The matrix can be row reduced to the following form:
\[ \begin{bmatrix} 1 & 0 & -4 & 1 \\ 0 & 1 & \frac{3}{2} & - \frac{1}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix} \]Which represents the following system of equations
\[ \begin{aligned} 1x + 0y + -4 z + 1w &= 0 \\ 0x + 1y + \frac{3}{2}z - \frac{1}{2}w &= 0 \\ 0x + 0y + 0z + 0w &= 0 \end{aligned} \]Since \( z \) and \( w \) are free variables, we let \( s, t \in \mathbb{R} \) and set \(z = s\) and \(w = t\) , now using back substitution, we can deduce the following
\[ \begin{aligned} 0x + 1y + \frac{3}{2}z - \frac{1}{2}t = 0 &\implies y = \frac{1}{2}t - \frac{3}{2}s \\ 1x + 0y + -4 z + 1w = 0 &\implies x = 4s - t\\ \end{aligned} \]Therefore the solution set is given by
\[ \left \{ \begin{bmatrix} 4s - t \\ - \frac{3}{2}s + \frac{1}{2}t \\ s \\ t \end{bmatrix} : s, t \in \mathbb{R} \right\} = \left \{ s \begin{bmatrix} 4 \\ - \frac{3}{2} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} -1 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix} : s, t \in \mathbb{R} \right\} \]It can be confirmed that the two vectors \( \vec{v_1} = \begin{bmatrix} -1 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix}\) and \( \vec{v_2} = \begin{bmatrix} 4 \\ - \frac{3}{2} \\ 1 \\ 0 \end{bmatrix} \) are linearly independent due to the non-matching zero in the third component, thus the solution space can be written as \( \operatorname{span} ( \{ \vec{v_1}, \vec{v_2} \} ) \), therefore a possible basis for the solution set is \( ( \vec{v_1}, \vec{v_2} ) \).