If $A =\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ and $A ^2-4 A=k I _3$
Here $I_3$ has a order of $3 \times 3$ matrix$
\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] \\
\text { now } A^2-4 A & =\left[\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right] \\
& =\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] \\
A^2-4 A & \left.=5 \begin{array}{lll}
9 & 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]=5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$
It is prove that $k$ will be 5 .

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