Proving formally that a set is subspace. I want to prove formally that  $U = (x,y,z\in \mathbb{R^3})$
  $| \det\begin{bmatrix}a_{11}&a_{12} & x\\
a_{21}&a_{22}&y\\ a_{31}&a_{32}&z  \end{bmatrix} = 0$ is a subspace.
I know this is true but the technique I learnt, I think I can't apply it here. 
To prove that they are closed under vector addition, suppose for example for some $f =(x,y,z) $ and $g = (x',y',z')$ determinant is $0$ then the I learnt I have to add both vectors $f,g$ and put it in the original expression but then I get $\det(A + B) = \det(A) + \det(B)$ and this way I can't prove that this is a vector space because in general $\det(A+B) \neq \det(A) + \det(B)$
 A: Hint
$$
\det\begin{bmatrix}a_{11}&a_{12} & cx\\
a_{21}&a_{22}&cy\\ a_{31}&a_{32}&cz  \end{bmatrix}
= c\det\begin{bmatrix}a_{11}&a_{12} & x\\
a_{21}&a_{22}&y\\ a_{31}&a_{32}&z  \end{bmatrix} 
$$
for all $c\in\mathbb{R}$ and
$$
\det\begin{bmatrix}a_{11}&a_{12} & x+u\\
a_{21}&a_{22}&y+v\\ a_{31}&a_{32}&z+w  \end{bmatrix}
= \det\begin{bmatrix}a_{11}&a_{12} & x\\
a_{21}&a_{22}&y\\ a_{31}&a_{32}&z  \end{bmatrix} +
\det\begin{bmatrix}a_{11}&a_{12} & u\\
a_{21}&a_{22}&v\\ a_{31}&a_{32}&w  \end{bmatrix} .
$$
In general the determinant is a multi-linear function of its columns.
A: Useful observation:
$$\det\begin{bmatrix}a_{11}&a_{12} & x\\
a_{21}&a_{22}&y\\ a_{31}&a_{32}&z  \end{bmatrix} = 0 \iff \left\{\begin{pmatrix}a_{11} \\ a_{21} \\a_{31}  \end{pmatrix}, \begin{pmatrix}a_{12} \\ a_{22} \\a_{32}  \end{pmatrix}, \begin{pmatrix}x \\ y \\z  \end{pmatrix}\right\} \text{ is linearly dependent}$$
So if $\begin{pmatrix}a_{11} \\ a_{21} \\a_{31}  \end{pmatrix}$ and $\begin{pmatrix}a_{12} \\ a_{22} \\a_{32}  \end{pmatrix}$ are linearly dependent, then $U = \mathbb{R}^3$ so it's a subspace.
Otherwise, $\begin{pmatrix}x \\ y \\z  \end{pmatrix} \in U$ if and only if $\begin{pmatrix}x \\ y \\z  \end{pmatrix} \in \operatorname{span}\left\{\begin{pmatrix}a_{11} \\ a_{21} \\a_{31}  \end{pmatrix}, \begin{pmatrix}a_{12} \\ a_{22} \\a_{32}  \end{pmatrix}\right\}$.
Thus, $U = \operatorname{span}\left\{\begin{pmatrix}a_{11} \\ a_{21} \\a_{31}  \end{pmatrix}, \begin{pmatrix}a_{12} \\ a_{22} \\a_{32}  \end{pmatrix}\right\}$ so it is certanly a subspace of $\mathbb{R}^3$.
A: Hint:
Determinants are linear w.r.t. columns, hence $U$  is the kernel of a linear map.
