Question

Show that \[A = \begin{bmatrix}6 & 5 \\ 7 & 6\end{bmatrix}\] satisfies the equation \[x^2 - 12x + 1 = O\]. Thus, find A⁻¹.

Answer 1

\[A = \begin{bmatrix} 6 & 5 \\7 & 6 \end{bmatrix} \]
\[ \therefore A^2 = \begin{bmatrix} 71 & 60 \\84 & 71 \end{bmatrix} \]
\[\text{ If }I_2\text{ is the identity matrix of order 2, then}\]
\[ A^2 - 12A + I_2 = \begin{bmatrix} 71 & 60 \\84 & 71 \end{bmatrix} - 12\begin{bmatrix} 6 & 5 \\7 & 6 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\0 & 1 \end{bmatrix}\]
\[ \Rightarrow A^2 - 12A + I_2 = \begin{bmatrix} 71 - 72 + 1 & 60 - 60 + 0 \\84 - 84 + 0 & 71 - 72 + 1 \end{bmatrix}\]
\[ \Rightarrow A^2 - 12A + I_2 = 0\]
\[\text{ Thus, A satisfies }x^2 - 12x + 1 = 0 . \]
Now, 
\[ A^2 - 12A + I_2 = 0\]
\[ \Rightarrow I_2 = 12A - A^2 \]
\[ \Rightarrow A^{- 1} I_2 = A^{- 1} \left( 12A - A^2 \right) \left[\text{ Pre - multiplying both sides by }A^{- 1} \right]\]
\[ \Rightarrow A^{- 1} = 12 I_2 - A\]
\[ \Rightarrow A^{- 1} = 12 \begin{bmatrix} 1 & 0 \\0 & 1 \end{bmatrix} - \begin{bmatrix} 6 & 5 \\7 & 6 \end{bmatrix} \]
\[ \Rightarrow A^{- 1} = \begin{bmatrix} 12 - 6 & 0 - 5\\ 0 - 7 & 12 - 6 \end{bmatrix} \]
\[ \Rightarrow A^{- 1} = \begin{bmatrix} 6 & - 5 \\ - 7 & 6 \end{bmatrix} \]