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

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

$\therefore$ $(A + B) = \left[ {\begin{array}{llllllllllllllllllll}{0 + 0}&{1 - 1}\\{1 + 1}&{1 + 0}\end{array}} \right] = {\left[ {\begin{array}{llllllllllllllllllll}0&0\\2&1\end{array}} \right]_{2 \times 2}}$

and $(A - B) = \left[ {\begin{array}{llllllllllllllllllll}{0 - 0}&{1 + 1}\\{1 - 1}&{1 - 0}\end{array}} \right] = {\left[ {\begin{array}{llllllllllllllllllll}0&2\\0&1\end{array}} \right]_{2 \times 2}}$

Since, $(A + B) \cdot (A - B)$ is defined,

if the number of columns of $(A + B)$ is equal to the number of rows of $(A - B)$,

so here multiplication of matrices $(A + B) \cdot (A - B)$ is possible.

Now, ${(A + B)_{2 \times 2}} \cdot {(A - B)_{2 \times 2}} = \left[ {\begin{array}{llllllllllllllllllll}{0 + 0}&{0 + 0}\\{0 + 0}&{4 + 1}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}0&0\\0&5\end{array}} \right]$

….(i)
Hence, ${A^2} = A \cdot A$
$= \left[ {\begin{array}{llllllllllllllllllll}0&1\\1&1\end{array}} \right] \cdot \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1\\1&1\end{array}} \right]$
$= \left[ {\begin{array}{llllllllllllllllllll}{0 + 1}&{0 + 1}\\{0 + 1}&{1 + 1}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&1\\1&2\end{array}} \right]$

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

$\therefore$ ${A^2} - {B^2} = \left[ {\begin{array}{llllllllllllllllllll}1&1\\1&2\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&0\\0&{ - 1}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}2&1\\1&3\end{array}} \right]$

….(ii)
Thus, we see that
$(A + B) \cdot (A - B) \ne {A^2} - {B^2}$

[using Eqs. (i) and (ii)]
$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}0&0\\0&5\end{array}} \right] \ne \left[ {\begin{array}{llllllllllllllllllll}2&1\\1&3\end{array}} \right]$