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Question
Find the inverse of the following matrix and verify that \[A^{- 1} A = I_3\]
Solution
\[ A = \begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}4 & 3 \\ 3 & 4\end{vmatrix} = 7, C_{12} = - \begin{vmatrix}1 & 3 \\ 1 & 4\end{vmatrix} = - 1\text{ and }C_{13} = \begin{vmatrix}1 & 4 \\ 1 & 3\end{vmatrix} = - 1\]
\[ C_{21} = - \begin{vmatrix}3 & 3 \\ 3 & 4\end{vmatrix} = - 3, C_{22} = \begin{vmatrix}1 & 3 \\ 1 & 4\end{vmatrix} = 1\text{ and }C_{23} = - \begin{vmatrix}1 & 3 \\ 1 & 3\end{vmatrix} = 0\]
\[ C_{31} = \begin{vmatrix}3 & 3 \\ 4 & 3\end{vmatrix} = - 3, C_{32} = - \begin{vmatrix}1 & 3 \\ 1 & 3\end{vmatrix} = 0\text{ and }C_{33} = \begin{vmatrix}1 & 3 \\ 1 & 4\end{vmatrix} = 1\]
\[adjA = \begin{bmatrix}7 & - 1 & - 1 \\ - 3 & 1 & 0 \\ - 3 & 0 & 1\end{bmatrix}^T = \begin{bmatrix}7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1\end{bmatrix}\]
\[\text{ and }\left| A \right| = 1\]
\[ A^{- 1} = \begin{bmatrix}7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1\end{bmatrix}\]
\[\text{ Now, }A^{- 1} A = \begin{bmatrix}7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = I_3 \]
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