Question

If \[A = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\] , then show that \[A - 3I = 2 \left( I + 3 A^{- 1} \right) .\]

Answer 1

\[\text{ We have, A }= \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\]
Now, 
\[adj(A) = \begin{bmatrix}1 & - 5 \\ - 2 & 4\end{bmatrix}\]
\[\text{ and }\left| A \right| = - 6\]
\[ \therefore A^{- 1} = - \frac{1}{6}\begin{bmatrix}1 & - 5 \\ - 2 & 4\end{bmatrix}\]
\[\text{ Now, }A - 3I = I + 3 A^{- 1} \]
\[\text{ LHS }= A - 3I = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix} - 3\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 5 \\ 2 & - 2\end{bmatrix}\]
\[\text{ RHS }= 2\left( I + 3 A^{- 1} \right) = 2\left\{ \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} - 3 \times \frac{1}{6}\begin{bmatrix}1 & - 5 \\ - 2 & 4\end{bmatrix} \right\} = 2\begin{bmatrix}0 . 5 & 2 . 5 \\ 1 & - 1\end{bmatrix} = \begin{bmatrix}1 & 5 \\ 2 & - 2\end{bmatrix} =\text{ LHS }\]
Hence proved .