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 $A^2 -4A-5I=0$?
 A: $$A^2 = \left(
\begin{array}{ccc}
 9 & 8 & 8 \\
 8 & 9 & 8 \\
 8 & 8 & 9 \\
\end{array}
\right)$$
$$-4A = \left(
\begin{array}{ccc}
 -4 & -8 & -8 \\
 -8 & -4 & -8 \\
 -8 & -8 & -4 \\
\end{array}
\right)$$
$$A^2  - 4A = \left(
\begin{array}{ccc}
 5 & 0 & 0 \\
 0 & 5 & 0 \\
 0 & 0 & 5 \\
\end{array}
\right)$$
This is just as easy as any other way for this problem.
A: $A=2J-I$, where $J$ is the all-one matrix with $J^2=3J$. Therefoer
$$
A^2-4A-5I=(2J-I)^2-4(2J-I)-5I=4J^2-12J=0.
$$
A: 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.
A: Another way to do this is to note that $x^2 - 4x - 5 = (x-5)(x+1)$ so if we wanted to calculate this matrix $A^2 - 4A + 5I$ we can just calculate $(A-5I)(A+I)$. 
This might not be much easier, especially just for a quadratic, but it does reduce the number of times you have to calculate a product of matrices, and the number of times you have to add them up.
The other ways given are perfectly good too, I just happen to prefer solving this sort of problem this way.
