Question

Show that \[A = \begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix}\] satisfies the equation \[x^2 - 3x - 7 = 0\]. Thus, find A⁻¹.

Answer 1

\[A = \begin{bmatrix} 5 & 3\\- 1 & - 2 \end{bmatrix} \]
\[ A^2 = \begin{bmatrix} 22 & 9\\ - 3 & 1 \end{bmatrix}\]
\[\text{ If } I_2\text{ is the identity matrix of order 2, then}\]
\[ A^2 - 3A - 7 I_2 = \begin{bmatrix} 22b & 9\\ - 3 & 1 \end{bmatrix} - 3\begin{bmatrix} 5 & 3\\ - 1 & - 2 \end{bmatrix} - 7\begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} \]
\[ \Rightarrow A^2 - 3A - 7 I_2 = \begin{bmatrix} 22 - 15 - 7 & 9 - 9 - 0\\ - 3 + 3 + 0 & 1 + 6 - 7 \end{bmatrix} = \begin{bmatrix} 0 & 0\\0 & 0 \end{bmatrix} = 0\]
\[ \Rightarrow A^2 - 3A - 7 I_2 = 0\]
\[\text{ Thus, A satisfies }x^2 - 3x - 7 = 0 . \]
Now, 
\[ A^2 - 3A - 7 I_2 = 0\]
\[ \Rightarrow A^2 - 3A = 7 I_2 \]
\[ \Rightarrow A^{- 1} \left( A^2 - 3A \right) = A^{- 1} \times 7 I_2 \left[\text{ Pre - multiplying both sides by } A^{- 1} \right]\]
\[ \Rightarrow A - 3 I_2 = 7 A^{- 1} \]
\[ \Rightarrow \begin{bmatrix} 5 & 3 \\ - 1 & - 2 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =  7 A^{- 1} \]
\[ \Rightarrow A^{- 1} = \frac{1}{7} \begin{bmatrix} 5 - 3 & 3 - 0\\- 1 - 0 & - 2 - 3 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 2 & 3\\- 1 & - 5 \end{bmatrix} \]