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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
\[ D = \begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}1 & 0 \\ 1 & 3\end{vmatrix} = 3, C_{12} = - \begin{vmatrix}5 & 0 \\ 1 & 3\end{vmatrix} = - 15 \text{ and }C_{13} = \begin{vmatrix}5 & 1 \\ 1 & 1\end{vmatrix} = 4\]
\[ C_{21} = - \begin{vmatrix}0 & - 1 \\ 1 & 3\end{vmatrix} = - 1, C_{22} = \begin{vmatrix}2 & - 1 \\ 1 & 3\end{vmatrix} = 7\text{ and }C_{23} = - \begin{vmatrix}2 & 0 \\ 1 & 1\end{vmatrix} = - 2\]
\[ C_{31} = \begin{vmatrix}0 & - 1 \\ 1 & 0\end{vmatrix} = 1, C_{32} = - \begin{vmatrix}2 & - 1 \\ 5 & 0\end{vmatrix} = - 5 \text{ and }C_{33} = \begin{vmatrix}2 & 0 \\ 5 & 1\end{vmatrix} = 2\]
\[ \therefore adjD = \begin{bmatrix}3 & - 15 & 4 \\ - 1 & 7 & - 2 \\ 1 & - 5 & 2\end{bmatrix}^T = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 7 & - 5 \\ 4 & - 2 & 2\end{bmatrix}\]
\[(adjD)D = \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}\]
\[\text{ and }\left| D \right| = 2\]
\[ \therefore \left| D \right|I = \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}\]
\[\text{ and }D(adjD) = \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}\]
\[Thus, (adjA)A = \left| A \right|I = A(adjA)\]
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