VIMUKTA ACADEMY

IIT-JEE | NEET | CBSE | FOUNDATION 1

Matrices and Determinants

Level - 1
1.(A) The determinant f is symmetric in x, y, z and vanishes when x is replaced by y or z and also when y is replaced by z
so that f = (x – y) (y – z) (z – x) F (x, y, z). F (x, y, z) being a symmetric function in x, y, z.
The main diagonal gives x y3z and in the product, we have the term (x – y) (y – z) (z – x). It follows that:
F (x, y, z) = xy + yz + zx
f = (x – y) (y – z) (z – x) (xy + yz + zx) and g = (x – y) (y – z) (z – x)
f
 = xy + yz + zx
g

x
a x eloge a x2 a x ax x2
2.(A) g (x) = a 3 x eloge a
3x
x4 = a 3 x a3x x4  e log a x  a x 
a 5 x elog e a
5x
1 a 5 x a5 x 1

ax a x x2 a x ax x2
 Interchanging Ist 
g (– x) = a 3 x a 3 x x 4 = – a 3 x a3 x x4   = – g (x)  g (x) + g (– x) = 0
 and IInd columns 
 
a5 x a 5 x 1 a 5 x a5 x 1

1 1 1
1
1 x 1 1 x x x
1 1 1
3.(C) 1  y 1  2y 1 = x y z 1 2
y y y
1 z 1 z 1 3z
1 1 1
1 1 3
z z z
Use, R1  R1  R2 + R3

1 1 1
 1 1 1 1 1 1
= x y z 3     1 2
 x y z y y y
1 1 1
1 1 3
z z z
Use, C2  C2  C1 and C3  C3 – C1

1 0 0
 1 1 1 1  1 1 1
= x y z 3     1  1 1 = 2 x y z  3      0
 x y z y  x y z
1
1 0 2
z
 x–1 + y–1 + z–1 = –3

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IIT-JEE | NEET | CBSE | FOUNDATION 2

 0 1 0   0 1 0 0 0 1
2      
4.(A) A = AA =  0 0 1    0 0 1 =  p q r 
 p q r   p q r   
 pr p  qr q  r 2 
Again
0 0 1   0 1 0  p q r 
   
3
A =A A=  p2
q r    0 0 1  =  pr p  qr qr 2

   3
 pr p  qr q  r 2   p q r  2
 pq  r p pr  q 2  qr 2 p  2qr  r 

p 0 0  0 q 0  0 0 r 
   
=  0 p 
0   0 0 q  +  pr qr r 2

 0 
p   pq q 2   3
0 qr   pr 2 pr  qr 2 qr  r 

1 0 0   0 1 0 0 0 1 
 
= p 0 1 0   q  0 0 1  + r
 
p q r 
0 0 1   p q r   
 pr p  qr q  r 2 
= pI + qA + rA2  A3 – rA2 – qA = pI

3  3 0 4 1 5
5.(AB) 2x – y =   . . . (i) and x + 2y =   . . . (ii)
3 3 2  1 4  4
Multiply (i) by 2 and add to (ii) to get :
6  6 0   4 1 5  10  5 5  2 1 1 
5x =        x =  
6 6 4  1 4  4   5 10 0  1 2 0
Multiplying (ii) by 2 and subtract (i) from it to get:
 8 2 10  3 3 0   5 5 10  1 1 2
5y      y 
 2 8 8  3 3 2   5 5 10   1 1 2 
3 0 3  1 2 1
 x y    and x  y   
0 3 2  2 1 2 

 cos ( x  y )  sin ( x  y ) 0  cos x  sin x 0 cos y  sin y 0 

6.(C) f (x + y) =  sin ( x  y ) cos ( x  y ) 0 =  sin x cos x 0  sin y cos y 0  = f (x) . f (y)
 

 0 0 1   0 0 1  0 0 1 

7.(B) cos2  cos2   cos  sin  cos  sin   0


 cos  cos  cos  cos   sin  sin    0  cos  cos  cos(   )  0   
2

 2 1  
 
8.(A) A  B   1  2 
 
 1  2 
 
 2 1    1  0 
  
 
( A  B )C   1  2      0 
  
 
 1  2   2  0 
 

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IIT-JEE | NEET | CBSE | FOUNDATION 3

2
0 c  b   a ab ac  0 0 0
 
9.(A)  c 0 a   ba b 2 bc  = 0 0 0 = 0.
    
 b  a 0   ca cb c 2  0 0 0
 
x  y y   2  3 2x  y  3  2x  y  3  x  1
10.(A)  2x   3x  y   2     xy   5
 x  y   1 2    

3x  y  2  y 5

1 1 1
11.(B)    0  bc  ca  ab  0
a b c

1  a  (1  b )(1  c )  1  1  c  1  1  1  b   

 (1  a ) b  c  bc   c  b  

 b  c  bc  ab  ac  abc  c  b    bc  ab  ca  abc      abc

x 3  14 x 2 x 3x  
12.(C) we have, ax 4  bx 3  cx 2  50x  d  4x  1 3x x 4
3 4 0

On differentiating with respect to x.

3x 2  28x 1 3 x 3  14 x 2 x 3x  
4ax 3  3bx 2  2cx  50  4x  1 3x x  4  4 3 1
3 4 0 3 4 0

0 1 3 0 0 
50  1 0 4  4 3 1  50  25  2
3 4 0 3 4 0

1  sin 2  cos2  4 sin 4 1 1 0

  R1  R1  R2 
13.(B) sin 2  1  cos2  4 sin 4  0 1 1  
 R2  R2  R3 
sin 2  cos2  1  4 sin 4 sin 2  cos2  1  4 sin 4

1 7  11
 1  4 sin 4   cos2   sin 2   2  4 sin 4   0  sin 4      ,
2 24 24
3 2
14.(A) f ( x )  x  9 x  27 x  23  0

f '( x)  3x 2  18 x  27  D0

 cos 2  sin  cos    cos 2  sin  cos  

15.(B) AB     
sin  cos  sin 2   sin  cos  sin 2  

cos  cos  cos(   ) cos  sin  cos(   ) cos  cos  cos  sin    
=   = cos(   )   = O ,       .
 cos  sin  cos(   ) sin sin  cos(   )   cos  sin  sin sin    2

1 1 i 0 2i 1
R1  R 1  R 2
16.(B)   2  i  1 1  2i 1  i =  2  i 0 1  2i 2i by
R2  R2  R3
1 2 1 i 1 2 1 i
= (2  i) {4 i 2  (1  2i)}  (2  i) (4  1  2i) = (2  i) (3  2i)  4  7 i .

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IIT-JEE | NEET | CBSE | FOUNDATION 4

x 1  2 x  1    2  2
17.(D)   x  2 1 = x  1    2 x  2 1 , (C1  C1  C 2  C 3 )
2 2
 1 x x  1    1 x

1  2
= x 1 x  2 1 ( 1     2  0 )
1 1 x 

= x [1{(x   2 ) (x   )  1}  {1  (x   )}   2 {1  (x   2 )}]

= x (x 2  x   2 x   3  1    x   2   2   2 x   4 ) = x 3 , (  3  1) .

yz x y 2 1 1
18.(B) zx z x = (x  y  z ) z  x z x
x y y z x y y z
By R1  R1  R 2  R 3
1 1 1
 x  y  z x z x   
; by C 1  C1  C 2   x  y  z  z 2  xy  xz  x 2   xy  xz   
x y z
2
  x  y  z  x  z   k 1.

x 3 7 1 1 1
19.(A) 2 x 2  0  (x  9) 2 x 2  0 , By R1  R1  R 2  R 3
7 6 x 7 6 x

 2
( x  9 ) {( x  12 )  (2 x  14 )  (12  7 x )}  0  ( x  9 ) (x 2  9 x  14 )  0   x  9  x  2  x  7   0
Hence the other two roots are x  2, 7

20.(D) Fundamental concept.

2(a  b  c) 0 abc
21.(B)  ca b c b
ab ca c
By R1  R1  R 2  R 3
2 0 1
  (a  b  c) . c  a b  c b
ab ca c

On expanding,  (a  b  c) (a 2  b 2  c 2  ab  bc  ca) =  (a 3  b 3  c 3  3 abc )  3 abc  a 3  b 3  c 3 .

22.(B) Since    2  2 2   2 . Therefore 2   4   .

1 0 1 0 1 0
23.(B)  2 1    bd .
c d a b c  ad bd
1 1 01 1 0 2 3 1 2 3 1 1 1 0  7 9 3
        
24.(B) A 2  AA   1 2 1   1 2 1= 5 6 2  A3  A2 A   5 6 2   1 2 1  =  15 19 6

 2 1 0   2 1 0   3 4 1   3 4 1   2 1 0   9 12 4 
1 0 0 
 
Here, A3  3 A 2   0 1 0  I  A3  3 A2  I  0
 0 0 1 

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IIT-JEE | NEET | CBSE | FOUNDATION 5

 1 1  a 1  1  a 0 
25.(B) Given, A , B     AB   
 2  1  b  1 2  b  2

 1 1  1 1  1 0  a 1  a 1  a 2  b a  1 
A2       ; B2      
 2  1  2  1  0  1  b  1 b  1 ab  b b  1 
 a 2  b  1 a  1
A2  B2   
 ab  b b 

 1  a 0  1  a 0 
Also, ( A  B)2    
 2  b  2  2  b  2
 (1  a)2 0
( A  B)2   
(2  b )(1  a)  2(2  b ) 4 

 (1  a)2 0   a 2  b  1 a  1
 ( A  B)2  A 2  B 2    
 (2  b )(a  1) 4   ab  b b 
By equating, a  1  0  a  1 and b  4 .
 3 5  3 5  29 25  15 25 
26.(B) A2  A.A       A2    and 5 A    20 10 
 4 2   4 2   20 24   
1 0 
 A2  5 A  14    14I .
0 1 

0 1 2 0 1 0 1 1 0 

27.(D) Given, Matrix A    . We know that A  A. A     .
1 0 1 0  1 0   0  1
8
 1 0  (1)8 0  1 0 
Therefore A16  (A 2 )8       
 0  1  0 (1)8  0 1 

 1 1 
28.(B) A  
 1 1 
 1 1   1 1   2 2   1 1 
A2      =    2 
 1 1   1 1   2 2   1 1 
 1 1   1 1  2 1 1 
A3  2    2  
 1 1   1 1   1 1 
 1 1  1 1
An  2n1    A100  299  .
 1 1  1 1
2
 a 0 0  a 0 0   a 0 0

29.(C) Since A2  A. A   0 b 0  0 b 0    0 b 2
0
 0 0 c   0 0 c   0 0

c2 
 
 a3 0 0
 
And A3   0 b3 0  ,....
 
 0 0 c3 


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IIT-JEE | NEET | CBSE | FOUNDATION 6

 a n 1 0 0  an 0 0
   a 0 0  
 An  An 1 . A   0 bn 1 0   0 b 0   0 bn 0.
     
 0 0 c n 1   0 0 c   0 0 cn 
 

2   4
30.(A) The given matrix A   1 3

4  is non singular, if | A |  0
 1  2  3 

2  4 1  3 0
A  1 3 4 = 1 3 4 ,  R1  R2  R1 
1 2 3 1 2 3
1 3 0
 R2  R2  R3 
= 0 1 1  
0  5  3  R3  R3  R1 

= 1 (3    5)  0   20    2.

x2  x x 1 x2
2
31.(A) 2x  3x  1 3x 3x  3
2
x  2x  3 2x  1 2x  1

x2  x x 1 x2
Use, R2  R2 – (R1 + R3) = gives 4 0 0 = 12 (2x – 1) = Ax + B
x2  2 x  3 2 x  1 2 x  1
  A = 24, B = – 12  A + 2B = 0
32.(D) We have
f  ( x ) g  ( x) h ( x) f  ( x) g  ( x) h ( x) f  ( x ) g  ( x ) h  ( x)
 (x) = f  ( x ) g  ( x) h ( x) + f  ( x) g  ( x ) h ( x ) + f  ( x) g  ( x) h ( x)
f  ( x) g  ( x ) h ( x ) f  ( x) g  ( x ) h ( x ) f iv ( x ) g iv ( x) hiv ( x )
f  ( x) g  ( x) h ( x)
= 0 + 0 + f  ( x) g  ( x ) h ( x) = 0 [ f, g, h are polynomials of degree 3, f iv (x) = giv (x) = hiv (x) = 0]
0 0 0
   (x) must be a constant.

1 0 2 1 0 1 0  1 0  1 0  1 0 1 0
33.(D) We have, A    A  1 1  1 1  =  2 1  A3 =    
1 1        2 1  1 1 3 1
 1 0
Similarly, An =   . Now, go option by option
n 1
Clearly A and C are ruled out.
1 0 n 0  n  1 0 
(B) n 1  n n    0 n  1
    
 n 0   n  1 0   1 0
(D) Now, nA – (n – 1) I =   =
n  1  n 1 
n n  0

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34.(C) Now, P3 = P (I – P)
= PI – P2 = PI – (I – P) = P – I + P = 2P – I
P4 = P · P3  P4 = P (2P – I)  P4 = 2 P2 – P
  P4 = 2I – 2P – P  P4 = 2I – 3P  P5 = P (2I – 3P)
  P5 = 2P – 3 (I – P)  P5 = 5P – 3I   P6 = P (5P – 3I)
 P6 = 5 P2 – 3P  P6 = 5 (I – P) – 3P  P6 = 5I – 8P, So, n = 6.

1 2 2  1 2 2  1 2 x
3 3 3  3 3 3  3 3 3
1 2 2      
2 1 2 2 1 2 2 1 2
35.(A) 3A =  2 1  2   A=   AAT = 
3 3 3 3 3 3 3 3 3
 x 2 y      
x 2 y  x 2 y  2

2 y
 3 3 3   3 3 3   3 3 3 
 x  2y  4 
 1 0 
 9 
 2x  2 y  2 
=  0 1 
9
 
 x  2y  4 2x  2 y  2 x2  4  y2 
 
 9 9 9 
 x + 2y + 4 = 0 . . . (i) 2x – 2y + 2 = 0 . . . (ii)
Solve (i) and (ii) to get : x = – 2 and y = – 1  x+y=–3

1 2 2  1 2 2  1 2 2 1 0 0
36.(ABD) A  4 A  5I3   2 1 2   2 1 2  4  2 1 2  5 0 1 0
2      
 2 2 1   2 2 1   2 2 1  0 0 1

9 8 8   4  8  8    5 0 0 0 0 0 
= 8 9 8    8  4  8    0  5 0  =
     0 0 0   0 A2 – 4A – 5I = 0
  3
8 8 9   8  8  4  0 0  5 0 0 0 
  A–1 A2 – 4A–1 A – 5A–1 I3 = 0  (A–1 A) A – 4 I3 – 5A–1 = 0
1
  IA – 4 I3 – 5A–1 = 0  A1  ( A  4I 3 )
5
9 8 8
2
Also, | A | = 8 9 8  25  0  A2 is invertible
8 8 9
and A3 = A · A2 = A · (4A + 5 I3) = 4A2 + 5A
36 32 32  5 10 10   41 42 42
= 32 36 32   10 5 10    42 41 42
32 32 36  10 10 5   42 42 41
  | A3 |  0  A3 is invertible.

37.(A) PTP = PPT = I  Q2 = PAPT PAPT = PA2 PT  Qn = PAn PT

1 2 1 3 1 n  1 2005
A2 =   , A3 =    An =    PT Q2005 P = PT PA2005 PT P = A2005 = 
0 1 0 1  0 1  0 1 

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eiA

i C  A  
i B  A 
e e eiA e iB eiC
i C  B  
i A  B 
38.(BC) z  eiA  e iB  eiC e eiB e   1  eiA eiB eiC

e

i B C  e

i A C  eiC eiA e iB e iC

Because e 
i A  B C 
 e i   cos   i sin    1 , etc.
1 1 1
 e 
i A  B C 
1 1 1   4
1 1 1

1 1 1
2 2 2
39.(A)  2x  2 x   3x  3 x   5x  5 x 
2 2 2
 2x  2 x   3x  3 x   5x  5 x 
Operate R2  R2  R3  4 R1 to get:

1 1 1
 0 0 0 0
x 2 x 2 x 2
 2x  2   3x  x   5x  5 
 1 1 1 
 
40.(D) Since, A  0 2 3 
 2 1 0 

 3 1 1
 
 B  adj A   6 2 3
 4 3 2

 5 5 5  5 5 5
 
 adj B   0 10 15   adj B  0 10 15  625
10 5 0  10 5 0
Given that C  5 A
1 1 1
adj B
 C  53 A  53 0 2 3  625  1
C
2 1 0

x a x2 1 1
41.(A) f x   x  b 2x 2  1 1 (differentiating column by column)
x c 3x 2  2 1

1 x2 1 1
x a 2x 1 x  a x2 1 0x a 1 1
 R2  R2  R1 
f '  x   1 2x 2  1 1  x  b 4x 1  x  b 2
 1 0  0  2x b  a 1 0  0
2x  
 R3  R3  R2 
1 3x 2  2 1 x  c 6x 1 x c 3x 2  2 0 c b 1 0

 2x  2b  a  c   0 ( a ,b,c are in A.P.)

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x 2 2 1 x 2 4
42.(CD) Given: 3 x 2  2 4x 8  33
3 3 x 4 8 16  x
 1   2  33 … (i)
x 2 2
1  3 x 2  
 x x 2  6  2  3x  6   2  9  3x   x 3  18x  30
3 3 x
1 x 2 4 1 2 4 x 2 4
2  2 4x 8  2 4x 8  0 4x 8
4 8 16  x 4 8 16  x 0 8 16  x
Operate C2  C2  2 C1;C 3  C3  4C1 in first determinant
1 0 0 1 2 4
 2 x 0 x 0 4x 8  21 x 2  x 3
4 0 x 0 8 16  x
 From equation (i), we get
x 3  18x  30  21x 2  x 3  33
 1 
 21x 2  18x  3  0  3  7 x  1 x  1  0  x   ,   1,  
 7 
n n 1 n 2
43.(AC) f n   n ! n  1 ! n  2  !
1 1 1

n 1 1
 C3  C3  C2 
 n ! nn ! n  1n  1 !  using : 
C2  C2  C1 
1 0 0 


 n ! n2  n 1 
Thus, f n  is divisible by n ! and n 2  n  1

44.(D) 0

1 
 f  x  f    f x   f    0
x
  x 


1 1
f x  f    f x   f   
x  x 
a  b x  a  xb   a  b x   a  xb 
n
n
n
n

Comparing coeffs. Of x n on both sides,

ab  b  a  1
Equating constant terms on both sides.
a 2  b 2  2a

 1  b 2  2  b2  1  b  1
n
 f x   1  x
4
f  2   17  1   2   n 4
4
 f 5  1  5  626

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1 3 2  1 
  
45.(B) We have 1 x 1 0 5 1  1   0
0 3 2   x 

1 
  1  5x  6  x 2  4 x   0
 1 5x  6 x  4   1   0 
 
 x 

9  53
 x 2  9x  7  0  x 
2
1 1 1
46.(C) 1 1 1  0  1[1  3]  1[1  3]  4  4  8  0  rank = 3
3 1 1

47.(C) R3  R3  3R1 R2  R 2  R1

1 1 1 1
 
0 1 3 5 
0 1 2 5 
 
R3  R 3  R2

1 1 1 1
 
0 1 3 5   rank = 3
0 0 1 0 
 

48.(C) R2  R3

1 3 4 3 
 
1 3 4 1 
3 9 12 9 
 
R2  R2  R1 R3  R3  3R1

1 3 4 3 
 
0 0 0 2 
0 0 0 0 
 

1 2 3
49.(A)  2 4 0
2 3 1

18
 14  2(   8)  3( 3  4)  0  18  11  0   
11
50.(A) R3  R3  3R1

R2  R2  2R1

1 2 3 
 
0 0 0   rank = 1
0 0 0 
 

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adj A
51.(B) Since A 1  and if  is eigen value of A, then  1 is eigen value of A 1 .
|A|
Thus for adj ( A )X  ( A 1X )| A || A |  1I
Thus, eigen value corresponding to   3 is 4/3 and corresponding to   2 is 4/–2 = – 2.

52.(A) 1  P  P 2  ..........  P n  O
Pre-multiplying both sides by P 1
 P 1  I  IP  ............  P n 1I  O .P 1  P 1  I (1  P  P 2  .........  P n 1 )  O
 P 1  (1  P  P 2  .........  P n 1 )I  P 1  (  P n )  P n

1 2 3   1 2 
     4 5 6 
53.(A) P  2 3 4   2 0   
3 4 5   0 4   0 0 1 23
  33   32
 3 14 
   4 5 6 
P   8 20   
 11 26   0 0 1  23
  32
12 15 4 
 
P   32 40 28   P22  40
44 55 40 
  33

54.(A)   1(3n  1)  n (2n  2n )  2n (n  4n )

  [( 3 )n  1]  0  2n [n  ( 3 )n .n ]

  1  1  0  2n [n  n ]  0

55.(A) Given that, I  P  P 2  ...  P n  0

Pre – multiplying both side by P 1 , we get:

  
P 1  P 1P  P 1P P  ...  P 1P P n 1  0
 
P 1  I  IP  ...  IP n 1  0  P 1  0  I I  P  P 2 ...P n 1    P
P 1   I  P n 1  P n

56.(C) D  diagonal d1, d2 ,d3 ,..., dn 

d1 0 0  0 
0 d 0  0 
 2
  0 0 d3  0 
 
     
0 0 0  dn 

As DD 1  In

1 
 0 0  0 
 d1 
 1 
1
0 0  0 
 D  d2 
 
      
 1 
0 0 0  
 dn 

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57.(C) A  A T  In  A  In  A  AA T  A In  A T  
 
A  In  A In  A T   A In  A T  A In  A

58.(C) If A is a square matrix, then A  A T is a skew –symmetric matrix. Then A  A T is ‘0’ or a perfect square

as A is of odd order or even order.

1 a b
59.(A) in ABC , given 1 c a  0
1 b c

  
1 c 2  ab  a c  a   b b  c   0  a 2  b 2  c 2  ab  bc  ca  0

1
 a  b 2  b  c 2  c  a 2   0
2 
Here, sum of squares of three members can be zero, if only, if a  b  c
 ABC is equilateral  A  B  C  60
9
 sin 2 A  sin 2 B  sin 2 C  3 sin 2 60 
4
a b c
60.(D) By the law of sines    k (say)
sin A sin B sin C
 a  k sin A,b  k sin B, c  k sin C

a2 ab / k ac / k 1 sin B sin C
Now   ab / k 1 cos  B  C   a 2 sin B 1 cos  B  C 
ac / k cos  B  C  1 sin C cos  B  C  1

1 sin  A  C  sin  A  B  sin A cos A 0 sin A cos A 0

 a 2 sin  A  C  1 cos  B  C   a 2 cos C sin C 0 cos C sin C 0  0
sin  A  B  cos  B  C  1 cos B sin B 0 cos B sin B 0

61.(A) 62.(C) 63.(B)

64.(C) For n  2

  A 1BA  A 1BA   A 1B2 A

3 2
 A 1BA    A 1BA   A 1BA    A 1B2A  A 1BA   A 1B3A and so on…
n
Thus  A 1BA   A 1B n A

 0 2b c  0 a a
65. (C) Let A   a b  c  . Now AT =  2b b  b 
 
 a  b c   c  c c 

As A is orthogonal  AAT = I 
 4 b2  c2 2 b2  c 2  2 b 2  c2 
 0 2b c   0 a a  1 0 0
  1 0 0
   a b  c   2b b  b   0 1 0   2 b 2  c 2 a2  b2  c2 a  b  c  = 0 1 0
2 2 2
 
 a  b c   c  c c  0 0 1  2
  2 b  c
2
a 2  b 2  c2 a 2  b2  c2  0 0 1 

Equating the corresponding elements, we get

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4 b2 + c2 = 1 . . . (i) , 2 b2 – c2 = 0. . . (ii) and a2 + b2 + c2 = 1 . . . (iii)

1 1 1
On solving (i), (ii) and (iii), we get a   ;b  c
2 6 3
66. (B) Solve 2x + y = 5 and x + 3y = 5 to get x = 2 and y = 1. As (x, y)  (2, 1) satisfies the equation x – 2y = 0, we have
unique solution for given system of equations.

67-70
67.(D) 68.(D) 69.(ABC) 70.(ABC)
bc  bd d  c 0 
 
Cofactor matrix of A is  c ac 1  ab 
 d ad ab  1

bc  bd c d 
 
( Adj A )   d  c ac ad 
 0 1  ab ab  1

bc  bd c d   f  bcf  bdf  cg  dh 
    
( Adj A )U   d  c ac ad   g   df  fc  acg  agd 
 0 1  ab ab  1 h   g  abg  abh  h 

hd  gc fc  fd 
 
Cofactor matrix of B is  g  h ah  f f  ag 
 c d ac ad 

a 2hd  a 2 gc 
hd  gc g  h c  d  a 2   
  
( Adj B )V   fc ah  f ac   0    a 2 fc 
   
  fd f  ag ad   0  2
    a fd 
 
From 67
 0 
 
  a 2 fc 
 2 
 a fc 
Also | A |  a (bc  bd )  (c  d )  abc  abd  c  d

t 2  3t  4
71.(A) 
t 2  3t  4
    1 t 2  3    1 t  4    1  0
Since, t is real
2 2
 9    1  16    1  0
1
  3  3  4   4  3  3  4  4   0  7    7  1  0  7
7
3 1 4
Now,   1 2 3
6 5 
[Determinant of coefficients of equations]
 1 
 7   5  0     7
 7 

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Hence the given system of equations has a unique solution.

 1 cos 
72.(A) Determinant of coefficients  3 1 2  cos   cos2   6
cos  1 2
And this is positive for all  since cos   1 .
The only solution is therefore the trivial solution.
73.(BC) Using R3  R3  R2 , R2  R2  R1, we get
x a x b x a c
 k k 2k  1 ,
k k 1  2k
x a k c
 C2  C2  C1 
Where k = common difference   k 0 k 1  
and C3  C3  C1 
k 0 k 1 
  k k k  1  k k  1   2k 2
2 2
2
dx   4k 2   16

   x  dx 
 2k  k  2, 2
0 0

74.(AB) The number of third-order determinants = the number of arrangements of nine different numbers in nine
places = 9!
Corresponding to each determinant made, there is a determinant obtained by interchanging two
consecutive rows or columns). So, the sum of this pair will be 0.
 the sum of all the determinants.
9!
 0  0  0  ... to times  0
2
p b c
  R1  R1  R2 
75.(D) a q c 0  
a b r  R2  R2  R3 

[ system has non –zero solution also, D1  D 2  D3  0 So, D must be equal to zero]

p q b q 0 1 1 0
0 q b c r  0   p  a q  b r  c  0 1 1 0
a b r a b r
p a q b r c

0 1 0
As p  a , q  b,r  c, we have  0 1 1 0
a b r b r
 
p a q b r c q b r c
 C1  C1  C2  C3 
a b r  a   b   r  p q c
   0  1    1    1    1
p a q b r c  p a  q b  r c  p a q b r c
     

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x b b
d
76.(B) 1  a x b  x 3  3 abx  1  3 (x 2  ab)
dx
a a x

x b d
and  2   x 2  ab  (1 )  3 ( x 2  ab)  3  2 .
a x dx

77.(A) Ratio of cofactor to its minor of the element –3, which is in the 3rd row and 2nd column = (1)3  2  1 .

2
78.(D)  c   n1  31   2  11  121 .
But we have to find the value of the square of the determinant, so required value is (121 )2  14641 .

  d 1 b1 c1 
 
 d 2 b 2 c 2 
D  d b c   (bcd )
79.(B) From Cramer’s rule, x  x   3 3 3   x .
D a
 1 b 1 c1  (abc )
 
a 2 b 2 c 2 
a 3 b 3 c 3 

10 ! 11! 12 ! 1 11 11  12
80.(C) 11! 12 ! 13 !  10 ! 11! 12! 1 12 12  13
12! 13 ! 14 ! 1 13 13  14

1 11 11  12
Applying R 2  R 2  R1 and R 3  R 3  R1 = 10 !11!12! 0 1 24  2(10 ! 11! 12!) .
0 2 50

1 3 1
rd 23
81.(B) The cofactor of element 4, in the 2 row and 3 nd
column is   1 8 0 1 = {1(–2) –3 (8 – 0) + 1.16} = 10.
0 2 1
a1 b1 c1 A1 B1 C1 a1 A1 0 0  0 0
82.(B) We know that    a 2 b 2 c 2 . A 2 B2 C2  0 a 2 A 2 0  0  0  3
a3 b3 c3 A3 B3 C3 0 0 a 3 A 3 0 0 
2
    .

83.(D) Fundamental concept

a1 c1 a1 b1
84.(A) B2   a1 c 3  c 1 a 3 ; C2    (a1 b 3  a 3 b1 )
a3 c3 a3 b3

a1 c1 a1 b1
B3    (a1 c 2  a 2 c 1 ) ; C3   a1 b 2  a 2 b 1
a2 c2 a2 b2

B2 C2 a1 c 3  a 3 c1 (a1 b 3  a 3 b1 ) a1 c 3  a1 b 3 a1 c 3 a 3 b1 a3 c1 a1b3 a3 c1 a3 b1

    
B3 C3  (a1 c 2  a 2 c 1 ) a1 b 2  a 2 b 1  a1 c 2 a1 b 2  a1 c 2  a 2 b1 a2 c1 a1b2 a2 c1  a2 b1

 a12 (b 2 c 3  b 3 c 2 )  a 1 b 1 (c 3 a 2  a 3 c 2 )  a1 c 1 ( a 3 b 2  a 2 b 3 )  c 1 b 1 (a 3 a 2  a 2 a 3 )  a1  .

85.(B) Fundamental concept

1 k 1
86.(C) It has a non-zero solution if 3  k  1 = 0   6k  6  0  k  1 .
1 3 1

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1 1 1
87.(D) For the system of given homogeneous equations   3  1  1  1(1  3)  1(3  1)  1(9  1)
1 3 1
 4  4  8  0
 There are infinite number of solutions.
88.(D) The given system of homogeneous equations has a non-zero solution if,   0
1 1 1
i.e., 3   3  2  6  0 ,i.e. if   3 .
1 3 1

dn dn dn  n   n 
xn sin x cos x n ! sin  x   cos  x  
n n n 2  2 
dx dx dx  
dn  n   n   n   n 
89.(B)   x   n! sin   cos    n! sin   cos  
n  2   2 
dx  2   2 
a a2 a3 a a2 a3

 n   n 
n ! sin  0   cos  0  
 2   2 
 n   n 
  n  x x 0  n! sin  
 2 
cos  
 2 
 0 . {Since R1  R 2 }.

a a2 a3

3 5 7
90.(D) Let A   2  3 1  , | A |  3 (7)  5(3)  7(5)  1
1 1 2 

 7  3 26 
 
Adj (A)   3  1 11 
 5 5  19 
Adj ( A)
A 1 
| A|
7 3  26 
1  
A 3 1  11 
 5 5 19 

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