9599624426

9773774546
Mukherjee Nagar, Batra Complex

MATRICES & DETERMINANT

1. If A is a singular matrix, then A.adj(A) is–
6 –9  8 –10 
7. If 3A – 2B =   and 2A – B = 8 –2  , then
;fn A ,d vO;qRØe.kh; vkO;wg gks] rks gS&
A.adj(A) 6 –3   
(a) an identity matrix/,d rRled vkO;wg A is–

(b) a null matrix/,d 'kwU; vkO;wg 6 –9  8 –10 

;fn 3A – 2B = 6 –3  vkSj2A – B = 8 –2  , rksA gS&
(c) a scalar matrix/,d vfn'k vkO;wg    

(d) N.O.T  –10 11 10 –11

(a)   (b)  
2. If a, b, c are in arithmetic progression, then  –10 1  10 –1 

x 1 x  2 x  a  –10 11

H
the value of x  2 x  3 x  b is–
x3 x4 xc

C
(c) 
 10 1 


5 8 6 
(d) N.O.T

;fn

O A E
x 1 x  2 x  a
a, b, c lekUrj Js.kh esa gSa]
x  rks
2 x3 xb
x3 x4 xc
dk
8. If a matrix 3 2 4  is expressed as A + B,
1 7 9 

where A is symmetric and B is skew-

C NC
symmetric, then B is equal to–
eku gS&
(a) 0 (b) 1 (c) 2 (d) abc 5 8 6 

E E Y
 cos x sin x 0 

H
;fn vkO;wg
3 2 4 
 
1 7 9 
dksA + B ds :i esa O;Dr fd;k tk;s]

T E F EM
3. If A = f(x) =  – sin x cos x 0  then A–1 is–
tgk¡A lefer vkSj B fo"ke lefer gS] rks
B dk eku gS&
 0 0 1 
 5 11/2 7 /2   0 5/2 5/2 

D AD

(a) 11/2 2  
11/2 (b)  –5 /2 0 –3 /2
 cos x sin x 0 
;fn A = f(x) =  – sin x cos x 0 
  rksA–1 gS&  7 /2 11/2 9   –5 /2 3 /2 0 
 0 0 1 

C
 0 –5 / 2 –5 / 2  5 11/ 2 7 /2
(a) f(x) (b) –f(x) (c) f(–x) (d) –f(–x) (c) 5 /2 0 3 /2  (d) 11/2 2 11 

A
4. If each element of a 3×3 matrix A is mutliplied 5 /2 –3 /2 0   7 /2 11 9 
by 3 then the determinant of the newly formed
matrix is– 9. If A and B are two matrices such that AB = B
and BA = A, then A2 + B2 is equal to
;fn ,d 3×3 vkO;wgA ds çR;sd vo;o dks3 ls xq.kk fd;k
;fn A rFkkB nks vkO;wg bl çdkj gSa
ABfd
= B rFkkBA = A,
x;k gS] rks ubZ cuh vkO;wg dh lkjf.kd gS& rksA + B cjkcj gS
2 2

(a) 3|A| (b) 9|A| (c) (|A|)3 (d) 27|A| (a) 2AB (b) 2BA (c) A + B (d) AB
2
5. If B is a matrix such that B = B and A = I – B, 10. If x 1 and x 2 are the roots of the equation
then which of the following is not correct?
1 4 20
;fn B ,d vkO;wg bl çdkj gS fdB2 = B vkSjA = I – B, 1 –2 5 = 0, then the value of x12  x 22 is
rks fuEufyf[kr esa ls dkSu lgh ugha gS\ 1 2x 5x 2
(a) A2 = A (b) A2 = I (c) AB = 0 (d) BA = 0
6. If A is a square matrix of order n, then value of 1 4 20
adj(adj A) is– ;fn x1 rFkkx2 lehdj.k 1 –2 5 =0 ds ewy gksa] rks
2
;fn A dksbZ oxZ vkO;wg gS ftldh
n gS]
dksfV
rks
adj(adj A) 1 2x 5x

dk eku gS& x12  x 22 dk eku gS

n n–1 n–2
(a) |A| A (b) |A| A (c) |A| A (d) N.O.T (a) 1 (b) 3 (c) 5 (d) 0

/thecoach /thecoach thecoachsir@gmail.com By : NEERAJ SINGH BAISLA SIR

 9599624426
9773774546
Mukherjee Nagar, Batra Complex

16. A square matrix A is said to be orthogonal, if:

1 2 
11. The inverse of the matrix   is
2 –1 ,d oxZ vkO;wg
A ledks.kh; vkO;wg dgykrk gS] ;fn

(a) AAT = A (b) AAT = A (c) AAT = A1 (d) AAT = 1

1 2 
vkO;wg
2 –1 dk çfrykse gS 17. If matrix A is of order of 3 × 3, then determinant
 
|KA| is equal to
1/5 2 /5  1/5 2 /5 
(a)   (b)   ;fn A, 3 × 3 dksfV dk vkO;wg gS] rks|KA|
lkjf.kd
dk eku gksxk%
2 /5 –1/5  2 /5 1/5 
(a) |KA| (b) K2|A| (c) K3|A| (d) 3K|A|
1/5 –2 /5   1/5 2 /5 
(c)   (d)   3 0 0 
2 /5 –1/5   –2 /5 –1/5 
18. If A = 0 3 0  , then A5 =
1 a  0 0 3 
n
12. If A =   , then A (where nN) is equal to:
0 1 
3 0 0 
;fn A =
1 a 

1 na 
(a) 
1 an2 
(b) 
H
0 1  , rksA (tgk¡ nN) cjkcj gS%
 
n

C 1 an 
 (c) 
n na 
 (d)  0 n 
;fn A = 0 3 0  , rksA5 =

(a) 243A
0 0 3 

(b) 81A (c) 27A (d) 9A

A
0 1  0 0
  0 1      19. Let P be a 4 × 4 matrix whose determinant is
13. If A be a matrix of order 3 such that |A| = 4 and 10. The determinant of the matrix –3P is:

;fn A, dksfV
O
C NC
B = adj(A), C = 2A, then
E adj  B 
C
3 dk ,d vkO;wg bl çdkj gS fd
is equal to

|A| = 4 rFkk
eku yhft, P, 4 × 4 vkO;wg gS ftldh lkjf.kd dk eku
rks vkO;wg
(a) 810
–3P dh lkjf.kd dk eku gS%
(b) 30 (c) –810
10 gS]

(d) –108

HE
B = adj(A), C = 2A, rks

E Y
adj  B 
C
dk eku gS
20. A pair of values of  and  for which the matrix

  1 2

T E F EM
A =  0 2   is invertible, is
(a) 8 (b) 6 (c) 4 (d) 1
 1 3 6
1 1 1 
 vkSj ds ekuksa dk ,d ;qXe ftlds fy, vkO;wg

D AD
14. If A = 1 2 –3 and A3 – 6A2 + 5A + 11I = 0,
2 –1 3 
  1 2
where I is a 3 × 3 identity matrix, then A–1 is A =  0 2   O;qRØe.kh; gS] gS%
equal to

C
 1 3 6

1 1 1 
;fn
A
A = 1 2 –3
2 –1 3 
rFkkA3 – 6A2 + 5A + 11I = 0 gS]

tgk¡ ij I, 3 × 3 ij dh rRled vkO;wg gS] A

rks
–1
cjkcj gS%
(a)  =
1
3
,=6

(c)  = 1,  = 5
(b)  = 1,  = 4

(d)  =
1
3
,=4

 –3 4 5   –3 4 5  21. Every square matrix can be expressed as

1  1 
(a) 9 –1 –4 (b) 0 –1 4 
11  11  çR;sx oxZ vkO;wg dks O;ÙkQ fd;k tk ldrk gS
 5 –3 –1  –5 3 0 
(a) a Hermitian matrix/,d gfeZVh vkO;wg ds :i esa
 –3 4 5   –3 4 5  (b) a skew-symmetric matrix/ ,d fo"ke&lefer
1  1 
(c) 8 6 –2 (d) 6 –9 –4 vkO;wg ds :i esa
11  11 
 2 0 –1  0 8 –1 (c) sum of symmetric and skew-symmetric
matrices/lefer rFkk fo"ke&lefer vkO;wgksa ds ;ksx ds :i e
2 3  –1
15. If A =   , such that A = kA, then k = (d) N.O.T
5 –2 
22. If A is a 3 × 3 non-singular matrix, then det(adj A)
is equal to
2 3 
;fn A = 5 –2 bl çdkj gS fd A–1 = kA, rksk = ;fn A ,d 3 × 3 O;qRØe.kh; vkO;wgdet(adj
gS] rksA) cjkcj gS
 
(a) 19 (b) 1/19 (c) –19 (d) –1/19 (a) 2 det A (b) 3 det A (c) (det A)2 (d) (det A)3

/thecoach /thecoach thecoachsir@gmail.com By : NEERAJ SINGH BAISLA SIR

 9599624426
9773774546
Mukherjee Nagar, Batra Complex

1 x x2 1 –3 2
28. If A =   and A – 4A + 10I = A, then K is
23. If f(x) = x x2 1 , then the value of f  33
is 2 K 
x2 1 x equal to–

1 –3
1 x x2 ;fn A = 2 K  vkSjA2 – 4A + 10I = A, rksK cjkcj gS
 
;fn f(x) = x x2 1 , rksf  3  dk eku gS
3

x2 1 x (a) 1 or 4 (b) 4 and not 1

(c) –4 (d) 0
(a) –6 (b) 6 (c) 4 (d) –4
24. If A is a 2 × 2 matrix such that a = 6, |A| = 12, 29. If A is skew symmetric matrix, then A2 is a–
then trace (A–1) is ;fn ;g fo"ke lefer vkO;wg
A gS] rks
A2 gS
;fn A ,d 2 × 2 vkO;wg bl çdkj gS fd vuqjs[k
a = 6, |A| = 12, (a) Null matrix/'kwU; vkO;wg
rks vuqjs[k
(A–1) gS
(b) Unitary matrix/,sfdd vkO;wg
(a)
1
2
(b)
1
3

CH (c)
1
6
25. If  is the unit cube root then what will be the
(d) 1 (c) Skew symmetric/fo"ke
(d) Symmetric/lefer
lefer

A
30. If A and B are square matrices of the same
x 1  2 order, then (A + B)(A – B) is–
 1  2

O
roots of the equation 1 = 0
2

C NC
1

E
obtained from the following determinant?
;fn  bdkbZ ?kuewy gks rc fuEu lkjf.kd ls çkIr lehdj.k
x
;fn A vkSjB leku Øe ds oxZ vkO;wg gSa]
blds cjkcj gS
(a) A2 – B2
(A + rks
B)(A – B)

(b) A2 – BA – AB – B2

E Y
(c) A2 – B2 + BA – AB (d) A2 – BA + B2 + AB
2
x 1  

E
31. If A and B are invertible matrices, then which
 1  2
2 1

(a) x = 1
H
1 = 0 ds ewy D;k gksaaxs\

T E F EM
x

(b) x =  (c) x = 2 (d) x = 0

of the following is not correct?
;fn A vkSjB O;qRØe.kh;
ugha gS\
vkO;wg gSa] rks fuEu esa ls D;k lg

D AD
26. Consider the following statements: (a) adj A = |A|A–1 (b) det(A)–1 = [det(A)]–1
fuEufyf[kr dFkuksa ij fopkj dhft,% (c) (AB)–1 = B–1A–1 (d) (A + B)–1 = B–1 + A–1
I. If A is skew-symmetric matrix, then A 2 is
4 – 3i i 
32. For a given matrix  where i = –1 .

C
symmetric.
 –i 4  3i 
;fn ,d fo"ke&lefer vkO;wg
A gS] rks
A lefer gksxkA 2
The inverse of the matrix is–
II.

A
Trace of a skew-symmetric matrix of an odd
order is always zero.
fn, x, vkO;wg esa
,d fo"ke dksfV okys fo"ke&lefer vkO;wg dk vuqjs[k lnSo 'kwU;
gksrk gSA
 –i
 4  3i 

rks vkO;wg dk çfrykse gS&

4 – 3i
ds fy, tgk¡ i =
i 
–1 gS]

Which of the following statements is/are true?

mi;qZÙkQ dFkuksa esa ls dkSu&lk@ls lR; gSa\ 1 4 – 3i i  1  i 4 – 3i 
(a) (b)
24  –i 4  3i  25 4  3i –i 
(a) Only I (b) Only II
(c) Both I and II (d) Neither I nor II
1 4  3i –i  1 4  3i –i 
(c) (d)
27. The system of equations x + 2y + 3z = 1, 24  i 4 – 3i  25  i 4 – 3i 
2x + y + 3z = 2, x + y + 2z = 3 has
lehdj.k fudk; x + 2y + 3z = 1, 2x + y + 3z = 2,  2 
33. If A =  3
 and |A | = 125, then the value of
x + y + 2z = 3 dk 2 

(a) no solution/dksbZ gy ugha gS  is–

(b) unique solution/vf}rh; gy gS  2 

;fn A = 2  vkSj|A3| = 125 gSa] rks
 dk eku gS&
(c) infinite solution/vuUr gy gS  
(d) N.O.T (a) ±1 (b) ±2 (c) ±3 (d) ±5

/thecoach /thecoach thecoachsir@gmail.com By : NEERAJ SINGH BAISLA SIR

 9599624426
9773774546
Mukherjee Nagar, Batra Complex

43 44 45 3 2 4
1
34. The value of the determinant 44 45 46 is 40. If A = 1 2 –1 and A–1 = adj(A), then the
k
45 46 46 0 1 1

equal to– value of k is

3 2 4
43 44 45 1
lkjf.kd 44 45 46 dk eku cjkcj gksxk& ;fn A = 1 2 –1 vkSjA–1 = adj(A), rksk dk eku gksxk
k
0 1 1
45 46 46

(a) 0 (b) –1 (c) 1 (d) 2 1

(a) 7 (b) –7 (c) (d) 11
7
35. The square matrix A is orthogonal, then
determinant of A is– 41. If ,  are the roots of the equation 2x2 + 3x + 5 = 0,
;fn oxZ vkO;wg
A yEcdks.kh; gS]Arks
dk lkjf.kd gS 0  
(a) ±1 (b) 0

CH
x 1
(c) ±2

x2
(d) N.O.T

x4
then the value of the determinants  0  is

;fn  vkSj ewy gS lehdj.k2x2 + 3x + 5 = 0 ds rks

  0

O A
36. The value of determinant x  3

E
x 5
x  7 x  10 x  14
x  8 is:

lkjf.kd
0  
 0  dk eku gksxk

C NC
x 1 x2 x4   0
lkjf.kd x3 x 5 x 8 dk eku gS%
3 15 3 14

E Y
x  7 x  10 x  14 (a) – (b) – (c) (d)
5 4 5 5
(a) –2 (b) x2 + 2

H E
T E F EM
(c) 2

x 2 x 3 x 5
(d) N.O.T

37. The value of determinant x  4 x  6 x  9 is:

0 1 
42. If A = 

;fn A =
0 1 
100
 , then A is equal to
1 0 

1 0  , rksA cjkcj gksxk

D AD
x  8 x  11 x  15 100

 

x 2 x3 x 5 1 0  0 1  1 1  0 0 
(a)  (b)  (d)  (d) 
lkjf.kd x4 x6 x9 dk eku gS%    

C
0 1  1 0  0 0  1 0 
x  8 x  11 x  15

A
2 3 1 
(a) 2 (b) –2 (c) 3 (d) x – 1
43. If A = 7 1 5  is equal to B + C, where B is
38. The value of  and µ for which the system of 1 9 8 
equations x + y + z = 6, x + 2y + 3z = 10 and
symmetric and C is skew symmetric, then B is
x + 2y + z = µ have no solution are equal to
 vkSjµ dk eku Kkr dfj;s tcfd fuEufyf[kr lehdj.kksa
x + y + z = 6, x + 2y + 3z = 10 and x + 2y + z = µ 2 3 1 
;fn A = 7 1 5  cjkcj gksB + C ds tgk¡B ,d lefer
dk dksbZ gy u gksa
1 9 8 
(a)  = 3, µ = 10 (b)  = 3, µ  10
rFkkC ,d fo"ke lefer vkO;wg gS] B
rkscjkcj gksxk
(c)   3, µ = 10 (d)   3, µ  10
39. For what value of k the equations 2x – 3y + 2z = a, 2 5 1  2 3 1 
5x + 4y – 2z = –3 and x – 13y + ky = 9 will not (a) 5 1 7  (b) 3 1 5 
have a unique solution?
1 7 8  1 5 8 
k ds fdl eku ds fy, lehdj.kksa2x – 3y + 2z = a,
5x + 4y – 2z = –3 vkSjx – 13y + ky = 9 dk ,dkadh eku 0 –2 0  0 5 1 
ugha gksxk (c) 2 0 –2 (d) 5 7 1 
0 2 0  1 1 8 
(a) 8 (b) 3 (c) 2 (d) 6

/thecoach /thecoach thecoachsir@gmail.com By : NEERAJ SINGH BAISLA SIR

 9599624426
9773774546
Mukherjee Nagar, Batra Complex

–a 2 ab ac  1 –1 3
49. If A =   , then A is equal to:
44. The value of the determinant ab –b2 bc is:  –1 1 
ac bc –c 2
 1 –1
;fn A =  –1 1  , rksA cjkcj gS%
3

 
–a 2 ab ac
(a) A (b) 4A (c) 3A (d) 2A
lkjf.kd ab –b2 bc dk eku gS%
50. Let A and B have 4 and 8 elments respectively.
ac bc –c 2 Then what can be maximum and minimum
number of elements in A × B?
(a) 0 (b) –(a2 + b2 + c2)
ekukA vkSjB esa Øe'k%
4 vkSj8 vo;o gSaA rks fiQj
A×B
(c) 4a2b2c2 (d) 2(ab + bc + ca)
esa rRoksa dh vf/dre vkSj U;wure la[;k D;k gks ldrh gS\
1 –1 1 4 2 2 (a) 16 and 64 (b) 32 and 32
1
45. Let A = 2 1 –3 and B = –5 0  . If B is (c) 32 and 64 (d) 64 and 64

H
10
1 1 1 1 –2 3
1 a 1 1

1 –1 1
ekukA = 2 1 –3
AC
the inverse of A, then  is:

rFkkB = 1
4 2 2
–5 0  ;fn
51. If 1
1
ab
1
equal to
1
1 c
= 0, then a–1 + b–1 + c–1 is

O
10

vkO;wg
1

B, vkO;wg
(a) –2
1

C NC
1

E
A dk O;qRØe gS] rks
(b) –1
gS%
(c) 2
1 –2 3

(d) 5
;fn
1 a
1
1
1
ab
1
1
1
1 c
= 0, rksa–1 + b–1 + c–1 cjkcj gksxk

E Y
46. If a, b, c are positive and unequal, then the value

a b c

H E
(a) 1 (b) –1 (c) abc
52. If all the elements of a 3 × 3 matrix P are 1,
(d) N.O.T

T E F EM
then P2 – 3P is
of  = b c a is:
c a b
;fn 3 × 3 vkO;wg
P ds lHkh vo;o 1 gSa] rks
P2 – 3P gS
(a) a null matrix/'kwU; vkO;wg

D AD
a b c (b) an unit matrix/bdkbZ vkO;wg
;fn a, b, c /ukRed o vleku gSa] rks
= b c a dk (c) a column matrix/LrEHk vkO;wg
c a b
(d) a diagonal matrix/fod.kZ vkO;wg
eku gksxk%
(a) 0

A C
(b) < 0

 sin  cos 
47. The matrix A =   is
(c) 1 (d) >1
0 1 
53. If A =   , then the matrix A is
1 0 

;fn A =
0 1 
 – cos  sin   1 0  , rks vkO;wg
A gS
 

 sin  cos  (a) idempotent matrix/oxZle vkO;wg

vkO;wg
A=  – cos  sin   gS%
  (b) nilpotent matrix/'kwU;Hkkoh vkO;wg
(a) Symmetric/lefer (c) involutary matrix/vuSfPNd vkO;wg

(b) Skew-symmetric/fo"ke&lefer (d) singular matrix/vO;qRØe.kh; vkO;wg

(c) Orthogonal/ykafcd cos x – sin x 0

(d) Singular/vO;qRØe.kh; 54. If f(x) = sin x cos x 0 , then f(x)f(y) is equal to
0 0 1
48. If Q is non-singular matrix and P is a square
matrix is such that det(Q–1P2Q) = 4, then det P is
equal to cos x – sin x 0

;fn Q ,d O;qRØe.kh; vkO;wgPrFkk,d oxZ vkO;wg bl çdkj ;fn f(x) = 0 0 1 , rc f(x)f(y) cjkcj gS
cos x – sin x 0
gS fddet(Q–1P2Q) = 4, rksdet P cjkcj gS
(a) 0 (b) 1 (c) 2 (d) 4 (a) f(x + y) (b) f(x – y) (c) f(x) + f(y) (d) f(x) – f(y)

/thecoach /thecoach thecoachsir@gmail.com By : NEERAJ SINGH BAISLA SIR

 9599624426
9773774546
Mukherjee Nagar, Batra Complex

3 –4  61. If A and B are matrices of the same order, then

n
55. If A =   , then A is equal to (A + B)2 = A2 + B2 + 2AB is possible if:
1 –1
;fn A rFkkB leku Øe ds vkO;wg gSa rks
3 –4 
;fn A=   , rksA cjkcj gS
n
(A + B)2 = A2 + B2 + 2AB rHkh vkSj dsoy rHkh lEHko gS] ;fn
1 –1
(a) AB = I (b) BA = I (c) AB = BA (d) N.O.T
n
 3n  –4   1  2n –4n 
(a)   (b)   1 2 3 
4
 1  –1   1  n 1 – 2n  62. If A = 1 3 4 , then |Adj A| is equal to:
1 4 3 
1  3n 1 – 4n  1  2n –4n 
(c)   (d) 
 1 n 1 – n   n 1 – 2n 
1 2 3 
56. If A is square matrix of order n, then |adj(adj A)| = ;fn A = 1 3 4 , rks|Adj A| gS
 
;fn A, n Øe dh oxZ vkO;wg gS] rks A)| =
|adj(adj 1 4 3 

(a) |A|(n – 1) (b) A

1 a 
57. If A =  n
 , then A equals:
CH
 n–12
(c) |A|(n + 1) (d) A
 n12
(a) –4

6 8 5 
(b) 4 (c) –2 (d) 2

A
 0 1  63. If A =  4 2 3  is the sum of a symmetric
9 7 1 

O E
1 a 
;fn A = 0 1  , rksA cjkcj gS%
n
matrix B and a skew-symmetric matrix C, then
 

C NC
B is–
1 na  1 a n  1 na  1 a n 
(a)   (b)   (c)   (d)   6 8 5 

E Y
0 n  0 1  0 1  0 n  ;fn A = 4 2 3
  ,d lefer vkO;wg B vkSj ,d

E
58. If A and B are square matrices of same order,
9 7 1 

lgh gS\ H
T E F EM
then which one of the following is true–
;fn A vkSjB ,d Øe ds oxZ vkO;wg gSa] rks fuEu esa ls dkSufo"ke&lefer
lk

(b) (AB)–1 = A–1B–1

vkO;wg
C dk ;ksx gS] rks
B gS&

6 6 7   0 2 –2

D AD
(a) (AB)' = A'B' (a)  4 2 3  (b)  –2 0 –2
(c) (A–1)' = (A')–1 (d) B'AB = BA'B 9 7 1   2 2 0 

 0 0 0
59. The matrix  1 0 0  is:

C
 6 –6 7   0 2 –2
 –2 0 0  (c)  –6 2 –5  (d)  2 0 –2

A
 7 –5 1   –2 –2 0 
0 0 0
vkO;wg
1
 0 0  gS% 1 3  1 0 
64. If a matrix X satisfies   X=   then X
 –2 0 0  0 1  0 1 

(a) Lower triangular matrix/fuEu f=kHkqth; vkO;wg is equal to

(b) Upper triangular matrix/mifj f=kHkqth; vkO;wg 1 3  1 0 
;fn dksbZ vkO;wg
0 1  X = 0 1  dks larq"V djrk gS] rks
(c) Diagonal matrix/fod.kZ vkO;wg    
(d) Column matrix/LrEHk vkO;wg X cjkcj gS

1 1 1 1 3  1 0  1 –3   1 0
(a)   (b)   (c)   (d)  –3 1 
60. The value of the determinant a b c is: 0 1  3 1   0 1   
a2 b2 c 2 65. If A and B are square matrices of order 3 such
1 1 1 that det A = –1 and det B = 3 then det(3AB) is
lkjf.kd a b c dk eku gS% equal to
a 2 2
b c 2
;fn A vkSjB dksfV3 ds oxZ vkO;wg ,sls gS
detfdA = –1

(a) abc (b) (a – b)(b – c)(c – a)

vkSjdet B = 3, rksdet(3AB) cjkcj gS
2
(c) a b c 2 2
(d) (a – b)(b – c)(a – c) (a) –9 (b) –27 (c) –81 (d) 81

/thecoach /thecoach thecoachsir@gmail.com By : NEERAJ SINGH BAISLA SIR

 9599624426
9773774546
Mukherjee Nagar, Batra Complex

66. If A is non-singular matrix of order 3 × 3, then

 0 
|adj A| is equal to 70. If A =  3
 and |A | = 125, then  equals
0  
;fn A ,d 3 × 3 dksfV dk vO;qRØe.kh; vkO;wg
|adjgS]
A|rks
fdlds cjkcj gS  0 
(a) |A|3 (b) |A|2 (c) |A| (d) 0 ;fn A = 0   vkSj|A3| = 125, rc  cjkcj gksxk%
 
67. Let A be a square matrix and At be its transpose
matrix. Then A – At, is: (a) 0 (b) ±2 (c) ±3 (d) ±5
ekuk fdA ,d oxZ vkO;wg gS A
rFkk
t
mldk ifjorZ vkO;wg gS] xa b c
rksA – A gS%
t
71. If determinant a xb c = 0, then x
(a) Symmetric matrix/lefer vkO;wg a b xc
(b) Skew-symmetric matrix/fo"ke lefer vkO;wg equals to
(c) Zero matrix/'kwU; vkO;wg
xa b c
(d) Identity matrix/bdkbZ vkO;wg

,d oxZ vkO;wg

CH
68. A square matrix A is idempotent if:
A fuf"Ø; gksxk ;fn%
;fn lkjf.kd a

(a) a + b + c
a
xb
b xc
c = 0, gks rcx cjkcj gksxk

(b) –(a + b + c)

A
(a) A2 = 0 (b) A2 = 1 (c) A = AT (d) A2 = A
(c) 0, a + b + c (d) 0, –(a + b + c)
1 a a2

O E
72. If A and B are two matrices such that A + B and
69. The ratio of the determinants 1 b b2 and AB are both defined, then

ab a  b 1 C NC 1 c c2
;fn vkO;wg
gks] rc
A vkSjB bl çdkj gSa fdA + B vkSjAB ifjHkkf"kr

bc b  c 1 :
ca ca 1

HE E Y (a) A and B are two matrices not necessarily of

same order/A vkSjB nksuksa vkO;wg vko';d ugha fd

T E F EM
,d gh Øe ds gks
2
1 a a ab a  b 1 (b) A and B are two matrices of same order/A
lkjf.kd 1 b b2 rFkkbc b  c 1 dk vuqikr gS% vkSjB oxZ vkO;wg vkSj ,d leku Øe ds gS

D AD
1 c c2 ca ca 1
(c) Number of columns of A = number of rows of B/
(a) 2:1 (b) 1:2 A ds LrEHk dh la[;k
= B ds dkye (iafÙkQ) dh la[;k
(c) 1:1 (d) abc : (a + b + c) (d) N.O.T

A C
Answer Key
1. (b) 9. (c) 17. (c) 25. (d) 33. (c) 41. (b) 48. (c) 55. (d) 62. (b) 69. (c)
2. (a) 10. (c) 18. (b) 26. (c) 34. (c) 42. (a) 49. (b) 56. (b) 63. (a) 70. (c)
3. (c) 11. (a) 19. (a) 27. (a) 35. (a) 43. (a) 50. (b) 57. (c) 64. (c) 71. (d)
4. (d) 12. (a) 20. (c) 28. (b) 36. (a) 44. (c) 51. (b) 58. (c) 65. (c) 72. (b)
5. (b) 13. (a) 21. (c) 29. (d) 37. (b) 45. (d) 52. (a) 59. (a) 66. (b)
6. (c) 14. (a) 22. (c) 30. (c) 38. (c) 46. (b) 53. (c) 60. (b) 67. (b)
7. (b) 15. (b) 23. (d) 31. (d) 39. (a) 47. (c) 54. (a) 61. (c) 68. (d)
8. (b) 16. (d) 24. (a) 32. (c) 40. (d)

/thecoach /thecoach thecoachsir@gmail.com By : NEERAJ SINGH BAISLA SIR