EXAMPLE 2. Find the general solution of the non-homogeneous system of linear equations

$$
\left\{\begin{aligned}
x_{2}+2 x_{3}-3 x_{4} & =-1 \\
2 x_{1}-x_{2}+3 x_{3}+4 x_{5} & =5 \\
2 x_{1}+5 x_{3}-3 x_{4}+4 x_{5} & =4
\end{aligned}\right.
$$
Solution.

1. We write the augmented matrix of the system and use elementary row transformations to reduce the matrix $A_{\text {ext. }}$ to its reduced form:

$$
\left(\begin{array}{rrrrr|r}
0 & 1 & 2 & -3 & 0 & -1 \\
2 & -1 & 3 & 0 & 4 & 5 \\
2 & 0 & 5 & -3 & 4 & 4
\end{array}\right) \sim\left(\begin{array}{rrrrr|r}
1 & 0 & 5 / 2 & -3 / 2 & 2 & 2 \\
0 & 1 & 2 & -3 & 0 & -1 \\
0 & 0 & 0 & 0 & 0 & 0
\end{array}\right) .
$$

2. Since $\operatorname{Rg} A_{\text {ext. }}=\operatorname{Rg} A=2$, the system is consistent. Since $n-r=5-2=3$, the general solution of the non-homogeneous system of linear equations is given by the formula

$$
X_{\text {g.n. }}=X_{\text {p.n. }}+C_{1} X_{1}+C_{2} X_{2}+C_{3} X_{3},
$$

where $X_{\text {p.n. }}$ is any particular solution of the non-homogeneous system, $X_{1}, X_{2}, X_{3}$ are the fundamental system of solutions of the corresponding homogeneous system, and $C_{1}, C_{2}, C_{3}$ are arbitrary constants.

3. We write the corresponding homogeneous system of equations

$$
\left\{\begin{array}{r}
x_{2}+2 x_{3}-3 x_{4}=0 \\
2 x_{1}-x_{2}+3 x_{3}+4 x_{5}=0 \\
2 x_{1}+5 x_{3}-3 x_{4}+4 x_{5}=0
\end{array}\right.
$$

This system coincides with the system given in Example 1. (If the homogeneous system of equations does not coincide with the system given in Example 1, then to find the fundamental system of solutions, we repeat the operations used in solving Example 1.)

In solving Example 1, the fundamental system of solutions of the homogeneous system of equations was found:

$$
X_{1}=\left(\begin{array}{c}
-5 / 2 \\
-2 \\
1 \\
0 \\
0
\end{array}\right), \quad X_{2}=\left(\begin{array}{c}
3 / 2 \\
3 \\
0 \\
1 \\
0
\end{array}\right), \quad X_{3}=\left(\begin{array}{c}
-2 \\
0 \\
0 \\
0 \\
1
\end{array}\right)
$$

4. We find any particular solution of the non-homogeneous system. The column of free terms $B$ of the augmented matrix is a linear combination of the basis columns of the matrix $A$, i.e., columns $A_{1}$ and $A_{2}$:

$$
B=2 \cdot A_{1}-1 \cdot A_{2}
$$

Adding to this expression the remaining columns with zero coefficients, we get

$$
B=2 \cdot A_{1}-1 \cdot A_{2}+0 \cdot A_{3}+0 \cdot A_{4}+0 \cdot A_{5}
$$

The coefficients in this expansion form a particular solution of the non-homogeneous system

$$
X_{\text {p.n. }}=\left(\begin{array}{r}
2 \\
-1 \\
0 \\
0 \\
0
\end{array}\right)
$$

We perform a check by substituting $X_{\text {p.n. }}$ into the original system of equations.

Answer. The general solution of the system is

$$
X_{\text {g.n. }}=\left(\begin{array}{r}
2 \\
-1 \\
0 \\
0 \\
0
\end{array}\right)+C_{1}\left(\begin{array}{c}
-5 / 2 \\
-2 \\
1 \\
0 \\
0