Question

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$, then compute $A^2 - 4A - 5I$.

Hint:

Notice the highly symmetric structure of matrix $A$. Because it is a symmetric matrix where all diagonal values are identical and all off-diagonal values are identical, its square $A^2$ preserves this exact pattern. This helps verify your arithmetic symmetry quickly.

Answer 1

Concept:
To evaluate the expression \(A^2-4A-5I\), we first find the square of the matrix \(A\), then compute the scalar multiples \(4A\) and \(5I\), and finally subtract the matrices element-wise. 

Matrix Square: \[ A^2=A\cdot A \] Identity Matrix: \[ I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \] 

Step 1: Compute \(A^2\).
Given, \[ A= \begin{bmatrix} 1 & 2 & 2\\ 2 & 1 & 2\\ 2 & 2 & 1 \end{bmatrix} \] Therefore, \[ A^2= \begin{bmatrix} 1 & 2 & 2\\ 2 & 1 & 2\\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2\\ 2 & 1 & 2\\ 2 & 2 & 1 \end{bmatrix} \] Computing each entry, First row: \[ \begin{aligned} c_{11}&=(1)(1)+(2)(2)+(2)(2)=9,\\ c_{12}&=(1)(2)+(2)(1)+(2)(2)=8,\\ c_{13}&=(1)(2)+(2)(2)+(2)(1)=8. \end{aligned} \] Second row: \[ \begin{aligned} c_{21}&=(2)(1)+(1)(2)+(2)(2)=8,\\ c_{22}&=(2)(2)+(1)(1)+(2)(2)=9,\\ c_{23}&=(2)(2)+(1)(2)+(2)(1)=8. \end{aligned} \] Third row: \[ \begin{aligned} c_{31}&=(2)(1)+(2)(2)+(1)(2)=8,\\ c_{32}&=(2)(2)+(2)(1)+(1)(2)=8,\\ c_{33}&=(2)(2)+(2)(2)+(1)(1)=9. \end{aligned} \] Hence, \[ A^2= \begin{bmatrix} 9 & 8 & 8\\ 8 & 9 & 8\\ 8 & 8 & 9 \end{bmatrix}. \] 

Step 2: Compute \(4A\) and \(5I\).
\[ 4A= 4 \begin{bmatrix} 1 & 2 & 2\\ 2 & 1 & 2\\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 8 & 8\\ 8 & 4 & 8\\ 8 & 8 & 4 \end{bmatrix}. \] \[ 5I= 5 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 5 \end{bmatrix}. \] 

Step 3: Evaluate \(A^2-4A-5I\).
\[ \begin{aligned} A^2-4A-5I &= \begin{bmatrix} 9 & 8 & 8\\ 8 & 9 & 8\\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8\\ 8 & 4 & 8\\ 8 & 8 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 5 \end{bmatrix} \\[1ex] &= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}. \end{aligned} \] Therefore, \[ \boxed{A^2-4A-5I=O}, \] where \(O\) is the zero matrix.