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Question
For the matrix
Solution
\[A = \begin{bmatrix}1 & - 1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}3 & 0 \\ 2 & 10\end{vmatrix} = 30 C_{12} = - \begin{vmatrix}2 & 0 \\ 18 & 10\end{vmatrix} = - 20 C_{13} = \begin{vmatrix}2 & 3 \\ 18 & 2\end{vmatrix} = - 50\]
\[ C_{21} = - \begin{vmatrix}- 1 & 1 \\ 2 & 10\end{vmatrix} = 12 C_{22} = \begin{vmatrix}1 & 1 \\ 18 & 10\end{vmatrix} = - 8 C_{23} = - \begin{vmatrix}1 & - 1 \\ 18 & 2\end{vmatrix} = - 20\]
\[ C_{31} = \begin{vmatrix}- 1 & 1 \\ 3 & 0\end{vmatrix} = - 3 C_{32} = - \begin{vmatrix}1 & 1 \\ 2 & 0\end{vmatrix} = 2 C_{33} = \begin{vmatrix}1 & - 1 \\ 2 & 3\end{vmatrix} = 5\]
\[adj A = \begin{bmatrix}30 & - 20 & - 50 \\ 12 & - 8 & - 20 \\ - 3 & 2 & 5\end{bmatrix}^T = \begin{bmatrix}30 & 12 & - 3 \\ - 20 & - 8 & 2 \\ - 50 & - 20 & 5\end{bmatrix}\]
\[ \therefore A(adj A) = \begin{bmatrix}1 & - 1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{bmatrix}\begin{bmatrix}30 & 12 & - 3 \\ - 20 & - 8 & 2 \\ - 50 & - 20 & 5\end{bmatrix} = \begin{bmatrix}30 + 20 - 50 & 12 + 18 - 20 & - 3 - 2 + 5 \\ 60 - 60 - 0 & 24 - 24 - 0 & - 6 + 6 + 0 \\ 540 - 40 - 500 & 216 - 16 - 200 & - 54 + 4 + 50\end{bmatrix} = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\]
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