If $A = \left[ {\begin{array}{cccccccccccccccccccc}1&2&{ - 3}\\5&0&2\\1&{ - 1}&1\end{array}} \right],B = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 1}&2\\4&2&5\\2&0&3\end{array}} \right]$ and $C = \left[ {\begin{array}{cccccccccccccccccccc}4&1&2\\0&3&2\\1&{ - 2}&3\end{array}} \right]$, then compute (A + B) and (B$-$C).
Also, verify that A + (B$-$C) = (A + B)$-$C.

.:

Here, A, B and C is a 3 × 3 matrix. So, A, B and C are comparable. So, (A + B), (B$-$C), A + (B$-$C) and (A + B )$-$C are defined and each one is 3 x 3 matrix.

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

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

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

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

Hence, A + (B$-$C) = (A + B)$-$C.