Find $A^{-1}$(W) of linear manifold W Given linear map $A:\mathbb{R}^2\to \mathbb{R}^4$ defined as 
$$A = \begin{pmatrix}
 1 & 1 \\ 
 1 & -1 \\
 0 & 2 \\
 3 & 1 
 \end{pmatrix}$$ 
and linear manifold $ W \subset \mathbb{R}^4$  $$ W = \begin{pmatrix}
 1 \\ 
 -2 \\
 0 \\
 0 
 \end{pmatrix} + \left[ \begin{pmatrix}
 1 \\ 
 -2 \\
 1 \\
 0 
 \end{pmatrix} , \begin{pmatrix}
 1 \\ 
 -1 \\
 0 \\
 1 
 \end{pmatrix} \right]_\lambda$$
($\left[ \dots \right]_\lambda$ denotes linear span). How do I find $A^{-1}$(W)? 
All I know is that generally $A^{-1}(W) \neq A^{-1}(a) + A^{-1}(P)$ ($P$ is the subspace spanned by those 2 vectors). The result has to be manifold in $\mathbb{R}^2$ right?
Are non-parametric equations of W needed here?
 A: There are most likely many different ways of doing this. Your affine subspace $W$ can be parametrised by ${\bf X} : \mathbb{R}^2 \to \mathbb{R}^4$, given by:
$${\bf X}(\lambda,\mu) = (1+\lambda+\mu,-2-2\lambda-\mu,\lambda,\mu) \, . $$
If we use $(x,y,z,t)$ as coordinates for $\mathbb{R}^4$ then $W$ can also be given by the simultaneous equations:
$$\begin{array}{ccc} x & = & 1+z+t \, , \\
y & = & -2 - 2z - t \, . \end{array}$$
If we use $(u,v)$ as coordinates for $\mathbb{R}^2$ then $A : (u,v) \mapsto (u+v,u-v,2v,3u+v).$ We see that $A(u,v)$ belongs to $W$ if $x=u+v,$ $y=u-v$, $z=2v$ and $t=3u+v$ satisfy the simultaneous equations $x=1+z+t$ and $y=-2-2z-t.$ From this we get:
$$\begin{array}{ccc}
u+v &=& 1+3u + 3v \, , \\
u-v &=& -2-3u-5v \, . 
\end{array}$$
Thus, the set of point $(u,v) \in \mathbb{R}^2$ for which $A(u,v) \in W$ satisfy:
$$2u+2v+1=0 \, . $$
Note that $A^{-1}(W)$ is just shorthand for $\{(u,v) \in \mathbb{R}^2 : A(u,v) \in W \}.$
A: $A^{-1}(W)=A^{-1}\left((\operatorname{range}A)\cap W\right)$ and in turn it is equal to the set of vectors $(x,y)^T\in\mathbb{R}^2$ such that
$$
\begin{bmatrix}
1& 1& 1& 1\\
1&-1&-2&-1\\
0& 2& 1& 0\\
3& 1& 0& 1
\end{bmatrix}
\begin{bmatrix}x\\y\\z\\w\end{bmatrix}
=\begin{bmatrix}1\\-2\\0\\0\end{bmatrix}.
$$
It is not hard to see that $(-\frac14, -\frac14, \frac12, 1)^T$ is a solution to the above equation and the kernel of the $4\times 4$ matrix on the left is $\operatorname{span}\{(1, -1, 2, -2)^T\}$. Therefore $A^{-1}(W)$ is the line $(-\frac14, -\frac14)^T+s(1, -1)^T$, or $2x+2y+1=0$.
