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Question
If \[A = \begin{bmatrix}- 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{bmatrix}\], show that adj A = A.
Solution
\[A = \begin{bmatrix}- 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}0 & 1 \\ 4 & 3\end{vmatrix} = - 4, C_{12} = - \begin{vmatrix}1 & 1 \\ 4 & 3\end{vmatrix} = 1\text{ and }C_{13} = \begin{vmatrix}1 & 0 \\ 4 & 4\end{vmatrix} = 4\]
\[ C_{21} = - \begin{vmatrix}- 3 & - 3 \\ 4 & 3\end{vmatrix} = - 3, C_{22} = \begin{vmatrix}- 4 & - 3 \\ 4 & 3\end{vmatrix} = 0\text{ and }C_{23} = - \begin{vmatrix}- 4 & - 3 \\ 4 & 4\end{vmatrix} = 4\]
\[ C_{31} = \begin{vmatrix}- 3 & - 3 \\ 0 & 1\end{vmatrix} = - 3, C_{32} = - \begin{vmatrix}- 4 & - 3 \\ 1 & 1\end{vmatrix} = 1\text{ and }C_{33} = \begin{vmatrix}- 4 & - 3 \\ 1 & 0\end{vmatrix} = 3\]
\[ \therefore adj A = \begin{bmatrix}- 4 & 1 & 4 \\ - 3 & 0 & 4 \\ - 3 & 1 & 3\end{bmatrix}^T = \begin{bmatrix}- 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{bmatrix} = A\]
Hence proved .
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