If , then find A2 – 5A – 14I.

Given, $A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$$\ldots$ (i)

Now,

$A^{2}=A \cdot A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]$

$A^{2}-5 A-14 I=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

So,

$A^{2}-5 A-14 I=O$

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

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

$A^{3}=5 A^{2}+14 A$

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

$=\left[\begin{array}{cc}145 & -125 \\ -100 & 120\end{array}\right]+\left[\begin{array}{cc}42 & -70 \\ -56 & 28\end{array}\right]$

Hence,

$A^{3}=\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]$