Parents can use ISC Maths Class 12 Solutions Chapter 3 Matrices Ex 3.2 to provide additional support to their children.

## 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 = $$\left[\begin{array}{rr} -2 & -10 / 3 \\ 4 & 14 / 3 \\ -31 / 3 & -7 / 3 \end{array}\right]$$