NCERT Solutions Class-12-Chapter-3-Matrices
Excercise-3.3
Note: The transpose of a matrix \( A \), denoted by \( A' \) or \( A^T \), is obtained by interchanging its rows and columns.
A matrix is Symmetric if \( A' = A \).
A matrix is Skew-Symmetric if \( A' = -A \).
Q1
Find the transpose of each of the following matrices:
(i) \( \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \)
(iii) \( \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix} \)
(i) \( \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \)
(iii) \( \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix} \)
Solution:
(i) Let \( A = \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix} \).
Then \( A' = \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix} \).
(ii) Let \( A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \).
Then \( A' = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \).
(iii) Let \( A = \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix} \).
Then \( A' = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix} \).
Q2
If \( A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix} \), then verify that
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
Solution:
(i) Verify \( (A+B)' = A' + B' \)
\[ A+B = \begin{bmatrix} -1-4 & 2+1 & 3-5 \\ 5+1 & 7+2 & 9+0 \\ -2+1 & 1+3 & 1+1 \end{bmatrix} = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix} \]
\[ (A+B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} \]
\[ A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}, \quad B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix} \]
\[ A' + B' = \begin{bmatrix} -1-4 & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3-5 & 9+0 & 1+1 \end{bmatrix} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} \]
Thus, LHS = RHS.
(ii) Verify \( (A-B)' = A' - B' \)
Similarly, calculate \( A-B \), find its transpose, and compare with \( A' - B' \). Both will yield \( \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} \).
Q3
If \( A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} \), then verify that
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
Solution:
First, find matrix A by transposing \( A' \):
\[ A = (A')' = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix} \]
(i) Verify \( (A+B)' = A' + B' \)
\[ A+B = \begin{bmatrix} 3-1 & -1+2 & 0+1 \\ 4+1 & 2+2 & 1+3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix} \]
\( (A+B)' = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix} \).
\( B' = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \).
\[ A' + B' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix} \]
Thus, verified.
Q4
If \( A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} \), then find \( (A+2B)' \).
Solution:
We know that \( (A+2B)' = A' + (2B)' = A' + 2B' \).
\( A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} \).
\( B' = \begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix} \Rightarrow 2B' = \begin{bmatrix} -2 & 2 \\ 0 & 4 \end{bmatrix} \).
\[ (A+2B)' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 2 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix} \]
Q5
For the matrices A and B, verify that \( (AB)' = B'A' \), where
(i) \( A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}, B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} \)
(ii) \( A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} \)
(i) \( A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}, B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} \)
(ii) \( A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} \)
Solution:
(i)
\[ AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \]
\( (AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \).
\[ B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \]
Verified.
Q6
If (i) \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \), then verify that \( A'A = I \).
(ii) If \( A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} \), then verify that \( A'A = I \).
(ii) If \( A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} \), then verify that \( A'A = I \).
Solution:
(i)
\( A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \).
\[ \begin{aligned} A'A &= \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \\ &= \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \end{aligned} \]
Q7
(i) Show that the matrix \( A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} \) is a symmetric matrix.
(ii) Show that the matrix \( A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} \) is a skew symmetric matrix.
(ii) Show that the matrix \( A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} \) is a skew symmetric matrix.
Solution:
(i)
\( A' = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} \). Since \( A' = A \), it is a symmetric matrix.
(ii)
\( A' = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} \).
\( -A = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} \). Since \( A' = -A \), it is a skew-symmetric matrix.
Q8
For the matrix \( A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} \), verify that
(i) \( (A + A') \) is a symmetric matrix
(ii) \( (A - A') \) is a skew symmetric matrix
(i) \( (A + A') \) is a symmetric matrix
(ii) \( (A - A') \) is a skew symmetric matrix
Solution:
\( A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} \).
(i) \( A + A' = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} \).
Since \( (A+A')' = A+A' \), it is symmetric.
(ii) \( A - A' = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \).
Since \( (A-A')' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -(A-A') \), it is skew-symmetric.
Q9
Find \( \frac{1}{2}(A + A') \) and \( \frac{1}{2}(A - A') \), when \( A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \)
Solution:
\( A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} \).
\[ A + A' = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \frac{1}{2}(A+A') = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
\[ A - A' = \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix} \Rightarrow \frac{1}{2}(A-A') = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \]
Q10
Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
(i) \( \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} \)
... [Other parts]
(i) \( \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} \)
... [Other parts]
Solution:
Any matrix \( A \) can be written as \( A = P + Q \), where \( P = \frac{1}{2}(A+A') \) is symmetric and \( Q = \frac{1}{2}(A-A') \) is skew-symmetric.
(i) Let \( A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} \). \( A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \).
\( P = \frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} \).
\( Q = \frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \).
\( P + Q = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = A \).
Q11
If A, B are symmetric matrices of same order, then \( AB - BA \) is a- Skew symmetric matrix
- Symmetric matrix
- Zero matrix
- Identity matrix
- Skew symmetric matrix
- Symmetric matrix
- Zero matrix
- Identity matrix
Answer: (A)
Solution:
Given \( A' = A \) and \( B' = B \).
Let \( P = AB - BA \).
\( P' = (AB - BA)' = (AB)' - (BA)' = B'A' - A'B' \).
\( P' = BA - AB = -(AB - BA) = -P \).
Since \( P' = -P \), it is a skew-symmetric matrix.
Q12
If \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \), and \( A + A' = I \), then the value of \( \alpha \) is- \( \frac{\pi}{6} \)
- \( \frac{\pi}{3} \)
- \( \pi \)
- \( \frac{3\pi}{2} \)
- \( \frac{\pi}{6} \)
- \( \frac{\pi}{3} \)
- \( \pi \)
- \( \frac{3\pi}{2} \)
Answer: (B)
Solution:
\( A' = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \).
\( A + A' = \begin{bmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{bmatrix} \).
Given \( A + A' = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
\( 2\cos \alpha = 1 \Rightarrow \cos \alpha = \frac{1}{2} \).
\( \alpha = \frac{\pi}{3} \).