We have, $A=\left[\begin{array}{ccc}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]$
$$
\begin{aligned}
&\therefore|\mathrm{A}|=1(-3)-2(-2)+0=1 \neq 0 \\
&\text { Now, } \mathrm{A}_{11}=-3, \mathrm{~A}_{12}=2, \mathrm{~A}_{13}=2, \mathrm{~A}_{21}=-2, \\
&\mathrm{~A}_{22}=1, \mathrm{~A}_{23}=1, \mathrm{~A}_{31}=-4, \mathrm{~A}_{32}=2 \text { and } \mathrm{A}_{33}=3 \\
&\therefore \quad \operatorname{adj}(\mathrm{A})=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}
-3 & -2 & -4 \\
2 & 1 & 2 \\
2 & 1 & 3
\end{array}\right]
\end{aligned}
$$



$$
\begin{aligned}
&\text { Now } A^{-1}=\frac{\operatorname{adj} A}{|A|} \\
&\Rightarrow A^{-1}=\left[\begin{array}{ccc}
-3 & -2 & -4 \\
2 & 1 & 2 \\
2 & 1 & 3
\end{array}\right]
\end{aligned}
$$
Also, we have the system of linear equations as
$$
\begin{aligned}
&x-2 y=10, \\
&2 x-y-z=8 \\
&\text { and }-2 y+z=7 \\
&\text { In the form of } C X=D \text {, } \\
&\qquad\left[\begin{array}{ccc}
1 & -2 & 0 \\
2 & -1 & -1 \\
0 & -2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
10 \\
8 \\
7
\end{array}\right]
\end{aligned}
$$


where, $C=\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{array}\right], X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$ and $D=\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]$ We know that, $\left(\mathrm{A}^{\mathrm{T}}\right)^{-1}=\left(\mathrm{A}^{-1}\right)^{\mathrm{T}}$
$\therefore C^{\mathrm{T}}=\left[\begin{array}{ccc}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]=\mathrm{A}$
[using Eq. (i)] $\therefore \mathrm{X}=\mathrm{C}^{-1} \mathrm{D}$
$$
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right]\left[\begin{array}{c}
10 \\
8 \\
7
\end{array}\right]=\left[\begin{array}{c}
0 \\
-5 \\
-3
\end{array}\right]
$$