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