Question

If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$, then show that

$A^2-4A-5I=0$, and hence find $A^{-1}$.

Answer 1

$A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$
$\therefore A^2=A.A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$
$=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]$
Now, $A^2-4A-5I=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-4\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]-5\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]-\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]$
$=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=0$

$A^2-4A-5I=0$
Multiplying by $A^{-1}$
$\Rightarrow A^{-1}{A}^2-4A^{-1}A-5A^{-1}I=0$
$\Rightarrow A-4I-5A^{-1}=0$
$\Rightarrow 5A^{-1}=A-4I$
$\Rightarrow A^{-1}=\frac{1}{5}[A-4I]$
$\Rightarrow A^{-1}=\frac{1}{5}\left\{\begin{array}{c}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]-\left[\begin{array}{ccc}4& 0& 0\\ 0& 4& 0\\ 0& 0& 4\end{array}\right]\end{array}\right\}$
$=\frac{1}{5}\left[\begin{array}{ccc}-3& 2& 2\\ 2& -3& 2\\ 2& 2& -3\end{array}\right]$