Question

If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$ then show that $A^2-4A-5I=0.$

Answer 1

Given

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

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

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

$-4A=\left[\begin{array}{ccc}-4& -8& -8\\ -8& -4& -8\\ -8& -8& -4\end{array}\right]$

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

$=5I$

$\therefore A^2-4A-5I=0$.