# Find and, when

Question:

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

Solution:

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, $A-A^{\prime}=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]-\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right]=\left[\begin{array}{ccc}0 & 2 a & 2 b \\ -2 a & 0 & 2 c \\ -2 b & -2 c & 0\end{array}\right]$

$\therefore \frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]$