Matrices — Class 12 Maths Solution

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$A = \left[ {\begin{array}{cccccccccccccccccccc}1&5\\6&7\end{array}} \right] \Rightarrow A' = \left[ {\begin{array}{cccccccccccccccccccc}1&6\\5&7\end{array}} \right]$

(i) $A + A' = \left[ {\begin{array}{cccccccccccccccccccc}1&5\\6&7\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}1&6\\5&7\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{1 + 1}&{5 + 6}\\{6 + 5}&{7 + 7}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}2&{11}\\{11}&{14}\end{array}} \right]$

$\Rightarrow$ $(A + A')' = \left[ {\begin{array}{cccccccccccccccccccc}2&{11}\\{11}&{14}\end{array}} \right] = (A + A')$

So, A + A’ is a symmetric matrix.

(ii) $A - A' = \left[ {\begin{array}{cccccccccccccccccccc}1&5\\6&7\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}1&6\\5&7\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{1 - 1}&{5 - 6}\\{6 - 5}&{7 - 7}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 1}\\1&0\end{array}} \right]$
$\Rightarrow$ $(A - A')$

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

So, A$-$A’ is a skew$-$symmetric matrix.