\(A=\begin{bmatrix}6&5\\7&6\end{bmatrix}\)
We have A2 – 12A + I = 0
A2 = \(\begin{bmatrix}6&5\\7&6\end{bmatrix}\)\(\begin{bmatrix}6&5\\7&6\end{bmatrix}\)\(=\begin{bmatrix}36+35&30+30\\42+42&35+36\end{bmatrix}\)
\(=\begin{bmatrix}71&60\\84&71\end{bmatrix}\)
Now, A2 – 12A + 1 = 0