Example 2. Solve the linear system

$$
\left\{\begin{aligned}
-x_{1}-2 x_{2}-6 x_{3}+3 x_{4} & =-1 \\
2 x_{1}+5 x_{2}+14 x_{3}-7 x_{4} & =3 \\
3 x_{1}+7 x_{2}+20 x_{3}-10 x_{4} & =4 \\
-x_{2}-2 x_{3}+x_{4} & =-1
\end{aligned}\right.
$$
Solution. Each time, we will take the simplest coefficient as the leading one, i.e., either 1 or -1. Let's emphasize that the goal of the transformations is to obtain zeros in the leading column. How to get zero coefficients in the unit column is clear from the solution of example 1. For this, the leading row needs to be multiplied by appropriate numbers (sometimes by 1 or -1) and added to the other rows that do not contain 0 in this leading column. Therefore, we will limit ourselves to highlighting the leading coefficient in each block, without commenting on the transformations themselves and without indicating the corresponding numbers with arrows. The results of the calculations will be placed in a single Gaussian table, which has the following form:

| $x_{1}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ | CB. ch. |
| ---: | ---: | ---: | ---: | ---: |
| -1 | -2 | -6 | 3 | -1 |
| 2 | 5 | 14 | -7 | 3 |
| 3 | 7 | 20 | -10 | 4 |
| 0 | -1 | -2 | 1 | -1 |
| 1 | 2 | 6 | -3 | 1 |
| 0 | 1 | 2 | -1 | 1 |
| 0 | 1 | 2 | -1 | 1 |
| 0 | -1 | -2 | 1 | -1 |
| 1 | 0 | 2 | -1 | -1 |
| 0 | 1 | 2 | -1 | 1 |

The last two rows are removed as zero (they correspond to trivial equations).

From the last block of the table, we obtain the system

$$
\left\{\begin{array}{l}
x_{1}=-1-2 x_{3}+x_{4} \\
x_{2}=1-2 x_{3}+x_{4}
\end{array}\right.
$$

expressing the "almost" general solution of the original system. The meaning of the word "almost" lies in the unequal participation of the unknowns.

Let $x_{3}=\alpha, x_{4}=\beta$ ( $\alpha$ and $\beta$ - arbitrary constants or parameters).

Then the system

$$
\left\{\begin{array}{l}
x_{1}=-1-2 \alpha+\beta \\
x_{2}=1-2 \alpha+\beta \\
x_{3}=\alpha \\
x_{4}=\beta
\end{array}\right.
$$

represents the general solution of the system in parametric form. All unknowns are expressed (equally) through two parameters $\alpha, \beta \in \mathbb{R}$.

Solutions obtained from the general for fixed values of the parameters $\alpha$ and $\beta$ are called particular.

For example, when $\alpha=1, \beta=2$, we get: $x_{1}=-1, x_{2}=1, x_{3}=1$, $x_{4}=2$.

When $\alpha=-1, \beta=1$, we get $x_{1}=2, x_{2}=4, x_{3}=-1, x_{4}=1$. The basic solution corresponds to the zero set of free variables: if $\alpha=0, \beta=0$, then $x_{1}=-1, x_{2}=1, x_{3}=0, x_{4}=0$.

We will write the answer as follows: $X_{0}=(-1-2 \alpha+\beta, 1-2 \alpha+\beta, \alpha, \beta)$, $X_{ch 1}=(-1,1,1,2), X_{ch 2}=(2,4,-1,1), X_{6}=(-1,1,0,0)$.