If $A = \left[ {\begin{array}{cccccccccccccccccccc}1&0&{ - 1}\\2&1&3\\0&1&1\end{array}} \right]$, then verify that ${A^2} + A = (A + I)$, where $I$ is $3 \times 3$ unit matrix.

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}1&0&{ - 1}\\2&1&3\\0&1&1\end{array}} \right]$

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

$= \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}&{ - 2}\\4&4&4\\2&2&4\end{array}} \right]$

$\therefore$ ${A^2} + A = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}&{ - 2}\\4&4&4\\2&2&4\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}1&0&{ - 1}\\2&1&3\\0&1&1\end{array}} \right]$

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

…..(i)
Now, $A + I = \left[ {\begin{array}{cccccccccccccccccccc}1&0&{ - 1}\\2&1&3\\0&1&1\end{array}} \right] + \left[ {\begin{array}{llllllllllllllllllll}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\2&2&3\\0&1&2\end{array}} \right]$

and $A(A + I) = \left[ {\begin{array}{cccccccccccccccccccc}1&0&{ - 1}\\2&1&3\\0&1&1\end{array}} \right] \cdot \left[ {\begin{array}{cccccccccccccccccccc}2&0&{ - 1}\\2&2&3\\0&1&2\end{array}} \right]$

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

…..(ii)
Thus, we see that ${A^2} + A = A(A + I)$

[using Eqs. (i) and (ii)]