Find (A+ A’) and $\cfrac{1}{2}(A - A')$, when , $A = \left[ {\begin{array}{cccccccccccccccccccc}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right]$

.:

$A = \left[ {\begin{array}{cccccccccccccccccccc}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right]$ $\Rightarrow$ $A' = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - a}&{ - b}\\a&0&{ - c}\\b&c&0\end{array}} \right]$

Now, A + A’$= \left[ {\begin{array}{cccccccccccccccccccc}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right]$ $+ \left[ {\begin{array}{cccccccccccccccccccc}0&{ - a}&{ - b}\\a&0&{ - c}\\b&c&0\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{0 + 0}&{a - a}&{b - b}\\{ - a + a}&{0 + 0}&{c - c}\\{ - b + b}&{ - c + c}&{0 + 0}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\0&0&0\\0&0&0\end{array}} \right]$

$\therefore$ $\cfrac{1}{2}(A + A') = \left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\0&0&0\\0&0&0\end{array}} \right]$

Now, $A - A' = \left[ {\begin{array}{cccccccccccccccccccc}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}0&{ - a}&{ - b}\\a&0&{ - c}\\b&c&0\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{0 - 0}&{a + a}&{b + b}\\{ - a - a}&{0 - 0}&{c + c}\\{ - b - b}&{ - c - c}&{0 - 0}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}0&{2a}&{2b}\\{ - 2a}&0&{2c}\\{ - 2b}&{ - 2c}&0\end{array}} \right]$

$\therefore$ $\cfrac{1}{2}(A - A') = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}0&{2a}&{2b}\\{ - 2a}&0&{2c}\\{ - 2b}&{ - 2c}&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right]$