Example 1.5. Solve the system of linear equations using the Gaussian method.

\[
\begin{aligned}
& 2 x_{1}+7 x_{2}+13 x_{3}=0 \\
& 3 x_{1}+14 x_{2}+12 x_{3}=18 \\
& 5 x_{1}+25 x_{2}+16 x_{3}=39
\end{aligned}
\]
Solution. Let's take the first equation of the system as the first leading equation, and the first leading unknown as $x_{1}$; the first leading element will be $a_{11}=2$. We will eliminate $x_{1}$ from the second and third equations by adding the leading equation, multiplied by $-\frac{3}{2}$, to the second equation, and the leading equation, multiplied by $-\frac{5}{2}$, to the third equation. We obtain:

$$
\begin{aligned}
& 2 x_{1}+7 x_{2}+13 x_{3}=0 \\
& \frac{7}{2} x_{2}-\frac{15}{2} x_{3}=18 \\
& \frac{15}{2} x_{2}-\frac{33}{2} x_{3}=39
\end{aligned}
$$

The first step is completed. The second and third equations form the first subsystem. Let's take the second equation of the system as the second leading equation, and the second leading unknown as $x_{2}$; the second leading element will be $\frac{7}{2}$. We will eliminate $x_{2}$ from the third equation. We obtain:

$$
\begin{aligned}
& 2 x_{1}+7 x_{2}+13 x_{3}=0, \\
& \frac{7}{2} x_{2}-\frac{15}{2} x_{3}=18, \\
& -\frac{3}{7} x_{3}=\frac{3}{7}
\end{aligned}
$$

The second step is completed. The second subsystem consists of one third equation. The forward pass of the Gaussian method is completed. By the backward pass, we get:

$$
\begin{gathered}
x_{3}=-1 \\
x_{2}=\frac{2}{7} 18+\frac{15}{2} x_{3}=\frac{2}{7} 18+\frac{15}{2}(-1)=3 \\
x_{1}=-\frac{1}{2}\left(7 x_{2}+13 x_{3}\right)=-\frac{1}{2}[7 \cdot 3+13(-1)]=-4
\end{gathered}
$$

Thus, the solution of the given system will be: $x_{1}=-4, x_{2}=3, x_{3}=-1$.