If $A = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right]$and $B = \left[ {\begin{array}{cccccccccccccccccccc}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right]$, then verify that

(i) $(A + B)' = A' + B',$

(ii) $(A - B)' = A' - B'$

.:

(i) $A + B = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 - 4}&{2 + 1}&{3 - 5}\\{5 + 1}&{7 + 2}&{9 + 0}\\{ - 2 + 1}&{1 + 3}&{1 + 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 5}&3&{ - 2}\\6&9&9\\{ - 1}&4&2\end{array}} \right]$

Now, $A' = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&5&{ - 2}\\2&7&1\\3&9&1\end{array}} \right]$ and $B' = \left[ {\begin{array}{cccccccccccccccccccc}{ - 4}&1&1\\1&2&3\\{ - 5}&0&1\end{array}} \right]$
$\Rightarrow$ $(A + B)' = \left[ {\begin{array}{cccccccccccccccccccc}{ - 5}&6&{ - 1}\\3&9&4\\{ - 2}&9&2\end{array}} \right]$

and $A' + B' = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&5&{ - 2}\\2&7&1\\3&9&1\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{ - 4}&1&1\\1&2&3\\{ - 5}&0&1\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 - 4}&{5 + 1}&{ - 2 + 1}\\{2 + 1}&{7 + 2}&{1 + 3}\\{3 - 5}&{9 + 0}&{1 + 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 5}&6&{ - 1}\\3&9&4\\{ - 2}&9&2\end{array}} \right]$
$\therefore$ $(A + B)' = A' + B'$

(ii) $A - B = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 + 4}&{2 - 1}&{3 + 5}\\{5 - 1}&{7 - 2}&{9 - 0}\\{ - 2 - 1}&{1 - 3}&{1 - 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}3&1&8\\4&5&9\\{ - 3}&{ - 2}&0\end{array}} \right]$

$\Rightarrow$ $(A - B)'$

% =${\left[ {\begin{array}{cccccccccccccccccccc}3&1&8\\4&5&9\\{ - 3}&{ - 2}&0\end{array}} \right]^{\prime}}$

$= \left[ {\begin{array}{cccccccccccccccccccc}3&4&{ - 3}\\1&5&{ - 2}\\8&9&0\end{array}} \right]$

and $A' - B' = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&5&{ - 2}\\2&7&1\\3&9&1\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}{ - 4}&1&1\\1&2&3\\{ - 5}&0&1\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1 + 4}&{5 - 1}&{ - 2 - 1}\\{2 - 1}&{7 - 2}&{1 - 3}\\{3 + 5}&{9 - 0}&{1 - 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}3&4&{ - 3}\\1&5&{ - 2}\\8&9&0\end{array}} \right]$
$\therefore$ $(A - B)' = A' - B'$.