Question

$\mathrm{If}A=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right],\mathrm{find}{A}^{-1}\mathrm{and}\mathrm{show}\mathrm{that}{A}^{-1}=\frac{1}{2}\left(A^2-3I\right).$

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right] \\ {A}^{-1}=\frac{1}{\left|A\right|}\mathrm{Adj}.A \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right| \\ =-1\left(-1\right)+1\left(1\right) \\ =2 \\ \mathrm{Now},\mathrm{to}\mathrm{find}\mathrm{Adj}.A \\ \begin{array}{ccc}{A}_{11}={\left(-1\right)}^{1+1}\left(-1\right)=-1 \\ {A}_{12}={\left(-1\right)}^{1+2}\left(-1\right)=1 \\ {A}_{13}={\left(-1\right)}^{1+3}\left(1\right)=1& A_{21}={\left(-1\right)}^{2+1}\left(-1\right)=1 \\ {A}_{22}={\left(-1\right)}^{2+2}\left(-1\right)=-1 \\ {A}_{23}={\left(-1\right)}^{2+3}\left(-1\right)=1& A_{31}={\left(-1\right)}^{3+1}\left(1\right)=1 \\ {A}_{32}={\left(-1\right)}^{3+2}\left(-1\right)=1 \\ {A}_{33}={\left(-1\right)}^{3+3}\left(-1\right)=-1\end{array} \\ \mathrm{Therefore}, \\ \mathrm{Adj}.A=\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right] \\ \mathrm{Thus}, \\ {A}^{-1}=\frac{1}{2}\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]. \\ \mathrm{Now}, \\ {A}^2=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right] \\ =\left[\begin{array}{ccc}0+1+1& 0+0+1& 0+1+0\\ 0+0+1& 1+0+1& 1+0+0\\ 0+1+0& 1+0+0& 1+1+0\end{array}\right] \\ =\left[\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right] \\ \mathrm{Now},\mathrm{to}\mathrm{show}{A}^{-1}=\frac{1}{2}\left(A^2-3I\right) \\ \mathrm{RHS} \\ =\frac{1}{2}\left(A^2-3I\right) \\ =\frac{1}{2}\left(\left[\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right]-3\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\right) \\ =\frac{1}{2}\left[\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right] \\ =A^{-1} \\ =\mathrm{LHS} \\ \mathrm{Hence},A^{-1}=\frac{1}{2}\left(A^2-3I\right). \end{gathered}$$