Solve this

Solution:

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

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

$\therefore A^{2}=\left[\begin{array}{ll}22 & 27 \\ 18 & 31\end{array}\right]$

Now,

$A^{2}-x A+y I=O$

$\Rightarrow\left[\begin{array}{ll}22 & 27 \\ 18 & 31\end{array}\right]-\left[\begin{array}{ll}4 x & 3 x \\ 2 x & 5 x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 0 & y\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}22-4 x-y & 27-3 x \\ 18-2 x & 31-5 x-y\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

Thus, we have

$22-4 x+y=0,27-3 x=0,18-2 x=0$ and $31-5 x+y=0$

$\Rightarrow-3 x=-27$

$\Rightarrow x=9$

On putting $x=9$ in $22-4 x+y=0$, we get

$22-36+y=0$

$\Rightarrow-14=-y$

$\Rightarrow y=14$

Now,

$A^{2}-9 A+14 I=0$

$\Rightarrow A^{2}-9 A=-14 I$

$\Rightarrow A^{-1} A^{2}-9 A A^{-1}=-14 I A^{-1} \quad\left[\right.$ Pre $-$ multiplying both sides by $\left.A^{-1}\right]$

$\Rightarrow A-9 I=-14 A^{-1}$

$\Rightarrow A^{-1}=-\frac{1}{14}(A-9 I)$

$\Rightarrow A^{-1}=-\frac{1}{14}\left\{\left[\begin{array}{ll}4 & 3 \\ 2 & 5\end{array}\right]-\left[\begin{array}{ll}9 & 0 \\ 0 & 9\end{array}\right]\right\}=\frac{1}{14}\left[\begin{array}{cc}5 & -3 \\ -2 & 4\end{array}\right]$