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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?)

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    @Mariano Suárez-Alvarez The idea is simply to interpose another step to reduce the risk of making a mistake. I could always move out of matrix notation, but going out of it and then back into it means that to be more explicit I am actually doing more work.2011-09-20

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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:

  1. It is not a matrix operation at all (Henning Makholm)
  2. 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.