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