[Math] Is the row calculation of row echelon form correct

linear algebramatrices

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 3×3 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 ?

    Best Answer

    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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