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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
\[B = \begin{bmatrix}1 & 2 & 5 \\ 2 & 3 & 1 \\ - 1 & 1 & 1\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}3 & 1 \\ 1 & 1\end{vmatrix} = 2, C_{12} = - \begin{vmatrix}2 & 1 \\ - 1 & 1\end{vmatrix} = - 3\text{ and } C_{13} = \begin{vmatrix}2 & 3 \\ - 1 & 1\end{vmatrix} = 5\]
\[ C_{21} = - \begin{vmatrix}2 & 5 \\ 1 & 1\end{vmatrix} = 3, C_{22} = \begin{vmatrix}1 & 5 \\ - 1 & 1\end{vmatrix} = 6\text{ and }C_{23} = - \begin{vmatrix}1 & 2 \\ - 1 & 1\end{vmatrix} = - 3\]
\[ C_{31} = \begin{vmatrix}2 & 5 \\ 3 & 1\end{vmatrix} = - 13, C_{32} = - \begin{vmatrix}1 & 5 \\ 2 & 1\end{vmatrix} = 9\text{ and }C_{33} = \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = - 1\]
\[ \therefore adjB = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 6 & - 3 \\ - 13 & 9 & - 1\end{bmatrix}^T = \begin{bmatrix}2 & 3 & - 13 \\ - 3 & 6 & 9 \\ 5 & - 3 & - 1\end{bmatrix}\]
\[(adjB)B = \begin{bmatrix}21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21\end{bmatrix}\]
\[\text{ and }\left| B \right| = 21\]
\[ \therefore \left| B \right|I = \begin{bmatrix}21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21\end{bmatrix}\]
\[\text{ and }B(adjB) = \begin{bmatrix}21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21\end{bmatrix}\]
\[\text{ Thus, }(adjA)A = \left| A \right|I = A(adjA)\]
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