# Solve this

Question:

If $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$, then prove that $A^{2}-4 A-5 /=0$

Solution:

Given : $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$

Now.

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{lll}1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]$

$A^{2}-4 A-5 I$

$\Rightarrow A^{2}-4 A-5 I=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-4\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]-5\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-4 A-5 I=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-\left[\begin{array}{lll}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]$

$\Rightarrow A^{2}-4 A-5 I=\left[\begin{array}{lll}9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5\end{array}\right]$

$\Rightarrow A^{2}-4 A-5 I=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=0$

Hence proved.