1.21. Solve the system from example 1.17 using the Gaussian method.
Solution. As a result of elementary transformations over the augmented matrix of the system, we obtained the following system at the end of the forward pass of the Gaussian method:

$$
\left\{\begin{array}{r}
x_{1}+2 x_{2}+x_{3}-x_{4}=1 \\
-x_{2}-5 x_{3}-x_{4}=-1 \\
6 x_{3}+5 x_{4}=1
\end{array}\right.
$$

Here, the free variable is $x_{4}$. We will move it to the right-hand sides, leaving the basic variables $x_{1}, x_{2}, x_{3}$ in the left-hand sides of the equations:

$$
\left\{\begin{array}{r}
x_{1}+2 x_{2}+x_{3}=x_{4}+1 \\
-x_{2}-5 x_{3}=x_{4}-1 \\
6 x_{3}=-5 x_{4}+1
\end{array}\right.
$$

Now, we will find the expression for $x_{3}$ from the last equation and, substituting it into the second equation, determine $x_{2}$. We have

$$
x_{3}=-\frac{5}{6} x_{4}+\frac{1}{6}, \quad x_{2}=1-5\left(\frac{5}{6} x_{4}+\frac{1}{6}\right)-x_{4}=\frac{1}{6}-\frac{31}{6} x_{4}
$$

Now, substituting $x_{3}$ and $x_{2}$ into the first equation, we will find $x_{1}$:

$$
x_{1}=-2\left(\frac{1}{6}-\frac{31}{6} x_{4}\right)-\left(-\frac{5}{6} x_{4}+\frac{1}{6}\right)+x_{4}+1=\frac{1}{2}+\frac{73}{6} x_{4}
$$

Thus, the answer can be written as:

$$
\left\{\begin{array}{l}
\forall x_{4} \in \mathbb{R} \\
x_{1}=\frac{1}{2}+\frac{73}{6} x_{4} \\
x_{2}=\frac{1}{6}-\frac{31}{6} x_{4} \\
x_{3}=-\frac{5}{6} x_{4}+\frac{1}{6}
\end{array}\right.
$$