Example 3. Solve the system of equations

$$
\left\{\begin{aligned}
& x_{1}+2 x_{2}+3 x_{3}+x_{4}=1 \\
& 3 x_{1}+13 x_{2}+13 x_{3}+5 x_{4}=3 \\
& 3 x_{1}+7 x_{2}+7 x_{3}+2 x_{4}=12 \\
& x_{1}+5 x_{2}+3 x_{3}+x_{4}=7 \\
& 4 x_{1}+5 x_{2}+6 x_{3}+x_{4}=19
\end{aligned}\right.
$$
Solution. Instead of Gauss's table, we will use another, more compact interpretation of its blocks. The vertical line in the blocks corresponds to the equality signs in the equations of the system. The sign (read as "tilde") between two adjacent blocks means that the systems corresponding to these blocks are equivalent. We have:

$$
\left(\begin{array}{rrrr|r}
(1) & 2 & 3 & 1 & 1 \\
3 & 13 & 1 & 5 & 3 \\
3 & 7 & 7 & 2 & 12 \\
1 & 5 & 3 & 1 & 7 \\
4 & 5 & 6 & 1 & 19
\end{array}\right) \sim\left(\begin{array}{rrrr|r}
1 & 2 & 3 & 1 & 1 \\
0 & 7 & 4 & 2 & 0 \\
0 & 1 & -2 & -1 & 9 \\
0 & 3 & 0 & 0 & 6 \\
0 & -3 & -6 & -3 & 15
\end{array}\right):(-3) \sim\left(\begin{array}{rrrr|r}
1 & 2 & 3 & 1 & 1 \\
0 & 7 & 4 & 2 & 0 \\
0 & 1 & -2 & -1 & 9 \\
0 & 1 & 0 & 0 & 2 \\
0 & 1 & 2 & 1 & -5
\end{array}\right) \sim
$$

$\left\{\begin{array}{l}\text { the unit column of the second block is obtained by multiplying the first row by } \\ -3,-3,-1,-4 \text { and subsequently adding to the second, third, fourth, and fifth }\end{array}\right.$ rows, respectively; in the second block, the fourth and fifth rows were divided term by term by 3 and -3, i.e., the equations were simplified.

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-009.jpg?height=183&width=885&top_left_y=640&top_left_x=194)

The second and third rows of the fourth block are discarded as proportional to the fifth. Note that the selection of the leading (pivotal) element uniquely determines the operations to zero out the elements in the leading column, so we have abandoned the use of numbers and arrows denoting row operations in the block.

The last block represents a system consisting of three equations $(r=3, m=5)$ with four unknowns ( $n=4$ ). Solving this system for the three basic unknowns, we get

$$
\left\{\begin{array}{l}
x_{1}=4-x_{3} \\
x_{2}=2 \\
x_{4}=-7-2 x_{3}
\end{array}\right.
$$

Let $x_{3}=\alpha$, then $\alpha=0$. The general and basic solutions then take the form respectively

$$
X_{0}=(4-\alpha, 2, \alpha,-7-2 \alpha), \quad X_{6}=(4,2,0,-7)
$$

Note that the variable $x_{2}$ cannot be obtained among the free variables (a free variable can take any value, whereas $\left.x_{2}=2\right)$).