Find $\frac{1}{2}\left(A+A^{\prime}\right)$ and $\frac{1}{2}\left(A-A^{\prime}\right)$, when $A=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]$
The given matrix is $A=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]$
$\therefore A^{\prime}=\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right]$
$A+A^{\prime}=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]+\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right]=\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$\therefore \frac{1}{2}\left(A+A^{\prime}\right)=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
Now, $A-A^{\prime}=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]-\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right]=\left[\begin{array}{ccc}0 & 2 a & 2 b \\ -2 a & 0 & 2 c \\ -2 b & -2 c & 0\end{array}\right]$
$\therefore \frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]$
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