I have a question about an exercise I am working on for a linear algebra class. The exercise is as follows
Let $\vec{v}_1=(1,1,1,1)$ and $\vec{v}_2=(3,5,2,1)$, and let $V$ be the subspace of $\mathbb{R}^4$ spanned by $\vec{v}_1$ and $\vec{v}_2$. Find two equations, each of the form $ax+by+cz+dw=0$, such that the common solution to these equations is the subspace $V$.
So, if $\vec{v}_1=(1,1,1,1)$ and $\vec{v}_2=(3,5,2,1)$ span $V$, then every vector in $V$ can be expressed as a linear combination of $\vec{v}_1=(1,1,1,1)$ and $\vec{v}_2=(3,5,2,1)$. That is, for any two scalars $c$ and $d$,
$$c\begin{bmatrix} 1\\ 1\\ 1\\ 1\\ \end{bmatrix} + d\begin{bmatrix} 3\\ 5\\ 2\\ 1\\ \end{bmatrix} = \begin{bmatrix} x\\ y\\ z\\ w\\ \end{bmatrix} .$$
From here I use an augmented matrix
$$\begin{bmatrix} 1& 3|& x\\ 1& 5|& y\\ 1& 2|& z\\ 1& 1|& w \end{bmatrix}$$
which, after putting into RREF form, yields
$$\begin{bmatrix} 1& 0|& 2w-z\\ 0& 1|& z-w\\ 0& 0|& x-3w-2z\\ 0& 0|& y-5w-4z \end{bmatrix}$$
Does this give me the two equations I am asked to provide? It would be
$$x-3w-2z=0$$ $$y-5w-4z=0$$
