### 1.16. Investigate the system of equations

$$
\left\{\begin{array}{l}
3 x_{1}+4 x_{2}+3 x_{3}=0 \\
x_{1}-x_{2}+x_{3}=0 \\
x_{1}+3 x_{2}-x_{3}=-2 \\
x_{1}+2 x_{2}+3 x_{3}=2
\end{array}\right.
$$
$$
\begin{aligned}
& \tilde{A}-\left(\begin{array}{ccc|c}
3 & 4 & 3 & 0 \\
1 & -2 & 1 & 0 \\
1 & 3 & -1 & -2 \\
1 & 2 & 3 & 2
\end{array}\right) \sim\left[\begin{array}{l}
\text { swap } \\
1 \text { and } 2 \text { rows }
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & -2 & 1 & 0 \\
3 & 4 & 3 & 0 \\
1 & 3 & -1 & -2 \\
1 & 2 & 3 & 2
\end{array}\right) \sim \\
& \sim\left[\begin{array}{l}
\text { row } 2 - \text { row } 1 \times 3 ; \\
\text { row } 3 - \text { row } 1 ; \\
\text { row } 4 - \text { row } 1
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & -2 & 1 & 0 \\
0 & 10 & 0 & 0 \\
0 & 5 & -2 & -2 \\
0 & 4 & 2 & 2
\end{array}\right) \sim \\
& \sim\left[\begin{array}{l}
\text { divide row } 2 \text { by 10; } \\
\text { divide row } 4 \text { by 10 }
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & -2 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 5 & -2 & -2 \\
0 & 2 & 1 & 1
\end{array}\right) \sim \\
& -\left[\begin{array}{l}
\text { row } 3 - \text { row } 2 \times 5 ; \\
\text { row } 4 - \text { row } 2 \times 2
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & -2 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1
\end{array}\right) \sim \\
& \sim\left[\begin{array}{l}
\text { row } 4 - \\
\text { row } 3
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & -2 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{array}\right)-\left[\begin{array}{l}
\text { remove } \\
\text { zero } \\
\text { fourth } \\
\text { row }
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & -2 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1
\end{array}\right) .
\end{aligned}
$$

The obtained result shows that matrix $A$ and the augmented matrix $\tilde{A}$ have the same rank:

$$
\operatorname{rank} A=\operatorname{rank} \tilde{A}=3
$$

Thus, the system of equations is consistent and has only one solution.