Solve this

Solution:

Given: $A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$

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

$=\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right]$

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

$A^{2}-5 A-14 I=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-14\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$=\left[\begin{array}{ll}29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14\end{array}\right]$

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

Therefore, $A^{2}-5 A-14 l=0$     ..(1)

Premultiplying the (1) by $A$, we get

$A\left(A^{2}-5 A-14 l\right)=A \cdot 0$

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

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

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

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

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