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

Given, $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$
$A^2=A\cdot A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]\cdot \left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$
$A^2=\left[\begin{array}{ccc}1+4+4& 2+2+4& 2+4+2\\ 2+2+4& 1+1+4& 4+2+2\\ 2+4+2& 4+2+2& 4+4+1\end{array}\right]$
$A^2=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-4\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]-5\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]-\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]$
$\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$⟹A^2-4A+5I=0$
Pre multiply both sides with $A^{-1},$ we get $\left(A^{-1}A\right)A-4A^{-1}A+5A^{-1}=0$
$IA-4I+5A^{-1}=0⟹A^{-1}=[4I-A]\frac{1}{5}$
$A^{-1}=\frac{1}{5}\left[\begin{array}{ccc}3& -2& -2\\ -2& 3& -2\\ -2& -2& 3\end{array}\right]$