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Show that \(A= \begin{bmatrix}5 &3\\-1&-2\end{bmatrix}\) satisfies the equation x2 – 3A – 7 = 0. 

Thus, find A–1.

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\(A= \begin{bmatrix}5 &3\\-1&-2\end{bmatrix}\)

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

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