If $A = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 5}\\{ - 4}&2\end{array}} \right]$, then find ${A^2} - 5A - 14I$. Hence, obtain ${A^3}$.

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 5}\\{ - 4}&2\end{array}} \right]$
……(i)
$\therefore$ ${A^2} = A \cdot A$

$= \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 5}\\{ - 4}&2\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 5}\\{ - 4}&2\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{29}&{ - 25}\\{ - 20}&{24}\end{array}} \right]$

…..(ii)
$\therefore$ ${A^2} - 5A - 14I = \left[ {\begin{array}{cccccccccccccccccccc}{29}&{ - 25}\\{ - 20}&{24}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}{15}&{ - 25}\\{ - 20}&{10}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}{14}&0\\0&{14}\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}0&0\\0&0\end{array}} \right]$

Now, ${A^2} - 5A - 14I = 0$

$\Rightarrow$ $A \cdot {A^2} - 5A \cdot A - 14AI = 0$

$\Rightarrow$ ${A^3} - 5{A^2} - 14A = 0$

$\Rightarrow$ ${A^3} = 5{A^2} = 14A$

$= 5\left[ {\begin{array}{cccccccccccccccccccc}{29}&{ - 25}\\{ - 20}&{24}\end{array}} \right] + 14\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 5}\\{ - 4}&2\end{array}} \right]$

[using Eqs. (i) and (ii)]
$= \left[ {\begin{array}{cccccccccccccccccccc}{145}&{ - 125}\\{ - 100}&{120}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{42}&{ - 70}\\{ - 56}&{28}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{187}&{ - 195}\\{ - 156}&{148}\end{array}} \right]$