boldsymbol{A}=left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{y}end{array}right|= )
A . ( x y )
в. ( x+y )
c ( . x-y )
D. ( x^{2} y^{2} )
12
817The value of which of the following
determinants can be non-zero?
( mathbf{A} cdotleft|begin{array}{lll}a_{1}+a_{2} & a_{2} & a_{1} \ a_{4}+a_{5} & a_{5} & a_{4} \ a_{7}+a_{8} & a_{8} & a_{7}end{array}right| )
( mathbf{B} cdotleft|begin{array}{lll}a_{1}+2 a_{2} & a_{2} & a_{3} \ a_{4}+2 a_{5} & a_{5} & a_{6} \ a_{7}+2 a_{8} & a_{8} & a_{9}end{array}right| )
( mathbf{C} cdotleft|begin{array}{lll}k a_{4} & k a_{5} & k a_{6} \ a_{4} & a_{5} & a_{6} \ a_{7} & a_{8} & a_{9}end{array}right| )
D. None of these
12
818Solve
( left|begin{array}{ccc}mathbf{1} & boldsymbol{b} boldsymbol{c} & boldsymbol{a}(boldsymbol{b}+boldsymbol{c}) \ mathbf{1} & boldsymbol{c} boldsymbol{a} & boldsymbol{b}(boldsymbol{c}+boldsymbol{a}) \ mathbf{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{c}(boldsymbol{a}+boldsymbol{b})end{array}right| )
( mathbf{A} cdot mathbf{0} )
B.
( c cdot a b c )
D. ( a b+b c+c a )
12
819( mathbf{f} boldsymbol{z}=left|begin{array}{ccc}mathbf{3}+mathbf{3} boldsymbol{i} & mathbf{5}-boldsymbol{i} & mathbf{7}-mathbf{3} i \ boldsymbol{i} & mathbf{2} boldsymbol{i} & -mathbf{3} i \ boldsymbol{3}-mathbf{2} boldsymbol{i} & mathbf{5}+boldsymbol{i} & mathbf{7}+mathbf{3} boldsymbol{i}end{array}right| ), then
A. z is purely real
B. z is purely imaginary
( c cdot 0 )
D. none of these
12
820Using properties of determinants, prove
[
text { the following: }left|begin{array}{ccc}
mathbf{1} & mathbf{1} & mathbf{1} \
boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \
boldsymbol{a}^{mathbf{3}} & boldsymbol{b}^{mathbf{3}} & boldsymbol{c}^{3}
end{array}right|=(boldsymbol{a}-
]
( b)(b-c)(c-a)(a+b+c) )
12
8212.
If (71) is a cube root of unity, then
1 1+i+02 02
(1995
-i
-i +0-1
-1
(a) 0
(6) 1
(c)
i
(d) w
12
822( operatorname{Let} boldsymbol{A}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right] ) and ( boldsymbol{f}(boldsymbol{lambda})=operatorname{det}(boldsymbol{A}- )
( boldsymbol{lambda} boldsymbol{I}), ) then
This question has multiple correct options
A ( cdot f(lambda)=lambda^{2}-(a+d) lambda+a d-b c )
в. ( f(A)=O )
c. ( A^{2}=O ) implies ( A^{r}=O forall r geq 2 )
D. none of these
12
8237. The number of distinct real roots of
sin x
cos x
cos x
VI
s
cos x cos x
sin x cos x = 0 in the interval
cos x sin x
b. 2
d. 3
(IIT-JEE 2001)
a. 0
c. 1
12
824Consider the determinant ( Delta= ) ( left|begin{array}{lll}a_{1} & a_{2} & a_{3} \ b_{1} & b_{2} & b_{3} \ c_{1} & c_{2} & c_{3}end{array}right| )
( M_{i j}= ) Minor of the element of ( i^{t h} ) row ( & ) ( j^{t h} ) column ( C_{i j}= ) Cofactor of element of ( i^{t h} ) row ( & ) ( j^{t h} ) column
( a_{2} cdot C_{12}+b_{2} cdot C_{22}+c_{2} cdot C_{32} ) is equal to
A.
B. ( Delta )
( c cdot 2 Delta )
D. ( Delta^{2} )
12
825( operatorname{Let} boldsymbol{D}_{boldsymbol{r}}=left|begin{array}{ccc}mathbf{2}^{boldsymbol{r}-mathbf{1}} & mathbf{2} mathbf{.} boldsymbol{3}^{boldsymbol{r}-mathbf{1}} & boldsymbol{4} mathbf{.} mathbf{5}^{boldsymbol{r}-mathbf{1}} \ boldsymbol{alpha} & boldsymbol{beta} & boldsymbol{gamma} \ mathbf{2}^{boldsymbol{n}}-mathbf{1} & boldsymbol{3}^{boldsymbol{n}}-mathbf{1} & mathbf{5}^{boldsymbol{n}}-mathbf{1}end{array}right| )
Then, the value of ( sum_{r=1}^{n} D_{r} ) is
( mathbf{A} cdot alpha beta mathbf{gamma} )
B . ( 2^{n} alpha+3^{n} beta+4^{n} gamma )
( c cdot 2 alpha+3 beta+4 gamma )
D. None of these
12
826( f(x)=left|begin{array}{lll}a^{-x} & e^{x ln a} & x^{2} \ a^{-3 x} & e^{3 x ln a} & x^{4} \ a^{-5 x} & e^{5 x ln a} & 1end{array}right|, ) then
A. ( f(x) cdot f(-x)=0 )
B. ( f(x)-f(-x)=0 )
c. ( f(x)+f(-x)=0 )
D. None of these
12
827Prove that
[
left|begin{array}{ccc}
boldsymbol{x}+boldsymbol{y}+mathbf{2} boldsymbol{z} & boldsymbol{x} & boldsymbol{y} \
boldsymbol{z} & boldsymbol{y}+boldsymbol{z}+mathbf{2} boldsymbol{x} & boldsymbol{y} \
boldsymbol{z} & boldsymbol{x} & boldsymbol{z}+boldsymbol{x}+boldsymbol{2} boldsymbol{y}
end{array}right|
]
( 2(x+y+z)^{3} )
12
828( left|begin{array}{ccc}1 ! & 2 ! & 3 ! \ 2 ! & 3 ! & 4 ! \ 3 ! & 4 ! & 5 !end{array}right|=2016 K )
then value of ( boldsymbol{K} ) is
A . 24
B. 84
c. ( frac{1}{24} )
D. ( frac{1}{84} )
12
829If the lines ( 3 x+2 y-5=0,2 x-5 y+ )
( mathbf{3}=mathbf{0}, mathbf{5 x}+mathbf{b y}+mathbf{c}=mathbf{0} ) are concurrent
then ( mathbf{b}+mathbf{c}= )
A. 7
B. – –
( c cdot 6 )
D.
12
830Find the values of the following
determinants
( left|begin{array}{cc}mathbf{2} boldsymbol{i} & -mathbf{3} i \ boldsymbol{i}^{3} & -mathbf{2} boldsymbol{i}^{5}end{array}right| ) where ( boldsymbol{i}=sqrt{-mathbf{1}} )
12
831Evaluate the following:
( left|begin{array}{ccc}1 & a & b c \ 1 & b & c a \ 1 & c & a bend{array}right| )
12
832[
mathbf{f} boldsymbol{a}_{mathbf{1}} boldsymbol{f}_{mathbf{1}}(boldsymbol{x})+boldsymbol{a}_{mathbf{2}} boldsymbol{f}_{mathbf{2}}(boldsymbol{x})+boldsymbol{a}_{mathbf{3}} boldsymbol{f}_{mathbf{3}}(boldsymbol{x})=mathbf{0}
]
where ( a_{1}, a_{2}, a_{3} ) are constants (not all
zero) and ( f_{1}, f_{2}, f_{3} ) are twice
differentiable functions.Then ( D= )
( left|begin{array}{ccc}boldsymbol{f}_{1}(boldsymbol{x}) & boldsymbol{f}_{2}(boldsymbol{x}) & boldsymbol{f}_{3}(boldsymbol{x}) \ boldsymbol{D} boldsymbol{f}_{1}(boldsymbol{x}) & boldsymbol{D} boldsymbol{f}_{2}(boldsymbol{x}) & boldsymbol{D} boldsymbol{f}_{3}(boldsymbol{x}) \ boldsymbol{D}^{2} boldsymbol{f}_{1}(boldsymbol{x}) & boldsymbol{D}^{2} boldsymbol{f}_{2}(boldsymbol{x}) & boldsymbol{D}^{2} boldsymbol{f}_{3}(boldsymbol{x})end{array}right| ) qual to ( D f_{1}(x)=frac{d}{d x} f_{1} )
12
833The value of ( frac{1}{x-y}left|begin{array}{ccc}1 & 0 & 0 \ 3 & x^{3} & 1 \ 5 & y^{3} & 1end{array}right| ) is
( mathbf{A} cdot x+y )
B . ( x^{2}-x y+y^{2} )
c. ( x^{2}+x y+y^{2} )
D. ( x^{3}-y^{3} )
12
834( fleft(a_{1}, a_{2}, a_{3}, dots, a_{n}, dots ) are in GP, then right.
the value of the determinant ( begin{array}{|ccc|}log a_{n} & log a_{n+1} & log a_{n+2} \ log a_{n+3} & log a_{n+4} & log a_{n+5} \ log a_{n+6} & log a_{n+7} & log a_{n+8}end{array} mid ), is
A . 0
B. 1
( c cdot 2 )
( D .-2 )
12
835If ( boldsymbol{A}=left[begin{array}{ll}2 & 5 \ 2 & 1end{array}right] ) and ( B=left[begin{array}{cc}4 & -3 \ 2 & 5end{array}right], ) verify
that ( |boldsymbol{A B}|=|boldsymbol{A}||boldsymbol{B}| )
12
836Find the value of ( lambda ) for which the points
( (6,-1,2),(8,-7, lambda) ) and (5,2,4) are
collinear.
12
837( left|begin{array}{ccc}mathbf{1}+boldsymbol{i} & mathbf{1}-boldsymbol{i} & mathbf{1} \ mathbf{1}-boldsymbol{i} & boldsymbol{i} & mathbf{1}+boldsymbol{i} \ boldsymbol{i} & mathbf{1}+boldsymbol{i} & mathbf{1}-boldsymbol{i}end{array}right| ) is a
A. real number
B. irrational number
c. complex member
D. Purely imaginary
12
838If ( A B C ) is a triangle, then the vectors
( (-1, cos C, cos B),(cos C,-1, cos A) )
and ( (cos B, cos C,-1) ) are
A. linearly independent for all triangles
B. linearly dependent for all triangles
c. linearly independent for all isosceles triangles
D. none of these
12
839Using properties of determinant, prove
[
text { that }left|begin{array}{lll}
boldsymbol{b}+boldsymbol{c} & boldsymbol{a}-boldsymbol{b} & boldsymbol{a} \
boldsymbol{c}+boldsymbol{a} & boldsymbol{b}-boldsymbol{c} & boldsymbol{b} \
boldsymbol{a}+boldsymbol{b} & boldsymbol{c}-boldsymbol{a} & boldsymbol{c}
end{array}right|=mathbf{3} boldsymbol{a} boldsymbol{b} boldsymbol{c}-boldsymbol{a}^{3}-
]
( b^{3}-c^{3} )
12
840( mathbf{f}left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{a} boldsymbol{alpha}+boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b} boldsymbol{alpha}+boldsymbol{c} \ boldsymbol{a} boldsymbol{alpha}+boldsymbol{b} & boldsymbol{b} boldsymbol{alpha}+boldsymbol{c} & boldsymbol{0}end{array}right|=mathbf{0} . ) Prove
that ( a, b, c ) are in G.P. or ( alpha ) is a root of
( boldsymbol{a} boldsymbol{x}^{2}+boldsymbol{2} boldsymbol{b} boldsymbol{x}+boldsymbol{c}=mathbf{0} )
12
841Find the value of following determinant. ( left|begin{array}{cc}-1 & 7 \ 2 & 4end{array}right| )12
842Prove that: ( 2left|begin{array}{cc}8 & -5 \ -2 & 6end{array}right|=left|begin{array}{cc}14 & -2 \ -4 & 6end{array}right| )12
843If ( boldsymbol{A}=left|begin{array}{lll}boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{a}end{array}right|, ) then the value of
( |boldsymbol{A}||boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{A})| ) is
A ( cdot a^{3} )
в. ( a^{6} )
( c cdot a^{9} )
D. ( a^{2} )
12
844The points (2,-3),(4,3) and ( (5, k / 2) ) are on the same straight line. The value(s) of k is (are):
A . 12
B. -12
( c .pm 12 )
D. 12 or 6
12
845( f(x)=left|begin{array}{ccc}a^{-x} & e^{x log _{e} a} & x^{2} \ a^{-3 x} & e^{3 x log _{e} a} & x^{4} \ a^{-5 x} & e^{5 x log _{e} a} & 1end{array}right|, ) then
A ( cdot g(x)+g(-x)=0 )
B . ( g(x)-g(-x)=0 )
C. ( g(x) times g(-x)=0 )
D. none of these
12
846If ( A ) is a square matrix such that ( A^{2}= )
( A, ) then ( |A| ) equals
( mathbf{A} cdot 0 ) or 1
B. – 2 or 2
( c .-3 ) or 3
D. None of these
12
847h non-zero entries and let A2=I,
(d) less wall
Let A be a 2 x 2 matrix with non-zero entries
where I is 2 x 2 identity matrix. Define
TT(A) = sum of diagonal elements of A and
A=determinant of matrix A.
Statement-1: Tr(A)=0.
Statement-2: A=1.
(a) Statement -1 is true, Statement -2 is true,
is 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-
is a correct explanation for Statement-1.
[2010]
ment-2 is true ; Stctement-2
12
848If ( A ) is an idempotent matrix satisfying, ( (I-0.4 A)^{-1}=I-alpha A, ) where ( I ) is the
unit matrix of the same order as that of
( A, ) then the value of ( |9 alpha| ) is equal to
12
849( f x neq 0 ) and ( left|begin{array}{ccc}1 & x & 2 x \ 1 & 3 x & 5 x \ 1 & 3 & 4end{array}right|=0, ) then ( x= )
( A )
B. –
( c cdot 2 )
( D ldots-2 )
12
850Find the value of the determinant ( left|begin{array}{ccc}1 & 0 & 0 \ 2 & cos x & sin x \ 3 & sin x & cos xend{array}right| )
( mathbf{A} cdot cos 2 x )
B.
( c cdot 0 )
D. ( sin 2 x )
12
851( mathbf{A}=left[begin{array}{ccc}mathbf{1}^{2} & mathbf{2}^{mathbf{2}} & mathbf{3}^{2} \ mathbf{2}^{mathbf{2}} & mathbf{3}^{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
852The adjoint of the matric ( boldsymbol{A}= ) ( left[begin{array}{lll}1 & 0 & 2 \ 2 & 1 & 0 \ 0 & 3 & 1end{array}right] ) is
A. ( left[begin{array}{ccc}-1 & 6 & 2 \ -2 & 1 & -4 \ 6 & 3 & 1end{array}right] )
B.
[
left[begin{array}{ccc}
1 & 6 & -2 \
-2 & 1 & 4 \
6 & -3 & 1
end{array}right]
]
( c )
[
left[begin{array}{ccc}
6 & 1 & 2 \
4 & -1 & 2 \
6 & 3 & -1
end{array}right]
]
D.
[
left[begin{array}{ccc}
-6 & 2 & 1 \
4 & -2 & 1 \
3 & 1 & -6
end{array}right]
]
12
853The straight lines ( imath_{1}, imath_{2} ) and ( imath_{3} ) are parallel and lie in the same plane. A
total of ( mathrm{m} ) points are taken on the line ( imath_{1} )
n points on ( imath_{2}, ) and ( mathrm{k} ) points on ( imath_{3} . ) How many triangles are there whose vertices are at these points?
12

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