1.20. Solve the system of equations from example 1.18 using the Gaussian method.
Solution. We have

$$
\begin{aligned}
& \tilde{A}=\left(\begin{array}{ccc|c}
1 & 2 & 4 & 5 \\
2 & 1 & 5 & 7 \\
3 & 2 & 6 & 9
\end{array}\right) \sim\left[\begin{array}{l}
(2)-(1) \times 2 ; \\
(3)-(1) \times 3
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & 2 & 4 & 5 \\
0 & -3 & -3 & -3 \\
0 & -4 & -6 & -6
\end{array}\right) \sim[(2):(-3)] \sim \\
& \sim\left(\begin{array}{ccc|c}
1 & 2 & 4 & 5 \\
0 & 1 & 1 & 1 \\
0 & -4 & -6 & -6
\end{array}\right) \sim[(3)-(2) \times(-4)] \sim\left(\begin{array}{ccc|c}
1 & 2 & 4 & 5 \\
0 & 1 & 1 & 1 \\
0 & 0 & 2 & 2
\end{array}\right)
\end{aligned}
$$

The forward pass of the Gaussian method is now complete, and for the backward pass, we form the system corresponding to the last matrix, which is equivalent to the original augmented matrix of the system:

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

From the last equation, we find $x_{3}=1$, substituting the found $x_{3}$ into the second equation, we get $x_{2}=1-x_{3}=0$. From the first equation, we find $x_{1}=5-2 \cdot x_{2}-4 \cdot x_{3}=1$.