Chapter 3 Matrices EXERCISE 3.3
EXERCISE 3.3
1. 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) $\begin{bmatrix}-1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1\end{bmatrix}$
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Solution
(i) Let $A= \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1\end{bmatrix} $, then $A^{T}= \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix} $
(ii) Let $A= \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix} $, then $A^{T}= \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^{T}= \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix} $
2. 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) $(A+B)^{\prime}=A^{\prime}+B^{\prime}$, (ii) $(A-B)^{\prime}=A^{\prime}-B^{\prime}$
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Solution
We have: $A^{\prime}= \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix} , B^{\prime}= \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix} $
(i) $A+B= \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix} + \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{bmatrix} = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix} $
$\therefore(A+B)^{\prime}= \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} $
$A^{\prime}+B^{\prime}= \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix} + \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} $
Hence, we have verified that $(A+B)^{\prime}=A^{\prime}+B^{\prime}$
(ii) $A-B= \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix} - \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{bmatrix} = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix} $
$\therefore(A-B)^{\prime}= \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} $
$A^{\prime}-B^{\prime}= \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix} - \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix} = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} $
Hence, we have verified that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$.
3. If $A^{\prime}=\begin{bmatrix}3 & 4 \\ -1 & 2 \\ 0 & 1\end{bmatrix}$ and $B=\begin{bmatrix}-1 & 2 & 1 \\ 1 & 2 & 3\end{bmatrix}$, then verify that (i) $(A+B)^{\prime}=A^{\prime}+B^{\prime}$ (ii) $(A-B)^{\prime}=A^{\prime}-B^{\prime}$
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Solution
(i) It is known that $A=(A^{\prime})^{\prime}$
Therefore, we have:
$ \begin{aligned} & A= \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix} \\ & B^{\prime}= \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \\ & A+B= \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix} \\ & \therefore(A+B)^{\prime}= \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix} \\ & A^{\prime}+B^{\prime}= \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} \end{aligned} $
Thus, we have verified that $(A+B)^{\prime}=A^{\prime}+B^{\prime}$.
(ii) $A-B= \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1\end{bmatrix} - \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3\end{bmatrix} = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix} $
$ \begin{aligned} & \therefore(A-B)^{\prime}= \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix} \\ & A^{\prime}-B^{\prime}= \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix} \end{aligned} $
Thus, we have verified that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$.
4. If $A^{\prime}=\begin{bmatrix}-2 & 3 \\ 1 & 2\end{bmatrix}$ and $B=\begin{bmatrix}-1 & 0 \\ 1 & 2\end{bmatrix}$, then find $(A+2 B)^{\prime}$
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Solution
We know that $A=(A^{\prime})^{\prime}$
$ \begin{aligned} & \therefore A= \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} \\ & \therefore A+2 B= \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} +2 \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix} \\ & \therefore(A+2 B)^{\prime}= \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix} \end{aligned} $
5. For the matrices $A$ and $B$, verify that $(A B)^{\prime}=B^{\prime} A^{\prime}$, where (i) $A=\begin{bmatrix}1 \\ -4 \\ 3\end{bmatrix}, B=\begin{bmatrix}-1 & 2 & 1\end{bmatrix}$ (ii) $A=\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}, B=\begin{bmatrix}1 & 5 & 7\end{bmatrix}$
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Solution
$ \begin{aligned} & \text{ (i) } A B= \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} \\ & \therefore(A B)^{\prime}= \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \\ & \text{ Now, } A^{\prime}= \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} , B^{\prime}= \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \\ & \therefore B^{\prime} A^{\prime}= \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} \end{aligned} $
Hence, we have verified that $(A B)^{\prime}=B^{\prime} A^{\prime}$.
(ii) $A B= \begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix} \begin{bmatrix} 1 & 5 & 7\end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix} $
$\therefore(A B)^{\prime}= \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix} $
Now, $A^{\prime}= \begin{bmatrix} 0 & 1 & 2\end{bmatrix} , B^{\prime}= \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} $
$\therefore B^{\prime} A^{\prime}= \begin{bmatrix} 1 \\ 5 \\ 7\end{bmatrix} \begin{bmatrix} 0 & 1 & 2\end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix} $
Hence, we have verified that $(A B)^{\prime}=B^{\prime} A^{\prime}$.
6. If (i) $A=\begin{bmatrix}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix}$, then verify that $A^{\prime} A=I$
(ii) If $A=\begin{bmatrix}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{bmatrix}$, then verify that $A^{\prime} A=I$
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Solution
(i)
$ \begin{aligned} & A= \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \\ & \therefore A^{\prime}= \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \\ & A^{\prime} 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 \alpha)(\cos \alpha)+(-\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\sin \alpha)+(-\sin \alpha)(\cos \alpha) \\ (\sin \alpha)(\cos \alpha)+(\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha)+(\cos \alpha)(\cos \alpha) \end{bmatrix} \\ & = \begin{bmatrix} \cos 2 \alpha+\sin ^{2} \alpha & \sin \alpha \cos \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha-\sin \alpha \cos \alpha & \sin ^{2} \alpha+\cos ^{2} \alpha \end{bmatrix} \\ & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =I \end{aligned} $
Hence, we have verified that $A^{\prime} A=I$.
(ii) $\quad A= \begin{cases} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{cases} $
$ \begin{aligned} & \therefore A^{\prime}= \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \\ & A^{\prime} A= \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} \end{aligned} $
$ \begin{aligned} & { \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} } \\ & = \begin{bmatrix} (\sin \alpha)(\sin \alpha)+(-\cos \alpha)(-\cos \alpha) & (\sin \alpha)(\cos \alpha)+(-\cos \alpha)(\sin \alpha) \\ (\cos \alpha)(\sin \alpha)+(\sin \alpha)(-\cos \alpha) & (\cos \alpha)(\cos \alpha)+(\sin \alpha)(\sin \alpha) \end{bmatrix} \\ & = \begin{bmatrix} \sin ^{2} \alpha+\cos ^{2} \alpha & \sin \alpha \cos \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha-\sin \alpha \cos \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha \end{bmatrix} \\ & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =I \end{aligned} $
Hence, we have verified that $A^{\prime} A=I$.
7. (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.
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Solution
(i) We have:
$ A^{\prime}= \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} =A $
$\therefore A^{\prime}=A$
Hence, $A$ is a symmetric matrix.
(ii) We have:
$ A^{\prime}= \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} =- \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} =-A $
$\therefore A^{\prime}=-A$
Hence, $A$ is a skew-symmetric matrix.
8. For the matrix $A=\begin{bmatrix}1 & 5 \\ 6 & 7\end{bmatrix}$, verify that
(i) $(A+A^{\prime})$ is a symmetric matrix
(ii) $(A-A^{\prime})$ is a skew symmetric matrix
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Solution
$ \begin{aligned} & A^{\prime}= \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} \\ & \text{ (i) } A+A^{\prime}= \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} \\ & \therefore(A+A^{\prime})^{\prime}= \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} =A+A^{\prime} \end{aligned} $
Hence, $(A+A^{\prime})$ is a symmetric matrix.
(ii) $A-A^{\prime}= \begin{bmatrix} 1 & 5 \\ 6 & 7\end{bmatrix} - \begin{bmatrix} 1 & 6 \\ 5 & 7\end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $ $(A-A^{\prime})^{\prime}= \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix} =- \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} =-(A-A^{\prime})$
Hence, $(A-A^{\prime})$ is a skew-symmetric matrix.
9. Find $\frac{1}{2}(A+A^{\prime})$ and $\frac{1}{2}(A-A^{\prime})$, when $A=\begin{bmatrix}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{bmatrix}$
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Solution
The given matrix is $A= \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} $
$\therefore A^{\prime}= \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} $
$A+A^{\prime}= \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{bmatrix} + \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
$\therefore \frac{1}{2}(A+A^{\prime})= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $
Now, $A-A^{\prime}= \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{bmatrix} - \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{bmatrix} = \begin{bmatrix} 0 & 2 a & 2 b \\ -2 a & 0 & 2 c \\ -2 b & -2 c & 0 \end{bmatrix}$
$\therefore \frac{1}{2}(A-A^{\prime})= \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} $
10. 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) $\begin{bmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix}$ (iii) $\begin{bmatrix}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{bmatrix}$ (iv) $\begin{bmatrix}1 & 5 \\ -1 & 2\end{bmatrix}$
Choose the correct answer in the Exercises 11 and 12.
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Solution
(i) Let $A= \begin{bmatrix} 3 & 5 \\ 1 & -1\end{bmatrix} $, then $A^{\prime}= \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} $
$ \begin{aligned} & \text{ Now, } A+A^{\prime}= \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} = \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} \\ & \text{ Let } P=\frac{1}{2}(A+A^{\prime})=\frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} \\ & \text{ Now, } P^{\prime}= \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} =P \end{aligned} $
Thus, $P=\frac{1}{2}(A+A^{\prime})$ is a symmetric matrix.
Now, $A-A^{\prime}= \begin{bmatrix} 3 & 5 \\ 1 & -1\end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 5 & -1\end{bmatrix} = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} $
Let $Q=\frac{1}{2}(A-A^{\prime})=\frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0\end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} $
Now, $Q^{\prime}= \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} =-Q$
Thus, $Q=\frac{1}{2}(A-A^{\prime})$ is a skew-symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :
$P+Q= \begin{bmatrix} 3 & 3 \\ 3 & -1\end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0\end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} =A$
(ii) Let $A= \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix} $, then $A^{\prime}= \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} $
Now, $A+A^{\prime}= \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix} + \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix} = \begin{bmatrix} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{bmatrix} $
Let $P=\frac{1}{2}(A+A^{\prime})=\frac{1}{2} \begin{bmatrix} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{bmatrix} = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} $
Now, $P^{\prime}= \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} =P$
Thus, $P=\frac{1}{2}(A+A^{\prime})$ is a symmetric matrix.
Now, $A-A^{\prime}= \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix} + \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $
Let $Q=\frac{1}{2}(A-A^{\prime})= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $
Now, $Q^{\prime}= \begin{cases} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{cases} =-Q$
Thus, $Q=\frac{1}{2}(A-A^{\prime})$ is a skew-symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :
$P+Q= \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} =A$
(iii) Let $A= \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{bmatrix} $, then $A^{\prime}= \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} $
Now, $A+A^{\prime}= \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{bmatrix} + \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{bmatrix} = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} $
Let $P=\frac{1}{2}(A+A^{\prime})=\frac{1}{2} \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} $
Now, $P^{\prime}= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} =P$
Thus, $P=\frac{1}{2}(A+A^{\prime})$ is a symmetric matrix.
Now, $A-A^{\prime}= \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{bmatrix} - \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{bmatrix} = \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} $
Let $Q=\frac{1}{2}(A-A^{\prime})=\frac{1}{2} \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{bmatrix} = \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix} $
Now, $Q^{\prime}= \begin{bmatrix} 0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix} =-Q$
Thus, $Q=\frac{1}{2}(A-A^{\prime})$ is a skew-symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :
$P+Q= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{bmatrix} + \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{bmatrix} = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} =A$
(iv) Let $A= \begin{bmatrix} 1 & 5 \\ -1 & 2\end{bmatrix} $, then $A^{\prime}= \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix} $
Now $A+A^{\prime}= \begin{bmatrix} 1 & 5 \\ -1 & 2\end{bmatrix} + \begin{bmatrix} 1 & -1 \\ 5 & 2\end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} $
Let $P=\frac{1}{2}(A+A^{\prime})= \begin{cases} 1 & 2 \\ 2 & 2 \end{cases} $
Now, $P^{\prime}= \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} =P$
Thus, $P=\frac{1}{2}(A+A^{\prime})$ is a symmetric matrix.
Now, $A-A^{\prime}= \begin{bmatrix} 1 & 5 \\ -1 & 2\end{bmatrix} - \begin{bmatrix} 1 & -1 \\ 5 & 2\end{bmatrix} = \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} $
Let $Q=\frac{1}{2}(A-A^{\prime})= \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix} $
Now, $Q^{\prime}= \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} =-Q$
Thus, $Q=\frac{1}{2}(A-A^{\prime})$ is a skew-symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :
$ P+Q= \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} =A $
11. If $A, B$ are symmetric matrices of same order, then $A B-B A$ is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
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Solution
The correct answer is A.
$A$ and $B$ are symmetric matrices, therefore, we have:
$$ \begin{equation*} A^{\prime}=A \text{ and } B^{\prime}=B \tag{1} \end{equation*} $$
$$ \begin{matrix} \text{ Consider }(A B-B A)^{\prime} & =(A B)^{\prime}-(B A)^{\prime} & & {[(A-B)^{\prime}=A^{\prime}-B^{\prime}]} \\ & =B^{\prime} A^{\prime}-A^{\prime} B^{\prime} & & {[(A B)^{\prime}=B^{\prime} A^{\prime}]} \\ & =B A-A B & {[\text{ by (1)] }} \tag{1}\\ & =-(A B-B A) & \end{matrix} $$
$\therefore(A B-B A)^{\prime}=-(A B-B A)$
Thus, $(A B-B A)$ is a skew-symmetric matrix.
12. If $A=\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}$, and $A+A^{\prime}=I$, then the value of $\alpha$ is (A) $\frac{\pi}{6}$ (B) $\frac{\pi}{3}$ (C) $\pi$ (D) $\frac{3 \pi}{2}$
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Solution
The correct answer is $B$.
$ \begin{aligned} & A= \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \\ & \Rightarrow A^{\prime}= \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \end{aligned} $
Now, $A+A^{\prime}=I$
$ \begin{aligned} & \therefore \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} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{aligned} $
Comparing the corresponding elements of the two matrices, we have:
$ \begin{aligned} & 2 \cos \alpha=1 \\ & \Rightarrow \cos \alpha=\frac{1 \pi}{2}=\cos \frac{\pi}{3} \\ & \therefore \alpha=\frac{\pi}{3} \end{aligned} $
