Is it legal to move the vertical bar in an augmented matrix? Say I am solving a system of linear equations using Gauss reduction. I have got it to this point:
$$\left[\begin{array}{ccc|c}
 1 & 0 & -1 & 1 \\
 0 & 1 & 1 & 0 \\
 0 & 0 & 0 & 0
 \end{array}\right]$$
It seems intuitive to me from basic arithmetic that it should be legal to do:
$$\left[\begin{array}{cc|cc}
 1 & 0 & 1 & 1 \\
 0 & 1 & -1 & 0 \\
 0 & 0 & 0 & 0
 \end{array}\right]$$
as a step to writing a line equation in 3-space:
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix}
= z\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}
+ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} $$
However, is moving the vertical bar considered a legal matrix operation? (And is there a more technical name for the vetical bar that I ought to be using?)
 A: I have thot about this more, with the help of the comments above, which I will cite in my answer.
The answer is no, moving the vertical bar is not a valid matrix operation, because: 
*

*It is not a matrix operation at all (Henning Makholm)

*If it were to be defined as a matrix operation, it would be rather ambiguous (Ted)
However, the underlying matrix  operations that were behind the idea are valid. The idea is just as
$$\left[\begin{array}{ccc|c}
 1 & 0 & -1 & 1 \\
 0 & 1 & 1 & 0 \\
 0 & 0 & 0 & 0
 \end{array}\right]$$
implies
$$\left[\begin{array}{ccc}
 1 & 0 & -1 \\
 0 & 1 & 1 \\
 0 & 0 & 0
 \end{array}\right]
\left[\begin{array}{c}
x \\ y \\ z
\end{array}\right] = 
\left[\begin{array}{c}
1 \\ 0 \\ 0
\end{array}\right]$$
So
$$\left[\begin{array}{cc|cc}
 1 & 0 & 1 & 1 \\
 0 & 1 & -1 & 0 \\
 0 & 0 & 0 & 0
 \end{array}\right]$$
would imply
$$\left[\begin{array}{ccc}
 1 & 0 \\
 0 & 1 \\
 0 & 0
 \end{array}\right]
\left[\begin{array}{c}
x \\ y
\end{array}\right] = 
z \left[\begin{array}{c}
1 \\ -1 \\ 0
\end{array}\right]
+ \left[\begin{array}{c}
1 \\ 0 \\ 0
\end{array}\right]$$
which is legal, and solves the problem of switching in and out of the matrix notation and should provide a good intermediate step, as desired, even if it is not quite as succinct. It  also makes the next steps much more clear: do the matrix multiplication on the LHS, add
$$ \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} $$
to both sides of the equation, and simplify.
