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If \(A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\), find the value of λ so that A2 = λA – 2I. 

Hence, find A–1.

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\(A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)

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

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