A question regarding $\,3 \times 4$ matrices Good day, I'm currently studying for an exam and need to learn about matrices too. Well, since I'm not good at English I'll just write what I've done so far. Below is a photo showing the full sheet of paper with the steps I did so far.
The thing I'm wondering about is, that I'd have four unknown variables $(w, x, y, z)$ after writing down the equations, but my matrix has got 3 rows only. Therefor I'm wondering how I'm supposed to find a generic solution for A*x = b

$$A=\begin{bmatrix}0&2&-2&3\\ 1&3&-1&2\\ 2&3&1&0 \end{bmatrix} , b=\begin{bmatrix}-1\\ -2\\ 0 \end{bmatrix}$$ 
$$A=\begin{bmatrix}1&3&-1&2&-2\\ 0&1&-1&3/2&-1/2\\ 0&0&0&1/2&-3/2 \end{bmatrix} $$ 
$$z=-3 $$ 
$$\operatorname{rank}(A)=3 $$ 

So I actually solved for $z$ but I can't seem to solve for the other unknown variables (or can I?). The thing that confuses me is the fact this matrix is $3 \times 4$ now instead of $3 \times 3$. Any tips would be very much appreciated!

Click image with middle-mouse-button for full size.

We can write this as an augmented matrix:
$$\begin{bmatrix}0&2&-2&3&-1\\ 1&3&-1&2&-2\\ 2&3&1&0&0 \end{bmatrix}$$ 
Using Gaussian Elimination (Row-Reduced-Echelon-Form), we arrive at:
$$A=\begin{bmatrix}1&0&2&0&12 \\ 0&1&-1&0&-8\\ 0&0&0&1&5 \end{bmatrix}$$ 
This means our solution can be written as:


*

*$z = 5$

*$x = -8 + y$

*$w = 12 - 2y$


$y$ is a free variable and this system has an infinite number of solutions.
You can easily test this solution by picking random values for $y$, substituting the values back into the system and validating equality. 
