Question

If $A=\left|\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right|$, then prove that $A^2-4A-5I_3=0$ and hence obtain $A^{-1}$.

Answer 1

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

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

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

$A^2-4A-5I_3=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]$$-4\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$$-$$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$⟹A^2-4A-5I_3=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

Multiplying both sides by $A^{-1}-$

$A^{-1}(A^2-4A-5I_3)=0$

$⟹A-4I_3-5A^{-1}=0$

$⟹A^{-1}=\frac{1}{5}\left[\begin{array}{ccc}-3& -2& -2\\ -2& -3& -2\\ -2& -2& -3\end{array}\right]$

$⟹A^{-1}=\frac{-1}{5}\left[\begin{array}{ccc}3& 2& 2\\ 2& 3& 2\\ 2& 2& 3\end{array}\right]$