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Question
Find the inverse of the following matrix.
\[\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{bmatrix}\]
Solution
\[A = \begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}3 & 1 \\ 1 & 2\end{vmatrix} = 5, C_{12} = - \begin{vmatrix}2 & 1 \\ 3 & 2\end{vmatrix} = - 1\text{ and }C_{13} = \begin{vmatrix}2 & 3 \\ 3 & 1\end{vmatrix} = - 7\]
\[ C_{21} = - \begin{vmatrix}2 & 3 \\ 1 & 2\end{vmatrix} = - 1, C_{22} = \begin{vmatrix}1 & 3 \\ 3 & 2\end{vmatrix} = - 7\text{ and }C_{23} = - \begin{vmatrix}1 & 2 \\ 3 & 1\end{vmatrix} = 5\]
\[ C_{31} = \begin{vmatrix}2 & 3 \\ 3 & 1\end{vmatrix} = - 7, C_{32} = - \begin{vmatrix}1 & 3 \\ 2 & 1\end{vmatrix} = 5\text{ and }C_{33} = \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = - 1\]
\[adjA = \begin{bmatrix}5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1\end{bmatrix}^T = \begin{bmatrix}5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1\end{bmatrix}\]
\[\text{ and }\left| A \right| = - 18\]
\[ \therefore A^{- 1} = - \frac{1}{18}\begin{bmatrix}5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1\end{bmatrix}\]
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