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Row echelon form?

  1. May 15, 2015 #1
    For some reason I just can't seem to wrap my head around the idea of reducing a Matrix to row echelon form. I'm familiar with the steps that the textbooks and tutorials use and how it's done but when I try practicing on my own I feel lost. e.g. all I end up with are just a bunch of random entries that don't bear any resemblance to row echelon form.
    How would I practice better for this?
    Last edited: May 15, 2015
  2. jcsd
  3. May 15, 2015 #2
    Perhaps try working a simple problem, say a two or three variable system with integer solutions, in tandem with the linear combination method? If you understand linear combination then you know the mechanics of how to reduce a matrix to ref. Often times mistakes come from the arithmetic; make sure you double check each calculation.
  4. May 15, 2015 #3


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    While it might be possible to simplify the calculations for special matrices, in general "row-reducing" is a very "mechanical" procedure.
    Here is the idea with a 3 by 3 general matrix:
    [tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}[/tex]

    I see that the "first column, first row" is "a" and I know I want "1" there so divide every number in the first column by a
    [tex]\begin{bmatrix}1 & b/a & c/a \\ d & e & f \\ g & h & i\end{bmatrix}[/tex]
    Now, I see that the "first column, second row" and "first column third row" are "d" and "g" respectively and I want "0" there. So subtract the first row times d from the second row and subtract the first row times g from the third row. That gives
    [tex]\begin{bmatrix}1 & b/a & c/a \\ 0 & e- bd/a & f- bd/a \\ 0 & h- bg/a & i- bg/a\end{bmatrix}[/tex]

    That completes the first column. Now look at the "second column, second row". It is "e- bd/a= (ae- bd)/a" and I want "1" there. So divide every number in the second row by (ae- bd/a)
    [tex]\begin{bmatrix} 1 & b/a & c/a \\ 0 & 1 & \frac{af- bd}{ae- bd} \\ 0 & \frac{ah- bg}{a} & {ai- bg}{a}\end{bmatrix}[/tex]
    There is now [itex]\frac{ah- bg}{a}[/itex] in the "second column, third row" and we want "0" there. So subtract [itex]\frac{ah- bg}{a}[/itex] times the second row from the third row.
    [tex]\begin{bmatrix} 1 & b/a & c/a \\ 0 & 1 & \frac{af- bd}{ae- bd} \\ 0 & 0 & {ai- bg}{a}- \frac{ai- bg}{a}\frac{ah- bg}{a}\end{bmatrix}[/tex]

    Start at the upper left and work down and to the right, doing one column at a time. That way, the "1"s and "0"s you already have won't be changed by further work.
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