If $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right]$ , show that ${A^2} - 5A + 7I = O.$ Hence, find ${A^{ - 1}}.$

$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right]$

L.H.S. $= {A^2} - 5A + 7I$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right] - 5\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right] + 7\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&1\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}8&5\\{ - 5}&3\end{array}} \right] - \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{15}&5\\{ - 5}&{10}\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7&0\\0&7\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{8 - 15 + 7}&{5 - 5 + 0}\\{ - 5 + 5 + 0}&{3 - 10 + 7}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\0&0\end{array}} \right] = O$

$\Rightarrow$ ${A^2} - 5A + 7I = O$

Hence, proved.
Now, multiplying by ${A^{ - 1}}$ on both sides,

we get
$({A^{ - 1}}A)A - 5A{A^{ - 1}} - 7I{A^{ - 1}} = O \Rightarrow IA - 5I + 7{A^{ - 1}} = O$

$\Rightarrow$ $A - 5I + 7{A^{ - 1}} = O \Rightarrow 7{A^{ - 1}} = 5I - A$

$\Rightarrow$ $7{A^{ - 1}} = 5\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&1\end{array}} \right] - \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right] \Rightarrow 7{A^{ - 1}}$

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

$\Rightarrow$ $7{A^{ - 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}\\1&3\end{array}} \right] \Rightarrow {A^{ - 1}} = \cfrac{1}{7}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}\\1&3\end{array}} \right]$