I have a question:
For the following system of linear equations, using Gaussian elimination, decide whether it has at least one solution. If it does, represent the general solution as an affine map in matrix/vector form.
w + 3x - y + 2z = 1
-w - 3x + 2y - z = 2
Using Gaussian elimination i got:
w + 3x + 3z = 4
y + z = 3
I think you actually meant (Gauss reduction):
$$\begin{pmatrix}\;1&\;\,3&\!\!-1&\;2&\;\;1\\ \!\!-1&\!\!-3&\,2&\!\!-1&\;\;2\end{pmatrix}\stackrel{R_2+R_1}\longrightarrow\begin{pmatrix}1&3&\!\!-1&2&\;\;1\\ 0&0&\,1&1&\;\;3\end{pmatrix}$$
The above reduced form mean:
$$\begin{align*}\text{Second line:}&\;\;y+z=3\implies y=3-z\\\text{First line:}&\;\;w+3x-(3-z)+2z=1\implies w+3x=4-3z\end{align*}$$
So you have two degrees of freedom: (choose freely the values of $\;z,x\;$ and then the values of all the variables get uniquely and completely determined.