Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 3 Matrices Ex 3.3 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3
Question 1.
Find the transpose of each of the following matrices:
(i) \(\left[\begin{array}{c}
5 \\
\frac{1}{2} \\
-1
\end{array}\right]\)
(ii) \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
(iii) \(\left[\begin{array}{ccc}
-1 & 5 & 6 \\
\sqrt{3} & 5 & 6 \\
2 & 3 & -1
\end{array}\right]\)
Solution.
(i) Let A = \(=\left[\begin{array}{c}
5 \\
\frac{1}{2} \\
-1
\end{array}\right]_{3 \times 1}\), then A’ = \(\left[5 \frac{1}{2}-1\right]_{1 \times 3}\).
(ii) Let A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\), then A’ = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right]\)
(iii) Let A = \(\left[\begin{array}{ccc}
-1 & 5 & 6 \\
\sqrt{3} & 5 & 6 \\
2 & 3 & -1
\end{array}\right]\), then A’ = \(\left[\begin{array}{ccc}
-1 & \sqrt{3} & 2 \\
5 & 5 & 3 \\
6 & 6 & -1
\end{array}\right]\).
Question 2.
If A = \(\) and B = \(\), then verify that
(i) (A + B) = A’ + B’
(ii) (A – B)’ = A’ – B’
Solution.
(i) 
Hence, we have verified that (A + B)’ = A’ + B’
(ii) 
Hence, we have verified that (A – B)’ = A’ – B’.
Question 3.
If A’ = \(\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\), then verify that
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’
Solution.
It is known that A = (A’)’
Therefore, we have
A = \(\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right]\)
B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right]\)
(i) A + B = \(\left[\begin{array}{ccc}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]=\left[\begin{array}{lll}
2 & 1 & 1 \\
5 & 4 & 4
\end{array}\right]\)
Thus, we have verified that (A + B)’ = A’ + B’
(ii) 
Thus, we have verified that (A – B)’ = A’ – B’.
Question 4.
If A’ = \(\left[\begin{array}{cc}
-\mathbf{2} & \mathbf{3} \\
\mathbf{1} & \mathbf{2}
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
-1 & 0 \\
1 & 2
\end{array}\right]\), then find (A + 2B)’.
Solution.
We know that A = (A’)’
Question 5.
For the matrices A and B, verify that (AB)’ = B’A’, where
(i) A = \(\left[\begin{array}{c}
1 \\
-4 \\
3
\end{array}\right]\), B = [- 1 2 1]
(ii) A = \(\left[\begin{array}{l}
\mathbf{0} \\
\mathbf{1} \\
\mathbf{2}
\end{array}\right]\), B = [1 5 7].
Solution.
(i) 
B’A’ = \(\left[\begin{array}{c}
-1 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
1 & -4 & 3
\end{array}\right]=\left[\begin{array}{ccc}
-1 & 4 & -3 \\
2 & -8 & 6 \\
1 & -4 & 3
\end{array}\right]\)
Hence, we have verified that (AB)’ = B’A’.
(ii) 
Hence, we have verified that (AB)’ = B’A’.
Question 6.
(i) If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), then verify that A’A = I.
(ii) If A = \(\left[\begin{array}{cc}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]\), then verify that A’A = I.
Solution.
Hence, we have verified that A’A = I.
(ii) Given, A = \(\left[\begin{array}{cc}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]\)
∴ A’ = \(\left[\begin{array}{cc}
\sin \alpha & -\cos \alpha \\
\cos \alpha & \sin \alpha
\end{array}\right]\)
Hence, we have verified that A’A = I.
Question 7.
(i) Show that the matrix A = \(\left[\begin{array}{rrr}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\) is a symmetric matrix.
(ii) Show that the matrix A = \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]\) is a skew symmetric matrix.
Solution.
We have A’ = \(=\left[\begin{array}{rrr}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\) = A
Hence, A is a symmetric matrix.
(ii) We have, A’ = \(\left[\begin{array}{rrr}
0 & -1 & 1 \\
1 & 0 & -1 \\
-1 & 1 & 0
\end{array}\right]\)
= – \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]\) = – A
∴ A’ = – A
Hence, A’ is a symmetric matrix.
Question 8.
For the matrix A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), verify that
(i) (A + A’)’ is a symmetric matrix.
(ii) (A – A’) is a skew symmetric matrix.
Solution.
We have, A’ = \(\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\)
(i) A + A’ = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]+\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]=\left[\begin{array}{cc}
2 & 11 \\
11 & 14
\end{array}\right]\)
(A + A’)’ = \(\left[\begin{array}{cc}
2 & 11 \\
11 & 14
\end{array}\right]\) = A + A’
Hence, (A + A’)’ is a symmetric matrix.
(ii) A – A’ = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]-\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]=\left[\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right]\)
(A – A’)’ = \(\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right]=-\left[\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right]\) = -(A – A’).
Thus, (A – A’) is a skew symmetric matrix.
Question 9.
Find \(\frac{1}{2}\) (A + A’) and \(\frac{1}{2}\) (A – A’) when A = \(\left[\begin{array}{ccc}
0 & \boldsymbol{a} & \boldsymbol{b} \\
-\boldsymbol{a} & \mathbf{0} & \boldsymbol{c} \\
-\boldsymbol{b} & -\boldsymbol{c} & \mathbf{0}
\end{array}\right]\).
Solution.
Question 10.
Express the following matrices as the sum of a symmetric and a skew-symmetric matrix.
(i) \(\left[\begin{array}{rr}
\mathbf{3} & \mathbf{5} \\
\mathbf{1} & -\mathbf{1}
\end{array}\right]\)
(ii) \(\left[\begin{array}{rrr}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\)
(iii) \(\left[\begin{array}{rrr}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
(iv) \(\left[\begin{array}{cc}
1 & 5 \\
-1 & 2
\end{array}\right]\)
Solution.
Let, Q = \(\frac{1}{2}\) (A – A’)
= \(\frac{1}{2}\) \(\left[\begin{array}{cc}
0 & 4 \\
-4 & 0
\end{array}\right]=\left[\begin{array}{cc}
0 & 2 \\
-2 & 0
\end{array}\right]\)
Now, Q’ = \(\left[\begin{array}{cc}
0 & -2 \\
2 & 0
\end{array}\right]\) = – Q
Thus, Q = \(\frac{1}{2}\) (A – A’) is a skew symmetric matrix.
Representing A as the sum of P and Q
P + Q = \(\left[\begin{array}{rr}
3 & 3 \\
3 & -1
\end{array}\right]+\left[\begin{array}{rr}
0 & 2 \\
-2 & 0
\end{array}\right]=\left[\begin{array}{rr}
3 & 5 \\
1 & -1
\end{array}\right]\) = A
(ii) 
(iii) 
Thus, Q = \(\frac{1}{2}\) (A – A’) is a skew-symmetric matrix.
P + Q = \(\left[\begin{array}{ccc}
3 & \frac{1}{2} & -\frac{5}{2} \\
\frac{1}{2} & -2 & -2 \\
-\frac{5}{2} & -2 & 2
\end{array}\right]+\left[\begin{array}{ccc}
0 & \frac{5}{2} & \frac{3}{2} \\
-\frac{5}{2} & 0 & 3 \\
-\frac{3}{2} & -3 & 0
\end{array}\right]=\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\) = A.
(iv) Let A = \(\left[\begin{array}{rr}
1 & 5 \\
-1 & 2
\end{array}\right]\), then A’ = \(\left[\begin{array}{cc}
1 & -1 \\
5 & 2
\end{array}\right]\)
Now, A + A’ = \(\left[\begin{array}{cc}
1 & 5 \\
-1 & 2
\end{array}\right]+\left[\begin{array}{cc}
1 & -1 \\
5 & 2
\end{array}\right]=\left[\begin{array}{cc}
2 & 4 \\
4 & 4
\end{array}\right]\)
Let P = \(\frac{1}{2}\) (A + A’) = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 2
\end{array}\right]\)
Now, p’ = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 2
\end{array}\right]\) = p
Thus, P = \(\frac{1}{2}\) (A + A’) is a symmetric matrix.
Now, A – A’ = \(\left[\begin{array}{cc}
1 & 5 \\
-1 & 2
\end{array}\right]-\left[\begin{array}{cc}
1 & -1 \\
5 & 2
\end{array}\right]=\left[\begin{array}{cc}
0 & 6 \\
-6 & 0
\end{array}\right]\)
Let Q = \(\frac{1}{2}\) (A – A’) = \(\left[\begin{array}{cc}
0 & 3 \\
-3 & 0
\end{array}\right]\)
Now, Q’ = \(\left[\begin{array}{cc}
0 & -3 \\
3 & 0
\end{array}\right]\) = – Q
Thus, Q = \(\frac{1}{2}\) (A – A’) is a skew symmetric matrix.
Representing A as the sum of P and Q:
P + Q = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 2
\end{array}\right]+\left[\begin{array}{cc}
0 & 3 \\
-3 & 0
\end{array}\right]=\left[\begin{array}{cc}
1 & 5 \\
-1 & 2
\end{array}\right]\) = A.
Direction (11 – 12) : Choose the correct answer in the following questions.
Question 11.
If A and 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
Solution.
A and B are symmetric matrices, therefore we have
A’ = A and B’ = B …………..(i)
Consider (AB – BA)’ = (AB)’ – (BA)’ [∵ (A – B)’ = A’ – B’]
= B’A’ – A’B’ [v (AB/= B’A’] [∵ (AB)’ = B’A’]
= BA – AB [FromEq. (i)]
= – (AB – BA)
∴ (AB – BA) = – (AB – BA)
Thus, (AB – BA) is a skew-symmetric matrix.
Hence, the correct answer is (A).
Question 12.
If A = \(\), then A + A’ = I, if the value of α is
(A) \(\frac{\pi}{6}\)
(B) \(\frac{\pi}{3}\)
(C) π
(D) \(\frac{3 \pi}{2}\)
Solution.
We have, A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
⇒ A’ = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\)
Given, A + A’ = I
∴ \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
On comparing the corresponding elements of these two matrices, we have
2 cos α = 1
⇒ cos α = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
∴ α = \(\frac{4}{4}\)
Hence, the correct answer is (B).