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Question
Find the inverse of the following matrix.
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Solution
\[C = \begin{bmatrix}2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}2 & - 1 \\ - 1 & 2\end{vmatrix} = 3, C_{12} = - \begin{vmatrix}- 1 & - 1 \\ 1 & 2\end{vmatrix} = 1\text{ and }C_{13} = \begin{vmatrix}- 1 & 2 \\ 1 & - 1\end{vmatrix} = - 1\]
\[ C_{21} = - \begin{vmatrix}- 1 & 1 \\ - 1 & 2\end{vmatrix} = 1, C_{22} = \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = 3\text{ and }C_{23} = - \begin{vmatrix}2 & - 1 \\ 1 & - 1\end{vmatrix} = 1\]
\[ C_{31} = \begin{vmatrix}- 1 & 1 \\ 2 & - 1\end{vmatrix} = - 1, C_{32} = - \begin{vmatrix}2 & 1 \\ - 1 & - 1\end{vmatrix} = 1\text{ and }C_{33} = \begin{vmatrix}2 & - 1 \\ - 1 & 2\end{vmatrix} = 3\]
\[adjC = \begin{bmatrix}3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3\end{bmatrix}^T = \begin{bmatrix}3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3\end{bmatrix}\]
\[\text{ and }\left| C \right| = 4\]
\[ \therefore C^{- 1} = \frac{1}{4}\begin{bmatrix}3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3\end{bmatrix}\]
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