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Question
`A=[[1,2,2],[2,1,2],[2,2,1]]`, then prove that A2 − 4A − 5I = 0
Solution
\[Given: \hspace{0.167em} A = \begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}1 + 4 + 4 & 2 + 2 + 4 & 2 + 4 + 2 \\ 2 + 2 + 4 & 4 + 1 + 4 & 4 + 2 + 2 \\ 2 + 4 + 2 & 4 + 2 + 2 & 4 + 4 + 1\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{bmatrix}\]
\[ A^2 - 4A - 5I\]
\[ \Rightarrow A^2 - 4A - 5I = \begin{bmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{bmatrix} - 4\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix} - 5\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^2 - 4A - 5I = \begin{bmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{bmatrix} - \begin{bmatrix}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{bmatrix} - \begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{bmatrix}\]
\[ \Rightarrow A^2 - 4A - 5I = \begin{bmatrix}9 - 4 - 5 & 8 - 8 - 0 & 8 - 8 - 0 \\ 8 - 8 - 0 & 9 - 4 - 5 & 8 - 8 - 0 \\ 8 - 8 - 0 & 8 - 8 - 0 & 9 - 4 - 5\end{bmatrix}\]
\[ \Rightarrow A^2 - 4A - 5I = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} = 0\]
Hence proved
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