Question

Show that

\[A = \begin{bmatrix}- 8 & 5 \\ 2 & 4\end{bmatrix}\] satisfies the equation \[A^2 + 4A - 42I = O\]. Hence, find A⁻¹.

Answer 1

\[A = \begin{bmatrix}- 8 & 5 \\ 2 & 4\end{bmatrix}\]
\[ \therefore A^2 = \begin{bmatrix}74 & - 20 \\ - 8 & 26\end{bmatrix}\]
and
\[ A^2 + 4A - 42I = \begin{bmatrix}74 & - 20 \\ - 8 & 26\end{bmatrix} + \begin{bmatrix}- 32 & 20 \\ 8 & 16\end{bmatrix} - \begin{bmatrix}42 & 0 \\ 0 & 42\end{bmatrix}\]
\[ \Rightarrow A^2 + 4A - 42I = \begin{bmatrix}74 - 32 - 42 & - 20 + 20 - 0 \\ - 8 + 8 - 0 & 26 + 16 - 42\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = O\]
Now, 
\[ A^2 + 4A - 42I = 0\]
\[ \Rightarrow A^2 + 4A = 42I\]
\[ \Rightarrow A^{- 1} A^2 + 4 A^{- 1} A = 42I A^{- 1} \left[\text{ Pre - multiplying both sides by }A^{- 1} \right]\]
\[ \Rightarrow A + 4I = 42 A^{- 1} \]
\[ \Rightarrow A^{- 1} = \frac{1}{42}\left( A + 4I \right)\]
\[ \Rightarrow A^{- 1} = \frac{1}{42}\left\{ \begin{bmatrix}- 8 & 5 \\ 2 & 4\end{bmatrix} + \begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix} \right\} = \frac{1}{42}\begin{bmatrix}- 4 & 5 \\ 2 & 8\end{bmatrix}\]