(a) \(A^{-1} = \begin{bmatrix}3&0\\2&1 \end{bmatrix}, |A| = 3 \ne0\)
so, A-1 will exist.
\(\because A^2x= b\)
⇒ \(x = A^{-2}b = A^{-1}(A^{-1})b\)
⇒ \(x = \left(\begin{bmatrix}3&0\\2&1\end{bmatrix}\begin{bmatrix}3&0\\2&1\end{bmatrix}\right) \begin{bmatrix}-1\\2\end{bmatrix}\)
\(= \begin{bmatrix}9&0\\8&1 \end{bmatrix} \begin{bmatrix}-1\\2 \end{bmatrix} \)
\(= \begin{bmatrix}-9\\-6 \end{bmatrix} \)
Hence, \(x= \begin{bmatrix}-9\\-6\end{bmatrix} = -3\begin{bmatrix}3\\2\end{bmatrix} \)
