Example 4. Solve the system of equations

$$
\left\{\begin{aligned}
x_{1}+x_{3}+2 x_{4}+3 x_{5}-4 x_{6}= & 20 \\
2 x_{1}+x_{2}-3 x_{3}+x_{5}+2 x_{6}= & -13 \\
5 x_{1}-x_{2}+x_{3}+2 x_{4}+6 x_{5} & =20 \\
2 x_{1}-2 x_{2}+3 x_{3}+2 x_{5}+2 x_{6} & =13
\end{aligned}\right.
$$
Solution. In the previous examples of Jordan-Gauss transformations, the operations reduced to actions on the equations of the system, or the rows of the table, because all the leading coefficients were equal to 1. If the leading coefficients are different from 1, then operations on the rows can cause difficulties, and in such cases, one should use the Jordan-Gauss transformation formulas, i.e., the rule of the rectangle.

To save space, we will also present the solution of this system in a block notation:

$$
\begin{aligned}
& \left(\begin{array}{rrrrrr|r}
1 & 0 & 1 & 2 & 3 & -4 & 20 \\
2 & 1 & -3 & 0 & 1 & 2 & -13 \\
5 & -1 & 1 & 2 & 6 & 0 & 20 \\
2 & -2 & 3 & 0 & 2 & 2 & 13
\end{array}\right) \sim\left(\begin{array}{rrrrrr|r}
1 & 0 & 1 & 2 & 3 & -4 & 20 \\
2 & 1 & -3 & 0 & 1 & 2 & -13 \\
7 & 0 & -2 & 2 & 7 & 2 & 7 \\
6 & 0 & -3 & 0 & 4 & 6 & -13
\end{array}\right) \sim \\
& \sim\left(\begin{array}{rrrrrr|r}
-6 & 0 & 3 & 0 & -4 & -6 & 13 \\
2 & 1 & -3 & 0 & 1 & 2 & -13 \\
7 / 2 & 0 & -1 & 1 & 7 / 2 & 1 & 7 / 2 \\
6 & 0 & -3 & 0 & 4 & 6 & -13
\end{array}\right) \sim\left(\begin{array}{rrrrrr|r}
1 & 0 & -1 / 2 & 0 & 2 / 3 & 1 & -13 / 6 \\
0 & 1 & -2 & 0 & -1 / 3 & 0 & -26 / 3 \\
5 / 2 & 0 & -1 / 2 & 1 & 17 / 6 & 0 & 17 / 3
\end{array}\right)
\end{aligned}
$$

(The last row is proportional to the first, so it is removed). Let us emphasize that the goal of our transformations is to obtain unit columns.

Let us provide examples of applying the rule of the rectangle in the third block. In this case, one vertex of each rectangle should coincide with the leading element \(a_{16}=-6\), and the opposite vertex with the element to be recalculated:

$$
\begin{aligned}
& \text { } \\
& 3-\overparen{-} \\
& \left\lvert\, \begin{array}{l}
\mid \\
-3
\end{array} a_{23}^{\prime}=-3-\frac{3 \cdot 2}{-6}=-2\right. \\
& -6)-13 \\
& 1-7 / 2 \\
& b_{3}^{\prime}=\frac{7}{2}-\frac{13 \cdot 1}{-6}=\frac{7}{2}+\frac{13}{6}=\frac{17}{3}
\end{aligned}
$$

From the last block, we obtain the general solution of the system in the basis \((x_{2}, x_{4}, x_{6}):\)

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x _ { 2 } = - \frac { 2 6 } { 3 } + 2 x _ { 3 } + \frac { 1 } { 3 } x _ { 5 } , } \\
{ x _ { 4 } = \frac { 1 7 } { 3 } - \frac { 5 } { 2 } x _ { 1 } + \frac { 1 } { 2 } x _ { 3 } - \frac { 1 7 } { 6 } x _ { 5 } , } \\
{ x _ { 6 } = - \frac { 1 3 } { 6 } - x _ { 1 } + \frac { 1 } { 2 } x _ { 3 } - \frac { 2 } { 3 } x _ { 5 } , }
\end{array} \text { i.e. } \left\{\begin{array}{l}
x_{1}=t_{1}, \\
x_{2}=-\frac{26}{3}+2 t_{2}+\frac{1}{3} t_{3}, \\
x_{3}=t_{2}, \\
x_{4}=\frac{17}{3}-\frac{5}{2} t_{1}+\frac{1}{2} t_{2}-\frac{17}{6} t_{3} \\
x_{5}=t_{3}, \\
x_{6}=\frac{13}{6}-t_{1}+\frac{1}{2} t_{2}-\frac{2}{3} t_{3},
\end{array}\right.\right. \\
& t_{1}, t_{2}, t_{3} \in \mathbb{R}
\end{aligned}
$$

For \(t_{1}=1, t_{2}=-1, \quad t_{3}=2\), we obtain a particular solution \(X_{4}=(1,-10,-1,-3,2,-5)\). The basis solution has the form \(X_{6}=(0,-26 / 3,0,17 / 3,0,-13 / 6)\).

Note. The Gauss method (truncated Jordan-Gauss method) allows obtaining a column in the next block of the Gauss table that is different from the unit column, i.e., the unknown does not necessarily have to be eliminated from all equations except one. In this case, we speak of reducing the system of equations to a stepwise form. This is important in terms of saving time when the coefficients of the system are "inconvenient," especially if the system turns out to be unsolvable.