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Question
Find A (adj A) for the matrix \[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & 2 & - 1 \\ - 4 & 5 & 2\end{bmatrix} .\]
Solution
\[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & 2 & - 1 \\ - 4 & 5 & 2\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}2 & - 1 \\ 5 & 2\end{vmatrix} = 9, C_{12} = - \begin{vmatrix}0 & - 1 \\ - 4 & 2\end{vmatrix} = 4\text{ and }C_{13} = \begin{vmatrix}0 & 2 \\ - 4 & 5\end{vmatrix} = 8\]
\[ C_{21} = - \begin{vmatrix}- 2 & 3 \\ 5 & 2\end{vmatrix} = 19, C_{22} = \begin{vmatrix}1 & 3 \\ - 4 & 2\end{vmatrix} = 14\text{ and }C_{23} = - \begin{vmatrix}1 & - 2 \\ - 4 & 5\end{vmatrix} = 3\]
\[ C_{31} = \begin{vmatrix}- 2 & 3 \\ 2 & - 1\end{vmatrix} = - 4, C_{32} = - \begin{vmatrix}1 & 3 \\ 0 & - 1\end{vmatrix} = 1\text{ and }C_{33} = \begin{vmatrix}1 & - 2 \\ 0 & 2\end{vmatrix} = 2\]
\[adj A = \begin{bmatrix}9 & 4 & 8 \\ 19 & 14 & 3 \\ - 4 & 1 & 2\end{bmatrix}^T = \begin{bmatrix}9 & 19 & - 4 \\ 4 & 14 & 1 \\ 8 & 3 & 2\end{bmatrix}\]
\[ \therefore A(adj A) = \begin{bmatrix}25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25\end{bmatrix}\]
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