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I was directed by a community member to a resource on how to calculate the row echelon form of a matrix here. The resource says:

First we wish to put A into reduced row echelon form. There are several ways to do those (and
thus several matrices P), but here is one possible way: (calculation next)  

So, I understand that a given matrix can have multiple row echelon forms.

To continue my self-study of linear algebra further, I looked at the example on wikihow.
It gives a simple 3x3 matrix and shows how to calculate the row echelon form. Fair enough.

As in wikihow, the given matrix is:
$$ \begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 4 & 5 \end{bmatrix} $$
and the row echelon form is this:
$$ \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \\ \end{bmatrix} $$
However, my answer is different and I am not sure if it is correct

My calculation on the same matrix:

  • Attempting to get all zeros under $A_1_1$ as:
  • $R_2$ - $R_1$ -> $R_2$
    $(3\times R_1$) - $R_3$ -> $R_3$
    So, the matrix is:
    $$ \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \\ \end{bmatrix} $$

  • Attempting to get all zeroes under $A_2_2$ as:
  • $R_2$ + $R_3$ -> $R_3$
    So the matrix is:
    $$ \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \\ \end{bmatrix} $$

    The only difference is that I have a $2$ in bottom right and wikihow has $-2$.
    Is it correct ?

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      $\begingroup$ yes, there are multiple echelon forms. For example you can continue multiplying third row by (-1) and get the other answer. Or you can add any multiple of a lower row to an upper row and still keep the echelon form. (What is unique is the reduced row echelon form where you insist pivots must be 1 and all other elements a pivot column should be zero.) $\endgroup$
      – Maesumi
      Jun 4, 2013 at 8:12
    • $\begingroup$ @Maesumi is there a rule that $3XR_3 - R_1$ -> $R_3$ because $R_3$ was the first operand ???? $\endgroup$
      – An SO User
      Jun 4, 2013 at 8:13
    • $\begingroup$ You can use the addition of rows $m R_n \pm R_k$ however you like (so long as $n\ne k$). Here to get zero you can use $3R_1-R_3$ or $R_3-3R_1$. The typical standard notation in text books is to use $mR_n + R_k$. $\endgroup$
      – Maesumi
      Jun 4, 2013 at 8:20
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      $\begingroup$ @Maesumi so my answer is correct :) Plus, there was a mistake in WIkiHow which I had to correct $\endgroup$
      – An SO User
      Jun 4, 2013 at 8:28

    1 Answer 1

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    You're getting different answers because you're subtracting $R_3$ from $3 R_1$. Usually, you subtract $3 R_1$ from $R_3$ (you can add or subtract rows, but you're changing the sign on $R_3$ here: we often choose to do it this way because adding or subtracting a multiple of a row keeps the determinant the same).

    So, to get Wikihow's answer, you should have taken $R_3 - 3R_1 \to R_3$ instead. This also explains the sign flip, since $R_3 - 3R_1 = - ( 3R_1 - R_3 )$.

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