Confused about a linear equation So I am working through some notes on Linear Algebra and I cant seem to follow this one part. The question asks to 
Solve:
$x+y-z+2w=-20$
$2x-y+z+w=11$
$3x-2y+z-2w+27$
I don't have a problem with putting the equation into matrix form and even reducing it. The way the notes explains it is what I don't understand at all. So first once its in echelon,
$$
\begin{bmatrix} 
1 & 1 & -1 & 2 & -20\\ 
0 & 1 & -1 & 1 & -17\\ 
0 & 0 & 1 & 3 &-2\\ 
\end{bmatrix} 
$$the notes states: "The solution is therefore
$$
X=\begin{pmatrix}
-w-3\\
-4w-19\\
-3w-2\\
2\\
\end{pmatrix}
$$
(This is the first place I am confused? Why are there $4$ rows now? And how did they get these numbers?).
Then, we can reduce and have 
$$
\begin{bmatrix}
1&0&0&1&-3\\
0&1&0&4&-19\\
0&0&1&3&-2
\end{bmatrix}
$$
Now for the second confusing part, it then states: "Note, the solution is 
$$
X=\begin{pmatrix}
-3\\
-19\\
-2\\
0\\
\end{pmatrix} +w\begin{pmatrix}
-1\\
-4\\
-3\\
1\\
\end{pmatrix}
$$
Note that the rank is $3$ and there is $1$ parameter.
Thanks a lot for the help guys. Hopefully I can understand it.
 A: You have four unknowns, so a solution is a vector with four components.
The system, after the final reduction, can be written as
$$
\begin{cases}
x+w=-3\\
y+4w=-19\\
z+3w=-2
\end{cases}
$$
Since $w$ is a free variable, you can set it to any value you want; therefore the solutions are
$$
\begin{cases}
x=-3-w\\
y=-19-4w\\
z=-2-3w\\
w=w
\end{cases}
$$
that in vector form can be written as
$$
\begin{bmatrix}
-3-w\\
-19-4w\\
-2-3w\\
w
\end{bmatrix}=
\begin{bmatrix}
-3\\
-19\\
-2\\
0
\end{bmatrix}+
w\begin{bmatrix}
-1\\
-4\\
-3\\
1
\end{bmatrix}
$$
where $w$ is any number.
A: Assuming your columns are x,y,z,w, your echelon form translates to
z + 3w - -2 or z = 2- 3w, the next row up says y + 4 w = -10 or y = -10 -4w, continuing
 x + w = -3, So solutions look like they have a freely chosen w from which says solutions look like: (x, y,z,w) = (-3-w, -10-4w, 2-3w, w) or w(-1,-4,-3 1) + (-3, -10, 2,0), a one-parameter family solutions with parameter w. Of course, you could write it as a column just as well. The solution is always of this form, an element of the kernel*
 + a particular solution (Here, the particular solution is (x,y,z,w) = ( -3, -19, -2, 0).
*The kernel is the set of vectors which are mapped to zero by the original matrix.
Because you have infinitely many solutions the original matrix has det = 0. You should't have been surprised that your solutions are 4-tuples, because you have 4 unknowns. This is actually incomplete because I do not know the precise statement of the original problem, I.E. Was it a homogeneous system? That requires a slightly different interpretation.
Hope this helps!
