Question

Find the inverse of the following matrix.

\[\begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]

Answer 1

\[D = \begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]
Now, 
\[ C_{11} = \begin{vmatrix}1 & 0 \\ 1 & 3\end{vmatrix} = 3, C_{12} = - \begin{vmatrix}5 & 0 \\ 0 & 3\end{vmatrix} = - 15\text{ and }C_{13} = \begin{vmatrix}5 & 1 \\ 0 & 1\end{vmatrix} = 5\]
\[ C_{21} = - \begin{vmatrix}0 & - 1 \\ 1 & 3\end{vmatrix} = - 1, C_{22} = \begin{vmatrix}2 & - 1 \\ 0 & 3\end{vmatrix} = 6\text{ and }C_{23} = - \begin{vmatrix}2 & 0 \\ 0 & 1\end{vmatrix} = - 2\]
\[ C_{31} = \begin{vmatrix}0 & - 1 \\ 1 & 0\end{vmatrix} = 1, C_{32} = - \begin{vmatrix}2 & - 1 \\ 5 & 0\end{vmatrix} = - 5\text{ and }C_{33} = \begin{vmatrix}2 & 0 \\ 5 & 1\end{vmatrix} = 2\]
\[adjD = \begin{bmatrix}3 & - 15 & 5 \\ - 1 & 6 & - 2 \\ 1 & - 5 & 2\end{bmatrix}^T = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\]
\[\text{ and }\left| D \right| = 1\]
\[ \therefore D^{- 1} = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\]