Show that $A = \left[ {\begin{array}{cccccccccccccccccccc}5&3\\{ - 1}&{ - 2}\end{array}} \right]$

satisfies the equation ${A^2} - 3A - 7I = 0$ and hence find the value of ${A^{ - 1}}$.

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}5&3\\{ - 1}&{ - 2}\end{array}} \right]$

$\therefore$ ${A^2} = A \cdot A = \left[ {\begin{array}{cccccccccccccccccccc}5&3\\{ - 1}&{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{cccccccccccccccccccc}5&3\\{ - 1}&{ - 2}\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}{25 - 3}&{15 - 6}\\{ - 5 + 2}&{ - 3 + 4}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}{22}&9\\{ - 3}&1\end{array}} \right]$

$3A = 3\left[ {\begin{array}{cccccccccccccccccccc}5&3\\{ - 1}&{ - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{15}&9\\{ - 3}&{ - 6}\end{array}} \right]$

and $7I = 7\left[ {\begin{array}{llllllllllllllllllll}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}7&0\\0&7\end{array}} \right]$
$\therefore$ ${A^2} - 3A - 7I = \left[ {\begin{array}{llllllllllllllllllll}{22}&9\\{ - 3}&1\end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll}{15}&9\\{ - 3}&{ - 6}\end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll}7&0\\0&7\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}{22 - 15 - 7}&{9 - 9 - 0}\\{ - 3 + 3 - 0}&{1 + 6 - 7}\end{array}} \right]$
$= \left[ {\begin{array}{llllllllllllllllllll}0&0\\0&0\end{array}} \right]$
$= 0$

Hence proved.

Since, ${A^2} - 3A - 7I = 0$
$\Rightarrow$ ${A^{ - 1}}\left[ {\left( {{A^2}} \right) - 3A - 7I} \right] = {A^{ - 1}}0$

$\Rightarrow {A^{ - 1}}A \cdot A - 3{A^{ - 1}}A - 7{A^{ - 1}}I = 0$

$\Rightarrow$ $IA - 3I - 7{A^{ - 1}} = 0$

$\Rightarrow A - 3I - 7{A^{ - 1}} =$
$\Rightarrow$ $- 7{A^{ - 1}} = - A + 3I$

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

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