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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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    $\begingroup$ The vertical bar is not formally part of the matrix at all -- it is just there to help you keep track of what you're doing, namely: applying the same row operations to the coefficient matrix and the constant column vector. So moving the bar is not a "legal matrix operation" at, because it is not a matrix operation at all. $\endgroup$ Sep 20, 2011 at 3:55
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    $\begingroup$ Before you can ask whether it's legal, you have to specify what it means to have 2 columns on the right side of the vertical bar. Usually the augmented matrix [A|b], where A is a matrix and b is a column, is an abbreviated version of the equation Ax=b. But now you have matrices on both sides, so it's not clear what that means any more. The "equation" interpretation of Ax=b would mean, in your example above, that x has to be a 2 by 2 matrix. That is totally different from what you started with, in which x was a 3 by 1 matrix. $\endgroup$
    – Ted
    Sep 20, 2011 at 3:56
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    $\begingroup$ What do you expect to achieve by doing this? It seems to me that you are misunderstanding something... $\endgroup$ Sep 20, 2011 at 4:30
  • $\begingroup$ @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. $\endgroup$
    – Kazark
    Sep 20, 2011 at 5:05

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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.

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