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Question
Find the inverse of the following matrix.
Solution
\[F = \begin{bmatrix}0 & 0 & - 1 \\ 3 & 4 & 5 \\ - 2 & - 4 & - 7\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}4 & 5 \\ - 4 & - 7\end{vmatrix} = - 8, C_{12} = - \begin{vmatrix}3 & 5 \\ - 2 & - 7\end{vmatrix} = 11\text{ and }C_{13} = \begin{vmatrix}3 & 4 \\ - 2 & - 4\end{vmatrix} = - 4\]
\[ C_{21} = - \begin{vmatrix}0 & - 1 \\ - 4 & - 7\end{vmatrix} = 4, C_{22} = \begin{vmatrix}0 & - 1 \\ - 2 & - 7\end{vmatrix} = - 2\text{ and }C_{23} = - \begin{vmatrix}0 & 0 \\ - 2 & - 4\end{vmatrix} = 0\]
\[ C_{31} = \begin{vmatrix}0 & - 1 \\ 4 & 5\end{vmatrix} = 4, C_{32} = - \begin{vmatrix}0 & - 1 \\ 3 & 5\end{vmatrix} = - 3\text{ and }C_{33} = \begin{vmatrix}0 & 0 \\ 3 & 4\end{vmatrix} = 0\]
\[adjF = \begin{bmatrix}- 8 & 11 & - 4 \\ 4 & - 2 & 0 \\ 4 & - 3 & 0\end{bmatrix}^T = \begin{bmatrix}- 8 & 4 & 4 \\ 11 & - 2 & - 3 \\ - 4 & 0 & 0\end{bmatrix}\]
\[\text{ and }\left| F \right| = 4\]
\[ \therefore F^{- 1} = \frac{1}{4}\begin{bmatrix}- 8 & 4 & 4 \\ 11 & - 2 & - 3 \\ - 4 & 0 & 0\end{bmatrix}\]
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