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.