Question
If \[ A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \] prove that \[ A^2 – 4A + 5I = O. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} \]Step 2: Form Expression
\[ A^2 – 4A + 5I = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} – \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \]Step 3: Simplify
\[ = \begin{bmatrix} 9 – 4 + 5 & 8 – 8 + 0 & 8 – 8 + 0 \\ 8 – 8 + 0 & 9 – 4 + 5 & 8 – 8 + 0 \\ 8 – 8 + 0 & 8 – 8 + 0 & 9 – 4 + 5 \end{bmatrix} \] \[ = \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} \]Final Result
\[
A^2 – 4A + 5I = 10I \neq O
\]
Hence, the given statement is incorrect. The correct result is \(10I\), not \(O\).