<s>[INST] Prove the matrix satisfies the equation $A^2 -4A-5I=0$ How to prove that
$$
A=\begin{bmatrix}
  1 & 2 & 2 \\
  2 & 1 & 2 \\
  2 & 2 & 1
\end{bmatrix}
$$
satisfies the equation $A^2 -4A-5I=0$?
 [/INST] If you would like to sped up the calculation (supposing you are doing it on your own) you can use this approach.
Let's take the equation
$$ A^2 - 4A - 5I = 0.$$
And now forget about the matrices. What we have is simple quadratic equation in $x$ written as
$$ x^2 - 4x - 5 = 0,$$
now just find the roots
$$ (x-5)(x+1) = 0.$$
Let's get back to that matrix
$$(A-5I)(A+I) = 0.$$
Simple calculations give
$$
\begin{pmatrix}
-4 & 2  &2\\
2  & -4 &2\\
2  & 2  &-4
\end{pmatrix}\cdot
\begin{pmatrix}
2 & 2  &2\\
2  & 2 &2\\
2  & 2  &2
\end{pmatrix} = 0.
$$
And now it is very simple since you divide by 2
$$
\begin{pmatrix}
-4 & 2  &2\\
2  & -4 &2\\
2  & 2  &-4
\end{pmatrix}\cdot
\begin{pmatrix}
1 & 1 &1\\
1 & 1 &1\\
1 & 1 &1
\end{pmatrix} = 0.
$$
Multiplying those matrices is just adding all the elements in each row of the first matrix. So it is $0$ everywhere. Such tricks might come in handy in more complicated examples.
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