NCERT SOLUTIONS BOOK I DETERMINANTS & MATRICES
EXERCISE 3.2
13.
cos x  sin x 0 cos y  sin y 0
F  x  F  y    sin x cos x 0  sin y cos y 0
0
1   0
0
1 
 0
cos x cos y  sin x sin y  0  0  cos x sin y  sin x cos y  0 0  0  0 
  sin x cos y  cos x sin y  0
 sin x sin y  cos x cos y  0 0  0  0 
000
000
0  0  1 

cos  x  y   sin  x  y  0 


  sin  x  y  cos  x  y  0   F  x  y 

0
0
1 
16.
We can directly calculate, A2 , A3 ,7 A etc. by using A2  AA, A3  A2 . A  A. A2 and prove
A3  6 A2  7 A  2 I  0
At an advanced level we will study that every square matrix satisfies its characteristic
equation (or characteristic polynomial equation). The characteristic equation of a square
matrix A is given by det  A   I   0 which will be a polynomial in  . The values of 
are called eigen values. Let us observe in this case
1 0 2   0 0 
Indeed A   I   0 2 1    0  0 

 

 2 0 3   0 0  
1  
  0
 2
2 
2
1 
0
3   
0
 det  A   I   1    2    3     0   2  2  2    
  2    1    3     4 
  2     2  4  1   3  6 2  7  2
Thus the characteristic equation is  3  6 2  7  2  0
Replacing  by matrix A and attaching I to constant term 2 we get
A3  6 A2  7 A  2  0
From characteristic equation all other powers (including 1, related to A1 ) of A can be
found. For instance if we multiply by A we get A4  6 A3  7 A2  2 A  0
1

 6  6 A2  7 A  2 I   7 A2  2 A  29 A2  44 A  12 I
A4  6 A3  7 A2  2 A
If we pre multiply by A1 we get A1 A3  6 A1 A2  7 A1 A  2 A1 I  A1 0  0

A2  6 A  7 I  2 A1  0
A1 

1
 A2  6 A  7 I 

2
Note that if one of the given values is zero A1 will not exist. Equivalently if det A  0, one of
the given values will be 0 .
17.
 3 2  3 2
A2  


 4 2  4 2
 3k
kA  2 I  
 4k
1 2 


 4 4 
2k   2 0 

2k  0 2 
If A2  kA  2 I we must have,
2k 
3k  2

2k  2 
 4k
3k  2  1,
2k  2
4k  4,
2k  2  4
Since values of k coming out from all the equations is same  k  1 such a k exists
equals 1 .
18.

 0
A
 tan 

2



 1
2 ,
 I  A
 tan 
0 


2
 tan
cos 
Now RHS   I  A 
 sin 


 cos   tan 2 sin 

  cos  tan   sin 

2
Now cos   tan

2
 tan

2

1 


1
 sin   

cos   

 tan

2
 tan
2  cos 

sin 
1 



cos  
2


tan sin   cos  
2

 sin   tan
sin   cos  
sin

2 .2sin  cos 
2
2
cos
2

2

 sin  
cos  
and
 cos   2sin 2

 cos   1  cos   1
2
Again  sin   tan


2
cos   2sin

cos  
 sin  2 cos 

2
2 cos  

2



2
cos
2

2

cos

2 .cos 

2


2
  2cos 2  cos  
 sin 


2




2 
2 
  2 cos 2  2 cos 2  1
 sin 


2


cos
sin
cos
  tan


2



2

 1
We can similarly show  cos  tan  sin   tan .Thus RHS  
2
2
 tan 

2

19.


2

1 

Let the trust decide to invest amount x on first type of bond, and 30000  x on the second
type of bond. Let us write amount invested as a row vector, and interest as a

then we must have  x 30000  x  

20.
 tan
5
100
7
100

  1800

5
7
  30000  x 
 1800
100
100

x

5x  210000  7 x  180000

Amount invested on two types of bonds must be equal.
Subject CHEM
PHY.
10 12
Price
column vector
80

2x  30000
ECO
8 12
60
10 12
40
Total amount received
3

x  15000
 80 
 
 120 96 120   60 
 40 
 
21.
 120  80  96  60  120  40  9600  5760  4800  20160
PY will be defined if number of columns in P is equal to number of rows in Y
 p  3 . k can be any integer. WY will be defined for all n and k . But PY and WY must
have same order  p  n, k  3 .
22.
If n  p then 7X and 5Z will be of order 2  n and p  k we must have p  2, k  n

Order of 7 X  5Z will be 2  n 
( B ) is correct.
EXERCISE 3.3
11.
If A and B are symmetric matrices of the same order then A  A, B  B
Let C  AB  BA, then C    AB  BA    AB    BA 
 BA  AB  BA  AB    AB  BA  C  C is skew symmetric.
12.
If A  A  I then 2cos   1 or cos  
Among the given choices only  

3
1
2
satisfies.
NOTE:- A full solution shall require more conditions on  which must be satisfied in order
that A  A  I
 cos 
  sin 
Indeed A  
sin  
 
cos  
 2cos 
A  A  
 0
We must have 2 cos   1, 2 cos   1 0  0, 0  0 etc.
EXERCISE 3.4
Instructions for solving 1 to 16. 2  2 matrix
4
 1 0


2cos    0 1 
0
a b 
 by elementary row transformations write
c d 
To find the inverse of 2  2 matrix 
 a b  1 0   a b 
 c d   0 1   c d 

 


We have to apply elementary row operations on matrix on the left to convert it to
1 0 
0 1  . The same row operation will be applied to identity matrix on RHS .


Step I:- To make a unity either apply R1 
1
R1 or interchange with some other row or
a
apply R1  R1  k R2
Step II:- Apply New R2  New R2  (New a ) c .
Thus we will get 0 at a21 position . Repeat the same for a12
Step III:- Make a22 unity by R2  kR2 .
EXERCISE 3-4
5.
2 1
 2 1  1 0   2 1 
, Write 
A




7 4 
7 4  0 1  7 4 
Apply R1 

1
1
R1 we get 

2
7
1 1

0  2 1 


2  2
 
 7 4
4   0 1
Apply R2  R2  7 R1
1   1



0 
1


1
2
1


2
2
we get 


 or 
0 4  7  0  7 1  7  0  2 4 
0

2   2


1  1

0


2 1
2
2


1  7
7 4 



1

2   2
Apply R1  R1  R2 (may not be equally convenient in other cases) we get
1 0   4


0 1    7

2  2
1
 2 1


1  7 4 

5
1 0  4 1  2 1 
Finally applying R2  2 R2 . 


 
0 1   7 2  7 4
1
2 1
 4 1
7 4   7 2 




INVERSE IN 3x3 CASE
STEP I:- Some how try to make a11  1 STEP II:- Create zeros at a21 , a31 positions
STEP III;- Create zeros at a12 , a13 positions
STEP IV:- Some how try to make a22  1
STEP V:- Create 0 at a32 and a23 positions
STEP VI:- Make a33 as 1
NOTE:- It is advisable to apply transformation on one side only . Later repeat them on matrix
on RHS :
Q.15
 2 3 3 
 2 3 3 1 0 0  2 3 3


A   2 2 3  Write  2 2 3  0 1 0  2 2 3

 


 3 2 2 0 0 1   3 2 2
 3 2 2 
Advised rough work
 2 3 3  2  3 2 3 2
1
A   2 2 3   2
2
3  R1  R1
2
 3 2 2  3 2
2 

1 3 2

 0
5

9
0 2 

2

32

0 
9
2 
2
1 3 2 3 2 
 0
5
0 
0 5 2  5 2 
Now fresh a21  
and a32 
R2  R2  2 R1
R3  R3  3R1
1 3 2 3 2 
1
 0
1
0  R2  R2
5
0 5 2  5 2
3
3
which will become zero by R1  R1  R2
2
2
5
5
which will become zero by R3  R3  R2
2
2
6
1 0 3 2 

0 
Applying these row operations LHS  0 1

 0 0 5 2 
1 0 3 2 
2
 0 1 0  R3   R3
5
0 0 1 
1 0 0 
3


Now a13  3 2 . To make it zero. Apply R1  R1  R3 , LHS  0 1 0


2
 0 0 1 
Let us now list down all above operations and their effect on Identity matrix on RHS
Operation
R1 
Effect on I
1

 2 0 0


 0 1 0
 0 0 1




1
R1
2
 1

0 0
 2


 1 1 0 
 3 2 0 1 




R2  R2  2 R1
R3  R3  3R1
1
R2  R2
5
0 0
12
 1 5 1 5 0 


 3 2 0 1 
3
R1  R1  R2
2
5
R3  R3  R2
2
1 3
 2  10

 1 5
 3 1
 
 2 2
3

0
10

15
0

1
0
1
2

0
3
 1

0
 5
10


 1 5 1 5 0 


1
1
 1 
2


7
3
 1

0 
 5
10


 1 5 1 5 0 
 2
1
2
 

5
5
 5
2
R3   R3
5
3 2
1 3 2 3 3



0

.
 5 2 5 10 10
2 5


15
0 
 1 5

2
1
2 



5
5
5 

3
R1  R1  R3
2
35
 2 5 0

= 1 5 1 5
0 

 2 5 1 5  2 5
Now in the fair work apply all the eight operations in succession simultaneously
Finally write
3 
 2
1

0

35
 2 3 3 
 2 5 0
1 0 0 
5
5   2 3 3 




0 1 0    1 5 1 5

0 
0   2 2 3    2 2 3    1 5 1 5

 
0 0 1   2 5 1 5  2 5  3 2 2 
 3 2 2 
 2 5 1 5  2 5




MISCELLANEOUS EXERCISE
1.
0 1  2 0 1  0 1  0 0 
A
 , A  0 0  0 0   0 0 
0 0 


 

We will prove by induction  aI  bA   a n I  na n 1bA
n
The result is true for n  1
k
k 1
Let the result be true for n  k then  aI  bA   a I  ka bA
k
To prove the result for n  k 1
We try to show  aI  bA 
k 1
 a k 1 I   k  1 a k bA
LHS   aI  bA   aI  bA 
k
8
  a k I  ka k 1bA   aI  bA 
  a k .I  aI  a k I  bA    ka k 1bA   aI    ka k 1bA   bA 
 a k 1 I 2  a k bA  ka k bA  0

AI  IA  A, A2  0, I commutes with A, I 2  I etc.
 ak 1I   k  1 ak bA
2.
1 1 1 1 1 1 3 3 3
A  1 1 1 1 1 1  3 3 3
1 1 1 1 1 1 3 3 3
2
Thus result is true for n  2 .
3k 1 3k 1 3k 1 
3k 1 3k 1 3k 1  1 1 1





Let Ak  3k 1 3k 1 3k 1  then Ak 1  3k 1 3k 1 3k 1  1 1 1


3k 1 3k 1 3k 1 
3k 1 3k 1 3k 1  1 1 1




3k 1

 3k 1
3k 1

5.
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k 1
3k

 3k
3k

3k

The result is true for n  k 1 . Hence the result follows by induction.
3k
3k
3k 

3k 
3k 
3k 1 

3k 1 
3k 1 

3.3k 1  3k 
Let C  BAB then C    BAB   B. A  B  (reversal law)
 BAB
Now if A is symmetric A  A and thus C  C  C is symmetric
If A is skew symmetric then A   A and then C  C  C is skew symmetric.
6.
x
x  0 2 y z 
0

AA   2 y y  y   x y  z 
 z  z z   x  y z 
2 x2

 0
 0

If AA  I then 2 x 2  1,6 y 2  1,3z 2  1 
x
9
0
6 y2
0
0 

0 
3 z 2 
1
1
1
,y
,z  
2
6
3
7.
Note that LHS will reduce to 1 1 matrix which will easily yield x  1 .
10.
 2.50
10000 2000 18000  
A
1.50 


 6000 20000 8000  23 
1.00  31
1000  2.50  2000 1.50  18000 1.00 


 6000  2.50  20000 1.50  8000 1.0 
2
10000 2000 18000   
B
 1 
 6000 20000 8000   
1 2
 46000  Total revenue in market I


53000  Total revenue in market II
31000 


36000 
15000

17000
Profits are given by the matrix A  B  
11.
Let X be n 2 (Number of columns in X has to be 2 why?)
Also  n 2    2  3 must become n 3
a b 
 n should also be 2 . Thus we may take X  

c d 
 a b  1 2 3  7 8 9



 c d   4 5 6  2 4 6 
Now 

 a  4b 2a  5b 3a  6b   7 8 9
c  4d 2c  5d 3c  6d    2 4 6 

 

1 2
.
2 0 
On comparing we will easily get the values of a, b, c, d and will finally get X  
12.(i)
To prove AB n  B n A , The result is true for n  1 since AB  BA
Let ABk  Bk A for some k then AB k 1  AB k .B
 B k AB
(ii)

AB k  B k A   B k BA

AB  BA  B k 1 A
n n
To prove  AB   A B for n  N . The result is true for n  1
n
10
Let  AB   Ak .B k then  AB 
k
13.
k 1
  AB  . AB
(From first part B k A  AB k )
 Ak  AB k  B

A2  

 2  

 0
    
     
 2    1
If A2  I

If A  A, A   A then A   A 
15.
 I  A
  I  A  I  A 

 I  2A  A
 I  A
3
  I  A  I  3 A 
 I  A
3


   
0
2
2 A  0  A  0  A is a zero matrix.
 I 2  IA  AI  A2
A2  A, AI  IA  A   I  3A
I 2  4 AI  3 A  I  7 A

 Ak 1Bk 1
 c  is correct.
14.
2
 Ak .B k AB (By induction hypothesis)
k
 7A  I
 I 2  3IA  AI  3 A2


AI  A
 C  is correct
EXERCISE 4.2
x a
1.
D y b
z
c
xa
y  b , Applying C1  C1  C2
xa a
xa
D  y b b
y b
zc
zc
z c
which is equal to zero since Ist and 3rd columns are identical.
0 bc ca
2.
Applying C1  C1  C2  C3 , D  0
c a a b  0
0 a b b c
2 7 5
3.
Applying C3  C3  10C2 ,
D  3 8 5
5 9 4
11
c
2 5 5
Again applying C2  C2  C1
D  3 5 5
5 4 4
2 5 5

 3 5 5 0
IInd and IIIrd column are identical).
5 4 4
4.
5.
Apply C3  C3  C2 , After which ab  bc  ac is common from IIIrd column.
bc
qr
yz
D  ca
r p
z  x applying R1  R1  R2  R3
ab
pq x y
2c
2r
2z
D  ca
r p
zx
ab
pq
x y
r
z
2 ca
r p
zx
ab
pq
x y
c
r
z
c
r
z
2 a
p
x
p
pq
xz
R2  R2  R1  2 a
p
x R3  R3  R2
y
ab
a
p
x
a
p
x
 2 c
r
z   2  1 p
q
y  RHS
p
q
y
r
z
0
a b
0
6.
c
Let D  a
b
c
a b
0 c then D  a
c
q
c (Interchanging rows and columns)
0
b c 0
0
0 a b
 a
0
c (-common from IIIrd row)
b
c
0
0
a b
        a 0 c (-common from Ist and second row)
b
c
0
12
Thus D   D giving D  0 .
7.
Taking a, b, c common from R1 , R, R3
a
b
c
0
0
2c
D  abc a
b
c
 abc a b
c
a
b
c
a
c
b
 R1  R2   abc 2c  ab  ab 
8.(i)
R2  R1 , R3  R1 ,  b  a  ,  c  a  become common
(ii)
C2  C1 , C3  C1 ,  b  a  ,  c  a  become common.
9.
x
x2
y
y
2
zx
z
z2
xy
x2

x2
yz
xyz 2
y
xyz 2
z
1 1 1
 . . y2
x y z 2
z
z x
2
xyz
z3
xyz
y
y 3 1 (Taking xyz common from C3 )
z3 1
x3
 y x
xyz
3
x3 1
x2
2
x3
3
0
  y  x  z  x  y  x
2
0
zx
2
y x
2
z x
3
3
x2
1
x2
x3
  y  x  z  x  y  x
zy
2
z 2  x 2  zx
0
0  R3  R2 
2
z  zx  y  xy 0
2
2
  z  y  z  y   x  z  y 
Now z 2  zx  y 2  xy  z 2  y 2  zx  xy
x2

1
y  x  xy 0
2
1
y  x  xy
2
x3
D   y  x  z  x  z  y  y  x
x3
1
y  x  xy 0
2
2
z yx
1
0
  y  x  z  x  z  y   y  x  z  y  x    y 2  x 2  xy 
13
 4a 2b 2 c 2 .
2
  x  y  y  z  z  x   z  y  x    y  x   y 2  x 2  xy 


  x  y  y  z  z  x  xy  yz  zx 
10. (i) You may start with operation C1  C1  C2  C3 to get 5x  y common
(ii)
11.(i)
You may start with operation C1  C1  C2  C3
Let us be strategic. Since we have to bring a  b  c several times. We try to create
a  b  c . Applying C1  C1  C2 , C3  C3  C2
a  b  c
D  abc
2a
0
bca
abc
2c
 a  b  c
0
(You can not apply C1  C1  C2 , C3  C3  C1 simultaneously)
Taking ( a  b  c ) common from first and third column
1
D  a  b  c 1
2
0
2a
0
bca
1
2c
1
  a  b  c  1 b  c  a  2c   2a  1
2
  a  b  c  b  c  a  2c    a  b  c 
2
(ii)
12.
3
Start with C1  C1  C2  C3 to take  x  y  z  common.
Start with C1  C1  C2  C3 to take 1  x  x 2 common. Note that
1  x3  1  x  1  x  x 2  After which two zeros can be easily created along first column.
13.
1  a 2  b2
0
2ab
1 a  b
Applying C2  C2  aC3 , D 
2

2a
2a  a 1  a 2  b 2  1  a 2  b 2
2b

2b
2

Nothing that 2a  a 1  a 2  b 2   a 1  a 2  b 2
14

D  1  a  b
2
D  1  a  b
2
2
2

1  a 2  b2
0
2b
2ab
1
2a
2b
a 1  a 2  b 2
1  a 2  b2

0
b 1  a  b
2
 1  a  b
2

2 2
2

1
0
2b
0
1
2a
,Applying C1  C1  C3
0
2b
1
2a
a 1  a 2  b 2
b a 1  a 2  b 2
 1  a 2  b2  1  a 2  b2  2a 2  2b  b 
2
 1  a 2  b2 
3
14.
Taking a, b, c common from first row, second row and third row
a
D  abc
1
a
b
a
b
a
b
c
1
b
c
c
1
c
Multiplying a, b, c , in the first, second and third columns (Note)
a2  1
b2
c2
a2
b2  1
c2
D
a
1
D a
2
0
15.
 a1

A   a2
a
 3
2
b
1
2
0
b  1 c2
2
1
b1
b2
b3
,Now apply R1  R1  R2 , R3  R3  R2
c 1
2
 b 2  1  c 2  1 a 2  0 
 a 2  b2  c2  1
1
c1 

c2 
c3 
det A  A  a1b2c3  a2b3c1  a3b1c2  a1b3c2  a3b2c1  a2b1c3
15
 ka1

Now kA  ka2

 ka
 3

kb1
kb2
kb3
kc1 

kc2 
kc3 
det kA   ka1  kb2  kc3    ka2  kb3  kc1   ......
 k 3  a1b2c3  a2b3c1  .......
 k3 A
EXERCISE 4.5
17.
A  AdjA   det A I3 
det  A  AdjA     det A  det I 3 (From Q.16 of 4.2)
3
Now det I 3  1 and det  AB    det A det B 

18.
 det A  det  AdjA    det A
det AA1  det I 3 
3

det A det A1  1 
det  AdjA    det A  
2
det A1   det A  
1
 B
 B
MISCELLANEOUS EXERCISE
1.


Directly expanding D  x  x 2  1  sin    x sin   c cos    cos    sin   x cos  
 x   x 2  1  x sin 2   sin  cos   sin  cos   x cos 2   x   x 2  1  x   x 3
a2
2.
1 1 1
1 2
b
Take , , common from R1 , R2 and R3 respectively then D 
a b c
abc 2
c
a3
abc
b
3
abc
c
3
abc
Now take abc common from third column. Then C2  C3 and finally C1  C2 .
3.
Taking cos  common from first row, sin  common from second row and sin 
common from third row.
cos 
D  cos  sin  sin  1
cos 
sin 
 tan 
cot 
0
sin 
cot 
16
, Applying R3  R3  R2
cos 
D  cos  sin  sin  1
sin 
 tan 
cot 
0
0
cot   tan 
0
Expanding along third row
D   cos  sin  sin   cot   tan    cos  cot   sin  
 cos  sin  sin 
1
cos  sin 
 cos2   sin 2  


sin 


1
1 ca ab
4.
Applying C1  C1  C2  C3 , D  2  a  b  c  1 a  b
bc
1 bc ca
1 ca ab
 2a  b  c 0 b  c c  a
0 b a c b
 2  a  b  c    b  c    c  a  b  a 


2
   a  b  c   2a 2  2b 2  2c 2  2ab  2ac  2bc 
2
2
2
   a  b  c   a  b    b  c    c  a  


Since a, b, c are real numbers the bracketed expression is zero only when a  b  c .
1
5.

R1  R1  R2  R3
 3x  a  1
x
x
xa
x
x
xa
1
1 x

6.
0
x
a 0  0   3x  a  a2  0 .Since a  0, x  a 3
0 0 a
 3x  a  0
Taking a, b, c common from first, second and third column respectively
a
D  abc a  b
b
c
ac
b
a
bc
c
, Applying R2  R2   R1  R3 
17
c
ac
a
c
ac
2c
2c
  abc  2c  0
1
1
a
D  abc 0
b bc
b bc
c
 2abc 2  a  c  b  c   b  c  a  c  
( 2c common from R2 )
c
(Expanding along first column)  4a 2b 2 c 2
7.
 AB 
9.
2 x  2 y can be taken common after C1  C1  C2  C3 .
10.
Two zeros can be brought along first columns by R2  R1 , R3  R1
11.
C3  C3  C1
1
 B 1 A1 8.
Lengthy question. Be non lethargic.
 2    
D   2    
  2    
 2 1
         2 1
 2 1
1 
2
       1   2
1  2
Now create zeros at a21 , a31 positions etc.
12.
By splitting property,
1 x
1 y
1 z
x2
x
x 2 1  px3
y
y 2 1  py 3  y
z
z2
1 x
x2
y 2  pxyz 1 y
z2
1 z
x
1  pz 3
z
x2 1
x
x2
px 3
y2 1  y
y2
py 3
z2 1
z2
pz 3
z
1 x
y 2 (Give details)  1  pxyz  1 y
z2
1 z
a  b  c a  b a  c
13.
C1  C1  C2  C3 , D  a  b  c
x2
y 2 etc.
z2
1 a  b a  c
b  c   a  b  c  1
3b
a  b  c c  b
3c
1 c  b
Now create zeros at a21 and a31 positions by R2  R2  R1 , R3  R3  R1
1
14.
1 p
Applying R3  R3  R1  R2 , D  2 3  2 p
0
1 p  q
4  3 p  2q
52p
2
18
3b
b  c
3c
1
1 p
D  2 3  2 p 1  p  2q
Applying C3  C3  C2
0
Applying R2  R2  2 R1
3 2p
2
1 1 p
15.
q
q
D 0
1
1 p
0
2
3 2p
 1 3  2 p  2  2 p   1
sin 
cos 
cos  cos   sin  sin 
D  sin 
cos 
cos  cos   sin  sin  ,Apply C3  C3  sin  C1  cos  cos  C2
sin 
cos 
cos  cos   sin  sin 
sin 
cos 
0
D  sin 
cos 
0 0
sin 
cos 
0
1
1
1
 u ,  v,  w
x
y
z
16.
Put
17.
Since 2b  a  b, to create zero at a23 position we apply C2  C2 
x2
D  x 3
x3
1
 C1  C2  , then
2
x  2a
1
1
1
 x  2  x  4  x  4   x  3  x  3 x  2b   x  2a  x  2b 
2
2
2
x4
x5
x  2c
x  2 x  3 x  2a

0
0
0
0
x  4 x  5 x  2c
18.
We can easily notice that cofactor of elements in a11 , a22 , a23 positions are yz, zx and xy
respectively. Cofactors of other elements are zero. Also det A  xyz .
 yz
1 
1
Thus A 
0
xyz 
 0
19.
0
zx
0

0   x 1

0    0
xy   0
0
y
1
0

0

0
z 1 


Just expanding detA  1 1  sin 2   sin    sin   sin    1 sin 2   1
19
 2 1  sin 2   .Now 1  sin 2  lies between 1 and 2 .
2 1  sin 2   lies from 2 to 4  det A 2, 4 

 D  is correct.
PAST YEARS CBSE BOARD QUESTIONS UP TO 2013
cos15
sin15
1
x
3 4
Find the minor of the element of second row and third column in the following
determinant 6
1
If

3.
2 3
2.
x
Evaluate
sin 75 cos 75
.
x
1.
1 2
, write positive value of x.
5
0
4 .
5
7
4 a bc
2
4.
What is the value of 4 b c  a ?
4 c ab
6.
If A 
7.
For what value of x, the matrix
8.
If A 
5.
4
Write value of 5 6
8 .
6 x 9 x 12 x
1 2
, then find k, if | 2 A | k  | A | .
4 2
3
1
2 3
5 x
x 1
2
4
, then find | adjA | . 9.
If
is singular matrix?
x  1 x  1 4 1

, then write the value of x.
x3 x 2 1 3
2 3
10.
3
If Aij is the cofactor of the element aij of the determinant 6
1
5
0
4 , then write the value of
5
7
a32  A32 .
11.
Let A be a square matrix of order 3  3. Write the value of |2A|where,|A|=4.Ans: 32
5 3 8
12.
If   2 0 1 , write the minor of the element a23 .
1 2 3
20
102 18 36
13.
Write the value of the determinant 1
17
3
4 .
3
6
14.
If the determinant of matrix A of order 3  3 is of value 4, write the value of |3A|.
15
For what value of x, A 
16.
If A is a non-singular matrix of order 3 and | adjA || A |k , then what is value of k?
17.
Evaluate
18.
 a  b 2a  c   1 5 
Find the value of a, if 

.
 2a  b 3c  d   0 13
19.
 9 1 4 
1 2 1
 A 
If 

 , then find the matrix A.
 2 1 3 
0 4 9 
20.
 1 1
If matrix A  
and A2  kA, then write the value of k.

 1 1 
21.
 0 1 2 
For what value of x, is the matrix A   1 0 3  a skew-symmetric matrix?
 x 3 0 
22.
 cos θ sin θ 
 sin θ
 sin θ 
Simplify cos θ 

  sin θ cos θ 
cos θ
23.
x
Find the value of y-x from following equation 2 
7
24.
cos α  sin α 
If A  
 , then for what value of α , A is an identity matrix?
 sin α cos α 
25.
1 2   3 1   7 11 
If 


 , then write the value of k.
3 4   2 5  k 23
a  ib
c  id
c  id
a  ib
2( x  1)
2x
x
x2
is a singular matrix?
.
 cos θ 
sin θ 
21
5  3 4   7 6 


.
y  3 1 2  15 14 
26.
If A is a matrix of order 3  4 and B is a matrix of order 4  3 , find order of matrix (AB).
27.
3
If A   1
 0
28.
1 2 
If A  
 , find A  A.
3 4 
29.
3 4
If A  
 , find A  A where A =transpose of A.
2 3
30.
If matrix A  1 2 3, write A A .
31.
2 0 1
If A   2 1 3  , find value of A2  3 A  2I3
1 1 0 
32.
Express the following matrix as a sum of a symmetric and a skew-symmetric matrix and verify
T
4
 1 2 1 
, then find AT  BT .
2  and B  
1 2 3

1 
 3 2 4 
your result for  3 2 5 .
 1 1
2 
33.
Using elementary transformation, find inverse of following matrices
(i)
6 5 
5 4 .


3 2
 2 5
. (iii) A  
(ii) 

.
7 5 
 1 3
 1 1 2 
(iv).  1 2 3 
 3 1 1 
 2 0 1
(v).  5 1 0 
 0 1 3 
 1 3 2 
(vi).  3 0 1
 2 1 0 
34.
 4 4 4 
Determine the product of matrices  7 1 3  and
 5 3 1
1 1 1 
1 2 2  and then use the result


 2 1
3 
to solve the system of equations x  y  z  4, x  2 y  2 z  9 and 2 x  y  3z  1.
22
1
35.
2
3
1
Find A , where A  2 3 2 . Hence solve the system of equations, x  2 y  3z  4,
3 3 4
2 x  3 y  2 z  2 and 3x  3 y  4 z  11.
36.
37.
 2 1 1 
If A   3 0 1 , find A1 using A1 solve the following of equations 2 x  y  z  3, 3x  z  0
 2 6 0 
and 2 x  6 y  2  0.
Using properties of determinants, prove the following
x  y x  z
3x
(i)
(ii).
(iii).
x y
3y
xz
yz
z  y  3( x  y  z )( xy  yz  zx).
3z
a 2  (b  c)2
b 2  (c  a ) 2
c 2  ( a  b) 2
a2
b2
c2
x
y
x
2
y
2
x
3
y
3
bc
ca  (a  b)(b  c)(c  a )(a  b  c )(a 2  b 2  c 2 ).
ab
z
x2
2x  3
3x  4
2
z  xyz ( x  y )( y  z )( z  x ). (iv).
x4
2x  9
3x  16  0.
3
x  8 2 x  27 3x  64
z
ax ax ax
(v)
a  x a  x a  x  0.
(vi).
ax ax ax
(vii).
a
b
2
2
a
b
xa
x
x
x
xa
x
x
x
xa
 0.
c
c 2  (a  b)(b  c)(c  a)(ab  bc  ca).
bc ca ab
(viii).
α
β
γ
α
β
γ2
2
2
 (α  β )( β  γ)(α  β  γ ).
βγ αγ α β
(ix).
1 x
1
1
1 y
1
1
x y
1
1  xyz  xy  yz  zx.
(x).
1 z
5x  4 y
x
x
4 x 2 x  x3.
10 x  8 y 8 x 3x
23
a
(xi).
b
c
a  b b  c c  a  a 3  b3  c3  3abc.
bc ca ab
(xii).
x
Show that if x  y  z and y
z
x 2 1  x3
y 2 1  y 3  0, then 1  xyz  0.
z 2 1  z3
ANSWERS
1.
0
2.
2
3.
13
7.
x3
8.
11
9.
x2
13.
0
14.
108
15
x  2
18.
a 1
19.
 8 3 5 
 2 3 6 


α  0
24.
10.
16.
k2
k  17 26.
0
5.
0
6.
k4
110
11.
32
12.
7
k2
17.
a 2  b2  c 2  d 2
21.
x2
22.
27.
4 3
 3 0  28.


 1 2 
3 3
23.
7
29.
6 6 
6 6 


33.
Using elementary transformation, find inverse of following matrices
(i).
 4 5 
 5 6 


(iv)
 1 1 1 
 8 7 5


 5 4 3 
34.
x  3, y  2 and z  1 35.
30.
(ii)
14
25.
20.
4.
31.
 5 2 
 7 3 


(v)
unit matrix
 2 5
5 8


 1 1 1
 3 3 4 


 3 2 0 
(iii)
 3 5
 1 2 


1 1 
 3
 15 6 5


 5 2 2 
(vi)
x  3, y  2 and z  1
24
 1 2 3
 2 4 7 


 3 5 9 
36.
1
1
3
x   , y  ,z  
2
2
2