Question

Find the adjoint of the matrix \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\]  and hence show that \[A\left( adj A \right) = \left| A \right| I_3\]. 

Answer 1

\[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\]
\[\text{ Now, to find Adj . A}\]
\[A_{11} = \left( - 1 \right)^{1 + 1} \left( - 3 \right) = - 3\]
\[ A_{12} = \left( - 1 \right)^{1 + 2} \left( 6 \right) = - 6\]
\[ A_{13} = \left( - 1 \right)^{1 + 3} \left( - 6 \right) = - 6\]
\[ A_{21} = \left( - 1 \right)^{2 + 1} \left( - 6 \right) = 6\]
\[ A_{22} = \left( - 1 \right)^{2 + 2} \left( 3 \right) = 3 \]
\[ A_{23} = \left( - 1 \right)^{2 + 3} \left( 6 \right) = - 6\]
\[ A_{31} = \left( - 1 \right)^{3 + 1} \left( 6 \right) = 6\]
\[ A_{32} = \left( - 1 \right)^{3 + 2} \left( 6 \right) = - 6\]
\[ A_{33} = \left( - 1 \right)^{3 + 3} \left( 3 \right) = 3\]
Therefore, 
\[Adj . A = \begin{bmatrix}- 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3\end{bmatrix}\]
Now,
\[\left| A \right| = \begin{vmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{vmatrix}\]
\[ = - 1\left( 1 - 4 \right) - 2\left( - 2 - 4 \right) + 2\left( 4 + 2 \right)\]
\[ = 3 + 12 + 12\]
\[ = 27\]
\[\text{ To show: } A\left( adj A \right) = \left| A \right| I_3 \]
LHS = 
\[\begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\begin{bmatrix}- 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3\end{bmatrix}\]
\[ = \begin{bmatrix}3 + 12 + 12 & - 6 - 6 + 12 & - 6 + 12 - 6 \\ - 6 - 6 + 12 & 12 + 3 + 12 & 12 - 6 - 6 \\ - 6 + 12 - 6 & 12 - 6 - 6 & 12 + 12 + 3\end{bmatrix}\]
\[ = \begin{bmatrix}27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27\end{bmatrix}\]
\[ = 27 \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \]
\[ = \left| A \right| I_3 \]
= RHS
\[\text{ Hence, }A\left( adj A \right) = \left| A \right| I_3 .\]