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Question
If \[A = \begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\] , find \[A^{- 1}\] and prove that \[A^2 - 4A - 5I = O\]
Solution
\[A = \begin{bmatrix} 1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1 \end{bmatrix} \]
\[ \Rightarrow \left| A \right| = \begin{vmatrix} 1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1 \end{vmatrix} = 1\left( 1 - 4 \right) - 2\left( 2 - 4 \right) + 2\left( 4 - 2 \right) = - 3 + 4 + 4 = 5 \]
\[\text{ Since, }\left| A \right| \neq 0\]
Hence, A is invertible .
Now,
\[ A^2 = \begin{bmatrix} 1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 + 4 + 4 & 2 + 2 + 4 & 2 + 4 + 2\\2 + 2 + 4 & 4 + 1 + 4 & 4 + 2 + 2\\2 + 4 + 2 & 4 + 2 + 2 & 1 + 4 + 4 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8\\8 & 9 & 8\\8 & 8 & 9 \end{bmatrix}\]
\[\text{ Now, }A^2 - 4A - 5I = \begin{bmatrix} 9 & 8 & 8\\8 & 9 & 8\\8 & 8 & 9 \end{bmatrix} - 4\begin{bmatrix} 1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1 \end{bmatrix} - 5\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 - 4 - 5 & 8 - 8 - 0 & 8 - 8 - 0\\8 - 8 - 0 & 9 - 4 - 5 & 8 - 8 - 0\\8 - 8 - 0 & 8 - 8 - 0 & 9 - 4 - 5 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0 \end{bmatrix} = O \]
\[ \Rightarrow A^2 - 4A - 5I = O [\text{ Proved }]\]
\[\text{ Again,} A^2 - 4A - 5I = O\]
\[ \Rightarrow A^{- 1} \left( A^2 - 4A - 5I \right) = A^{- 1} O [\text{ Pre - multiplying with }A^{- 1} ]\]
\[ \Rightarrow A^{- 1} A^2 - 4 A^{- 1} A - 5 A^{- 1} = O\]
\[ \Rightarrow A - 4I = 5 A^{- 1} \]
\[ \Rightarrow 5 A^{- 1} = \begin{bmatrix} 1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1 \end{bmatrix} - 4\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 - 4 & 2 - 0 & 2 - 0\\2 - 0 & 1 - 4 & 2 - 0\\2 - 0 & 2 - 0 & 1 - 4 \end{bmatrix} = \begin{bmatrix} - 3 & 2 & 2\\ 2 & - 3 & 2\\ 2 & 2 & - 3 \end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{5}\begin{bmatrix} - 3 & 2 & 2\\ 2 & - 3 & 2\\ 2 & 2 & - 3 \end{bmatrix}\]
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