\(A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)
A2 =\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)\(=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}\)
\(=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)
Now, A2 = λA – 2I
= λA = A2 + 2I
