Solve this

Solution:

$A=\left[\begin{array}{ll}3 & -2\end{array}\right.$

$4-2]$

$\therefore \mathrm{A}^{2}=\left[\begin{array}{ll}1 & -2\end{array}\right.$

$4-4]$

Given :

$A^{2}=\lambda A-2 I$            ....(1)

$\Rightarrow[1-2$

$4-4]=\lambda\left[\begin{array}{ll}3 & -2\end{array}\right.$

$4-2]-2\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}1 & -2\end{array}\right.$

$4-4]=\left[\begin{array}{ll}3 \lambda & -2 \lambda\end{array}\right.$

$4 \lambda \quad-2 \lambda-2]$

On equating corresponding terms, we get

$\begin{aligned}-2 \lambda &=-2 \\ \Rightarrow \lambda &=1 \end{aligned}$

On substituting $\lambda=1$ in (1), we get

$A^{2}=A-2 I$

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

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

$\Rightarrow A^{-1}\left(A-A^{2}\right)=A^{-1} \times 2 I \quad\left(\right.$ Pre $-$ multiplying both sides with $\left.A^{-1}\right)$

$\Rightarrow I-A=2 A^{-1}$

$0 \quad 1]-\left[\begin{array}{ll}3 & -2\end{array}\right.$

$4-2]=\left[\begin{array}{ll}1-3 & 0+2\end{array}\right.$

$0-4 \quad 1+2]$

$\Rightarrow A^{-1}=\frac{1}{2}\left[\begin{array}{ll}-2 & 2\end{array}\right.$

$-4 \quad 3]$