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# If A=$\begin{bmatrix}1 & 2& 2\\2 & 1 & 2\\2 & 2&1\end{bmatrix}$ verify that $A^2-4A-5I=0.$

Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
• An identity matrix or unit matrix of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. An identity matrix of order 3, $I_{3}= \begin{bmatrix} 1 &0 &0 \\ 0&1&0\\0&0&1 \end{bmatrix}$
Step 1: $A^2=A.A=\begin{bmatrix}1 & 2& 2\\2 & 1 & 2\\2 & 2& 1\end{bmatrix}\begin{bmatrix}1 & 2& 2\\2 & 1 & 2\\2 & 2& 1\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}1(1)+2(2)+2(2) & 1(2)+2(1)+2(2)& 1(2)+2(2)+2(1)\\2(1)+1(2)+2(2) & 2(2)+1(1)+2(2) & 2(2)+2(2)+1(1)\\2(1)+2(2)+1(2) & 2(2)+2(1)+1(2)& 2(2)+2(2)+1(1)\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}1+4+4 & 2+2+4& 2+4+2\\2+2+4 & 4+1+4 & 4+4+1\\2+4+2 & 4+2+2& 4+4+1\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}9 & 8& 8\\8 & 9 & 8\\8 & 8& 9\end{bmatrix}$
Step 2: 4A=4$\begin{bmatrix}1 & 2 & 2\\2 & 1 &2\\2 & 2 &1\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}4 & 8& 8\\8 & 4 & 8\\8 & 8& 4\end{bmatrix}$
Step 3: 5I=5$\begin{bmatrix}1 & 0 & 0\\0 & 1 &0\\0 & 0 &1\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}5 & 0& 0\\0 & 5 & 0\\0 & 0& 5\end{bmatrix}$
Step 4: $A^2-4A-5I=\begin{bmatrix}9 & 8& 8\\8 & 9 & 8\\8 & 8& 9\end{bmatrix}+(-1)\begin{bmatrix}4 & 8& 8\\8 & 4 & 8\\8 & 8& 4\end{bmatrix}-\begin{bmatrix}5 & 0& 0\\0 & 5 & 0\\0 & 0& 5\end{bmatrix}$
$\qquad\qquad\qquad\qquad=\begin{bmatrix}9-4-5 & 8-8+0& 8-8-0\\8-8-0 & 9-4-5 & 8-8-0\\8-8-0 & 8-8-0& 9-4-5\end{bmatrix}$
$\qquad\qquad\qquad\qquad=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}=0$[zero matrix]