Sum of subspaces by setting determinant = 0 Given two subspaces of $R^4$
$W = \operatorname{Span}( (1,1,0,-1), (1,2,3,0) )$
$U = \operatorname{Span}( (1, 2, 2, -2), (2, 3, 2, -3) )$
Find the cartesian equations of the intersection.
I would solve this by putting all the vectors in a matrix as columns and the coordinates x,y,z,t on the side and doing gaussian elimination (this follows straight from the definition).
Instead I read a very strange (to me) solution to this problem.
Given that 
$$ \det \left(\begin{bmatrix}1&1&1\\1&2&2\\0&3&2\end{bmatrix}\right) = -1 \neq 0$$
and that the last vector is the sum of the first and the third, a base of the sum is given by the first 3 vectors, so we can find cartesian equations for the sum by setting:
$$ \det \left(\begin{bmatrix}x&y&z&t\\1&1&1&-1\\1&2&2&0\\0&3&2&-2\end{bmatrix}\right) = 0.$$
No more explanation is given could you please explain me how this procedure works?
 A: The proposed solution is simply wrong. As both $W$ and $U$ are two-dimensional, $W\cap U$ is at most two-dimensional. However, the determinantal equation in the proposed solution is a single linear equation in four unknowns. So, what it specifies is a three-dimensional subspace, which cannot possibly be equal to $W\cap U$.
In fact, since the first three vectors in the specifications of $W$ and $U$ are linearly independent and the fourth vector is a linear combination of the first and the third, we know right away that $W\cap U$ is the span of the first vector $(1,1,0,-1)^T$ (and you need a system of three linear equations to specify this span, such as $x=y,\,x=-t$ and $z=0$). However, $(1,1,0,-1)$ is not even a solution to the determinantal equation. So, you see how wrong that proposed solution is.
Edit. I've just noticed that there is a mismatch between your question title (which is about the sum $W+U$) and the question body (which is about the intersection $W\cap U$). Anyway, the proposed solution is still wrong when $W+U$ is concerned. We have $(1,1,0,-1)\in W\subseteq W+U$, but the LHS of the determinantal equation (which amounts to $2x+4y-5z+t=0$) is $5$ when $(x,y,z,w)=(1,1,0,-1)$.
