Example 1. Find the dimension $d$ of the solution space, its basis (fundamental system of solutions), and the general solution of the homogeneous system of linear equations

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

1. We write down the matrix of the system and transform it to the reduced form using elementary row operations:

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

It is clear that the first and second columns of the matrix $A_{\text {red. }}$ (and the original matrix $A$) are linearly independent, and the other columns are their linear combinations. Therefore, the first and second columns are the basis columns.

2. Since the number of linearly independent columns of the matrix $A_{\text {red. }}$ is two, then

$$
\operatorname{Rg} A=\operatorname{Rg} A_{\text {red. }}=2
$$

Thus, the dimension of the solution space is

$$
d=n-r=5-2=3
$$

and the fundamental system of solutions consists of three linearly independent solutions.

3. The unknowns $x_{1}$ and $x_{2}$, corresponding to the basis columns, are the basis unknowns, and the unknowns $x_{3}, x_{4}, x_{5}$ are free.

We write down the system of equations with the matrix $A_{\text {red. }}$ (this system is equivalent to the original one) and move the free unknowns to the right-hand sides of the equations:

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

For the first set of free unknowns $x_{3}=1, x_{4}=0, x_{5}=0$, we get $x_{1}=-5 / 2, x_{2}=-2$, i.e., the first solution is

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

For the second set of free unknowns $x_{3}=0, x_{4}=1, x_{5}=0$, we get $x_{1}=3 / 2, x_{2}=3$, i.e., the second solution is

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

For the third set of free unknowns $x_{3}=0, x_{4}=0, x_{5}=1$, we get $x_{1}=-2, x_{2}=0$, i.e., the third solution is

$$
X_{3}=\left(\begin{array}{r}
-2 \\
0 \\
0 \\
0 \\
1
\end{array}\right)
$$

We will check these solutions by substituting them into the original system of equations, and also verify that the solutions are linearly independent (the rank of the matrix formed from the columns $X_{1}, X_{2}, X_{3}$ is 3).

Therefore, the solutions $X_{1}, X_{2}, X_{3}$ form a basis in the solution space (the fundamental system of solutions).

Answer. The dimension of the solution space is $d=3$. The fundamental system of solutions is

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

and the general solution of the homogeneous system is

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

where $C_{1}, C_{2}$, and $C_{3}$ are arbitrary constants.