Students can track their progress and improvement through regular use of OP Malhotra Class 10 Solutions Chapter 8 Matrices Exercise 8(b).
S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices Exercise 8(b)
Question 1.
Given matrix A = \(\left[\begin{array}{l}
3 \\
2
\end{array}\right]\), B = \(\left[\begin{array}{l}
-2 \\
-1
\end{array}\right]\), C = \(\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\), find the matrix X in each of the following cases:
(i) X = A + B + C
(ii) X = A + B – C
(iii) X – A = B
(iv) A + X = B + C
(v) 2X = A – C.
Solution:
Question 2.
If 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 the matrix B – C.
Solution:
B = \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 5
\end{array}\right]\), C = \(\left[\begin{array}{cc}
1 & -3 \\
-4 & 4
\end{array}\right]\)
B – C = \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 5
\end{array}\right]-\left[\begin{array}{cc}
1 & -3 \\
-4 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2-1 & -3+3 \\
-4+4 & 5-4
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Question 3.
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 2A – 3B.
Solution:
Question 4.
Given A = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
-4 & -1 \\
-3 & -2
\end{array}\right]\)
(i) Find the matrix 2A + B;
(ii) Find a matrix C such that C + B = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
Question 5.
If P = \(\left[\begin{array}{cc}
4 & 6 \\
2 & -8
\end{array}\right]\) and Q = \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 1
\end{array}\right]\), find P + 2Q.
Solution:
P = \(\left[\begin{array}{cc}
4 & 6 \\
2 & -8
\end{array}\right]\), Q = \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 1
\end{array}\right]\)
∴ P + 2Q = \(\left[\begin{array}{cc}
4 & 6 \\
2 & -8
\end{array}\right]+2\left[\begin{array}{cc}
2 & -3 \\
-1 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4 & 6 \\
2 & -8
\end{array}\right]+\left[\begin{array}{cc}
4 & -6 \\
-2 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4+4 & 6-6 \\
2-2 & -8+2
\end{array}\right]=\left[\begin{array}{cc}
8 & 0 \\
0 & -6
\end{array}\right]\)
Question 6.
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.
Solution:
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 = \(\frac { 4 }{ 2 }\) = 2
and 8 +7 = 0 ⇒ y = – 8
∴ x = 2, y = – 8
Question 7.
If \(\left[\begin{array}{ll}
a & 3 \\
4 & 2
\end{array}\right]+\left[\begin{array}{cc}
2 & b \\
1 & -2
\end{array}\right]-\left[\begin{array}{cc}
1 & 1 \\
-2 & c
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]\), find the value of a, b and c.
Solution:
\(\left[\begin{array}{ll}
a & 3 \\
4 & 2
\end{array}\right]+\left[\begin{array}{cc}
2 & b \\
1 & -2
\end{array}\right]-\left[\begin{array}{cc}
1 & 1 \\
-2 & c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
a+2-1 & 3+b-1 \\
4+1+2 & 2-2-c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
a+1 & b+2 \\
7 & -c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]\)
Comparing the corresponding elements a + 1 = 5 ⇒ a = 5 – 1 = 4
b + 2 = 0 ⇒ b = – 2
– c = 3 ⇒ c = – 3
∴ a = 4, b = – 2, c = – 3
Question 8.
If \(\left[\begin{array}{cc}
2 & -1 \\
2 & 0
\end{array}\right]+2 A=\left[\begin{array}{cc}
-3 & 5 \\
4 & 3
\end{array}\right]\), find A.
Solution:
Question 9.
If 2\(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]+3\left[\begin{array}{ll}
1 & 3 \\
y & 2
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\), find the values of x, y and z.
Solution:
2\(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]+3\left[\begin{array}{ll}
1 & 3 \\
y & 2
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6 & 2 x \\
0 & 2
\end{array}\right]+\left[\begin{array}{cc}
3 & 9 \\
3 y & 6
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6+3 & 2 x+9 \\
0+3 y & 2+6
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
9 & 2 x+9 \\
3 y & 8
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\)
Comparing the corresponding elements
2x + 9 = – 7 ⇒ 2x + – 7 – 9 ⇒ 2x = – 16
⇒ x = \(\frac { – 16 }{ 2 }\) = – 8
3y = 15 ⇒ y = \(\frac { 15 }{ 3 }\) = 5
z = 9
Hence x = – 8, y = 5, z = 9
Question 10.
Given that M = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]\) and N = \(\left[\begin{array}{cc}
2 & 0 \\
-1 & 2
\end{array}\right]\), find M + 2N.
Solution:
M = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]\) and N = \(\left[\begin{array}{cc}
2 & 0 \\
-1 & 2
\end{array}\right]\)
∴ M + 2N = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]+2\left[\begin{array}{cc}
2 & 0 \\
-1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]+\left[\begin{array}{cc}
4 & 0 \\
-2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2+4 & 0+0 \\
1-2 & 2+4
\end{array}\right]=\left[\begin{array}{cc}
6 & 0 \\
-1 & 6
\end{array}\right]\)
Question 11.
(a) If A = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]\), C = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find A – B + C.
(b) Is the following statement true or false? You do not have to give any reason or proof for your answer. “If A and B are two ‘2 x 2’ matrices such that A – B = A + B, then B is a zero matrix.
Solution:
(a) A = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]\), C = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
∴ A – B + C = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]-\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3-1+1 & 0-3+0 \\
0-0+0 & 3-1+1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3 & -3 \\
0 & 3
\end{array}\right]\)
(b) ∵ A and B are ‘2 x 2’ matrices
A – B = A + B
⇒ – B = B
It is possible if B is a zero matrix.
Question 12.
If P = \(\left[\begin{array}{cc}
-3 & 1 \\
2 & 5
\end{array}\right]\), Q = \(\left[\begin{array}{cc}
1 & 6 \\
-4 & 0
\end{array}\right]\) and R = \(\left[\begin{array}{cc}
4 & -1 \\
2 & 3
\end{array}\right]\)
Find the value of :
(i) 2P + 3Q – R
(ii) 4p – 2Q + 3R
Solution:
Question 13.
Compute :
Solution:
Question 14.
If A = \(\left[\begin{array}{ll}
x & y \\
z & w
\end{array}\right]\), B = \(\left[\begin{array}{cc}
x & -y \\
-z & w
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
0 & y \\
2 z & 0
\end{array}\right]\)
Show that (A + B) + C = A + (B + C).
Solution:
Question 15.
If A = \(\left[\begin{array}{ll}
8 & 1 \\
5 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-4 & 6 \\
-2 & 12
\end{array}\right]\), find the matrix C such that 2A + 3B + 4C is a null matrix.
Solution:
Question 16.
If P = \(\left[\begin{array}{ll}
6 & -2 \\
4 & -6
\end{array}\right]\), Q = \(\left[\begin{array}{ll}
5 & 3 \\
2 & 0
\end{array}\right]\), find X such that 3P – 2Q + 3X = 0.
Solution:
Question 17.
Find X and Y if X + Y = \(\left[\begin{array}{cc}
9 & 7 \\
2 & 11
\end{array}\right]\) and X – Y = \(\left[\begin{array}{cc}
-5 & -6 \\
4 & 3
\end{array}\right]\).
Solution:
Question 18.
Solve for A and B, when
2A + B = \(\left[\begin{array}{cc}
3 & -4 \\
2 & 7
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ll}
4 & 3 \\
1 & 1
\end{array}\right]\).
Solution:
