NCERT Solutions for Class 12 Maths Chapter 3 Matrices

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Maths
Chapter : 3
Chapter Name : Matrices

[ ]
2 5 19 −7
5
Q1 In the matrix A = 35 −2 2
12 ,write

√3 1 −5 17

(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elementsa 13, a 21, a 33, a 24, a 23

Answer.
(i) In the given matrix, the number of rows is 3 and the number of columns is 4.
Therefore, the order of the matrix is 3 × 4 .
(ii) since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it.
5
(iii) a 13 = 19, a 21 = 35, a 33 = − 5, a 24 = 12, a 23 = 2

Page : 64 , Block Name : Exercise 3.1

Q2 If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13
elements?

Answer. We know that if a matrix is of the order m x n, it has mn elements. Thus, to nd all the
possible orders of a matrix having 24 elements, we have to nd all the ordered pairs of natural
numbers whose product is 24.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and
Hence, the possible orders of a matrix having 24 elements are:
1 x 24,24 x 1, 2 x 12, 12 x 2, 3 x 8, 8 x 3, 4 x 6, and 6 x 4
(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 x 13 and 13 x 1.

Page : 64 , Block Name : Exercise 3.1

Q3 If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5
elements?

Answer. We know that if a matrix is of the order m x n, it has mn elements. Thus, to nd all the
possible orders of a matrix having 18 elements, we have to nd all the ordered pairs of natural

numbers whose product is 18. The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are: 1 x 18, 18 x 1, 2 x 9, 9 x 2,3 x 6 and 6
x 3 (1, 5) and (5, 1) are the ordered pairs Of natural numbers whose product is 5. Hence, the
possible orders Of a matrix having 5 elements are 1 x 5 and 5 x 1.

Page : 64 , Block Name : Exercise 3.1

Q4 Construct a 2 × 2 matrix, A = a y , whose elements are given by: [ ]
( i + j )2
(A) a ij = 2
i i
(B)a ij = j a ij = j
( i + 2j ) 2
(C) a ij = 2

[ ]
a 11 a 12
Answer. In general, a 2 × 2 matrix is given by A =
a 21 a 22
( i + j )2
a ij = 2
; i, j = 1, 2
( 1 + 1 )2 4 ( 1 + 2 )2 9
∴ a 11 = 2
= 2 =2 a 12 = 2
= 2
( 2 + 1 )2 9 ( 2 + 2 )2 16
a 21 = 2
= 2 a 22 = 2
= 2 =8

[ ]
9
2 2
Therefore, the required matrix is A = 9
2 8

(ii)
i
a ij = j , i, j = 1, 2
1 1
∴ a 11 = 1 = 1 a 12 = 2
2 2
a 21 = 1 = 2 a 22 = 2 = 1

[ ]
1
1 2
Therefore, the required matrix is A =
2 1

(iii)
( i + 2j ) 2
a ii ) a y = 2
, i, j = 1, 2
( 1 + 2 )2 32 9 ( 1 + 4 )2 52 25
∴ a 11 = 2
= 2 = 2 a 12 = 2
= 2 = 2
( 2 + 2 )2 42 ( 2 + 4 )2 62
a 21 = 2
= 2 =8 a 22 = 2
= 2 = 18

[ ]
9 25
2 2
Therefore, the required matrix is A =
8 18

Page : 64 , Block Name : Exercise 3.1

Q5 Construct a 3 × 4 matrix, whose elements are given by:
1
(i) a ij = 2 | − 3i + j |
(ii) a ij = 2i − j

Answer. In general, a 3 x 4 matrix is given by

[ ]
a 11 a 12 a 13 a 14

A= a 21 a 22 a 23 a 24
a 31 a 32 a 33 a 34

(i)
1
a ij = 2 | − 3i + j | , i = 1, 2, 3 and j = 1, 2, 3, 4
1 1 1 2
∴ a 11 = 2 | − 3 × 1 + 1 | = 2 | − 3 + 1 | = 2 | − 2 | = 2 = 1

1
| 1
a 21 = 2 − 3 × 2 + 1 = 2 − 6 + 1 = 2 − 5 = 2
| | | |
1 5

1 1 1 8
a 31 = 2 | − 3 × 3 + 1 | = 2 | − 9 + 1 | = 2 | − 8 | = 2 = 4
1 1 1 1
a 12 = 2 | − 3 × 1 + 2 | = 2 | − 3 + 2 | = 2 | − 1 | = 2
1 1 1 4
a 22 = 2 | − 3 × 2 + 2 | = 2 | − 6 + 2 | = 2 | − 4 | = 2 = 2
1 1 1 7
a 32 = 2 | − 3 × 3 + 2 | = 2 | − 9 + 2 | = 2 | − 7 | = 2
1 1
a 13 = 2 | − 3 × 1 + 3 | = 2 | − 3 + 3 | = 0
1 1 1 3
a 23 = 2 | − 3 × 2 + 3 | = 2 | − 6 + 3 | = 2 | − 3 | = 2
1 1 1 6
a 33 = 2 | − 3 × 3 + 3 | = 2 | − 9 + 3 | = 2 | − 6 | = 2 = 3
1 1 1 1
a 14 = 2 | − 3 × 1 + 4 | = 2 | − 3 + 4 | = 2 | 1 | = 2
1 1 1 2
a 24 = 2 | − 3 × 2 + 4 | = 2 | − 6 + 4 | = 2 | − 2 | = 2 = 1
1 1 1 5
a 34 = 2 | − 3 × 3 + 4 | = 2 | − 9 + 4 | = 2 | − 5 | = 2

[ ]
1 1
1 2 0 2
5 3
Therefore, the required matrix is A = 2 2 2 1
7 5
4 2 3 2

(ii)
a ij = 2i − j, i = 1, 2, 3 and j = 1, 2, 3, 4
∴ a 11 = 2 × 1 − 1 = 2 − 1 = 1
a 21 = 2 × 2 − 1 = 4 − 1 = 3
a 31 = 2 × 3 − 1 = 6 − 1 = 5
a 12 = 2 × 1 − 2 = 2 − 2 = 0
a 22 = 2 × 2 − 2 = 4 − 2 = 2
a 32 = 2 × 3 − 2 = 6 − 2 = 4
a 13 = 2 × 1 − 3 = 2 − 3 = − 1
a 23 = 2 × 2 − 3 = 4 − 3 = 1
a 33 = 2 × 3 − 3 = 6 − 3 = 3
a 14 = 2 × 1 − 4 = 2 − 4 = − 2
a 24 = 2 × 2 − 4 = 4 − 4 = 0
a 34 = 2 × 3 − 4 = 6 − 4 = 2

[ ]
1 0 −1 −2
Therefore, the required matrix is A = 3 2 1 0
5 4 3 2

Page : 64 , Block Name : Exercise 3.1

Q6 Find the values of x, y and z from the following equations

(i)
[ ] [ ]
4
x
3
5
=
y
1 5
z

(ii)
[ x+y
5+z
2
xy ] [ ]
=
6 2
5 8

[ ] []
x+y+z 9
(iii) x+z = 5
y+z 7

Answer. (i) [ ] [ ]
4
x
3
5
=
y
1
z
5
As the given matrices are equal, their corresponding elements are also equal. Comparing the
corresponding elements, we get:
x = 1, y = 4, and z = 3

(ii) [ x+y
5+z
2
xy ] [ ]
=
6
5
2
8
As the given matrices are equal, their corresponding elements are also equal. Comparing the
corresponding elements, we get:
x + y = 6, xy = 8, 5 + z = 5
Now, 5 + z = 5 ⇒ z = 0
We know that:
(x − y) 2 = (x + y) 2 − 4xy
⇒ (x − y) 2 = 36 − 32 = 4
⇒ x − y = ± 2

Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2
When x − y = − 2 and x + y = 6, we get x = 2 and y = 4
∴ x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0

[ ] []
x+y+z 9
(iii) x+z = 5
y+z 7
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x + y + z = 9…(1)
x + z = 5…(2)
y + z = 7…(3)
From (1) and (2), we have:
y+5=9
⇒ y=4
Then, from (3), we have:
4+z=7
⇒ z=3
∴ x+z=5
∴ x=2
∴ x = 2, y = 4, and z = 3

Page : 64 , Block Name : Exercise 3.1

Q7 Find the value of a, b, c and d from the equation:

[ a−b
2a − b
2a + c
3c + d ] [
=
−1
0
5
13 ]
Answer. [ a−b
2a − b
2a + c
3c + d ] [
=
−1
0
5
13 ]
As the two matrices are equal, their corresponding elements are also equal. Comparing the
corresponding elements, we get:
a − b = − 1…(1)
2a − b = 0…(2)
2a + c = 5…(3)
3c + d = 13…(4)
From (2), we have:
b = 2a
Then, from (1), we have:
a − 2a = − 1
⇒ a=1
⇒ b=2
Now, from (3), we have:
2×1+c=5
⇒ c=3
From (4) we have:
3 × 3 + d = 13
⇒ 9 + d = 13 ⇒ d = 4
∴ a = 1, b = 2, c = 3, and d = 4

Page : 64 , Block Name : Exercise 3.1

Q8

[ ]
A = a ij
m×m
is a sqaure matrix, if
(A)m < n
(B)m > n
(C) m=n
(D) none of these

Answer.
The correct answer is C.
It is known that a given matrix is said to be a square matrix if the number of rows is
equal to the number of columns.
Therefore,A = a ij [ ]
m×n
is a square matrix, if m = n

Page : 64 , Block Name : Exercise 3.1

Q9 Which of the given values of x and y make the following pair of matrices equal

[ 3x + 7
y+1
5
2 − 3x ] [ =
0 y−2
8 4 ]
−1
(A) x = 3 , y = 7
(B)Not Possible to nd
−2
(c) y = 7, x= 3
−1 −2
(D) x = 3 , y= 3

Answer. The correct answer is B

It is given that
[ 3x + 7
y+1
5
2 − 3x ] [ =
0 y−2
8 4 ]
Equating the corresponding elements , we get,
7
3x + 7 = 0 ⇒ x = − 3

5=y−2⇒y=7
y+1=8⇒y=7
2
2 − 3x = 4 ⇒ x = − 3
We nd that on comparing the corresponding elements Of the two matrices, we get two
different values Of x, which is not possible.
Hence, it is not possible to nd the values Of x and y for which the given matrices are
equal.

Page : 64 , Block Name : Exercise 3.1

Q10 The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
27
28
81
512

Answer.
The correct answer is D.
The given matrix of the order 3 × 3 has 9 elements and each of these elements can be
either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Therefore, by the multiplication principle, the required number of possible matrices is 2 9
= 512

Page : 64 , Block Name : Exercise 3.1

Q1 Let A =
[ ] [ ] [ ]
2
3
4
2
,B =
1
−2
3
5
,C =
−2
3
5
4

Find each of the following:
(i) A+B
(ii) A-B
(iii) 3A-C
(iv) AB
(v) BA

Answer. (i) A + B =
[ ][ ] [
2
3
4
2
+
1
−2
3
5
=
2+1
3−2
4+3
2+5 ] [ ]
=
3
1
7
7

(ii) A − B =
[ ][ ] [
2
3
4
2
−
−2
1 3
5
=
2−1
3 − ( − 2)
4−3
2−5 ] [ ]
=
1
5
1
−3

(iii)

[ ][ ]
3A − C = 3
2
3
4
2
−
−2
3
5
4

[ =
][ ]3×2
3×3
3×4
3×2
−
−2
3
5
4

[ ][ ]
=
6
9
12
6
−
−2
3
5
4

=
[ 6+2
9−3] 12 − 5
6−4

=
[ ]8 7
6 2
(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B
Therefore, AB is defined as:

AB =
[ ][ ] [
2 4
3 2
1
−2
3
5
=
2(1) + 4( − 2)
3(1) + 2( − 2)
2(3) + 4(5)
3(3) + 2(5) ]
=
[
2−8
3−4 ] [ ] 6 + 20
9 + 10
=
−6
−1
26
19

(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A .
Therefore, BA is defined as:

BA =
[ ][ ] [
1
−2
3
5
2
3
4
2
= 11 − 2(2) + 5(3) − 2(4) + 5(2) ]

=
[ 2+9
] [
− 4 + 15
4+6
− 8 + 10
=
11
11
10
2 ]
Page : 64 , Block Name : Exercise 3.2

Q2 Compute the following:

(i)
[ ][ ]
−b
a b
a
+
a
b
b
a

[ ][ ]
→2
a2 + b b2 + c2 2ab 2bc
(ii) +
a2 + c2 a2 + b2 − 2ac − 2ab

[ ][ ]
−1 4 −6 12 7 6
(iii) 8 5 16 + 8 0 5
2 8 5 3 2 4

(iv)
[ cos 2x
sin 2x
sin 2x
cos 2x ][ +
sin 2x
cos 2x
cos 2x
sin 2x ]
Answer. (i)
[ ][ ] [
−b
a b
a
+
a
b
b
a
=
a+a
−b + b
b+b
a+a ] [
=
2a 2b
0 2a ]
(ii)
[ a2 + b2
a2 + c2
b2 + c2
a2 + b2 ][ +
2ab
− 2ac
2bc
− 2ab ]
=
[ a 2 + b 2 + 2ab
a 2 + c 2 − 2ac
b 2 + c 2 + 2bc
a 2 + b 2 − 2ab ]
=
[ (a + b) 2
(a − c) 2
(b + c) 2
(a − b) 2 ]
[ ][ ]
−1 4 −6 12 7 6
(iii) 8 5 16 + 8 0 5
2 8 5 3 2 4

[ ]
− 1 + 12 4+7 −6 + 6
= 8+8 5+0 16 + 5
2+3 8+2 5+4

[ ]
11 11 0
= 16 5 21
5 10 9

(iv)
[ cos 2x
sin 2x
sin 2x
cos 2x ][
+
sin 2x
cos 2x
cos 2x
sin 2x ]
=
[ cos 2x + sin 2x
sin 2x + cos 2x
sin 2x + cos 2x
cos 2x + sin 2x ]
=
[ ](1
1
1
1
∵ sin 2x + cos 2x = 1 )
Page : 64 , Block Name : Exercise 3.2

Q3 Compute the indicated products.

(i)
[ ][ ]
a
−b
b
a
a
b
−b
a

[]
1
(ii) 2 [2 3 4 ]
3

(iii)
[ ][
1
2
−2
3
1 2
2 3
3
1 ]

[ ][ ]
2 3 4 1 −3 5
(iv) 3 4 5 0 2 4
4 5 6 3 0 5

[ ][ ]
2 1
1 0 1
(v) 3 2
−1 2 1
−1 1

][ ]
2 −3
(vi)
[ 3
−1
−1
0
3
2
1
3
0
1

Answer. (i)

[ ][ ]
−b
a b
a
a −b
b a

=
[ a(a) + b(b)
− b(a) + a(b)
a( − b) + b(a)
− b( − b) + a(a) ]
=
[ a2 + b2
− ab + ab
− ab + ab
b2 + a2 ][ =
a2 + b2
0
0
a2 + b2 ]
[] [ ][ ]
1 1(2) 1(3) 1(4) 2 3 4
(ii) 2 [2 3 4 ] = 2(2) 2(3) 2(4) = 4 6 8
3 3(2) 3(3) 3(4) 6 9 12

(iii)
[ ][1
2
−2
3
1
2
2
3
3
1 ]
=
[ 1(1) − 2(2)
2(1) + 3(2)
1(2) − 2(3)
2(2) + 3(3)
1(3) − 2(1)
2(3) + 3(1) ]
=
[ 1−4
2+6
2−6
4+9
3−2
6+3 ] [ =
−3
8
−4
13
1
9 ]

[ ][ ]
2 3 4 1 −3 5
(iv) 3 4 5 0 2 4
4 5 6 3 0 5

[ ]
2(1) + 3(0) + 4(3) 2( − 3) + 3(2) + 4(0) 2(5) + 3(4) + 4(5)
= 3(1) + 4(0) + 5(3) 3( − 3) + 4(2) + 5(0) 3(5) + 4(4) + 5(5)
4(1) + 5(0) + 6(3) 4( − 3) + 5(2) + 6(0) 4(5) + 5(4) + 6(5)

[ ][ ]
2 + 0 + 12 −6 + 6 + 0 10 + 12 + 20 14 0 42
= 3 + 0 + 15 −9 + 8 + 0 15 + 16 + 25 = 18 −1 56
4 + 0 + 18 − 12 + 10 + 0 20 + 20 + 30 22 −2 70

[ ][
2 1
(v) 3
−1
2
1
1
−1
0
2
1
1 ]

[ ]
2(1) + 1( − 1) 2(0) + 1(2) 2(1) + 1(1)
= 3(1) + 2( − 1) 3(0) + 2(2) 3(1) + 2(1)
− 1(1) + 1( − 1) − 1(0) + 1(2) − 1(1) + 1(1)

[ ][ ]
2−1 0+2 2+1 1 2 3
= 3−2 0+4 3+2 = 1 4 5
−1 − 1 0+2 −1 + 1 −2 2 0

][ ]
2 −3
(vi)
[ 3
−1
−1
0
3
2
1
3
0
1

= [ 3(2) − 1(1) + 3(3)
− 1(2) + 0(1) + 2(3)
3( − 3) − 1(0) + 3(1)
− 1( − 3) + 0(0) + 2(1) ]
= [ 6−1+9
−2 + 0 + 6
−9 − 0 + 3
3+0+2 ] [ =
14
4
−6
5 ]
Page : 64 , Block Name : Exercise 3.2

[ ] [ ] [ ]
1 2 −3 3 −1 2 4 1 2
Q4 if A = 5 0 2 , (B = 4 2 5 andC = 0 3 2 , then compute
1 −1 1 2 0 3 1 −2 3
(A + B) and (B − C). Also, verify that A + (B − C) = (A + B) − C

Answer.

][ ] [
1 2 −3 3 −1 2
A+B= 5 0 2 + 4 2 5
1 −1 1 2 0 3

[ ][ ]
1+3 2−1 −3 + 2 4 1 −1
= 5+4 0+2 2+5 = 9 2 7
1+2 −1 + 0 1+3 3 −1 4

[ ][ ]
3 −1 2 4 1 2
B−C= 4 2 5 − 0 3 2
2 0 3 1 −2 3

[ ][ ]
3−4 −1 − 1 2−2 −1 −2 0
= 4−0 2−3 5−2 = 4 −1 3
2−1 0 − ( − 2) 3−3 1 2 0

[ ][ ]
1 2 −3 −1 −2 0
A + (B − C) = 5 0 2 + 4 −1 3
1 −1 1 1 2 0

[ ][ ]
1 + ( − 1) 2 + ( − 2) −3 + 0 0 0 −3
= 5+4 0 + ( − 1) 2+3 = 9 −1 5
1+1 −1 + 2 1+0 2 1 1

[ ][ ]
4 1 −1 4 1 2
(A + B) − C = 9 2 7 − 0 3 2
3 −1 4 1 −2 3

[ ][ ]
4−4 1−1 −1 − 2 0 0 −3
= 9−0 2−3 7−2 = 9 −1 5
3−1 − 1 − ( − 2) 4−3 2 1 1
Hence we have veri ed that A + (B − C) = (A + B) − C

Page : 64 , Block Name : Exercise 3.2

[ ][ ]
2 5 2 3
3 1 3 5 5 1
1 2 4 1 2 4
Q5 if A = 3 3 3 andB = 5 5 5 then compute 3A-5B.
7 2 7 6 2
3 2 3 5 5 5

Answer.

[ ][ ]
2 5 2 3
3 1 3 5 5 1
1 2 4 1 2 4
3A − 5B = 3 3 3 3 −5 5 5 5
7 2 7 6 2
3 2 3 5 5 5

[ ][ ][ ]
2 3 5 2 3 5 0 0 0
= 1 2 4 − 1 2 4 = 0 0 0
7 6 2 7 6 2 0 0 0

Page : 64 , Block Name : Exercise 3.2

Q6 Simplify cosθ
[ cosθ
− sinθ
sinθ
cosθ ] [ + sinθ
sinθ
cosθ
− cosθ
sinθ ]

Answer.

cosθ
[ cosθ
− sinθ
sinθ
cosθ ] [ + sinθ
sinθ
cosθ
− cosθ
sinθ ]
=
[ cos 2θ
− sinθcosθ
cosθsinθ
cos 2θ ][
+
sin 2θ
sinθcosθ
− sinθcosθ
sin 2θ ]
=
[ cos 2θ + sin 2θ
− sinθcosθ + sinθcosθ
cosθsinθ − sinθcosθ
cos 2θ + sin 2θ ]
=
[ ](1
0
0
1
∵ cos 2θ + sin 2θ = 1 )
Page : 64 , Block Name : Exercise 3.2

Q7 Find X and Y, if

(i)X + Y =
[ ] 7
2
0
5
and X − Y =
[ ] 3 0
0 3

(ii) 2X + 3Y =
[ ] 2
4
3
0
and 3X + 2Y =
[ −1
2 −2
5 ]
Answer. (i)

X+Y=
[ ]
7
2
0
5
…(1)
. . . . (2)

X−Y=
[ ]
3
0
0
3
Adding equation (1) and (2)

2X =
[ ][ ] [
7
2
0
5
+
3
0
0
3
=
7+3
2+0
0+0
5+3 ] [ ]
=
10
2
0
8

∴ X= 2
[ ] [ ]
1 10
2
0
8
=
5
1
0
4

Now, X + Y =
[ ] 7
2
0
5

⇒
[ ] [ ]
5 0
1 4
+Y=
7
2
0
5

⇒ Y=
[ ][ ] 7
2
0
5
−
5
1
5
4

⇒ Y= [ ]7−5
2−1
0−0
5−4

∴ Y= [ ] 2 0
1 1

(ii)2X + 3Y = [ ] 2 3
4 0
…(3)

3X + 2Y = [ ] −1
2 −2
5
…(4)

Multiplying equations (3) with (2), we get:

2(2X + 3Y) = 2
| ]2
4
3
0

⇒ 4X + 6Y =
[ ] 4
8
6
0
…(5)

Multiplying equation (4) and (3), we get :

3(3X + 2Y) = 3 [ −1
2 −2
5 ]
⇒ 9X + 6Y = [ 6
−3
−6
15 ]…(6)

From (5) and (6), we have

(4X + 6Y) − (9X + 6Y) = [ ][ ]
4
8
6
0
−
6
−3
−6
15

⇒ − 5X = [ 4−6
8 − ( − 3) ] [ ]
6 − ( − 6)
0 − 15
=
−2
11
12
− 15

[ ]
2 12
− 5
∴ X= − 5
1
[ −2
11
12
− 15 ] =
− 5
5
11
3

Now,2X + 3Y =
[ ] 2
4
3
0

[ ] []
2 12
− 5
5 2 3
⇒ 2 11
+ 3Y =
4 0
− 5 3

[ ] []
4 24
− 5
5 2 3
⇒
22
+ 3Y =
4 0
− 5 6

[ ][ ]
4 24
− 5
2 3 5
⇒ 3Y = − 22
4 0
− 5 6

[ ][ ]
4 24 6 39
2− 5 3+ 5 5 5
⇒ 3Y = 22
= 42
4+ 5 0−6 5 −6

[ ][ ]
6 39 2 13
1 5 5 5 5
∴ Y = 3 42 = 14
5 −6 5 −2

Page : 64 , Block Name : Exercise 3.2

Q8 Find X, if Y =
[ ] 3
1
2
4
and 2X + Y =
[ ]
1
−3
0
2

Answer. 2X + Y =
[ ] 1
−3
0
2

⇒ 2X +
[ 3 2
1 4 ] [ ]=
−3
1 0
2

⇒ 2X =
[ 1
−3
0
2][ ] [ −
3
1
2
4
=
1−3
−3 − 1
0−2
2−4 ]
⇒ 2X =
[ −2
−4
−2
−2]
∴
1
X= 2
[ −2
−4
−2
−2] [ ] =
−1
−2
−1
−1

Page : 64 , Block Name : Exercise 3.2

Q9 Find x and y, if 2
[ ][ ] [ ]
1
0
3
x
+
y
1 2
0
=
5
1
6
8

Answer.

2
[ ][ ] [ ]
1
0
3
x
+
y
1 2
0
=
5 6
1 8

⇒
[ ][ ] [ ]
2
0
6
2x
+
y
1
0
2
=
5
1
6
8

⇒
[ ] [ ]
2+y
1
6
2x + 2
=
5 6
1 8
Comparing the corresponding elements of these two matrices, we have
2+y=5
⇒ y=3
2x + 2 = 8
⇒ x=3
∴ x = 3 and y = 3

Page : 64 , Block Name : Exercise 3.2

Q10 Solve the equation for x, y, z and t, if 2
[ ] [ ] [ ]
x
y
z
t
+3
1
0
−1
2
=3
3
4
5
6

Answer.

2
[ ] [ ] [ ]
x
y
z
t
+3
1
0
−1
2
=3
3
4
5
6

⇒
[ ][ ] [ ]
2x
2y
2z
2t
+
3
0
−3
6
=
9
12
15
18

⇒
[ ] [ ]
2x + 3
2y
2z − 3
2t + 6
=
9
12
15
18
Comparing the corresponding elements of these two matrices, we get:
2x + 3 = 9
⇒ 2x = 6
⇒ x=3
2y = 12
⇒ y=6

2z − 3 = 15
⇒ 2z = 18
⇒ z=9
2t + 6 = 18
⇒ 2t = 12
⇒ t=6
∴ x = 3, y = 6, z = 9, and t = 6

Page : 64 , Block Name : Exercise 3.2

Q11 If x [] [ ] [ ]
2
3
+y
−1
1
=
10
5
nd the values of x and y.

Answer.

x [] [ ] [ ]
2
3
+y
−1
1
=
10
5

⇒
[ ][ ] [ ]
2x
3x
+
−y
y
=
10
5

⇒
[ ] [ ]
2x − y
3x + y
=
10
5
Comparing the corresponding elements of these two matrices, we get:
2x y = 10 and 3x+ y = 5
Adding these two equations, we have:
5x=15
x=3
Now, 3x+y=5
y=5-3x
⇒ y = 5 − 9 = − 4

∴ x = 3 and y = − 4

Page : 64 , Block Name : Exercise 3.2

Q12 Given3
[ | [
x
z
y
w
=
x
−1
6
2w ][+
z+w
4 x+y
3 ]
, nd the values of x, y, z and w.

Answer.

3 [ ] [ ][
x
z
y
w
=
−1
x 6
2w
+
z+w
4 x+y
3 ]
⇒
[ ] [
3x
3z
3y
3w
=
x+4
−1 + z + w
6+x+y
2w + 3 ]

Comparing the corresponding elements of these two matrices, we get:
3x = x + 4
⇒ 2x = 4
⇒ x=2
3y = 6 + x + y
⇒ 2y = 6 + x + y
⇒ y=6+x=6+2=8
⇒ y=4
3w = 2w + 3
⇒ w=3
3z = − 1 + z + w
⇒ 2z = − 1 + w = − 1 + 3 = 2
⇒ z=1
∴ x = 2, y = 4, z = 1, and w = 3

Page : 64 , Block Name : Exercise 3.2

[ ]
cosx − sinx 0
Q13 If F (x) = sinx cosx 0 show that F(x) F(y) = F(x + y).
0 0 1

Answer.

[ ] [ ]
cosx − sinx 0 cosy − siny 0
F(x) = sinx cosx 0 , F(y) = siny cosy 0
0 0 1 0 0 1

[ ]
cos(x + y) − sin(x + y) 0
F(x + y) = sin(x + y) cos(x + y) 0
0 0 1
F(x)F(y)

[ ][ ]
cosx − sinx 0 cosy − siny 0
= sinx cosx 0 siny cosy 0
0 0 1 0 0 1

[ ]
cosxcosy − sinxsiny + 0 − cosxsiny − sinxcosy + 0 0
= sinxcosy + cosxsiny + 0 − sinxsiny + cosxcosy + 0 0
0 0

[ ]
cos(x + y) − sin(x + y) 0
= sin(x + y) cos(x + y) 0
0 0 1

= F(x + y)
∴ F(x)F(y) = F(x + y)

Page : 64 , Block Name : Exercise 3.2

Q14 Show that(i)

[ ][ ] [ ][ ]
5
6
−1
7
2 1
3 4
≠
2
3
1
4
5
6
−1
7
(ii)

[ ][ ][ ][ ]
1 2 3 −1 1 0 −1 1 0 1 2 3
0 1 0 0 −1 1 ≠ 0 −1 1 0 1 0
1 1 0 2 3 4 2 3 4 1 1 0

Answer.
(i)

[ ][ ]
5
6
−1
7
2 1
3 4

[
=
]
5(2) − 1(3)
6(2) + 7(3)
5(1) − 1(4)
6(1) + 7(4)

[
=
] [ ]
10 − 3
12 + 21
5−4
6 + 28
=
7
33
1
34

[ ][ ]
2
3
1
4
5
6
−1
7

[
=
]
2(5) + 1(6)
3(5) + 4(6)
2( − 1) + 1(7)
3( − 1) + 4(7)

[
=
] [ ]
10 + 6
15 + 24
−2 + 7
− 3 + 28
=
16
39
5
25

[ ][ ] [ ][ ]
∴
5
6
−1
7
2
3
1
4
≠
2
3
1
4
5
6
−1
7
(ii)

[ ][ ]
1 2 3 −1 1 0
0 1 0 0 −1 1
1 1 0 2 3 4

[ ]
1( − 1) + 2(0) + 3(2) 1(1) + 2( − 1) + 3(3) 1(0) + 2(1) + 3(4)
= 0( − 1) + 1(0) + 0(2) 0(1) + 1( − 1) + 0(3) 0(0) + 1(1) + 0(4)
1( − 1) + 1(0) + 0(2) 1(1) + 1( − 1) + 0(3) 1(0) + 1(1) + 0(4)

[ ]
5 8 14
= 0 −1 1
−1 0 1

[ ][ ]
−1 1 0 1 2 3
0 −1 1 0 1 0
2 3 4 1 1 0

[ ]
− 1(1) + 1(0) + 0(1) − 1(2) + 1(1) + 0(1) − 1(3) + 1(0) + 0(0)
= 0(1) + ( − 1)(0) + 1(1) 0(2) + ( − 1)(1) + 1(1) 0(3) + ( − 1)(0) + 1(0)
2(1) + 3(0) + 4(1) 2(2) + 3(1) + 4(1) 2(3) + 3(0) + 4(0)

[ ]
−1 −1 −3
= 1 0 0
6 11 6

[ ][ ] [ ][ ]
1 2 3 −1 1 0 −1 1 0 1 2 3
∴ 0 1 0 0 −1 1 ≠ 0 −1 1 0 1 0
1 1 0 2 3 4 2 3 4 1 1 0

Page : 64 , Block Name : Exercise 3.2

[ ]
2 0 1
Q15 Find A 2 − 5A + 6I, if A = 2 1 3
1 −1 0

Answer. We have A 2 = A × A

[ ][ ]
2 0 1 2 0 1
A 2 = AA = 2 1 3 2 1 3
1 −1 0 1 −1 0

[ ]
2(2) + 0(2) + 1(1) 2(0) + 0(1) + 1( − 1)
= 2(2) + 1(2) + 3(1) 2(0) + 1(1) + 3( − 1)
1(2) + ( − 1)(2) + 0(1) 1(0) + ( − 1)(1) + 0( − 1) 1(1) + ( − 1)(3) + 0(0)

[ ]
4+0+1 0+0−1 2+0+0
= 4+2+3 0+1−3 2+3+0
2−2+0 0−1+0 1−3+0
∴ A 2 − 5A + 6I

[ ][ ][ ]
5 −1 2 2 0 1 1 0 0
= 9 −2 5 −5 2 1 3 +6 0 1 0
0 −1 −2 1 −1 0 0 0 1

[ ][ ][ ]
5 −1 2 10 0 5 6 0 0
= 9 −2 5 − 10 5 15 + 0 6 0
0 −1 −2 5 −5 0 0 0 6

[ ][ ]
5 − 10 −1 − 0 2−5 6 0 0
= 9 − 10 −2 − 5 5 − 15 + 0 6 0
0−5 −1 + 5 −2 − 0 0 0 6

[ ][ ]
−5 −1 −3 6 0 0
= −1 −7 − 10 + 0 6 0
−5 4 −2 0 0 6

[ ]
−5 + 6 −1 + 0 −3 + 0
= −1 + 0 −7 + 6 − 10 + 0
−5 + 0 4+0 −2 + 6

[ ]
1 −1 −3
= −1 −1 − 10
−5 4 4

[ ]
1 0 2
Q16 If A = 0 2 1 , prove that A 3 − 6A 2 + 7A + 2I = 0
2 0 3

[ ][ ]
1 0 2 1 0 2
Answer. A 2 = AA = 0 2 1 0 2 1
2 0 3 2 0 3

[ ][ ]
1+0+4 0+0+0 2+0+6 5 0 8
= 0+0+2 0+4+0 0+2+3 = 2 4 5
2+0+6 0+0+0 4+0+9 8 0 13
Now A 3 = A 2 ⋅ A

[ ][ ]
5 0 8 1 0 2
= 2 4 5 0 2 1
8 0 13 2 0 3

[ ]
5 + 0 + 16 0+0+0 10 + 0 + 24
= 2 + 0 + 10 0+8+0 4 + 4 + 15
8 + 0 + 26 0+0+0 16 + 0 + 39

[ ]
21 0 34
= 12 8 23
34 0 55
∴ A 3 − 6A 2 + 7A + 2I

[ ][ ][ ][ ]
21 0 34 5 0 8 1 0 2 1 0 0
= 12 8 23 − 6 2 4 5 +7 0 2 1 +2 0 1 0
34 0 55 8 0 13 2 0 3 0 0 1

[ ][ ]
21 + 7 + 2 0+0+0 34 + 14 + 0 30 0 48
= 12 + 0 + 0 8 + 14 + 2 23 + 7 + 0 12 24 30
34 + 14 + 0 0+0+0 55 + 21 + 2 48 0 78

[ ][ ]
30 0 48 30 0 48
= 12 24 30 − 12 24 30
48 0 78 48 0 78

[ ]
0 0 0
= 0 0 0 =O
0 0 0
∴ A 3 − 6A 2 + 7A + 2I = O

Page : 82 , Block Name : Exercise 3.2

Q17 If A =
[ ] 3
4
−2
−2
and I =
[ ] 1
0
0
1
, find k so that A 2 = kA − 2I?

Answer. A 2 = A ⋅ A =
[ ][ ]
3 −2
4 −2
3 −2
4 −2

=
[ 3(3) + ( − 2)(4)
4(3) + ( − 2)(4)
3( − 2) + ( − 2)( − 2)
4( − 2) + ( − 2)( − 2) ] [ ]
=
1
4
−2
−4
Now A 2 = kA − 2I

⇒
[ ] [ ] [ ]
1
4
−2
−4
=k
3
4
−2
−2
−2
1 0
0 1

⇒
[ ] [ ][ ]
1
4
−2
−4
=
3k
4k
− 2k
− 2k
−
2
0
0
2

⇒
[ ] [
1
4
−2
−4
=
3k − 2
4k
− 2k
− 2k − 2 ]
Comparing the corresponding elements, we have:
3k − 2 = 1
⇒ 3k = 3
⇒ k=1
Thus the value of k is 1

Page : 82 , Block Name : Exercise 3.2

[ ]
α
0 − tan 2
Q18 If A = α
and I is the identity matrix of order 2, , show that
tan 2 0

I + A = (I − A)
[ cosα
sinα
− sinα
cosα ]
Answer. On the L.H.S

[ ]
α
− tan 2
[ ] 1 0 0
I+A= + α
0 1
tan 2 0

[ ]
α
1 − tan 2
= α
......(1)
tan 2 1

On the R.H.S

(I − A)
[ cosα
sinα
− sinα
cosα ]

( [ ] [ ]) [
α
− tan 2
]
1 0 0 cosα − sinα
= − α
0 1 sinα cosα
tan 2 0

[ ][ ]
α
1 tan 2
cosα − sinα
= α sinα cosα
− tan 2 1

[ ]
α α
cosα + sinαtan 2 − sinα + cosαtan 2
= α α
......(2)
− cosαtan 2 + sinα sinαtan 2 + cosα

[( ) ( )
]
α α α α α α α α
1 − 2sin 2 2 + 2sin 2 cos 2 tan 2 − 2sin 2 cos 2 + 2cos 2 2 − 1 tan 2
=
α α α α α α α α α
− 2cos 2 2 − 1 tan 2 + 2sin 2 cos 2 2sin 2 2sin 2 cos 2 tan 2 + 1 − 2sin 2 2

[ ]
α α α α α α α
1 − 2sin 2 2 + 2sin 2 2 − 2sin 2 cos 2 + 2sin 2 cos 2 − tan 2
= α α α α α α α
− 2sin 2 cos 2 + tan 2 + 2sin 2 cos 2 2sin 2 2 + 1 − 2sin 2 2

[ ]
α
1 − tan 2
= α
tan 2 1

Thus, from (1) and (2) ,we get L.H.S=R.H.S

Page : 82 , Block Name : Exercise 3.2

Q19 A trust fund has ₹ 30,000 that must be invested in two different types of bonds.The rst bond
pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix
multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund
must obtain an annual total interest of:
(a) ₹ 1800
(b) ₹ 2000

Answer. (a) Let Rs x be invested in the rst bond. Then, the sum of money invested in the second
bond will be ₹ (30000 — x).
It is given that the rst bond pays 5% interest per year and the second bond pays 7% interest per
year.
Therefore, in order to obtain an annual total interest of ₹ 1800, we have:

[] [
5

[x (30000 − x) ]
100
7
100
= 1800 S.I. for 1 year =
Principal × Rate
100 ]

5x 7 ( 30000 − x )
⇒
100
+ 100
= 1800

⇒ 5x + 21000 − 7x = 180000
⇒ 210000 − 2x = 180000
⇒ 2x = 210000 − 180000
⇒ 2x = 30000
⇒ x = 15000
Thus, in order to obtain an annual total interest Of ₹ 1800, the trust fund should invest ₹ 15000 in
the rst bond and the remaining ₹ 15000 in the second bond.
(b) Let Rs x be invested in the rst bond. Then, the sum Of money invested in the second bond will
be ₹ (30000 — x).
Therefore, in order to obtain an annual total interest of ₹ 2000, we have:

[]
5
100
[x (30000 − x) ]
7
= 2000
100

5x 7 ( 30000 − x )
⇒
100
+ 100
= 2000

⇒ 5x + 210000 − 7x = 200000
⇒ 210000 − 2x = 200000
⇒ 2x = 210000 − 200000
⇒ 2x = 10000
⇒ x = 5000
Thus, in order to obtain an annual total interest of ₹ 2000, the trust fund should invest
₹ 5000 in the rst bond and the remaining ₹ 25000 in the second bond.

Page : 82 , Block Name : Exercise 3.2

Q20 The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10
dozen economics books. Their selling prices are ₹80, ₹60 and ₹40 each respectively. Find the total
amount the bookshop will receive from selling all the books using matrix algebra. Assume X, Y, Z,
W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct
answer in Exercises 21 and 22

Answer. The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen
economics books. The selling prices Of a chemistry book, a physics book, and an economics book
are respectively given as ₹ 80, ₹ 60, and ₹ 40. The total amount of money that will be received from
the sale of all these books can be represented in the form Of a matrix as:

[]
80
12 [ 10 8 10 ] 60
40

= 12[10 × 80 + 8 × 60 + 10 × 40]
= 12(800 + 480 + 400)
= 12(1680)
= 20160
Thus, the bookshop will receive ₹ 20160 from the sale Of all these books.

Page : 83 , Block Name : Exeercise 3.2

Q21 The restriction on n, k and p so that PY + WY will be de ned are:
(A) k = 3, p = n
(B) k is arbitrary, p = 2
(C) p is arbitrary, k = 3
(D) k = 2, p = 3

Answer.
Matrices P and Y are of the orders p × k and 3 × k respectively.
Therefore, matrix PY will be defined if k = 3. Consequently, PY will be of the order p × k .
Matrices W and Y are of the orders n × 3 and 3 × k respectively.
Since the number of columns in W is equal to the number of rows in Y, matrix WY is
well-defined and is of the order n × k .
Matrices PY and WY can be added only when their orders are the same.
However, PY is of the order p × k and WY is of the order n × k. Therefore, we must have
p = n.
Thus, k = 3 and p = n are the restrictions on n, k, and p so that PY + WY will be
defined.

Page : 83 , Block Name : Exercise 3.2

Q22 IIf n = p, then the order of the matrix 7X – 5Z is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

Answer. The correct answer is B.
Matrix X is of the order 2 x n.
Therefore, matrix 7X is also of the same order.
Matrix Z is of the order 2 x p, i.e., 2 x n [Since n = p]
Therefore, matrix 5Z is also of the same order.
Now, both the matrices 7X and 52 are of the order 2 x n.
Thus, matrix 7X — 52 is well-de ned and is of the order 2 x n.

Page : 83 , Block Name : Exercise 3.2

Q1 Find the transpose of each of the following matrices:

[]
5
1
(i) 2
−1

(ii)
[ ]
1
2
−1
3

[ ]
−1 5 6

(iii) √3 5 6
2 3 −1

[]
5

Answer. (i) LetA =
1
2
−1
, then A T =
[ 5
1
2
−1
]
(ii) Let A =
[ ]
1
2
−1
3
, then A T =
[ ] 1
−1
2
3

[ ] [ ]
−1 5 6 −1 √3 2
(iii) Let LetA = √3 5 6 , then A T = 5 5 3
2 3 −1 6 6 −1

Page : 88 , Block Name : Exercise 3.3

[ ] [ ]
−1 2 3 −4 1 −5
Q2 If A = 5 7 9 and B = 1 2 0 then verify that,
−2 1 1 1 3 1
(i)(A + B) ′ = A ′ + B ′
(ii)(A − B) ′ = A ′ − B ′

[
] [ ]
−1 5 −2 −4 1 1
Answer. A ′ = 2 7 1 , B′ = 1 2 3
3 9 1 −5 0 1

[ ][ ][ ]
−1 2 3 −4 1 −5 −5 3 −2
(i) A + B = 5 7 9 + 1 2 0 = 6 9 9
−2 1 1 1 3 1 −1 4 2

[ ]
−5 6 −1
∴ (A + B) ′ = 3 9 4
−2 9 2

[ ][ ][ ]
−1 5 −2 −4 1 1 −5 6 −1
A′ + B′ = 2 7 1 + 1 2 3 = 3 9 4
3 9 1 −5 0 1 −2 9 2
Hence, we have veri ed that (A + B) ′ = A ′ + B ′

[ ][ ][ ]
−1 2 3 −4 1 −5 3 1 8
(ii) A − B = 5 7 9 − 1 2 0 = 4 5 9
−2 1 1 1 3 1 −3 −2 0

[ ]
3 4 −3
∴ (A − B) ′ = 1 5 −2
8 9 0

[ ][ ][ ]
−1 5 −2 −4 1 1 3 4 −3
′ ′ 2 7 1 1 2 3 1 5 −2
A −B = − =
3 9 1 −5 0 1 8 9 0
Hence we have veri ed that (A − B) ′ = A ′ − B ′

Page : 88 , Block Name : Exercise 3.3

[ ]
3 4
Q3 If A ′ = −1
0
2 and B =
1
[ −1 2
1 2
1
3 ] , then verify that

(i) (A + B) ′ = A ′ + B ′
(i) (A − B) ′ = A ′ − B ′

Answer.

(i) It is known that A = A ′ ( ) ′

Therefore, we have:

A=
[ 3
4
−1
2
0
1 ]
[ ]
−1 1
B′ = 2 2
1 3

A+B=
[ 3
4
−1
2
0
1 ][ +
−1
1
2
2
1
3 ] [ =
2
5
1
4
1
4 ]
[ ]
2 5
∴ (A + B) ′ = 1 4
1 4

[ ][ ][ ]
3 4 −1 1 2 5
A′ + B′ = −1 2 + 2 2 = 1 4
0 1 1 3 1 4
Hence we have veri ed that (A + B) ′ = A ′ + B ′

(ii) A − B =
[ 3
4
−1
2
0
1 ][ −
−1
1
2
2
1
3 ] [
=
4
3
−3
0
−1
−2 ]
[ ]
4 3
∴ (A − B) ′ = −3 0
−1 −2

[ ][ ][ ]
3 4 −1 1 4 3
A′ − B′ = −1 2 − 2 2 = −3 0
0 1 1 3 −1 −2
Thus, we have veri ed that (A − B) ′ = A ′ − B ′

Page : 88 , Block Name : Exercise 3.3

Q4 If A ′ =
[ ] −2
1
3
2
and B =
[ ]
−1
1
0
2
, then find (A + 2B) ′

Answer.
We know that A = A ′ ( )
∴ A=
[ ]
−2
3
1
2

∴
[ ] [ ] [ ][ ] [ ]
A + 2B =
−2 1
3 2
+2
−1
1
0
2
=
−2
3
1
2
+
−2
2
0
4
=
−4
5
1
6

∴
[ ]
(A + 2B) ′ =
−4
1
5
6

Page : 88 , Block Name : Exercise 3.3

Q5 For the matrices A and B, verify that (AB)′ = B′A′, where

[]
1
(i) A= −4 , B = [−1 2 1]
3

[]
0
(ii) A = 1 , B = [1 5 7]
2

[] [ ]
1 −1 2 1
Answer. (i) AB = −4 [−1 2 1 ] = 4 −8 −4
3 −3 6 3

[ ]
−1 4 −3
∴ (AB) ′ = 2 −8 6
1 −4 3

[]
−1
Now A ′ = [ 1 − 4 3 ], B ′ = 2
1

[] [ ]
−1 −1 4 −3
∴ B ′A ′ = 2 [1 −4 3] = 2 −8 6
1 1 −4 3

[] [ ]
0 0 0 0
(ii) AB = 1 [1 5 7 ] = 1 5 7
2 2 10 14

[ ]
0 1 2
∴ (AB) ′ = 0 5 10
0 7 14

[]
1
A′ = [ 0 1 2 ], B ′ = 5
7

[] [ ]
1 0 1 2
∴ B ′A ′ = 5 [0 1 2 ] = 0 5 10
7 0 7 14
Hence, we have veri ed that (AB)' = B'A'

Page : 88 , Block Name : Exercise 3.3

Q6 If (i) A =
[ cosα
− sinα
sinα
cosα ] , then verify that A ′A = I

(ii) A =
[ sinα
− cosα
cosα
sinα ] , then verify that A ′A = I

Answer. (i) A =
[ cosα
− sinα
sinα
cosα ]
∴ A′ =
[ cosα
sinα cosα ]
− sinα

A ′A =
[ cosα
sinα
− sinα
cosα ][ cosα
− sinα
sinα
cosα ]
=
[ (cosα)(cosα) + ( − sinα)( − sinα)
(sinα)(cosα) + (cosα)( − sinα)
(cosα)(sinα) + ( − sinα)(cosα)
(sinα)(sinα) + (cosα)(cosα) ]
=
[ cos 2α + sin 2α
sinαcosα − sinαcosα
sinαcosα − sinαcosα
sin 2α + cos 2α ]
=
[ ] 1 0
0 1
=I

Hence, we have verified that A ′A = I

(ii)

A=
[ sinα
− cosα
cosα
sinα ]
∴ A′ =
[ sinα
cosα
− cosα
sinα ]
A ′A =
[ sinα
cosα
− cosα
sinα ][ sinα
− cosα
cosα
sinα ]
[ sinα
cosα
− cosα
sinα ][ sinα
− cosα
cosα
sinα ]
=
[ sin 2α + cos 2α
sinαcosα − sinαcosα
sinαcosα − sinαcosα
cos 2α + sin 2α ]
=
[ ] 1 0
0 1
=I=I

Hence, we have verified that A ′A = I

Page : 89 , Block Name : Exercise 3.3

[ ]
1 −1 5
Q7 (i) Show that the matrix A = −1 2 1 is a symmetric matrix.
5 1 3

[ ]
0 1 −1
(ii) Show that the matrix A = −1 0 1 is a skew symmetric matrix.
1 −1 0

Answer.
(i) We have:

[ ]
1 −1 5
A′ = −1 2 1 =A
5 1 3

∴ A′ = A
Hence, A is a symmetric matrix.

(ii) We have

[ ] [ ]
0 −1 1 0 1 −1
A′ = 1 0 −1 = − −1 0 1 = −A
−1 1 0 1 −1 0
∴ A′ = − A
Hence, A is a skew-symmetric matrix.

Page : 89 , Block Name : Exercise 3.3

Q8 For the matrix A =
[ ]
1
6
5
7
, verify that

(i) (A + A′) is a symmetric matrix
(ii) (A – A′) is a skew symmetric matrix

Answer.

A′ =
[ ] 1 6
5 7

[ ][ ] [
(i) A + A ′ =
1
6
5
7
+
1
5
6
7
=
2
11
11
14 ]
( ) [ ] ′ 2 11
∴ A + A′ = = A + A′
11 14

(ii) A − A ′ =
[ ][ ] [ ] 1
6
5
7
−
1
5
6
7
=
0
1
−1
0

) [ ] [ ] (
0 1 0 −1
(A − A′
′
=
−1 0
= −
1 0
= − A − A′ )
( )
Hence, A − A ′ is a skew-symmetric matrix.

Page : 89 , Block Name : Exercise 3.3

[ ]
0 a b

( ) ( )
1 1
Q9 Find 2 A + A ′ and 2 A − A ′ , when A = −a 0 c
−b −c 0

[ ]
0 a b
Answer. The given matrix is A = −a 0 c
−b −c 0

[ ]
0 −a −b
∴ A′ = a 0 −c
b c 0

[ ][ ][ ]
0 a b 0 −a −b 0 0 0
A + A′ = −a 0 c + a 0 −c = 0 0 0
−b −c 0 b c 0 0 0 0

( ) [ ]
0 0 0
1
∴
2
A + A′ = 0 0 0
0 0 0

[ ][ ][ ]
0 a b 0 −a −b 0 2a 2b
Now, A − A ′ = −a 0 c − a 0 −c = − 2a 0 2c
−b −c 0 b c 0 − 2b − 2c 0

( ) [ ]
0 a b
1
∴
2
A − A′ = −a 0 c
−b −c 0

Page : 89 , Block Name : Exercise 3.3

Q10 Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i)
[ ] 3
1 −1
5

[ ]
6 −2 2
(ii) − 2 3 −1
2 −1 3

[ ]
3 3 −1
(iii) − 2 −2 1
−4 −5 2

(iv) [ ] 1
−1
5
2
Choose the correct answer in the Exercises 11 and 12.

Answer. Let A = [ ] [ ] 3
1
5
−1
, then A ′ =
3
5
1
−1

Now , A + A ′ = [ ][ ] [ ] 3
1 −1
5
+
3
5 −1
=
6
6 −2
6

( ) [ ] [ ]
1 1 6 6 3
Let P = 2 A + A ′ = 2 =
6 −2 3 −1

Now , P ′ = [ ] 3
3
3
−1
=P

( )
1
Thus , P = 2 A + A ′ is a symmetric matrix.

Now, Q ′ = [ ] 0
−2
2
0
= −Q

( )
1
Thus , Q = 2 A − A ′ is a skew-symmetric matrix.
Representing A as the sum of P and Q:

P+Q= [ ][ ] [ ]
3
3 −1
3
+
0
−2
2
0
=
3
1
5
−1
=A

[ ] [ ]
6 −2 2 6 −2 2
(ii) Let A = −2 3 −1 , then A ′ = −2 3 −1
2 −1 3 2 −1 3

[ ][ ][ ]
6 −2 2 6 −2 2 12 −4 4
Now , A + A ′ = −2 3 −1 + −2 3 −1 = −4 6 −2
2 −1 3 2 −1 3 4 −2 6

( ) [ ][ ]
12 −4 4 6 −2 2
1 1
Let P = 2 A + A ′ = 2 − 4 6 −2 = −2 3 −1
4 −2 6 2 −1 3

[ ]
6 −2 2
Now, P ′ = −2 3 −1 =P
2 −1 3

( )
1
Thus, P = 2 A + A ′ is a symmetric matrix.

[ ]
0 0 0
Now, Q ′ = 0 0 0 = −Q
0 0 0

( )
1
Thus Q = 2 A − A ′ is a skew-symmetric matrix.
Representing A as the sum of P and Q:

[ ][ ][ ]
6 −2 2 0 0 0 6 −2 2
P+Q= −2 3 −1 + 0 0 0 = −2 3 −1 =A
2 −1 3 0 0 0 2 −1 3

[ ] [ ]
3 3 −1 3 −2 −4
(iii) let A = −2 −2 1 , then A ′ = 3 −2 −5
−4 −5 2 −1 1 2

[ ][ ][ ]
3 3 −1 3 −2 −4 6 1 −5
Now, A + A ′ = −2 −2 1 + 3 −2 −5 = 1 −4 −4
−4 −5 2 −1 1 2 −5 −4 4

[ ]
1 5
3 −2

[ ]
2
6 1 −5

( )
1 1 1
Let P = 2 A + A ′ = 2 1 −4 −4 = 2 −2 −2
−5 −4 4 5
−2 −2 2

[ ]
1 5
3 2 −2
1
Now, P ′ = 2 −2 −2 =P
5
−2 −2 2

( )
1
Thus, P = 2 A + A ′ is a symmetric matrix .

[ ][ ][ ]
3 3 −1 3 −2 −4 0 5 3
Now, A − A ′ = −2 −2 1 − 3 −2 −5 = −5 0 6
−4 −5 2 −1 1 2 −3 −6 0

[ ]
5 3
0

[ ]
2 2
0 5 3

( )
1 1 5
Let Q = 2 A − A ′ = 2 − 5 0 6 = −2 0 3
−3 −6 0 3
−2 −3 0

( )
1
Thus, Q = 2 A − A ′ is a skew-symmetric matrix.
Representing A as the sum of P and Q:

[ ][ ]
1 5 5 3
3 −2 0

[ ]
2 2 2
3 3 −1
1 5
P+Q= −2 −2 + −2 0 3 = −2 −2 1 =A
2
5 3 −4 −5 2
−2 −2 2 −2 −3 0

(iv)

Let A =
[ ] 1
−1 [ ] 5
2
, then A ′ =
1
5
−1
2

Now A + A ′ =
[ ][ ] [ ]−1
1 5
2
+
1
5
−1
2
=
2 4
4 4

( ) [ ]
1 1 2
Let P = 2 A + A ′ =
2 2

Now, P ′ =
[ ] 1
2
2
2
=P

( )
1
Thus,P = 2 A + A ′ is a symmetric matrix.

Now, A − A ′ =
[ ][ ] [ ] 1
−1
5
2
−
1
5
−1
2
=
0
−6
6
0

Now, Q ′ =
[ ] 0
3
−3
0
= −Q

( )
1
Thus Q = 2 A − A ′ is a skew-symmetric matrix.
Representing A as the sum of P and Q:

P+Q= [ ][ ] [ ]
1
2
2
2
+
−3
0 3
0
=
1
−1
5
2
=A

Page : 89 , Block Name : Exercise 3.3

Q11 If A, B are symmetric matrices of same order, then AB – BA is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix

Answer. The correct answer is A.
A and B are symmetric matrics ,therefore , we have :
A ′ = A and B ′ = B ....(1)

Consider (AB − BA) ′ = (AB) ′ − (BA) ′ [(A − B) = A − B ]
′ ′ ′

= B ′A ′ − A ′B ′ [(AB) = B A ]
′ ′ ′

= BA − AB [ by (1)]
= − (AB − BA)
′
∴ (AB − BA) = − (AB − BA)

Thus ,( AB - BA ) is a skew-symmetric matrix.

Page : 90 , Block Name : Exercise 3.3

Q12 If A =

π
[ cosα
sinα
− sinα
cosα ], and A + A ′ = I, then the value of α

(A) 6
π
(B) 3
(C) π
3π
(D) 2

Answer. The correct answer is B.

A=
[ cosα
sinα
− sinα
cosα ]
⇒ A′ =
[ cosα
− sinα
sinα
cosα ]
Now, A + A ′ = I

∴
[ cosα
sinα
− sinα
cosα ][ cosα
− sinα
sinα
cosα ] [ ]
=
1
0
0
1

⇒
[ 2cosα
0
0
2cosα ] [ ]
=
1 0
0 1
Comparing the corresponding elements of the two matrices, we have:
2cosα = 1
1
⇒ cosα = 2 = cos 3 )
π
∴ α= 3

Page : 90 , Block Name : Exercise 3.3

Q1 Using elementary transformations, nd the inverse of the matrix.if it is exist-

[ ]
1
2
−1
3

Answer.

Let A = [ ] 1
2
−1
3

We know that A = IA

∴
[ ] [ ]
1
2
−1
3
=
1
0
0
1
A

⇒
[ ] [ ] (
1
0
−1
5
=
−2
1 0
1
A R 2 → R 2 − 2R 1 )

[ ] [ ] (
1 0
⇒
1
0
−1
1
=
−5
2 1
5
A
1
R2 → 5 R2 )

[ ]
3 1

⇒
[ ] 1
0
0
1
=
5
2
−5
5
1
5
A (R1 → R1 + R2 )

[ ]
3 1
5 5
∴ A −1 = 2 1
−5 5

Page : 97 , Block Name : Exercise 3.4

Q2 Using elementary transformations, nd the inverse of the matrix.if it is exist-

[ ]
2
1
1
1

Answer.

Let A =
[ ] 2
1
1
1
We know that A = IA

∴
[ ] [ ]
2
1
1
1
=
1
0
0
1
A

⇒
[ ] [ ] (
1
1
0
1
=
1
0
−1
1
A R1 → R1 − R2 )
⇒
[ ] [ ] (
1
0
0
1
=
−1
1 −1
2
A R2 → R2 − R1 )
∴
[ ]
A −1 =
−1
1 −1
2

Page : 97 , Block Name : Exercise 3.4

Q3 Using elementary transformations, nd the inverse of the matrix.if it is exist-

[ ]
1
2
3
7

Answer.

Let A =
[ ] 1 3
2 7

We know that A = IA

∴
[ ] [ ]
1
2
3
7
=
1
0
0
1
A

⇒
[ ] [ ] (
1
0
3
1
=
1
−2
0
1
A R 2 → R 2 − 2R 1 )
⇒
[ ] [ ] (
1
0
0
1
=
−2
7 −3
1
A R 1 → R 1 − 3R 2 )
∴
[ ]
A −1 =
7
−2
−3
1

Page : 97 , Block Name : Exercise 3.4

Q4 Using elementary transformations, nd the inverse of the matrix.if it is exist-

[ ]
2
5
3
7

Answer.

Let A =
[ ] 2 3
5 7

We know that A = IA

∴
[ ] [ ]
2
5
3
7
=
1
0
0
1
A

[ ][ ](
3 1

⇒
1

5 7
2
=
0
2 0

1
A
1
R1 → 2 R1
)

[ ][ ]
3 1
1 2 2 0
⇒
1
= 5
A (R2 → R2 − 5R1 )
0 −2 −2 1

⇒
[ ][ ](
1

0 −2
0
1 =
−7

−2
5
3

1
A R 1 → R 1 + 3R 2 )

⇒
[ ] [ ] (
1
0
0
1
=
−7
5
3
−2
A R 2 → − 2R 1 )
∴
[ ]
A −1 =
−7
5
3
−2

Page : 97 , Block Name : Exercise 3.4

Q5 Using elementary transformations, nd the inverse of the matrix.if it is exist-

[ ]
2
7
1
4

Answer.

Let A =
[ ]2
7
1
4
We know that A = IA

∴
[ ] [ ]
2
7
1
4
=
1
0
0
1
A

[ ][ ] (
1 1

⇒
1

7 4
2
=
0
2 0

1
A
1
R1 → 2 R1 )

[ ][ ]
1 1
1 2 2 0
⇒
1
= 7
A (R2 → R2 − 7R1 )
0 2
−2 1

⇒
[ ][ ](
1

0
0
1
2
=
−2
4
7
−1

1
A R1 → R1 − R2 )

⇒
[ ] [ ] (
1
0
0
2
=
−7
4 −1
2
A R 2 → 2R 2 )
∴
[ ]
A −1 =
−7
4 −1
2

Page : 97 , Block Name : Exercise 3.4

Q6 Using elementary transformations, nd the inverse of the matrix.if it is exist-

[ ]
2
1
5
3

Answer.

Let A = [ ] 2
1
5
3
We know that A = IA

∴
[ ] [ ]
2
1
5
3
=
1
0
0
1
A

[ ][ ](
5 1

→
1

1 3
2
= 2

0
0

1
A
1
R1 → 2 R1 )

| ]| |
5 1
1 2 2 0
⇒
1
= 1
A (R 2 → R 2 − R 1 )
0 2
−2 1

⇒
[ ][ ](
1

0
0
1
2
=
3

−2
1
−5

1
A R 1 → R 2 − 5R 2 )

⇒
[ ] [
1
0
0
1
=
3
−1
−5
2 ] A (R2 → 2R2 )
Page : 97 , Block Name : Exercise 3.4

Q7 Using elementary transformations, nd the inverse of the matrix.if it is exist-

[ ]
3
5
1
2

Answer.

Let A =
[ ] 3 1
5 2

We know that A = Al

∴
[ ] [ ]
3
5
1
2
=A
1 0
0 1

⇒
[ ] [ ](
1
1
1
2
=A
−2
1 0
1
C 1 → C 1 − 2C 2 )
⇒
[ ] [ ](
1
1
0
1
=A
−2
1 −1
3
C2 → C2 − C1 )
⇒
[ ] [ ](
1
0
0
1
=A
−5
2 −1
3
C1 → C1 − C2 )
∴
[ ]
A −1 =
2
−5
−1
3

Page : 97 , Block Name : Exercise 3.4

Q8 Using elementary transformations, nd the inverse of the matrix.if it is exist-

[ ]
4
3
5
4

Answer.

Let A =
[ ] 4 5
3 4
We know that A = IA

∴
[ ] [ ]
4
3
5
4
=
1
0
0
1
A

⇒
[ ] [ ] (
1
3
1
4
=
1
0
−1
1
A R1 → R1 − R2 )
⇒
[ ] [ ] (
1
0
1
1
=
−3
1 −1
4
A R 2 → R 2 − 3R 1 )
⇒
[ ] [ ] (
1
0
0
1
=
−3
4 −5
4
A R1 → R1 − R2 )
∴
[ ]
A −1 =
−3
4 −5
4

Page : 97 , Block Name : Exercise 3.4

Q9 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ ] 3
2
10
7

Answer.

Let A =
[ ] 3
2
10
7
We know that A = IA

∴
[ 3
2
10
7 ] [ ]
=
1
0
0
1
A

⇒
[ 1
2
3
7 ] [ ] (
=
1
0
−1
1
A R1 → R1 − R2 )
⇒
[ 1
0
3
1 ] [ ] (
=
−2
1 −1
3
A R 2 → R 2 − 2R 1 )
⇒
[ 1
0
0
1 ] [ ] (
=
−2
7 − 10
3
A R 1 → R 1 − 3R 2 )
∴ A −1 =
[ ] −2
7 − 10
3

Page : 97 , Block Name : Exercise 3.4

Q10 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ −4
3 −1
2 ]

Answer.

Let A =
[ −4
3 −1
2 ]
We know that A = AI

∴
[ ] [ ]
−4
3 −1
2
=A
1
0
0
1

⇒
[ ] [ ](
1
0
−1
2
=A
1
2
0
1
C 1 → C 1 + 2C 2 )
⇒
[ ] [ ](
1
0
0
2
=A
1
2
1
3
C2 → C2 + C1 )

[ ](
1

[ ]
1
⇒
1
0
0
1
=A
2
2
3
2
1
C2 → 2 C2
)

[]
1
1 2
∴ A −1 = 3
2 2

Page : 97 , Block Name : Exercise 3.4

Q11 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ ]
2
1
−6
−2

Answer.

Let A =
[ ] 2
1
−6
−2
We know that A=AI

∴
[ ] [ ]
2
1
−6
−2
=A
1
0
0
1

⇒
[ ] [ ](
2
1
0
1
=A
1
0
3
1
C 2 → C 2 + 3C 1 )
⇒
[ ] [ ](
2
0
0
1
=A
−2
−1
3
1
C1 → C1 − C2 )

[ ] [ ](
−1 3
⇒
1
0
0
1
=A
−2
1
1
1
C1 → 2 C1 )
∴ A
−1 =
[ ] −1

−2
1
3

1

Page : 97 , Block Name : Exercise 3.4

Q12 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ 6
−2
−3
1 ]
Answer.

Let A =
[ −2
6 −3
1 ]]
We know that A = IA

∴
[ 6
−2
−3
1 ] [ ]
=
1
0
0
1
A

[ ][ ] (
1 1

⇒

−2
1 −2

1
= 6

0
0

1
A
1
R1 → 6 R1 )

[ ][ | (
1
1 0
1 −2 6
⇒

0 0
= 1
A R 2 → R 2 + 2R 1 )
3 1

Now, in the above equation, we can see all the zeros in the second row of the matrix on the L.H.S.
Therefore, A − 1 does not exist.

Page : 97 , Block Name : Exercise 3.4

Q13 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ 2
−1
−3
2 ]
Answer.

Let A =
[ 2
−1
−3
2 ]
We know that A = 1A

∴
[ ] [ ]
−1
2 −3
2
=
1
0
0
1
A

⇒
[ ] [ ] (
−1
1 −1
2
=
1
0
1
1
A R1 → R1 + R2 )
⇒
| | | | (
1
0
−1
1
=
1
1
1
2
A R2 → R2 + R1 )
⇒
[ ] [ ] (
1
0
0
1
=
2
1
3
2
A R1 → R1 + R2 )
∴
[ ]
A −1 =
2
1
3
2

Page : 97 , Block Name : Exercise 3.4

Q14 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ ]
2
4
1
2

Answer.

Let A =
[ ] 2 1
4 2

We know that A = IA

∴
[ ] [ ]
2
4
1
2
=
1
0
0
1
A

1
ApplyingR 1 → R 1 − 2 R 2, we have

[ ] [ ]
1
0 0 1 −2
= A
4 2 0 1

Now, in the above equation, we can see all the zeros in the rst row of the matrix on the L.H.S.
Therefore, A − 1 does not exist.

Page : 97 , Block Name : Exercise 3.4

Q15 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ ]
2 −3 3
2 2 3
3 −2 2

[ ]
2 −3 3
Answer. A = 2 2 3
3 −2 2
We know that A = IA

[ ][ ]
2 −3 3 1 0 0
∴ 2 2 3 = 0 1 0 A
3 −2 2 0 0 1

[ ][ ](
2 −3 3 1 0 0
⇒ 0 5 0 = −1 1 0 A R2 → R2 − R1 )
3 −2 2 0 0 1

[ ][ | (
1 0 0
2 −3 3
⇒ 0
3
1
−2
0
2
= −5
1

0
1
5
0
0 A
1
R2 → 5
1
)

[ ][ ]
1 0 −1
−1 −1 1
1 1
⇒ 0 1 0 = −5 5
0 A (R 1 → R 1 − R 3 )
3 −2 2
0 0 1

[ |
4 1
−1

[ ]
5 5
−1 0 1
1 1
⇒ 0 1 0 = −5 5 0 A (R1 → R1 + R2 and R3 → R3 + 2R2 )
3 0 2 2 2
−5 5 1

[ ]
4 1

[ ]
−1 0 1 5 5
−1

⇒ 0 1 0 = −5
1 1
5
0 A (R3 → R3 + 3R1 )
0 0 5
2 1 −2

[ ]
4 1
−1

[ ]
5 5
−1 0 1
⇒ 0
0
1
0
0
1
= −5
2
1 1
5
1
0
2
A
( 1
R3 → 5 R3
)
5 5 −5

[ ][ |
2 5
5 0 −5
−1 0 0
1 1
⇒ 0 1 0 = −5 5 0 A (R1 → R1 − R3 )
0 0 1 2 1
5 5 −2

[ ][ ]
2 3
−5 0 5
1 0 0
1 1
⇒ 0 1 0 = −5 5 0 A (R1 → ( − 1)R1 )
0 0 1 2 1 2
5 5 −5

[ ]
2 3
−5 0 5
1 1
∴ A −1 = −5 5 0
2 1 2
5 5 −5

Page : 97 , Block Name : Exercise 3.4

Q16 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ ]
1 3 −2
−3 0 −5
2 5 0

[ ]
1 3 −2
Answer. LetA = −3 0 −5
2 5 0
We know that A = IA

[ ][ ]
1 3 −2 1 0 0
∴ −3 0 −5 = 0 1 0 A
2 5 0 0 0 1
Applying R 2 → R 2 + 3R 1 and R 3 → R 3 − 2R 1, we have:

[ ][ ]
1 3 −2 1 0 0
0 9 − 11 = 3 1 0 A
0 −1 4 −2 0 1
Applying R 1 → R 1 + 3R 3 and R 2 → R 2 + 8R 3, we have

[ ][ ]
1 0 10 −5 0 3
0 1 21 = − 13 1 8 A
0 −1 4 −2 0 1
Applying R 3 → R 3 + R 2, we have:

[ ][ ]
1 0 10 −5 0 3
0 1 21 = − 13 1 8 A
0 0 25 − 15 1 9
1
Applying R 3 → 25 R 3, , we have:

[ ][ ]
−5 0 3
1 0 10
0 1 21 − 13 1 8
= A
3 1 9
0 0 1 −5 25 25

Applying R 1 → R 1 − 10R 3, and R 2 → R 2 − 21R 3, , we have:

[ ]
2 3
1 −5 −5

[ ]
1 0 0
2 4 11
0 1 0 = −5 25 25 A
0 0 1 3 1 9
−5 25 25

[ ]
2 3
1 −5 −5
2 4 11
∴ A −1 = −5 25 25
3 1 9
−5 25 25

Page : 97 , Block Name : Exercise 3.4

Q17 Using elementary transformations, nd the inverse of the matrix. if it is exist-

[ ]
2 0 −1
5 1 0
0 1 3

[ ]
2 0 −1
Answer. LetA = 5 1 0
0 1 3
We know that A = IA

[ ][ ]
2 0 −1 1 0 0
∴ 5 1 0 = 0 1 0 A
0 1 3 0 0 1
1
Applying R 1 → 2 R 1 , we have:

[ ][ ]
1 1
1 0 −2 2 0 0
5 1 0 = 0 1 0 A
0 1 3 0 0 1

Applying R 2 → R 2 − 5R 1 , we have:

[ ][ ]
1 1
1 0 −2 2 0 0
5 = 5
0 1 2
−2 1 0 A

0 1 3 0 0 1

Applying R 3 → R 3 − R 2 , we have:

[ ][ ]
1 1
1 0 −2 2 0 0
5 5
0 1 2 = −2 1 0 A
1 5
0 0 2 2 −1 1

Applying R 3 → 2R 3 , we have:

[ ][ ]
1 1
1 0 −2 2
0 0
5 = 5 A
0 1 2
−2 1 0
0 0 1 5 −2 2

1 5
Applying R 1 → R 1 + 2 R 3, and R 2 → R 2 − 2 R 3 , we have:

[ ][ ]
1 0 0 3 −1 1
0 1 0 = − 15 6 −5 A
0 0 1 5 −2 2

[ ]
3 −1 1
∴ A −1 = − 15 6 −5
5 −2 2

Page : 97 , Block Name : Exercise 3.4

Q18 Matrices A and B will be inverse of each other only if
A. AB = BA
C. AB = 0, BA = I
B. AB = BA = 0
D. AB = BA = I

Answer. D
We know that if A is a square matrix of order m, and if there exists another square matrix B of the
same order m, such that AB = BA = I, then B is said to be the inverse of A. In this case, it is clear
that A is the inverse of B.
Thus, matrices A and B will be inverses of each other only if AB = BA = I.

Page : 97 , Block Name : Exercise 3.4

Q1 A =
[ ]0
0
1
0
, show that (aI + bA) n = a nI + na n − 1bA, where I is the identity matrix of order 2 and

n ∈ N.

Answer.

It is given that A = [ ]0
0
1
0

To show : P(n) : (aI + bA) n = a nI + na n − 1bA, n ∈ N
We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:
P(1) : (al + bA) = aI + ba 0A = aI + bA
Therefore, the result is true for n = 1.
Let the result be true for n = k.
That is,
Now, we prove that the result is true for n = k + 1.
Consider
(aI + bA) k + 1 = (aI + bA) k(aI + bA)

( )
= a kI + ka k − 1bA (aI + bA)
.......(1)
k+1 k k k−1 2 2
=a I + ka bAI + a bLA + ka b A
k+1 k k−1 2 2
=a I + (k + 1)a bA + ka b A

A2 = [ ][ ] [ ]
0
0
1
0
0 1
0 0
=
0
0
0
0
=0

From (1), we have:
(aI + bA) k + 1 = a k + 1I + (k + 1)a kbA + O
= a k + 1I + (k + 1)a kbA
Therefore, the result is true for n = k + 1.
Thus, by the principle of mathematical induction, we have:

(aI + bA) n = a nI + na n − 1bA where A = [ ]
0
0
1
0
,n ∈ N

Page : 100 , Block Name : Miscellaneous Exercise

Q2

A=
[ ]
1 1
1 1
1 1
1
1 , prove that A n =
1 [ 3n − 1
3n − 1
3n − 1
3n − 1
3n − 1
3n − 1
3n − 1

]
3n − 1 , n ∈ N
3n − 1

Answer. It is given that

[ ]
1 1 1
A= 1 1 1
1 1 1

To show: P(n) : A = n
[ 3n − 1
3n − 1
3n − 1
3n − 1
3n − 1
3n − 1 ]
,n ∈ N

We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:

[ ][ ][ ]
31 − 1 31 − 1 31 − 1 30 30 30 1 1 1
P(1) : 3 −1 31 − 1 31 − 1 = 30 30 30 = 1 1 1 =A
3 −1 31 − 1 31 − 1 30 30 30 1 1 1

Therefore, the result is true for n = 1.
Let the result be true for n = k.

[ ]
3k − 1 3k − 1 3k − 1
That is P(k) : A k = 3k − 1 3k − 1 3k − 1
3k − 1 3k − 1 3k − 1

Now, we prove that the result is true for n = k + 1.
Now, A k + 1 = A ⋅ A k

[ ][ |
1 1 1 3k − 1 3k − 1 3k − 1
= 1 1 1 3k − 1 3k − 1 3k − 1
1 1 1 3k − 1 3k − 1 3k − 1

=
[ 3 ⋅ 3k − 1
3 ⋅ 3k − 1
3 ⋅ 3k − 1
3 ⋅ 3k − 1
3 ⋅ 3k − 1
3 ⋅ 3k − 1
3 ⋅ 3k − 1
3 ⋅ 3k − 1
3 ⋅ 3k − 1 ]
=
[ 3 (k+1) −1
3 (k+1) −1
3 (k+1) −1
3 (k+1) −1
3 (k+1) −1
3 (k+1) −1

Therefore, the result is true for n = k + 1.
3 (k+1) −1
3 (k+1) −1
3 (k+1) −1 ]
Thus by the principle of mathematical induction, we have:

An =
[ 3n − 1
3n − 1
3n − 1
3n − 1
3n − 1
3n − 1
3n − 1

]
3n − 1 , n ∈ N
3n − 1

Page : 100 , Block Name : Miscellaneous Exercise

Q3

A=
[ ]
3
1
−4
−1
, then prove that A n =
[ 1 + 2n
n
− 4n
1 − 2n ] , where n is any positive integer

Answer. It is given that

A=
[ ]
3
1
−4
−1

To prove: P(n) : A n =
[ 1 + 2n
n
− 4n
1 − 2n ] ,n ∈ N

We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:

P(1) : A 1 =
[ 1+2
1
−4
1−2 ] [ ]
=
3
1
−4
−1
=A

Therefore, the result is true for n = 1.
Let the result be true for n = k.
That is,

P(k) : A k =
[ 1 + 2k
k
− 4k
1 − 2k ] ,n ∈ N

Now, we prove that the result is true for n = k + 1.
Consider
Ak + 1 = Ak ⋅ A

=
[ 1 + 2k
k
− 4k
1 − 2k ][ ] 3
1
−4
−1

=
[ 3(1 + 2k) − 4k
3k + 1 − 2k
− 4(1 + 2k) + 4k
− 4k − 1(1 − 2k) ]
=
[ 3 + 6k − 4k
3k + 1 − 2k
− 4 − 8k + 4k
− 4k − 1 + 2k ]
=
[ 3 + 2k
1+k
− 4 − 4k
− 1 − 2k ]
=
[ 1 + 2(k + 1)
1+k
− 4(k + 1)
1 − 2(k + 1) ]
Therefore, the result is true for n = k + 1.
Thus, by the principle of mathematical induction, we have:

An =
[ 1 + 2n
n
− 4n
1 − 2n ] ,n ∈ N

Page : 100 , Block Name : Miscellaneous Exercise

Q4 If A and B are symmetric matrices, prove that AB − BA is a skew symmetric matrix.

Answer. It is given that A and B are symmetric matrices. Therefore, we have:

Now (AB − BA) ′ = (AB) ′ − (BA) ′ [(A − B) = A − B ]
′ ′ ′

= B ′A ′ − A ′B ′ [(AB) = B A ]
′ ′ ′

= BA − AB
[ Using (1)]
= − (AB − BA)
′
∴ (AB − BA) = − (AB − BA)

Thus, (AB − BA) is a skew-symmetric matrix.

Page : 100 , Block Name : Miscellaneous Exercise

Q5 Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or
skew symmetric

Answer. We suppose that A is a symmetric matrix, then A ′ = A…(1)
Consider

(B AB ) = {B (AB) }
′ ′ ′ ′

= (AB) (B )
[ (AB) = B A ]
′ ′ ′
′ ′ ′

′ ′
= B A (B)
[ (B ) = B ] ′ ′

= B (A B )
′ ′
[ Using (1)]
= B ′(AB)
∴
(B AB ) = B AB
′ ′ ′

Thus, if A is a symmetric matrix, then B′AB is a symmetric matrix.
Now, we suppose that A is a skew-symmetric matrix.
Then,
A′ = − A
Consider

(B AB ) = [B (AB) ] = (AB) (B )
′ ′ ′ ′ ′ ′

= (B A )B = B ( − A)B
′ ′ ′

= − B ′AB
∴
(B AB ) = − B AB
′ ′ ′

Thus, if A is a skew-symmetric matrix, then B′AB is a skew-symmetric matrix.
Hence, if A is a symmetric or skew-symmetric matrix, then B′AB is a symmetric or skew-symmetric
matrix accordingly.

Page : 100 , Block Name : Miscellaneous Exercise

[ ]
0 2y −z
Q6 Find the values of x, y, z if the matrix A = x y −z satisfy the equation
x −y z
A′A = I

Answer.

[ ]
0 2y z
It is given that A = x y −z
x −y z

[ ]
0 x x
∴ A′ = 2y y −y
z −z z
Now A ′A = I

[ ][ ][ ]
0 x x 0 2y z 1 0 0
⇒ 2y y −y x y −z = 0 1 0
z −z z x −y z 0 0 1

[ 0 + x2 + x2

][ ]
0 + xy − xy 0 − xz + xz 1 0 0
⇒ 0 + xy − xy 4y 2 + y 2 + y 2 2yz − yz = 0 1 0
0 − xz + zx 2yz − yz − yz z2 + z2 0 0 1

[ 2x 2

][ ]
0 0 1 0 0
⇒ 0 6y 2 0 = 0 1 0
0 0 3z 2 0 0 1

On comparing the corresponding elements, we have:
1
2x 2 = 1 ⇒ x = ±
√2
1
6y 2 = 1 ⇒ y = ±
√6
1
3z 2 = 1 ⇒ z = ±
√3

Page : 100 , Block Name : Miscellaneous Exercise

[ ][ ]
1 2 0 0
Q7 For what values of x : [ 1 2 1] 2 0 1 2 = O?
1 0 2 x

Answer. We have:

[ ][ ]
1 2 0 0
[1 2 1 ] 2 0 1 2 =0
1 0 2 x

[]
0
⇒ [1 + 4 + 1 2+0+0 0+2+2] 2 =0
x

[]
0
⇒ [6 2 4 ] 2 =0
x

⇒ [6(0) + 2(2) + 4(x)] = 0
⇒ [4 + 4x] = [0]
∴4 + 4x = 0
⇒ x = −1
Thus, the required value of x is −1.

Page : 100 , Block Name : Miscellaneous Exercise

Q8 If A =
[ ]
−1
3 1
2
, show that A 2 − 5A + 7I = 0

Answer.

It is given that A =
[ ] 3
−1
1
2

∴ A2 = A ⋅ A =
[ ][ ]
3
−1
1
2
3 1
−1 2

=
[ 3(3) + 1( − 1)
− 1(3) + 2( − 1) ] 3(1) + 1(2)
− 1(1) + 2(2)

=
[ 9−1
−3 − 2 ] [ ]
3+2
−1 + 4
=
8
−5
5
3
∴ L.H.S. = A 2 − 5A + 7I

=
[ ] [ ] [ ]
−5 3
8 5
−5
3
−1
1
2
+7
1
0
0
1

=
[ ][ ][ ]
8
−5
5
3
−
15
−5
5
10
+
7
0
0
7

=
[ ][ ]
−7
0 −7
0
+
7
0
0
7

=
[ ]
0 0
0 0
= 0 = R.H.S
2
∴ A − 5A + 7I = 0

Page : 100 , Block Name : Miscellaneous Exercise

[ ][ ]
1 0 2 x
Q9 Find x, if [ x −5 −1 ] 0 2 1 4 =0
2 0 3 1

Answer. We have:

[ ][ ]
1 0 2 x
[x −5 −1 ] 0 2 1 4 =0
2 0 3 1

[]
x
⇒ [x + 0 − 2 0 − 10 + 0 2x − 5 − 3 ] 4 =0
1

[]
x
⇒ [x − 2 − 10 2x − 8 ] 4 =0
1

⇒ [x(x − 2) − 40 + 2x − 8] = O

⇒
[x − 2x − 40 + 2x − 8 ] = [0]
2

⇒
[x − 48 ] = [0]
2

∴ x 2 − 48 = 0
⇒ x 2 = 48
⇒ x = ± 4√3

Page : 100 , Block Name : Miscellaneous Exercise

Q10 A manufacturer produces three products x, y, z which he sells in two markets. Annual sales
are indicated below:

(a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, nd the total
revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50 paise respectively.
Find the gross pro t.

Answer. (a) The unit sale prices of x, y, and z are respectively given as Rs 2.50, Rs 1.50, and Rs 1.00.
Consequently, the total revenue in market I can be represented in the form of a matrix as:

[ ]
2.50
[ 10000 2000 18000 ] 1.50
1.00

= 25000 + 2.50 + 2000 × 1.50 + 18000 × 1.0
= 25000 + 3000 + 18000
= 46000
The total revenue in market II can be represented in the form of a matrix as:

[ ]
2.50
[ 6000 20000 8000 ] 1.50
1.00

= 6000 × 2.50 + 20000 × 1.50 + 8000 × 1.00
= 15000 + 30000 + 8000
= 53000
Therefore, the total revenue in market I isRs 46000 and the same in market II isRs 53000.

(b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50 paise.
Consequently, the total cost prices of all the products in market I can be represented in the form of
a matrix as:

[ ]
2.00
[ 10000 2000 18000 ] 1.00
0.50

= 20000 × 2.00 + 2000 × 1.00 + 18000 × 0.50
= 20000 + 2000 + 9000
= 31000
Since the total revenue in market I isRs 46000, the gross pro t in this marketis (Rs 46000 − Rs
31000) Rs 15000.
The total cost prices of all the products in market II can be represented in the form of a matrix as:

[ ]
2.00
[ 6000 20000 8000 ] 1.00
0.50

= 6000 × 2.00 + 20000 × 1.00 + 8000 × 0.50

= 12000 + 20000 + 4000
= Rs36000
Since the total revenue in market II isRs 53000, the gross pro t in this market is
(Rs 53000 − Rs 36000) Rs 17000.

Page : 101 , Block Name : Miscellaneous Exercise

Q11

X so that X [ 1 2 3
4 5 6 ] [ =
−7
2
−8
4
−9
6 ]
Answer.It is given that:

X [ 1
4
2
5
3
6 ] [
=
−7
2
−8
4
−9
6 ]
The matrix given on the R.H.S. of the equation is a 2 × 3 matrix and the one given on the L.H.S. of
the equation is a 2 × 3 matrix. Therefore, X has to be a 2 × 2 matrix.

Now, let X = [ ] a
b
c
d
Therefore, we have:

[ ][a
b
c
d
1 2
4 5
3
6 ] [ =
−7
2
−8
4
−9
6 ]
⇒
[ α + 4c
b + 4d
2a + 5c
2b + 5d
3a + 6c
3b + 6d ] [ =
−7
2
−8
4
−9
6 ]
Equating the corresponding elements of the two matrices, we have:
a + 4c = − 7, 2a + 5c = − 8,
b + 4d = 2, 2b + 5d = 4, 3b + 6d = 6
Now, a + 4c = − 7 ⇒ a = − 7 − 4c
∴ 2a + 5c = − 8 ⇒ − 14 − 8c + 5c = − 8

⇒ − 3c = 6

⇒ c = − 2

∴ a = − 7 − 4( − 2) = − 7 + 8 = 1

Now , b + 4d = 2 ⇒ b = 2 − 4d
∴ 2b + 5d = 4 ⇒ 4 − 8d + 5d = 4
⇒ − 3d = 0
d=0 ⇒

∴ b = 2 − 4(0) = 2

Thus, a = 1, b = 2, c = −2, d = 0

Hence, the required matrix X is [ ]
1
2
−2
0

Page : 101 , Block Name : Miscellaneous Exercise

Q12 If A and B are square matrices of the same order such that AB = BA, then prove by induction

that tAB n = B nA. Further, prove that (AB) n = A nB n for all n ∈ N

Answer. A and B are square matrices of the same order such that AB = BA.
To prove: P(n) : AB n = B nA, n ∈ N
For n = 1, we have:
P(1) : AB = BA [Given]
⇒ AB 1 = B 1A
Therefore, the result is true for n = 1
Let the result be true for n = k .
P(k) : AB k = B kA .. . . . . . . . . (1)
Now, we prove that the result is true for n = k + 1
AB k + 1 = AB k ⋅ B

( )
= B kA B[ By(1) ]

= B k(AB)[ Associative law ]
= B k(BA) [AB = BA( Given )]

( )
= B kB A [ Associative law ]

= B k + 1A
Therefore, the result is true for n = k + 1
Thus, by the principle of mathematical induction, we have AB n = B nA, n ∈ N .
Now, we prove that (AB) n = A ′′B n for all n ∈ N
For n = 1, we have:
(AB) ′ = A ′B 1 = AB
Therefore, the result is true for n = 1
Let the result be true for n = k .
(AB) k = A kB k.. . (2)
Now, we prove that the result is true for n = k + 1.
AB k + 1 = AB k ⋅ B

( )
= B kA B [By(1)]

= B k(AB) [ Associative law ]
k
= B (BA) [AB = BA( Given )]

( )
= B kB A [ Associative law ]

= B k + 1A
Therefore, the result is true for n = k + 1
n n n
Thus, by the principle of mathematical induction, we have ( AB ) = A B , for all natural
numbers.

Page : 101 , Block Name : Miscellaneous Exercise

Q13 Choose the correct answer in the following questions:

If A =
[ α
γ
β
−α ] is such that A 2 = I, then

(A) 1 + α 2 + βγ = 0 (B) 1 − α 2 + βγ = 0
(C) 1 − α 2 − βγ = 0 (D) 1 + α 2 − βγ = 0

Answer.

A=
[ α
γ −α
β
]
∴ A2 = A ⋅ A =
[ α
γ
β
−α ][ α
γ
β
−α ]
=
[ α 2 + βγ
αγ − αγ
αβ − αβ
βγ + α 2 ][ =
α 2 + βγ
0
0
βγ + α 2 ]Now,

2
A =I⇒
[ α 2 + βγ
0
0
βγ + α 2 ][ ]=
1
0
0
1

On comparing the corresponding elements, we have:
α 2 + βγ = 1
⇒ α 2 + βγ − 1 = 0
⇒ 1 − α 2 − βγ = 0

Page : 101 , Block Name : Miscellaneous Exercise

Q14 If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix (B) A is a zero matrix
(C) A is a square matrix (D) None of these

Answer.B
If A is both symmetric and skew-symmetric matrix, then we should have
A ′ = A and A ′ = − A
⇒ A= −A
⇒ A+A=O
⇒ 2A = O
⇒ A=O
Therefore, A is a zero matrix.

Page : 101 , Block Name : Miscellaneous Exercise

Q15 If A is square matrix such that A 2 = A, then (I + A) 3 − 7A is equal to
(A) A (B) I – A (C) I (D) 3A

Answer.

(I + A) 3 − 7A = I 3 + A 3 + 3I 2A + 3A 2I − 7A
= I + A 3 + 3A + 3A 2 − 7A
= I + A 2 ⋅ A + 3A + 3A − 7A [A = A ]
2

=I+A⋅A−A
= I + A2 − A
=I+A−A
=I
∴ (I + A) 3 − 7A = I

Page : 101 , Block Name : Miscellaneous Exercise