Question
If $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$, show that $A^2-4 A$ is a scalar matrix.

Answer

$
\begin{aligned}
& A^2=A \cdot A=\left(\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right)\left(\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right) \\
& =\left(\begin{array}{lll}
1+4+4 & 2+2+4 & 2+4+2 \\
2+2+4 & 4+1+4 & 4+2+2 \\
2+4+2 & 4+2+2 & 4+4+1
\end{array}\right) \\
& =\left(\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right) \\
& \therefore A ^2-4 A =\left(\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right)-4\left(\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right) \\
& =\left(\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right)-\left(\begin{array}{lll}
4 & 8 & 8 \\
8 & 4 & 8 \\
8 & 8 & 4
\end{array}\right) \\
& =\left(\begin{array}{lll}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5
\end{array}\right) \\
&
\end{aligned}
$
which is a scalar matrix.

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