Since you are working relative to the basis $B$, you may as well imagine that you are working over the standard basis and that the vectors you have are, as lhf points out, $(1,-1,0,1)$ and $(-1,0,3,0)$.
What is a system of equations that determines these vectors? To get a subspace, you'll want a homogeneous system of equations. And these two vectors should generate all solutions. So, first of all, your system should have four unknowns and two equations, since you want two parameters. The solutions will all look like $\left(\begin{array}{c} x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r} 1\\-1\\0\\1\end{array}\right)r + \left(\begin{array}{r}-1 \\0\\3\\0\end{array}\right)s$ with $r$ and $s$ arbitrary.
What system is this? Well, \begin{align*} x_1 & = r - s\\ x_2 &= -r\\ x_3 &= 3s\\ x_4 &= r. \end{align*} Equivalently, plugging in $x_4$ for $r$ and $\frac{1}{3}x_3$ for $s$, we get \begin{align*} x_1 +\frac{1}{3}x_3 - x_4 &= 0\\ x_2 +x_4 &= 0. \end{align*} You'll note that the two vectors you have, $(1,-1,0,1)$ and $(-1,0,3,0)$, are solutions. And any solution is a linear combination of these two. So this gives the system you want in terms of coordinate vectors relative to $B$.