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Show that \(A=\begin{bmatrix}6&5\\7&6\end{bmatrix}\) satisfies the equation x2–12 x + 1 = 0. 

Thus, find A–1.

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\(A=\begin{bmatrix}6&5\\7&6\end{bmatrix}\)

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

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