Answer
$(i)\,Adj\,\mathbf{A} = \begin{bmatrix}0&1&0\\-5&-1&-5\\5&2&10\end{bmatrix}............Adjugate$
$(ii)\,\mathbf{A}^{-1}=\begin{bmatrix}0&\frac{1}{5}&0\\-1&-\frac{1}{5}&-1\\1&\frac{2}{5}&2\end{bmatrix}...........Inverse\,of\,the Matrix$
Work Step by Step
Given:
$\mathbf{A} = \begin{bmatrix}0&-2&-1\\5&0&0\\-1&1&1\end{bmatrix}$
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 cofactors of $\mathbf{A}$;
$C_{11}=(+)\begin{vmatrix}0&0\\1&1\end{vmatrix}\,,C_{12}=(-)\begin{vmatrix}5&0\\-1&1\end{vmatrix}\,,C_{13}=(+)\begin{vmatrix}5&0\\-1&1\end{vmatrix}$
$C_{21}=(-)\begin{vmatrix}-2&-1\\1&1\end{vmatrix}\,,C_{22}=(+)\begin{vmatrix}0&-1\\-1&1\end{vmatrix}\,,C_{23}=(-)\begin{vmatrix}0&-2\\-1&1\end{vmatrix}$
$C_{31}=(+)\begin{vmatrix}-2&-1\\0&0\end{vmatrix}\,,C_{32}=(-)\begin{vmatrix}0&-1\\5&0\end{vmatrix}\,,C_{33}=(+)\begin{vmatrix}0&-2\\5&0\end{vmatrix}$
Finding the Matrix of the cofactors we have,
Matrix of Co-factors =$\begin{bmatrix}0&-5&5\\1&-1&2\\0&-5&10\end{bmatrix}$
$Adj\,\mathbf{A} = \begin{bmatrix}0&-5&5\\1&-1&2\\0&-5&10\end{bmatrix}^T= \begin{bmatrix}0&1&0\\-5&-1&-5\\5&2&10\end{bmatrix}$
$\mathbf{A}^{-1}=\frac{1}{det\mathbf{A}}\times Adj\,\mathbf{A}................theorem\,8$
we first find $det\,\mathbf{A}$ using Co-factor expansion;
$det\,\mathbf{A}=0-(-2)(5-0)+(-1)(5-0)=5$
therefore,
$\mathbf{A}^{-1}=\frac{1}{5}\begin{bmatrix}0&1&0\\-5&-1&-5\\5&2&10\end{bmatrix}=\begin{bmatrix}0&\frac{1}{5}&0\\-1&-\frac{1}{5}&-1\\1&\frac{2}{5}&2\end{bmatrix}$
