Question

If \[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\], show that 

\[A^2 - 5A + 7I = O\].  Hence, find A⁻¹.

Answer 1

\[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix} = \begin{bmatrix}9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4\end{bmatrix} = \begin{bmatrix}8 & 5 \\ - 5 & 3\end{bmatrix}\]
and
\[ A^2 - 5A + 7I = \begin{bmatrix}8 & 5 \\ - 5 & 3\end{bmatrix} - \begin{bmatrix}15 & 5 \\ - 5 & 10\end{bmatrix} + \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}\]
\[ \Rightarrow A^2 - 5A + 7I = \begin{bmatrix}8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = O\]
Now, 
\[ A^2 - 5A + 7I = 0\]
\[ \Rightarrow A^2 - 5A = - 7I\]
\[ \Rightarrow A^{- 1} A^2 - 5 A^{- 1} A = - 7 A^{- 1} I \left[\text{ Pre - multiplying both sides by }A^{- 1} \right]\]
\[ \Rightarrow A - 5I = - 7 A^{- 1} \]
\[ \Rightarrow A^{- 1} = - \frac{1}{7}\left( A - 5I \right)\]
\[ \Rightarrow A^{- 1} = - \frac{1}{7}\left\{ \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix} - \begin{bmatrix}5 & 0 \\ 0 & 5\end{bmatrix} \right\} = \frac{1}{7}\begin{bmatrix}2 & - 1 \\ 1 & 3\end{bmatrix}\]