ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.2

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.2

Question 1.
If A and B are matrices of order 3 × n and m × 5 respectively, then find the order of the matrix 5A – 3B, given that it is defined.
Solution:
Given A be a matrix of order 3 × n.
and B be a matrix of order m × 5.
Now 5A be a matrix of order 3 × n
and 3B be a matrix of order m × 5
Since 5A – 3B is defined.
∴ order of 5A (matrix) = order of matrix 3B
⇒ 3 × n = m × 5
∴ m = 3 and n = 5
Thus, required orders of matrix A and B are 3 × 5.

Question 2.
If A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rr}
1 & 3 \\
-2 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
-2 & 5 \\
3 & 4
\end{array}\right]\), find the following:
(i) A + B
(ii) A – B
(iii) 3A – C (NCERT)
Solution:
Given A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) ;
B = \(\left[\begin{array}{rr}
1 & 3 \\
-2 & 5
\end{array}\right]\) and
C = \(\left[\begin{array}{rr}
-2 & 5 \\
3 & 4
\end{array}\right]\)

(i) A + B = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) + \(\left[\begin{array}{rr}
1 & 3 \\
-2 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2+1 & 4+3 \\
3-2 & 2+5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3 & 7 \\
1 & 7
\end{array}\right]\)

(ii) A – B = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) – \(\left[\begin{array}{rr}
1 & 3 \\
-2 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2-1 & 4-3 \\
3+2 & 2-5
\end{array}\right]\)
= \(\left[\begin{array}{rr}
1 & 1 \\
5 & -3
\end{array}\right]\)

(iii) 3 A – C = 3 \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) – \(\left[\begin{array}{rr}
-2 & 5 \\
3 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rr}
6 & 12 \\
9 & 6
\end{array}\right]-\left[\begin{array}{rr}
-2 & 5 \\
3 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
8 & 7 \\
6 & 2
\end{array}\right]\)

Question 3.
(i) \(\left[\begin{array}{rrr}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]\) + \(\left[\begin{array}{rrr}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\) (NCERT)
(ii) \(\left[\begin{array}{ll}
\cos ^2 x & \sin ^2 x \\
\sin ^2 x & \cos ^2 x
\end{array}\right]\) + \(\left[\begin{array}{ll}
\sin ^2 x & \cos ^2 x \\
\cos ^2 x & \sin ^2 x
\end{array}\right]\) (NCERT)
Solution:
\(\left[\begin{array}{rrr}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]\) + \(\left[\begin{array}{rrr}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-1+12 & 4+7 & -6+6 \\
8+8 & 5+0 & 16+5 \\
2+3 & 8+2 & 5+4
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
11 & 11 & 0 \\
16 & 5 & 21 \\
5 & 10 & 9
\end{array}\right]\)

(ii) \(\left[\begin{array}{ll}
\cos ^2 x & \sin ^2 x \\
\sin ^2 x & \cos ^2 x
\end{array}\right]\) + \(\left[\begin{array}{ll}
\sin ^2 x & \cos ^2 x \\
\cos ^2 x & \sin ^2 x
\end{array}\right]\)
= \(\left[\begin{array}{ll}
\cos ^2 x+\sin ^2 x & \sin ^2 x+\cos ^2 x \\
\sin ^2 x+\cos ^2 x & \cos ^2 x+\sin ^2 x
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]_{2 \times 2}\) [∵ cos² θ + sin² θ = 1 ∀ θ]

Question 3 (old).
Compute the following:

(i) \(\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\) (NCERT)
(ii) \(\left[\begin{array}{ll}
a^2+b^2 & b^2+c^2 \\
a^2+c^2 & a^2+b^2
\end{array}\right]+\left[\begin{array}{rr}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]\) (NCERT)
Solution:
(i) \(\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\)
= \(\left[\begin{array}{rr}
a+a & b+b \\
-b+b & a+a
\end{array}\right]\)
= \(\left[\begin{array}{rr}
2 a & 2 b \\
0 & 2 a
\end{array}\right]_{2 \times 2}\) (NCERT)

(ii) \(\left[\begin{array}{ll}
a^2+b^2 & b^2+c^2 \\
a^2+c^2 & a^2+b^2
\end{array}\right]+\left[\begin{array}{rr}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a^2+b^2+2 a b & b^2+c^2+2 b c \\
a^2+c^2-2 a c & a^2+b^2-2 a b
\end{array}\right]\)
= \(\left[\begin{array}{ll}
(a+b)^2 & (b+c)^2 \\
(a-c)^2 & (a-b)^2
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{rr}
1 & 5 \\
-2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
0 & -1 \\
4 & -7
\end{array}\right]\), find 2A – 3B.
Solution:
Given A = \(\left[\begin{array}{rr}
1 & 5 \\
-2 & 3
\end{array}\right]\) and
B = \(\left[\begin{array}{ll}
0 & -1 \\
4 & -7
\end{array}\right]\)
∴ 2A – 3B = 2 \(\left[\begin{array}{rr}
1 & 5 \\
-2 & 3
\end{array}\right]\) – 3 \(\left[\begin{array}{ll}
0 & -1 \\
4 & -7
\end{array}\right]\)
= \(\left[\begin{array}{rr}
2 & 10 \\
-4 & 6
\end{array}\right]-\left[\begin{array}{rr}
0 & -3 \\
12 & -21
\end{array}\right]\)
= \(\left[\begin{array}{rr}
2-0 & 10+3 \\
-4-12 & 6+21
\end{array}\right]\)
= \(\left[\begin{array}{rr}
2 & 13 \\
-16 & 27
\end{array}\right]\)

Question 4 (old).
If B = \(\left[\begin{array}{rr}
-1 & 5 \\
0 & 3
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{rr}
0 & 4 \\
-7 & 5
\end{array}\right]\) find the matrix A.
Solution:
Given B = \(\left[\begin{array}{rr}
-1 & 5 \\
0 & 3
\end{array}\right]\)
and A – 2B = \(\left[\begin{array}{rr}
0 & 4 \\
-7 & 5
\end{array}\right]\)
A = \(\left[\begin{array}{rr}
0 & 4 \\
-7 & 5
\end{array}\right]+2\left[\begin{array}{rr}
-1 & 5 \\
0 & 3
\end{array}\right]\)
A = \(\left[\begin{array}{rr}
0 & 4 \\
-7 & 5
\end{array}\right]+\left[\begin{array}{rr}
-2 & 10 \\
0 & 6
\end{array}\right]\)
= \(\left[\begin{array}{ll}
-2 & 14 \\
-7 & 11
\end{array}\right]\)

Question 5.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\), then find 2A – B. (NCERT)
Solution:
Given, A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\) ; B = \(\left[\begin{array}{rrr}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\)
Now 2A – B = 2 \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\) – \(\left[\begin{array}{rrr}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2 & 4 & 6 \\
4 & 6 & 2
\end{array}\right]-\left[\begin{array}{rrr}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-1 & 5 & 3 \\
5 & 6 & 0
\end{array}\right]\)

Question 6.
If \(\left[\begin{array}{rrr}
9 & -1 & 4 \\
-2 & 1 & 3
\end{array}\right]\) = A + \(\left[\begin{array}{rrr}
1 & 2 & -1 \\
0 & 4 & 9
\end{array}\right]\), then find matrix A.
Solution:
Given \(\left[\begin{array}{rrr}
9 & -1 & 4 \\
-2 & 1 & 3
\end{array}\right]\) = A + \(\left[\begin{array}{rrr}
1 & 2 & -1 \\
0 & 4 & 9
\end{array}\right]\)
⇒ A = \(\left[\begin{array}{rrr}
9 & -1 & 4 \\
-2 & 1 & 3
\end{array}\right]-\left[\begin{array}{rrr}
1 & 2 & -1 \\
0 & 4 & 9
\end{array}\right]\)
Thus A = \(\left[\begin{array}{rrr}
8 & -3 & 5 \\
-2 & -3 & -6
\end{array}\right]\)

Question 7.
If A = diagonal [1, – 2, 5], B = diagonal (3, 0, – 4] and C = diagonal [- 2, 7, 0], then find A + 2B – 3C.
Solution:
Given A = diagonal [1, – 2, 5] = \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 5
\end{array}\right]\)
B = diagonal [3, 0, – 4] = \(\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -4
\end{array}\right]\)
and C = diagonal [- 2, 7, 0] = \(\left[\begin{array}{rrr}
-2 & 0 & 0 \\
0 & 7 & 0 \\
0 & 0 & 0
\end{array}\right]\)
∴ A + 2B – 3C = \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 5
\end{array}\right]\) + 2 \(\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -4
\end{array}\right]\) – 3 \(\left[\begin{array}{rrr}
-2 & 0 & 0 \\
0 & 7 & 0 \\
0 & 0 & 0
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 5
\end{array}\right]+\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -8
\end{array}\right]-\left[\begin{array}{rrr}
-6 & 0 & 0 \\
0 & 21 & 0 \\
0 & 0 & 0
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
1+6+6 & 0 & 0 \\
0 & -2+0-21 & 0 \\
0 & 0 & 5-8-0
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
13 & 0 & 0 \\
0 & -23 & 0 \\
0 & 0 & -3
\end{array}\right]\)
= diagonal [13, – 23, – 3]

Question 8.
If A = \(\left[\begin{array}{rr}
2 & -3 \\
5 & 0
\end{array}\right]\) and kA = \(\left[\begin{array}{rr}
8 & 3 a \\
-2 b & c
\end{array}\right]\), find the values of k, a, b and c.
Solution:
Given A = \(\left[\begin{array}{rr}
2 & -3 \\
5 & 0
\end{array}\right]\) and kA = \(\left[\begin{array}{rr}
8 & 3 a \\
-2 b & c
\end{array}\right]\)
⇒ \(k\left[\begin{array}{rr}
2 & -3 \\
5 & 0
\end{array}\right]=\left[\begin{array}{rr}
8 & 3 a \\
-2 b & c
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
2 k & -3 k \\
5 k & 0
\end{array}\right]=\left[\begin{array}{rr}
8 & 3 a \\
-2 b & c
\end{array}\right]\)
Thus their corresponding elements are equal.
2k = 8
⇒ k = 4
– 3k = 3a
⇒ a = – k = – 4 ;
5k = – 2b
⇒ b = \(-\frac{5 k}{2}=-\frac{20}{2}\) = – 10 ;
0 = c
Thus, k = 4 ;
a = – 4 ;
b = – 10 and
c = 0.

Question 9.
If \(x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{r}
-1 \\
1
\end{array}\right]=\left[\begin{array}{r}
10 \\
5
\end{array}\right]\), find the values of x and y.
Solution:
Given, \(x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{r}
-1 \\
1
\end{array}\right]=\left[\begin{array}{r}
10 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 x \\
3 x
\end{array}\right]+\left[\begin{array}{r}
-y \\
y
\end{array}\right]=\left[\begin{array}{r}
10 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 x-y \\
3 x+y
\end{array}\right]=\left[\begin{array}{r}
10 \\
5
\end{array}\right]\)
∴ 2x – y = 10 …………….. (1)
and 3x + y = 5 ……………. (2)
On solving (1) and (2) ; we have
x = 3 ;
y = – 4.

Question 9 (old).
If \(2\left[\begin{array}{cc}
1 & 3 \\
0 & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\), find the values of x and y.
Solution:
Given, \(2\left[\begin{array}{cc}
1 & 3 \\
0 & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 & 6 \\
0 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2+y & 6 \\
1 & 2 x+2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ 2 + y = 5
⇒ y = 3 ;
2x + 2 = 8
⇒ x = 3.

Question 10.
If \(2\left[\begin{array}{rr}
x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{rr}
-3 & -4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rr}
7 & 6 \\
15 & 14
\end{array}\right]\), find the value of (x – y).
Solution:
Given \(2\left[\begin{array}{rr}
x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{rr}
-3 & -4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rr}
7 & 6 \\
15 & 14
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
2 x & 10 \\
14 & 2 y-6
\end{array}\right]+\left[\begin{array}{rr}
-3 & -4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rr}
7 & 6 \\
15 & 14
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
2 x-3 & 10-4 \\
14+1 & 2 y-4
\end{array}\right]=\left[\begin{array}{rr}
7 & 6 \\
15 & 14
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
2 x-3 & 6 \\
15 & 2 y-4
\end{array}\right]=\left[\begin{array}{rr}
7 & 6 \\
15 & 14
\end{array}\right]\)
Thus, their corresponding elements are equal.
∴ 2x – 3 = 7
x = 5 and
2y – 4 = 14
⇒ y = 9
Thus x – y = 5 – 9 = – 4.

Question 11.
Given \(2\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]+3\left[\begin{array}{rr}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\) find x, y, z, t. (NCERT)
Solution:
Given \(2\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]+3\left[\begin{array}{rr}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
2 x+3 & 2 z-3 \\
2 y & 2 t+6
\end{array}\right]=\left[\begin{array}{rr}
9 & 15 \\
12 & 18
\end{array}\right]\)
Thus, their corresponding elements are equal.
⇒ 2x + 3 = 9
⇒ 2x = 6
⇒ x = 3 ;
2y = 12
⇒ y = 6 ;
2z – 3 = 15
⇒ z = 9 ;
2t + 6 = 18
⇒ t = 6.

Question 12.
If \(3\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{rr}
a & 6 \\
-1 & 2 d
\end{array}\right]+\left[\begin{array}{cc}
4 & a+b \\
c+d & 3
\end{array}\right]\), find a, b, c and d. (NCERT)
Solution:
Given, \(3\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{rr}
a & 6 \\
-1 & 2 d
\end{array}\right]+\left[\begin{array}{cc}
4 & a+b \\
c+d & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
3 a & 3 b \\
3 c & 3 d
\end{array}\right]=\left[\begin{array}{cc}
a+4 & 6+a+b \\
-1+c+d & 2 d+3
\end{array}\right]\)
Thus, their corresponding elements are equal.
∴ 3a = a + 4
⇒ 2a = 4
⇒ a = 2
3b = 6 + a + b
⇒ 2b = 6 + 2 = 8
⇒ b = 4
3c = – 1 + c + d
⇒ c – d = – 1 …………..(1)
3d = 2d + 3
⇒ d = 3
∴ from (1) ; 2c – 3 = – 1
⇒ 2c = 2
⇒ c = 1
Thus, a = 2 ; b = 4, c = 1 and d = 3.

Question 13.
Find X if Y = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) and 2X + Y = \(\left[\begin{array}{rr}
1 & 0 \\
-3 & 2
\end{array}\right]\). (NCERT)
Solution:
2X + Y = \(\left[\begin{array}{rr}
1 & 0 \\
-3 & 2
\end{array}\right]\)
⇒ 2X + \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\) [∵ Y = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\)]
⇒ 2X = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
-4 & -2
\end{array}\right]\)
⇒ X = \(\left[\begin{array}{ll}
-1 & -1 \\
-2 & -1
\end{array}\right]\)

Question 14.
Find matrices X and Y, if X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\).
Solution:
Given X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) …………..(1)
and X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\) ……………..(2)
On adding eqn. (1) and (2) ; we have
2X = \(=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
= \(\left[\begin{array}{rr}
10 & 0 \\
2 & 8
\end{array}\right]\)
X = \(\frac{1}{2}\left[\begin{array}{rr}
10 & 0 \\
2 & 8
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]\)
From (1) ;
Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) – X
= \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]-\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]\)
⇒ Y = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right]\).

Question 15.
Find the matrices X and Y if 2X – Y = \(\left[\begin{array}{rrr}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) and X + 2Y = \(\left[\begin{array}{rrr}
3 & 2 & 5 \\
-2 & 1 & -7
\end{array}\right]\).
Solution:
Given, 2X – Y = \(\left[\begin{array}{rrr}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) ………………(1)
and X + 2Y = \(\left[\begin{array}{rrr}
3 & 2 & 5 \\
-2 & 1 & -7
\end{array}\right]\) ………………(2)
Multiplying eqn. (1) by 2 ; we have
4X – 2Y = \(\left[\begin{array}{rrr}
12 & -12 & 0 \\
-8 & 4 & 2
\end{array}\right]\) …………………(3)
On adding eqn. (2) and (3) ; we have

Question 15 (old).
If A = \(\left[\begin{array}{rr}
1 & 3 \\
2 & 1 \\
3 & -1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & 1 \\
1 & 2 \\
-1 & 0
\end{array}\right]\), find the matrix C such that A + B + C is a zero matrix.
Solution:
Given \(\left[\begin{array}{rr}
1 & 3 \\
2 & 1 \\
3 & -1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & 1 \\
1 & 2 \\
-1 & 0
\end{array}\right]\)
We want to find matrix C such taht
A + B + C = O
C = O – A – B
C = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}\right]-\left[\begin{array}{rr}
1 & 3 \\
2 & 1 \\
3 & -1
\end{array}\right]-\left[\begin{array}{rr}
2 & 1 \\
1 & 2 \\
-1 & 0
\end{array}\right]\)
= \(\left[\begin{array}{ll}
0-1-2 & 0-3-1 \\
0-2-1 & 0-1-2 \\
0-3+1 & 0+1-0
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-3 & -4 \\
-3 & -3 \\
-2 & 1
\end{array}\right]\)

Question 16.
Find a matrix X such that 3A – 2B + X = O, where A = \(\left[\begin{array}{ll}
4 & 2 \\
1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{rr}
-2 & 1 \\
3 & 2
\end{array}\right]\).
Solution:
Given, A = \(\left[\begin{array}{ll}
4 & 2 \\
1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{rr}
-2 & 1 \\
3 & 2
\end{array}\right]\)
Let X = \(\left[\begin{array}{ll}
x & y \\
a & b
\end{array}\right]\)
also, 3A – 2B + X = O
⇒ \(3\left[\begin{array}{ll}
4 & 2 \\
1 & 3
\end{array}\right]-2\left[\begin{array}{rr}
-2 & 1 \\
3 & 2
\end{array}\right]+\left[\begin{array}{ll}
x & y \\
a & b
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
12 & 6 \\
3 & 9
\end{array}\right]-\left[\begin{array}{rr}
-4 & 2 \\
6 & 4
\end{array}\right]+\left[\begin{array}{ll}
x & y \\
a & b
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
16+x & 4+y \\
-3+a & 5+b
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
On comparing ; we have
16 + x = 0
⇒ x = – 16 ;
4 + y = 0
⇒ y = – 4 ;
– 3 + a = 0
⇒ a = 3
5 + b = 0
⇒ b = – 5
Hence X = \(\left[\begin{array}{cc}
-16 & -4 \\
3 & -5
\end{array}\right]\).

Question 17.
If A = \(\left[\begin{array}{ll}
9 & 1 \\
7 & 8
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
1 & 5 \\
7 & 12
\end{array}\right]\), find matrix such that 5A + 3B + 2C is a null matrix. (NCERT Exampler).
Solution:
Given A = \(\left[\begin{array}{ll}
9 & 1 \\
7 & 8
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
1 & 5 \\
7 & 12
\end{array}\right]\)_{2 × 2} where O = \(\left[\begin{array}{rr}
0 & 0 \\
0 & 0
\end{array}\right]\)_{2 × 2}
⇒ 2C = O – 3B – 5A
= O + (- 3) B + (- 5) A
⇒ 2C = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]+(-3)\left[\begin{array}{rr}
1 & 5 \\
7 & 12
\end{array}\right]+(-5)\left[\begin{array}{ll}
9 & 1 \\
7 & 8
\end{array}\right]\)
2C = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]+\left[\begin{array}{rr}
-3 & -15 \\
-21 & -36
\end{array}\right]+\left[\begin{array}{ll}
-45 & -5 \\
-35 & -40
\end{array}\right]\)
2C = \(\left[\begin{array}{rr}
0-3-45 & 0-15-5 \\
0-21-35 & 0-36-40
\end{array}\right]\)
= \(\left[\begin{array}{ll}
-48 & -20 \\
-45 & -76
\end{array}\right]\)
C = \(\frac{1}{2}\left[\begin{array}{ll}
-48 & -20 \\
-56 & -76
\end{array}\right]\)
= \(\left[\begin{array}{ll}
-24 & -10 \\
-28 & -38
\end{array}\right]\)

Question 18.
If A = \(\left[\begin{array}{rr}
8 & 0 \\
4 & -2 \\
3 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & -2 \\
4 & 2 \\
-5 & 1
\end{array}\right]\), then find matrix X such that 2A + 3X = 5B. (NCERT)
Solution:
Given 2A + 3X = 5B
⇒ 3X = 5B – 2A
⇒ 3X = \(\left[\begin{array}{rr}
10 & -10 \\
20 & 10 \\
-25 & 5
\end{array}\right]+\left[\begin{array}{rr}
-16 & 0 \\
-8 & 4 \\
-6 & -12
\end{array}\right]\)
⇒ 3X = \(\left[\begin{array}{rr}
10-16 & -10 \\
20-8 & 10+4 \\
-25-6 & 5-12
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-6 & -10 \\
12 & 14 \\
-31 & -7
\end{array}\right]\)
X = \(\left[\begin{array}{rr}
-2 & -10 / 3 \\
4 & 14 / 3 \\
-31 / 3 & -7 / 3
\end{array}\right]\)