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Question
Given \[A = \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix}\], compute A−1 and show that \[2 A^{- 1} = 9I - A .\]
Solution
\[\text{ We have, }A = \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix}\]
\[\text{ Now, }adj(A) = \begin{bmatrix}7 & 3 \\ 4 & 2\end{bmatrix}\]
\[\text{ and }\left| A \right| = 2\]
\[ \therefore A^{- 1} = \frac{1}{2}\begin{bmatrix}7 & 3 \\ 4 & 2\end{bmatrix}\]
\[\text{ Now, 2 }A^{- 1} = 9I - A\]
\[\text{ LHS }= 2 A^{- 1} = \begin{bmatrix}7 & 3 \\ 4 & 2\end{bmatrix}\]
\[\text{ RHS =} 9I - A = 9\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} - \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix} = \begin{bmatrix}7 & 3 \\ 4 & 2\end{bmatrix} =\text{ LHS }\]
Hence proved .
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