Example 2 (to item $6^{\circ}$). Solve the system

$$
\left\{\begin{aligned}
3 x_{1}-x_{2}+3 x_{3} & =5 \\
2 x_{1}-x_{2}+4 x_{3} & =5 \\
x_{1}+2 x_{2}-3 x_{3} & =0
\end{aligned}\right.
$$
Solution.

$$
A=\left(\begin{array}{rrr}
3 & -1 & 3 \\
2 & -1 & 4 \\
1 & 2 & -3
\end{array}\right), \quad \operatorname{det} A=\left|\begin{array}{rrr}
3 & -1 & 3 \\
2 & -1 & 4 \\
1 & 2 & -3
\end{array}\right|=-10
$$

Therefore, $A$ is a non-singular matrix, so it has an inverse matrix $A^{-1}$.

We will compute 9 algebraic complements (see $\S 2$, point $3^{\circ}$ ):

$$
\begin{array}{ll}
A_{11}=\left|\begin{array}{rr}
-1 & 4 \\
2 & -3
\end{array}\right|=-5, \quad A_{12}=-\left|\begin{array}{rr}
2 & 4 \\
1 & -3
\end{array}\right|=10, \quad A_{13}=\left|\begin{array}{rr}
2 & -1 \\
1 & 2
\end{array}\right|=5, \\
A_{21}=-\left|\begin{array}{rr}
-1 & 3 \\
2 & -3
\end{array}\right|=3, \quad A_{22}=\left|\begin{array}{rr}
3 & 3 \\
1 & -3
\end{array}\right|=-12, \quad A_{23}=-\left|\begin{array}{rr}
3 & -1 \\
1 & 2
\end{array}\right|=-7, \\
A_{31}=\left|\begin{array}{rr}
-1 & 3 \\
-1 & 4
\end{array}\right|=-1, \quad A_{32}=-\left|\begin{array}{ll}
3 & 3 \\
2 & 4
\end{array}\right|=-6, \quad A_{33}=\left|\begin{array}{rr}
3 & -1 \\
2 & -1
\end{array}\right|=-1 .
\end{array}
$$

According to Theorem 1

$$
A^{-1}=-\frac{1}{10}\left(\begin{array}{rrr}
-5 & 3 & -1 \\
10 & -12 & -6 \\
5 & -7 & -1
\end{array}\right)
$$

We strongly recommend checking the equalities $A \cdot A^{-1}=$ $=A^{-1} \cdot A=E$.

Thus, by Theorem 5, considering the notation (8), we get

$$
X=\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right)=-\frac{1}{10}\left(\begin{array}{rrr}
-5 & 3 & -1 \\
10 & -12 & -6 \\
5 & -7 & -1
\end{array}\right)\left(\begin{array}{l}
5 \\
5 \\
0
\end{array}\right)=\left(\begin{array}{l}
1 \\
1 \\
1
\end{array}\right), \text { i.e. }\left\{\begin{array}{l}
x_{1}=1 \\
x_{2}=1 \\
x_{3}=1
\end{array}\right.
$$