\(A= \begin{bmatrix}5 &3\\-1&-2\end{bmatrix}\)
A2 =\(\begin{bmatrix}5 &3\\-1&-2\end{bmatrix}\)\(\begin{bmatrix}5 &3\\-1&-2\end{bmatrix}\)\(=\begin{bmatrix}25-3&15-6\\-5+12&-3+4\end{bmatrix}\)
\(=\begin{bmatrix}22&9\\-3&1\end{bmatrix}\)
Now, A2 – 3A – 7 = 0
So, A2 – 3A – 7I = 0
Multiply by A–1 both sides
= A.A. A–1 – 3A. A–1 – 7I. A–1 = 0
= A – 3I – 7A–1 = 0
= 7A–1 = A – 3I