Question

Find the inverse of the following matrix.

\[\begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\]

Answer 1

\[B = \begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\]
Now, 
\[ C_{11} = \begin{vmatrix}- 1 & - 1 \\ 3 & - 1\end{vmatrix} = 4, C_{12} = - \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = - 1\text{ and }C_{13} = \begin{vmatrix}1 & - 1 \\ 2 & 3\end{vmatrix} = 5\]
\[ C_{21} = - \begin{vmatrix}2 & 5 \\ 3 & - 1\end{vmatrix} = 17, C_{22} = \begin{vmatrix}1 & 5 \\ 2 & - 1\end{vmatrix} = - 11\text{ and }C_{23} = - \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = 1\]
\[ C_{31} = \begin{vmatrix}2 & 5 \\ - 1 & - 1\end{vmatrix} = 3, C_{32} = - \begin{vmatrix}1 & 5 \\ 1 & - 1\end{vmatrix} = 6\text{ and }C_{33} = \begin{vmatrix}1 & 2 \\ 1 & - 1\end{vmatrix} = - 3\]
\[adjB = \begin{bmatrix}4 & - 1 & 5 \\ 17 & - 11 & 1 \\ 3 & 6 & - 3\end{bmatrix}^T = \begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}\]
\[\text{  and }\left| B \right| = 27\]
\[ \therefore B^{- 1} = \frac{1}{27}\begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}\]