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

Given the matrix $A = \left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right].$
First,find the transpose $A^{\prime}$ by interchanging rows and columns:
$A^{\prime} = \left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right].$
Now,calculate $A+A^{\prime}$:
$A+A^{\prime} = \left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right] + \left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right].$
Therefore,$\frac{1}{2}(A+A^{\prime}) = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right].$
Next,calculate $A-A^{\prime}$:
$A-A^{\prime} = \left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right] - \left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right] = \left[\begin{array}{ccc}0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0\end{array}\right].$
Therefore,$\frac{1}{2}(A-A^{\prime}) = \left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right].$