Question

Find the inverse of the following matrix and verify that \[A^{- 1} A = I_3\]

\[\begin{bmatrix}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{bmatrix}\]

Answer 1

\[B = \begin{bmatrix}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{bmatrix}\]
Now, 
\[ C_{11} = \begin{vmatrix}4 & 1 \\ 7 & 2\end{vmatrix} = 1, C_{12} = - \begin{vmatrix}3 & 1 \\ 3 & 2\end{vmatrix} = - 3\text{ and }C_{13} = \begin{vmatrix}3 & 4 \\ 3 & 7\end{vmatrix} = 9\]
\[ C_{21} = - \begin{vmatrix}3 & 1 \\ 7 & 2\end{vmatrix} = 1, C_{22} = \begin{vmatrix}2 & 1 \\ 3 & 2\end{vmatrix} = 1\text{ and } C_{23} = - \begin{vmatrix}2 & 3 \\ 3 & 7\end{vmatrix} = - 5\]
\[ C_{31} = \begin{vmatrix}3 & 1 \\ 4 & 1\end{vmatrix} = - 1, C_{32} = - \begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix} = 1\text{ and }C_{33} = \begin{vmatrix}2 & 3 \\ 3 & 4\end{vmatrix} = - 1\]
\[adjB = \begin{bmatrix}1 & - 3 & 9 \\ 1 & 1 & - 5 \\ - 1 & 1 & - 1\end{bmatrix}^T = \begin{bmatrix}1 & 1 & - 1 \\ - 3 & 1 & 1 \\ 9 & - 5 & - 1\end{bmatrix}\]
\[\text{ and }\left| B \right| = 2\]
\[ B^{- 1} = \frac{1}{2}\begin{bmatrix}1 & 1 & - 1 \\ - 3 & 1 & 1 \\ 9 & - 5 & - 1\end{bmatrix}\]
\[\text{ Now, }B^{- 1} B = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = I_3\]