Question

If \[A = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\] , verify that \[A^2 - 4 A + I = O,\text{ where }I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\text{ and }O = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\] . Hence, find A⁻¹.

Answer 1

\[A = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]
\[ \therefore A^2 = \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix}\]
and
\[ A^2 - 4A + I = \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix} - \begin{bmatrix}8 & 12 \\ 4 & 8\end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^2 - 4A + I = \begin{bmatrix}7 - 8 + 1 & 12 - 12 + 0 \\ 4 - 4 + 0 & 7 - 8 + 1\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = O\]
\[ \Rightarrow A^2 - 4A + I = 0\]
\[ \Rightarrow A^2 - 4A = - I\]
\[ \Rightarrow A^{- 1} A^2 - 4A A^{- 1} = - I A^{- 1} \left[\text{ Pre - multiplying both sides by }A^{- 1} \right]\]
\[ \Rightarrow A - 4I = - A^{- 1} \]
\[ \Rightarrow A^{- 1} = 4I - A\]
\[ \Rightarrow A^{- 1} = \left\{ \begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix} - \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix} \right\} = \begin{bmatrix}2 & - 3 \\ - 1 & 2\end{bmatrix}\]