Question

Find $A^{-1}$, if $A=\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|$ and show that $A^{-1}=\frac{A^2-3I}{2}$.

Answer 1

We have, $A=\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|$
$\therefore A_{11}=-1,A_{12}=1,A_{13}=1,A_{21}=1,A_{22}=-1,A_{23}=1,A_{31}=1,A_{32}=1and{A}_{33}=-1\therefore adjA={\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|}^T=\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$
and $|A|=-1(-1)+1\left(1\right)=2$
$\therefore A^{-1}=\frac{adjA}{|A|}=\frac{1}{2}\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]$ .....(i)
and $A^2=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right].\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]=\left[\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right]$ ..... (ii)
$\therefore \frac{A^2-3I}{2}=\frac{1}{2}\left\{\begin{array}{c}\left|\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right|-\left|\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right|\end{array}\right\}=\frac{1}{2}\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$
$=A^{-1}$ [using Eq. (i)]
Hence proved.