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Question
If \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] , show that adj A = 3AT.
Solution
\[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}1 & - 2 \\ - 2 & 1\end{vmatrix} = - 3, C_{12} = - \begin{vmatrix}2 & - 2 \\ 2 & 1\end{vmatrix} = - 6\text{ and }C_{13} = \begin{vmatrix}2 & 1 \\ 2 & - 2\end{vmatrix} = - 6\]
\[ C_{21} = - \begin{vmatrix}- 2 & - 2 \\ - 2 & 1\end{vmatrix} = 6, C_{22} = \begin{vmatrix}- 1 & - 2 \\ 2 & 1\end{vmatrix} = 3\text{ and }C_{23} = - \begin{vmatrix}- 1 & - 2 \\ 2 & - 2\end{vmatrix} = - 6\]
\[ C_{31} = \begin{vmatrix}- 2 & - 2 \\ 1 & - 2\end{vmatrix} = 6, C_{32} = - \begin{vmatrix}- 1 & - 2 \\ 2 & - 2\end{vmatrix} = - 6 \text{ and }C_{33} = \begin{vmatrix}- 1 & - 2 \\ 2 & 1\end{vmatrix} = 3\]
\[adj A = \begin{bmatrix}- 3 & - 6 & - 6 \\ 6 & 3 & - 6 \\ 6 & - 6 & 3\end{bmatrix}^T = \begin{bmatrix}- 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3\end{bmatrix}\]
\[ A^T = \begin{bmatrix}- 1 & 2 & 2 \\ - 2 & 1 & - 2 \\ - 2 & - 2 & 1\end{bmatrix}\]
\[ \Rightarrow 3 A^T = \begin{bmatrix}- 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3\end{bmatrix}\]
\[ \Rightarrow 3 A^T = adj A\]
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