Find the general solution of the system Given:
\begin{cases} x- z = 1 \\ y + 2z - w = 3 \\ x + 2y + 3z - w = 7 \end{cases} 
My work:
Use Gauss method
Leading variables: x,y, and z
Free variables: w
Express the leading variables x,y,and z in terms the free variable w.
1) I perform $-2R_2+R_3$ and got $x-z+w=1$ which replaces the third row of the system.
2) Then I perform $-R_1+R_3$ and got $w = 0$ which replaces the third row of the system.
3) Finally I perform $R_2+R_3$ and got $y+2z=3$ which replaces the third row of the system.
The system at the end looks like this:
\begin{cases} x- z = 1 \\ y + 2z - w = 3 \\ y = 3 -2 z \end{cases} 
I am stuck as to how to get the leading variables in terms of the free variables since I know the value of the free variable now. But plugging in the value of w to the equations causes the free variable to disappear... 
 A: It appears you wrote the system as: $x, y, z, w$, so I followed suit.
The RREF gives:


*

*$w = 0$

*$y = 3 -2z$

*$x = 1 + z$


$z$ is a free variable.
A: As you can see, you have a linear system of three equations in four unknowns. So at least one of the variables $x, y, w, z$ will be a free variable and in this case, since it turns out that $w$ is identically zero, its value is fixed, and so it cannot be a free variable. 
What we end with, if you use an augmented coefficient matrix to row reduce, and with columns representing the coefficients of each variable, ordered as follows:
$x\;y\;w \; z$, , is 
$$\begin{pmatrix} 1 & 0 & 0 & -1 &\mid& 1\\ 0 & 1 & 0 & 2  &\mid& 3 \\ 0 & 0 & 1 & 0 &\mid& 0 \\ 0 & 0 & 0 & 0 &\mid& 0\end{pmatrix}$$
corresponding to the system of equations: $$\begin{align} x & = 1 + z \\ \\ y & = 3 - 2z \\ \\ w & = 0 \\ \\ z &= t\end{align}$$
where $z = t$ simply means that we have $z$ as the free variable, and we denote its value by the parameter $t$.
Then, the general solution is given by $$\begin{pmatrix} x \\ y \\ w \\ z \end{pmatrix}=\begin{pmatrix} 1 \\ 3 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -2 \\ 0 \\ 1\end{pmatrix}$$
For a particular solution, just let the parameter $z = t = 0$.
A: Note that $w$ is necessarily zero, since $(***) - 2(**) - (*) = w = 7 - 2\cdot 3 - 1 = 0$, where $(*)$ denotes the first equation, $(**)$ denotes the second and $(***)$ the third. Then look at the matrix over which you have to compute the inverse. It has rank $2$. Why? Do you see a problem? Hence, you can't simplify it beyond the parametric form you have it in with $x-z=1$ and $y=3-2z$.
