Answer
$Adj\,\mathbf{A} = \begin{bmatrix}8&4&-5\\2&0&-1\\-4&-2&2\end{bmatrix}$
$\mathbf{A}^{-1}=\frac{1}{-2}\begin{bmatrix}8&4&-5\\2&0&-1\\-4&-2&2\end{bmatrix}=\begin{bmatrix}-4&-2&\frac{5}{2}\\-1&0&\frac{1}{2}\\2&1&-1 \end{bmatrix}$
Work Step by Step
Given the Matrix:
$\mathbf{A} = \begin{bmatrix}1&-1&2\\0&2&1\\2&0&4\end{bmatrix}$
Using the following formula to find the adjugate;
$Adj\,\mathbf{A}=\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33}\end{bmatrix}^T$
$C_{ij}$ are the Matrix of minors;
$C_{11}=(+)\begin{vmatrix}2&1\\0&4\end{vmatrix}\,,C_{12}=(-)\begin{vmatrix}0&1\\2&4\end{vmatrix}\,,C_{13}=(+)\begin{vmatrix}0&2\\2&0\end{vmatrix}$
$C_{21}=(-)\begin{vmatrix}-1&2\\0&4\end{vmatrix}\,,C_{22}=(+)\begin{vmatrix}1&2\\2&4\end{vmatrix}\,,C_{23}=(-)\begin{vmatrix}1&-1\\2&0\end{vmatrix}$
$C_{31}=(+)\begin{vmatrix}-1&2\\2&1\end{vmatrix}\,,C_{32}=(-)\begin{vmatrix}1&2\\0&1\end{vmatrix}\,,C_{33}=(+)\begin{vmatrix}1&-1\\0&2\end{vmatrix}$
Finding the Matrix of the cofactors we have,
Matrix of Co-factors =$\begin{bmatrix}8&2&-4\\4&0&-2\\-5&-1&2\end{bmatrix}$
$Adj\,\mathbf{A} =\begin{bmatrix}8&2&-4\\4&0&-2\\-5&-1&2\end{bmatrix}^T= \begin{bmatrix}8&4&-5\\2&0&-1\\-4&-2&2\end{bmatrix}$
$\mathbf{A}^{-1}=\frac{1}{det\mathbf{A}}\times Adj\,\mathbf{A}................theorem\,8$
we first find $det\,\mathbf{A}$ using Co-factor expansion;
$det\,\mathbf{A}=1(2\times4-1\times0)+1(0\times4-1\times2)+2(0\times0-2\times2)=-2$
therefore,
$\mathbf{A}^{-1}=\frac{1}{-2}\begin{bmatrix}8&4&-5\\2&0&-1\\-4&-2&2\end{bmatrix}=\begin{bmatrix}-4&-2&\frac{5}{2}\\-1&0&\frac{1}{2}\\2&1&-1 \end{bmatrix}$
