I'm having a hard time understanding this question:
Determine the reduced echelon form of the homogeneous linear system of three equations in variables $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ such that $x_{1}, x_{2}, x_{4}$ are leading variables; $x_{3}, x_{5}$ are free variables and which has solutions
$$\begin{pmatrix}2 \\-1\\1\\0\\0 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix}-3 \\2\\0\\-4\\1 \end{pmatrix} $$
You have \begin{cases} x_1=2x_3-3x_5\\ x_2=-x_3+2x_5\\ x_4=-4x_5 \end{cases} which means \begin{cases} x_1-2x_3+3x_5=0\\ x_2+x_3-2x_5=0\\ x_4+4x_5=0 \end{cases} This system's matrix is what you're looking for.
If you can't see why, here's a different strategy: the matrix must be of the form \begin{bmatrix} 1 & 0 & a & 0 & b \\ 0 & 1 & c & 0 & d \\ 0 & 0 & 0 & 1 & e \end{bmatrix} and the two given vectors should be in its null space.