Example 1. Solve the linear system

$$
\left\{\begin{aligned}
x_{1}+2 x_{2}-3 x_{3}-x_{4} & =10 \\
-2 x_{1}-3 x_{2}+7 x_{3} & =-23 \\
2 x_{1}+6 x_{2}-5 x_{3}-5 x_{4} & =18 \\
-x_{1}+3 x_{3}-4 x_{4} & =-11
\end{aligned}\right.
$$
Solution. We have $m=4, n=4$.

The first block of the Gauss table for this system has the following form («free terms» means the free terms of the system of equations, the vertical line corresponds to the equality signs):

| $x_{1}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ | free terms |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| (1) | 2 | -3 | -1 | 10 | 12 | -2 |
| -2 | -3 | 7 | 0 | -23 | $\downarrow$ |  |
| 2 | 6 | -5 | -5 | 18 |  | $\downarrow$ |
| -1 | 0 | 3 | -4 | -11 |  |  |

1. Perform the first iteration, i.e., obtain the first unit column, choosing the leading coefficient $a_{11}=1$ (in the table it is circled). For this, perform the following actions on the rows of the table (on the equations of the system) (they are denoted to the right of the table):
1) copy the first row (rewrite it);
2) add the first row, multiplied by 2, to the second;
3) add the first row, multiplied by -2, to the third;
4) add the first row to the fourth.

We get the second block of the table:

| $x_{1}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ | free terms |
| ---: | ---: | ---: | ---: | ---: |
| 1 | 2 | -3 | -1 | 10 |
| 0 | 1 | (1) | -2 | -3 |
| 0 | 2 | 1 | -3 | -2 |
| 0 | 2 | 0 | -5 | -1 |

2. Reduce the third column to a unit column, in which there is already one zero. The leading coefficient $a_{23}=1$ is circled. Next:
1) add the second row, multiplied by 3, to the first and write it instead of the first row;
2) rewrite the second row without change;
3) add the second row, multiplied by -1, to the third;
4) rewrite the fourth row without change.

These actions are expressed by numbers and arrows shown to the right of the second block of the table. The third block of the table has the following form:

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-007.jpg?height=216&width=754&top_left_y=1355&top_left_x=1530)

3. The next iteration involves obtaining the third unit column. For this, we take the leading coefficient $a_{32}=1$ and perform the following actions: add the third row, multiplied by $-5, -1, -2$, to the first, second, and fourth rows, respectively. Rewrite the third row without changes. We get the fourth block:

| $x_{1}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ | free terms |
| ---: | ---: | ---: | ---: | ---: |
| 1 | 0 | 0 | -2 | -4 |
| 0 | 0 | 1 | -1 | -4 |
| 0 | 1 | 0 | -1 | 1 |
| 0 | 0 | 0 | -3 | -3 |

4. Finally, perform the last iteration, choosing the leading coefficient $a_{44}=-3$. Divide the fourth row by -3. The other actions are obvious. We get:

| $x_{\mathbf{I}}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ | free terms |
| ---: | ---: | ---: | ---: | ---: |
| 1 | 0 | 0 | 0 | -2 |
| 0 | 0 | 1 | 0 | -3 |
| 0 | 1 | 0 | 0 | 2 |
| 0 | 0 | 0 | 1 | 1 |

5. After four iterations, we obtained a table corresponding to the system resolved relative to all unknowns ($r=m=n=4$):

$$
x_{1}=-2, \quad x_{2}=2, \quad x_{3}=-3, \quad x_{4}=1
$$

We can also write this as: $X=(-2,2,-3,1)$. The system is definitively consistent.

Note. Substitute these values of the unknowns into the given system and verify that they yield correct numerical equalities.