OP Malhotra Class 10 Maths Solutions Chapter 8 Matrices Exercise 8(c)

Well-structured OP Malhotra Class 10 Solutions Chapter 8 Matrices Exercise 8(c) facilitate a deeper understanding of mathematical principles.

S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices Exercise 8(c)

Question 1.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
6 & 1 \\
1 & 1
\end{array}\right]\), C = \(\left[\begin{array}{cc}
-2 & -3 \\
0 & 1
\end{array}\right]\), find each of the following:
(a) AB, BA, AC, CA, BC, CB
(b) A², B², C²
(c) A(BC), (AB)C, B(CA), (BC)A.
Solution:

Question 2.
Answer true or false :
(i) \(\frac { 1 }{ 2 }\) is the identity matrix for addition of 2 x 2 matrices.
(ii) If A, B and C are any 2 x 2 matrices, then
A (B – C) = A. B – A. C
Solution:
(i) False : (∵ For addition it is false)
(ii) True.

Question 3.
If \(\left[\begin{array}{ll}
a & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{c}
2 \\
-1
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right]\), find the value of a.
Solution:
\(\left[\begin{array}{ll}
a & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{c}
2 \\
-1
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
a \times 2+3 \times(-1) \\
1 \times 2+2 \times(-1)
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
2 a-3 \\
2-2
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right] \Rightarrow\left[\begin{array}{c}
2 a-3 \\
0
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right]\)
Comparing the corresponding terms;
2a – 3 = 5 ⇒ 2a = 5 + 3 ⇒ 2a = 8
⇒ a = \(\frac { 8 }{ 2 }\)
⇒ a = 4
∴ a = 4

Question 4.
If A = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-3 & 2 \\
4 & 0
\end{array}\right]\), C = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]\), simplify A² + BC.
Solution:

Question 5.
Find x and y if \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\).
Solution:
\(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
1 \times 3+0 \times 4 \\
0 \times 3+(-1) \times 4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right] \Rightarrow\left[\begin{array}{l}
3+0 \\
0-4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
3 \\
-4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
Comparing the corresponding elements,
x = 3, y = – 4

Question 6.
Let M x \(\left[\begin{array}{ll}
1 & 1 \\
0 & 2
\end{array}\right]\) = [1 2], where M is a matrix,
(i) State the order of the matrix M.
(ii) Find the matrix M.
Solution:
(i) The order of the matrix will be 1 x 2

(ii) Let M = [x y], then
[x y] x \(\left[\begin{array}{ll}
1 & 1 \\
0 & 2
\end{array}\right]\) = [1 2]
⇒ [x x 1 + x o x × 1 + y x 2] = [1 2]
[x + 0 x + 2y] = [1 2]
⇒ [x x + 2y] = [1 2]
Comparing the corresponding elements
x = 1
x + 2y = 2 ⇒ 1 +2y = 2 ⇒ 2y = 2 – 1 = 1
⇒ y = \(\frac { 1 }{ 2 }\)
∴ M = [x y] = \(\frac { 1 }{ 2 }\)

Question 7.
Let A = \(\left[\begin{array}{ll}
4 & -2 \\
6 & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & 2 \\
1 & -1
\end{array}\right]\), C = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & -3
\end{array}\right]\), find
(i) A²
(ii) BC
(iii) A² – A + BC
Solution:

Question 8.
If P = \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\), Q = \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]\), then compute
(i) P² – Q²
(ii) (P + Q) (P – Q)
Is (P + Q) (P – Q) = P² – Q² true for matrix algebra?
Solution:

We see that (P + Q) (P – Q) ≠ P² – Q² for matrix.

Question 9.
If \(\left[\begin{array}{ll}
3 & 4 \\
2 & 5
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right]\) write down the values of a, b, c and d.
Solution:
\(\left[\begin{array}{ll}
3 & 4 \\
2 & 5
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a \times 1+b \times 0 & a \times 0+b \times 1 \\
c \times 1+d \times 0 & c \times 0+d \times 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a+0 & 0+b \\
c+0 & 0+d
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Comparing we get,
a = 3, b = 4, c = 2, d = 5

Question 10.
If A = \(\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\), and B = \(\left[\begin{array}{cc}
0 & 1 \\
-2 & 3
\end{array}\right]\), find (i) BA (ii) A².
Solution:
A = \(\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & 1 \\
-2 & 3
\end{array}\right]\)
(i) BA = \(\left[\begin{array}{cc}
0 & 1 \\
-2 & 3
\end{array}\right]\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
0 \times 2+1 \times(-3) & 0 \times 0+1 \times 1 \\
-2 \times 2+3 \times(-3) & -2 \times 0+3 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
0-3 & 0+1 \\
-4+(-9) & 0+3
\end{array}\right]=\left[\begin{array}{cc}
-3 & 1 \\
-13 & 3
\end{array}\right]\)

A² = A x A = \(\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right] \times\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2 \times 2+0 \times(-3) & 2 \times 0+0 \times 1 \\
-3 \times 2+1 \times(-3) & -3 \times 0+1 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4+0 & 0+0 \\
-6-3 & 0+1
\end{array}\right]=\left[\begin{array}{cc}
4 & 0 \\
-9 & 1
\end{array}\right]\)

Question 11.
If A = \(\frac { 1 }{ 2 }\) and B = \(\frac { 1 }{ 2 }\), find the value of x, given that A² = B.
Solution:

Question 12.
(i) Find the integers p and q such that [p q] \(\left[\begin{array}{l}
p \\
q
\end{array}\right]\) = [25].

(ii) \(\left[\begin{array}{ll}
1 & 3 \\
0 & 0
\end{array}\right]\left[\begin{array}{c}
2 \\
-1
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right]\), findp.
Solution:
[p q] \(\left[\begin{array}{l}
p \\
q
\end{array}\right]\) = [25]
[p² + q²] = [25]
∴ 25 = sum of two squares which can be
(±0)² + (±5)²; (±5)² + (±0)²; (±3)² + (±4)². (±4)² + (+3)²
∴ p = 0, and q = ± 5
p = ± 5, q = o
P = ±3, q = ± 4
P = ± 4, q = ± 3

(ii) \(\left[\begin{array}{ll}
1 & 3 \\
0 & 0
\end{array}\right]\left[\begin{array}{c}
2 \\
-1
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
1 \times 2+3 \times(-1) \\
0 \times 2+0 \times(-1)
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2-3 \\
0+0
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right] \Rightarrow\left[\begin{array}{c}
-1 \\
0
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right]\)
Comparing corresponding elements, P = – 1

Question 13.
If A and B are any two 2 x 2 matrices such that AB = B and B is not a zero matrix, what can you say about the matrix A?
Solution:
A and B are any two 2 x 2 matrices and AB = B
∵ B is not a zero matrix
∴ A matrix will be unit matrix or identity matric = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

Question 14.
If A = \(\left[\begin{array}{cc}
3 & 2 \\
4 & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & -3 \\
-4 & 4
\end{array}\right]\), find (i) BC (ii) A² + A.
Solution:

Question 15.
(i) Find a and b if
\(\left[\begin{array}{ll}
a-b & b-4 \\
b+4 & a-2
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)

(ii) Given \(\left[\begin{array}{cc}
8 & -2 \\
1 & 4
\end{array}\right] X=\left[\begin{array}{l}
12 \\
10
\end{array}\right]\), , write down
(a) the order of matrix X, (b) the matrix X.
Solution:
(i) \(\left[\begin{array}{ll}
a-b & b-4 \\
b+4 & a-2
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
(a-b) \times 2+(b-4) \times 0 & (a-b) \times 0+(b-4) \times 2 \\
(b+4) \times 2+(a-2) \times 0 & (b+4) \times 0+(a-2) \times 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 a-2 b+0 & 0+2 b-8 \\
2 b+8+0 & 0+2 a-4
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 a-2 b & 2 b-8 \\
2 b+8 & 2 a-4
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)
Comparing the corresponding elements
2b – 8 = – 2 ⇒ 2b = 8 – 2 = 6 ⇒ b = \(\frac { 6 }{ 2 }\) = 3
2a – 4 = 0 ⇒ 2a = 4 ⇒ a = \(\frac { 4 }{ 2 }\) = 2
Hence a = 2, 6 = 3

(ii) \(\left[\begin{array}{cc}
8 & -2 \\
1 & 4
\end{array}\right] X=\left[\begin{array}{l}
12 \\
10
\end{array}\right]\)
(a) Here matrix X is of the order 2 x 1
(b) Let X = \(\left[\begin{array}{l}
a \\
b
\end{array}\right]\), then
\(\left[\begin{array}{cc}
8 & -2 \\
1 & 4
\end{array}\right]\left[\begin{array}{l}
a \\
b
\end{array}\right]=\left[\begin{array}{l}
12 \\
10
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
8 a-2 b \\
a+4 b
\end{array}\right]=\left[\begin{array}{c}
12 \\
10
\end{array}\right]\)
Comparing the corresponding elements
8a – 26 = 12 … (i)
a + 46 = 10 … (ii)
Multiply (i) by 2 and (ii) by 1

From (ii) 2 + 4b = 10 ⇒ 4b = 10 – 2 = 8
= \(\frac { 8 }{ 4 }\) = 2
∴ X = \(\left[\begin{array}{l}
2 \\
2
\end{array}\right]\)

Question 16.
If A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\), show that A² – 4A + 5I = 0, where I is the unit matrix.
Solution:

Question 17.
If A = \(\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right]\), compute (AB) C and (CB) A.
Is (AB) C = (CB) A?
Solution:

Question 18.
(i) Write as a single matrix : \(\left[\begin{array}{cc}
6 & -2 \\
1 & 2
\end{array}\right]\) \(\left[\begin{array}{ll}
5 & 3 \\
2 & 4
\end{array}\right]\).
(ii) Find x and y if \(\left[\begin{array}{cc}
2 & 3 \\
-1 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right]\).
Solution:
(i) \(\left[\begin{array}{cc}
6 & -2 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
5 & 3 \\
2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
6 \times 5+(-2) \times 2 & 6 \times 3+(-2) \times 4 \\
1 \times 5+2 \times 2 & 1 \times 3+2 \times 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
30-4 & 18-8 \\
5+4 & 3+8
\end{array}\right]=\left[\begin{array}{cc}
26 & 10 \\
9 & 11
\end{array}\right]\)

(ii) \(\left[\begin{array}{cc}
2 & 3 \\
-1 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
2 x+3 y \\
-x+0
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right] \Rightarrow\left[\begin{array}{c}
2 x+3 y \\
-x
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right]\)
Comparing the corresponding elements,
– x = – 2 ⇒ x = 2
2x + 3y = 7 ⇒ 2 x 2 + 3y = 7
⇒ 4 + 3y = 7 ⇒ 3y = 7 – 4 = 3
⇒ y = \(\frac { 3 }{ 3 }\) = 1
∴ x = 2, y = 1

Question 19.
If A = \(\left[\begin{array}{ll}
3 & 1 \\
2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), verify if (A – B)² = A² – 2AB + B².
Solution:

Question 20.
Given that A = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
a & b \\
0 & c
\end{array}\right]\) and that AB = A + B, find the values of a, b and c.
Solution:
A = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
a & b \\
0 & c
\end{array}\right]\)
AB = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]\left[\begin{array}{ll}
a & b \\
0 & c
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3 a+0 & 3 b+0 \\
0+0 & 0+4 c
\end{array}\right]=\left[\begin{array}{cc}
3 a & 3 b \\
0 & 4 c
\end{array}\right]\)
A + B = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
0 & c
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3+a & 0+b \\
0+0 & 4+c
\end{array}\right]=\left[\begin{array}{cc}
3+a & b \\
0 & 4+c
\end{array}\right]\)
Now ∵ AB = A + B
∴ \(\left[\begin{array}{cc}
3 a & 3 b \\
0 & 4 c
\end{array}\right]=\left[\begin{array}{cc}
3+a & b \\
0 & 4+c
\end{array}\right]\)
Comparing the corresponding elements,
3a = 3 + a ⇒ 3a – a = 3 ⇒ 2a = 3 ⇒ a = \(\frac { 3 }{ 2 }\)
3b = b ⇒ 3b – b = 0 ⇒ 2b = 0 4c = 4 + c ⇒ 4c – c = 4 ⇒ 3c = 4
⇒ \(\frac { 4 }{ 3 }\)
Hence a = \(\frac { 3 }{ 2 }\), b = 0, c = \(\frac { 4 }{ 3 }\)

Question 21.
(i) Solve the matrix equation \(\left[\begin{array}{ll}
2 & 1 \\
5 & 0
\end{array}\right]\) – 3X = \(\left[\begin{array}{cc}
-7 & 4 \\
2 & 6
\end{array}\right]\).
Solution:

Question 22.
If A = \(\left[\begin{array}{ll}
9 & 1 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & 5 \\
7 & -11
\end{array}\right]\), find the matrix X such that 3A + 5B – 2X = 0.
Solution:

Question 23.
(i) Find x and y if, \(\left[\begin{array}{ll}
x & 3 x \\
y & 4 y
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right], B=\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right], C=\left[\begin{array}{ll}
5 & 1 \\
7 & 4
\end{array}\right]\).
Compute A (B + C) and (B + C) A.
Solution:

Question 24.
(i) If \(\left[\begin{array}{cc}
x-2 & 5 \\
3 & 3
\end{array}\right]=\left[\begin{array}{ll}
4 & 2 \\
y & 5
\end{array}\right]\) + \(\left[\begin{array}{cc}
-4 & 3 \\
-1 & -2
\end{array}\right]\), find the values of x and y.
(ii) If A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right]\), C = \(\left[\begin{array}{ll}
1 & 3 \\
3 & 1
\end{array}\right]\). Find the matrix C (B – A).
Solution:

Question 25.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 1 \\
2 & 1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 2 \\
1 & 1
\end{array}\right]\), write down the matrix AB. Would it be possible to find the product BA ? If so, compute it, and if not, give reasons.
Solution:

Hence it is possible.

Self Evaluation And Revision
(LATEST ICSE QUESTIONS)

Question 1.
Evaluate x, y if
Solution:
\(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\left[\begin{array}{c}
2 x \\
1
\end{array}\right]=\left[\begin{array}{c}
6 x-2 \\
-2 x+4
\end{array}\right]\)
∴ \(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\left[\begin{array}{c}
2 x \\
1
\end{array}\right]+2\left[\begin{array}{c}
-4 \\
5
\end{array}\right]=\left[\begin{array}{c}
6 x-2 \\
-2 x+4
\end{array}\right]\) + \(\left[\begin{array}{c}
-8 \\
10
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
6 x-2-8 \\
-2 x+4+10
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
6 x-10 \\
-2 x+14
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
Comparing the corresponding elements
6x – 10 = 8 ⇒ 6x = 8 + 10 = 18
⇒ x = \(\frac { 18 }{ 6 }\) = 3
and – 2x + 14 = 4y ⇒ – 2 x 3 + 14 = 4y
⇒ 4y = 14 – 6 = 8 ⇒ y = \(\frac { 8 }{ 4 }\) = 2
Hence x = 3, y = 2

Question 2.
Evaluate :

Solution:

We know that cos 60° = \(\frac { 1 }{ 2 }\), sin 30° = \(\frac { 1 }{ 2 }\)
tan 45° = 1, cos 0° = 1
cot 45° = 1 cosec 30° = \(\frac { 1 }{ sin 30° }\) = \(\frac { 2 }{ 1 }\),
sec 60° = \(\frac { 1 }{ cos 60° }\) = \(\frac { 2 }{ 1 }\), sin 90° = 1
∴ \(\left[\begin{array}{cc}
2 \times \frac{1}{2} & -2 \times \frac{1}{2} \\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & 2 \\
2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 \times 1+(-1) \times 2 & 1 \times 2+(-1)(1) \\
-1 \times 1+1 \times 2 & -1 \times 2+1 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1-2 & 2-1 \\
-1+2 & -2+1
\end{array}\right]=\left[\begin{array}{cc}
-1 & 1 \\
1 & -1
\end{array}\right]\)

Question 3.
Find the value of x and y; if
\(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\left[\begin{array}{ll}
x & 0 \\
0 & y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\left[\begin{array}{ll}
x & 0 \\
0 & y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
1 \times x+2 \times 0 & 1 \times 0+2 \times y \\
3 \times x+3 \times 0 & 3 \times 0+3 \times y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
0+x & 0+2 y \\
3 x+0 & 0+3 y
\end{array}\right]=\left[\begin{array}{cc}
x & 0 \\
9 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
x & 2 y \\
3 x & 3 y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]\)
Comparing the corresponding elements,
2y = 0 ⇒ y = 0, 3x = 9 ⇒ x = \(\frac { 9 }{ 3 }\) = 3
Hence x = 3, y = 0

Question 4.
Find the 2 x 2 matrix X which satisfies the equation.
\(\left[\begin{array}{ll}
3 & 7 \\
2 & 4
\end{array}\right]\left[\begin{array}{ll}
0 & 2 \\
5 & 3
\end{array}\right]+2 X=\left[\begin{array}{cc}
1 & -5 \\
-4 & 6
\end{array}\right]\)
Solution:

Comparing the corresponding elements,
35 + 2 a = 1 ⇒ 2a = 1 – 35 = -34
⇒ a = \(\frac { -34 }{ 2 }\) = -17
27 + 2b = – 5 ⇒ 2b = -5 – 27 = – 32
⇒ b = \(\frac { -32 }{ 2 }\) = – 16
20 + 2c = – 4 ⇒ 2c = – 4 – 20 = – 24
⇒ c = \(\frac { -24 }{ 2 }\) = – 12
16 + 2d = 6 ⇒ 2d = 6 – 16 = – 10
⇒ d = \(\frac { -10 }{ 2 }\) = – 5
⇒ x = \(\left[\begin{array}{cc}
-17 & -16 \\
-12 & -5
\end{array}\right]\)

Question 5.
Given A = \(\left[\begin{array}{ll}
1 & 1 \\
8 & 3
\end{array}\right]\), evaluate A² – 4A.
Solution:

Question 6.
Find x and y, if
\(\left[\begin{array}{cc}
-3 & 2 \\
0 & -5
\end{array}\right]\left[\begin{array}{l}
x \\
2
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
-3 & 2 \\
0 & -5
\end{array}\right]\left[\begin{array}{l}
x \\
2
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
-3 x+2 \times 2 \\
0 \times x+(-5) \times 2
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
-3 x+4 \\
0-10
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right] \Rightarrow\left[\begin{array}{c}
-3 x+4 \\
-10
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right]\)
Comparing the corresponding elements,
– 3x + 4 = – 5
⇒ – 3x = – 5 – 4
⇒ – 3x = – 9
⇒ x = \(\frac { -9 }{ -3 }\) = 3
y = – 10
Hence x = 3, y = – 10

Question 7.
Find x and y, if \(\left[\begin{array}{ll}
x & 3 x \\
y & 4 y
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right]\).
Solution:
\(\left[\begin{array}{ll}
x & 3 x \\
y & 4 y
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 x+3 x \\
2 y+4 y
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right] \Rightarrow\left[\begin{array}{l}
5 x \\
6 y
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right]\)
Comparing the corresponding elements, 5
5x = 5 ⇒ x = \(\frac { 5 }{ 5 }\) = 1
6y = 12 ⇒ y = \(\frac { 12 }{ 6 }\) = 2
Hence x = 1, y = 2

Question 8.
Find x and y, if:
\(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\left[\begin{array}{c}
2 x \\
1
\end{array}\right]+2\left[\begin{array}{c}
-4 \\
5
\end{array}\right]=4\left[\begin{array}{l}
2 \\
y
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\left[\begin{array}{c}
2 x \\
1
\end{array}\right]+2\left[\begin{array}{c}
-4 \\
5
\end{array}\right]=4\left[\begin{array}{l}
2 \\
y
\end{array}\right]\)
\(\left[\begin{array}{c}
6 x-2 \\
-2 x+4
\end{array}\right]+\left[\begin{array}{c}
-8 \\
10
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
\(\left[\begin{array}{c}
6 x-2-8 \\
-2 x+4+10
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
Now, comparing the corresponding elements of two equal matrices
6x – 10 = 8
∴ 6x = 18
∴ x = \(\frac { 18 }{ 6 }\) = 3
and 4y = – 2x + 14(Using x = 3)
4y = – 2 x 3 + 14
⇒ 4y = 14 – 6 = 8
y = \(\frac { 8 }{ 4 }\)
Hence x = 3, y = 2

Question 9.
Given A = \(\left[\begin{array}{cc}
2 & -1 \\
2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-3 & 2 \\
4 & 0
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]\), find the matrix X, such that A + X = 2B + C.
Solution:

Comparing the corresponding elements,
2 + a = – 5 ⇒ a = – 5 – 2 ⇒ a = – 7
– 1 + b = 4 ⇒ b = 4 + 1 ⇒ b = 5
2 + c = 8 ⇒ c = 8 – 2 ⇒ c = 6
d = 2
Hence matrix X = \(\left[\begin{array}{cc}
-7 & 5 \\
6 & 2
\end{array}\right]\)

Question 10.
Find the value of x given that
A = \(\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right], B=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right]\) and A² = B.
Solution:
A = \(\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right], B=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right]\)
A² = A x A = \(\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2 \times 2+12 \times 0 & 2 \times 12+12 \times 1 \\
0 \times 2+1 \times 0 & 0 \times 12+1 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4+0 & 24+12 \\
0+0 & 0+1
\end{array}\right]=\left[\begin{array}{cc}
4 & 36 \\
0 & 1
\end{array}\right]\)
But A² = B
∴ \(\left[\begin{array}{cc}
4 & 36 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right]\)
Comparing the corresponding elements x = 36

Question 11.
Let A = \(\left[\begin{array}{cc}
4 & -2 \\
6 & -3
\end{array}\right], B=\left[\begin{array}{cc}
0 & 2 \\
1 & -1
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & -1
\end{array}\right]\). Find A² – A + BC.
Solution:

Question 12.
(a) If 2\(\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\), find the values of x and y.
(b) Let A = Find A² + AB + B².
Solution:
(a) 2\(\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6 & 8 \\
10 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6+1 & 8+y \\
10+0 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
7 & 8+y \\
10 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
Comparing the corresponding elements, 2x + 1 = 5 ⇒ 2x = 5 – 1 = 4
⇒ x = \(\frac { 4 }{ 2 }\)
and 8 + y = 0 ⇒ y = – 8
Hence x = 2, y = – 8

(b) \(A=\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right], B=\left[\begin{array}{cc}
2 & 3 \\
-1 & 0
\end{array}\right]\)

Question 13.
(a) If \(\left[\begin{array}{cc}
1 & 4 \\
-2 & 3
\end{array}\right]\) + 2M = 3\(\left[\begin{array}{cc}
3 & 2 \\
0 & -3
\end{array}\right]\), find the matrix M.
(b) Given, \(\mathbf{A}=\left[\begin{array}{ll}
p & 0 \\
0 & 2
\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}
0 & -q \\
1 & 0
\end{array}\right]\), C = \(\left[\begin{array}{cc}
2 & -2 \\
2 & 2
\end{array}\right]\) and BA = C². Find the values of p and q.
Solution:
(a) \(\left[\begin{array}{cc}
1 & 4 \\
-2 & 3
\end{array}\right]+2 \mathrm{M}=3\left[\begin{array}{cc}
3 & 2 \\
0 & -3
\end{array}\right]\)
Let M = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\), then 2M = \(\left[\begin{array}{ll}
2 a & 2 b \\
2 c & 2 d
\end{array}\right]\)
∴ \(\left[\begin{array}{cc}
1 & 4 \\
-2 & 3
\end{array}\right]+\left[\begin{array}{cc}
2 a & 2 b \\
2 c & 2 d
\end{array}\right]=\left[\begin{array}{cc}
9 & 6 \\
0 & -9
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
1+2 a & 4+2 b \\
-2+2 c & 3+2 d
\end{array}\right]=\left[\begin{array}{cc}
9 & 6 \\
0 & -9
\end{array}\right]\)
Comparing the corresponding elements
1 + 2a = 9 ⇒ 2a = 9 – 1 = 8 ⇒ a = \(\frac { 8 }{ 2 }\) = 4
4 + 2b = 6 ⇒ 2b = 6 – 4 = 2 ⇒ b = \(\frac { 2 }{ 2 }\) = 1
– 2 + 2c = 0 ⇒ 2c = 2 ⇒ c = \(\frac { 1 }{ 2 }\) = 1
3 + 2d = – 9 ⇒ 2d = – 9 – 3 ⇒ 2d = – 12
⇒ d = \(\frac { -12 }{ 2 }\) = – 6
∴ M = \(\left[\begin{array}{cc}
4 & 1 \\
1 & -6
\end{array}\right]\) = – 6

(b) A = \(\left[\begin{array}{ll}
p & 0 \\
0 & 2
\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}
0 & -q \\
1 & 0
\end{array}\right]\), C = \(\left[\begin{array}{cc}
2 & -2 \\
2 & 2
\end{array}\right]\)

Comparing the corresponding elements,
p = 8, -2 q = – 8 ⇒ q = \(\frac { -8 }{ -2 }\) = 4
⇒ p = 8, q = 4

Question 14.
Find x and y, if \(\left[\begin{array}{cc}
2 x & x \\
y & 3 y
\end{array}\right]\left[\begin{array}{l}
3 \\
2
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]\)
Solution:
Given : \(\left[\begin{array}{cc}
2 x & x \\
y & 3 y
\end{array}\right]\left[\begin{array}{l}
3 \\
2
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]\)
= \(\left[\begin{array}{l}
2 x \times 3+2 \times x \\
3 \times y+3 y \times 2
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]\)
∴ \(\left[\begin{array}{l}
6 x+2 x \\
3 y+6 y
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]=\left[\begin{array}{l}
8 x \\
9 y
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]\)
⇒ 8x =16 and 9y = 9
⇒ x = 2 and y = 1

Question 15.
Given A = \(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\), B = \(\left[\begin{array}{l}
6 \\
1
\end{array}\right]\), C = \(\left[\begin{array}{c}
-4 \\
5
\end{array}\right]\) and D = \(\left[\begin{array}{l}
2 \\
2
\end{array}\right]\), find AB + 2C – 4D.
Solution:

Question 16.
Evaluate \(\left[\begin{array}{cc}
4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\
\sin 90^{\circ} & 2 \cos 0^{\circ}
\end{array}\right]\left[\begin{array}{ll}
4 & 5 \\
5 & 4
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\
\sin 90^{\circ} & 2 \cos 0^{\circ}
\end{array}\right]\left[\begin{array}{ll}
4 & 5 \\
5 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\
1 & 2 \times 1
\end{array}\right]\left[\begin{array}{ll}
4 & 5 \\
5 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
4 & 5 \\
5 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
8+5 & 10+4 \\
4+10 & 5+8
\end{array}\right]=\left[\begin{array}{cc}
13 & 14 \\
14 & 13
\end{array}\right]\)

Question 17.
If A = \(\left[\begin{array}{cc}
3 & 5 \\
4 & -2
\end{array}\right]\) and B = \(\left[\begin{array}{l}
2 \\
4
\end{array}\right]\), is the product AB possible? Give a reason. If yes, find AB.
Solution:
Yes, the product is possible because number of column in A = number of row in B i.e., (2 x 2). (2 x 1) = (2 x 1) is the order of matrix.
AB = \(\left[\begin{array}{cc}
3 & 5 \\
4 & -2
\end{array}\right]\left[\begin{array}{l}
2 \\
4
\end{array}\right]=\left[\begin{array}{c}
3 \times 2+5 \times 4 \\
4 \times 2+(-2) \times 4
\end{array}\right]\)
= \(\left[\begin{array}{c}
6+20 \\
8-8
\end{array}\right]=\left[\begin{array}{c}
26 \\
0
\end{array}\right]\)

Question 18.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find A² – 5A + 7I.
Solution:

Question 19.
Given A = \(\left[\begin{array}{cc}
2 & -6 \\
2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-3 & 2 \\
4 & 0
\end{array}\right]\), C = \(\left[\begin{array}{ll}
4 & 0 \\
0 & 2
\end{array}\right]\)
Find the matrix X such that A + 2X = 2B + C.
Solution:

Question 20.
Let A = \(\left[\begin{array}{cc}
2 & 1 \\
0 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
-3 & 2 \\
-1 & 4
\end{array}\right]\). Find A² + AC – 5B.
Solution:

Question 21.
If A = \(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
9 & 16 \\
0 & -y
\end{array}\right]\), find x and y when A² = B.
Solution:
Given : A = \(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
9 & 16 \\
0 & -y
\end{array}\right]\), find x and y when A² = B
Now, A² = A x A
= \(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right] \times\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
9 & 3 x+x \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
9 & 4 x \\
0 & 1
\end{array}\right]\)
We have A² = B
Two matrices are equal if each and every corresponding element is equal.
Thus, \(\left[\begin{array}{cc}
9 & 4 x \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
9 & 16 \\
0 & -y
\end{array}\right]\)
⇒ 4x = 16 and 1 = – y
⇒ x = 4 and y = – 1

Question 22.
Given A = \(\frac { 1 }{ 2 }\) and I = \(\frac { 1 }{ 2 }\) and A² = 9A + ml. Find m.
Solution:
A² = 9A + ml
⇒ A² – 9A = ml … (i)
A² = AA

Question 23.
Given matrix A = \(\left[\begin{array}{cc}
4 \sin 30^{\circ} & \cos 0^{\circ} \\
\cos 0^{\circ} & 4 \sin 30^{\circ}
\end{array}\right]\) and B = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
If AX = B
(i) Write the order of matrix ‘X’.
(ii) Find the matrix ‘X’.
Solution:
A = \(\left[\begin{array}{cc}
4 \sin 30^{\circ} & \cos 0^{\circ} \\
\cos 0^{\circ} & 4 \sin 30^{\circ}
\end{array}\right]\) and B = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
(i) Order of matrix A is 2 x 2
Order of matrix B is 2 x 1
∴ Order of matrix X is 2 x 1

(ii) Let the matrix X = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
AX = B
⇒ \(\left[\begin{array}{cc}
4 \sin 30^{\circ} & \cos 0^{\circ} \\
\cos 0^{\circ} & 4 \sin 30^{\circ}
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
4\left(\frac{1}{2}\right) & 1 \\
1 & 4\left(\frac{1}{2}\right)
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 x+y \\
x+2 y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ 2x + y = 4 … (i)
and x + 2y – 5 … (ii)
Subtracting (ii) from (i), we get
⇒ 2x+ y – (x + 2y) = 4 – 5
⇒ 2x + y – x – 2y = 4 – 5
x – y = – 1 … (iii)
Adding (i) and (ii), we get
⇒ 2x + y + x + 2y = 4 + 5
⇒ 3x + 3y = 9
⇒ x + y = 3
Adding (iii) and (iv), we get
2x = 2 ⇒ x – 1
Substitute x in (iv), we get = 2
Hence, the matrix X = \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)