\(A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)
A2 =\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)\(=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}\)
\(=\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)
Now, A2 – 5A + 7I\(=\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)\(-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)\(+7\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
