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Question
Find the inverse of the following matrix.
Solution
\[ E = \begin{bmatrix}0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}- 3 & 4 \\ - 3 & 4\end{vmatrix} = 0, C_{12} = - \begin{vmatrix}4 & 4 \\ 3 & 4\end{vmatrix} = - 4\text{ and }C_{13} = \begin{vmatrix}4 & - 3 \\ 3 & - 3\end{vmatrix} = - 3\]
\[ C_{21} = - \begin{vmatrix}1 & - 1 \\ - 3 & 4\end{vmatrix} = - 1, C_{22} = \begin{vmatrix}0 & - 1 \\ 3 & 4\end{vmatrix} = 3\text{ and }C_{23} = - \begin{vmatrix}0 & 1 \\ 3 & - 3\end{vmatrix} = 3\]
\[ C_{31} = \begin{vmatrix}1 & - 1 \\ - 3 & 4\end{vmatrix} = 1, C_{32} = - \begin{vmatrix}0 & - 1 \\ 4 & 4\end{vmatrix} = - 4\text{ and }C_{33} = \begin{vmatrix}0 & 1 \\ 4 & - 3\end{vmatrix} = - 4\]
\[adjE = \begin{bmatrix}0 & - 4 & - 3 \\ - 1 & 3 & 3 \\ 1 & - 4 & - 4\end{bmatrix}^T = \begin{bmatrix}0 & - 1 & 1 \\ - 4 & 3 & - 4 \\ - 3 & 3 & - 4\end{bmatrix}\]
\[and \left| E \right| = - 1\]
\[ \therefore E^{- 1} = - 1\begin{bmatrix}0 & - 1 & 1 \\ - 4 & 3 & - 4 \\ - 3 & 3 & - 4\end{bmatrix} = \begin{bmatrix}0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4\end{bmatrix}\]
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