If $A = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - x}\\x&0\end{array}} \right],B = \left[ {\begin{array}{cccccccccccccccccccc}0&1\\1&0\end{array}} \right]$ and ${x^2} = - 1$, then show that ${(A + B)^2} = {A^2} + {B^2}$.

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - x}\\x&0\end{array}} \right],B = \left[ {\begin{array}{cccccccccccccccccccc}0&1\\1&0\end{array}} \right]$ and ${x^2} = - 1$

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

and ${(A + B)^2} = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - x + 1}\\{x + 1}&0\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}0&{ - x + 1}\\{x + 1}&0\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{1 - {x^2}}&0\\0&{1 - {x^2}}\end{array}} \right]$

…….(i)
${A^2} = A \cdot A = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - x}\\x&0\end{array}} \right]{\left[ {\begin{array}{cccccccccccccccccccc}0&{ - x}\\x&0\end{array}} \right]_ - }$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - {x^2}}&0\\0&{ - {x^2}}\end{array}} \right]$

and ${B^2} = B \cdot B = \left[ {\begin{array}{llllllllllllllllllll}0&1\\1&0\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}0&1\\1&0\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&0\\0&1\end{array}} \right]$

Now, ${A^2} + {B^2} = \left[ {\begin{array}{cccccccccccccccccccc}{ - {x^2} + 1}&0\\0&{ - {x^2} + 1}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{1 - {x^2}}&0\\0&{1 - {x^2}}\end{array}} \right]$

[using Eq.(i)]
$= {(A + B)^2}$