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Question
Compute the adjoint of the following matrix:
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
Solution
\[ C = \begin{bmatrix}2 & - 1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & - 1\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}2 & 5 \\ 4 & - 1\end{vmatrix} = - 22, C_{12} = - \begin{vmatrix}4 & 5 \\ 0 & - 1\end{vmatrix} = 4\text{ and }C_{13} = \begin{vmatrix}4 & 2 \\ 0 & 4\end{vmatrix} = 16\]
\[ C_{21} = - \begin{vmatrix}- 1 & 3 \\ 4 & - 1\end{vmatrix} = 11, C_{22} = \begin{vmatrix}2 & 3 \\ 0 & - 1\end{vmatrix} = - 2\text{ and }C_{23} = - \begin{vmatrix}2 & - 1 \\ 0 & 4\end{vmatrix} = - 8\]
\[ C_{31} = \begin{vmatrix}- 1 & 3 \\ 2 & 5\end{vmatrix} = - 11, C_{32} = - \begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = 2\text{ and }C_{33} = \begin{vmatrix}2 & - 1 \\ 4 & 2\end{vmatrix} = 8\]
\[ \therefore adjC = \begin{bmatrix}- 22 & 4 & 16 \\ 11 & - 2 & - 8 \\ - 11 & 2 & 8\end{bmatrix}^T = \begin{bmatrix}- 22 & 11 & - 11 \\ 4 & - 2 & 2 \\ 16 & - 8 & 8\end{bmatrix}\]
\[(adjC)C = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\]
\[\text{ and }\left| C \right| = 0\]
\[ \therefore \left| C \right|I = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\]
\[\text{ and }C\left( adjC \right) = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\]
\[\text{ Thus, }(adjA)A = \left| A \right|I = A(adjA)\]
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