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Math Help - prove elementary row operations

  1. #1
    Junior Member Tclack's Avatar
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    prove elementary row operations

    the one where you can switch them is understandable, the one where you can multiply through by a constant is easy, but the one where you add the multiple of one row to another row to get a new row is weird. I've found that it's similar to solving linear equations by adding them together ex
    3x + 4y = 2
    -3x - 2y = 1
    ___________
    2y = 3


    but back when I was learning that I wasn't really concerned with proof. I guess I don't even see why the above works.

    Any insights?
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  2. #2
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    Re: prove elementary row operations

    the basic idea is this:

    if:

    A+B = C

    and

    D+E = F

    then C+F = (A+B) + (D+E) (since A+B is just C, and D+E is just F).

    now let's put x's and y's in this:

    ax + by = c
    dx + ey = f

    (here, a,b,c,d,e and f are numbers, and x and y are variables we want to find values for).

    so (a+d)x + (b+e)y = c+f

    now if it just so happens that a = -d, then a + d will be -d + d = 0, so that term will "disappear". but if not, we can MAKE that happen:

    d(ax + by) = cd (since the two sides were equal, they're still equal if we multiply both sides by the same number, d).
    (-a)(dx + ey) = -af

    that is:

    (ad)x + (bd)y = cd
    (-ad)x + (-ae)y = -af

    and adding these two equations, we get:

    (ad - ad)x + (bd - ae)y = cd - af

    that is:

    (bd - ae)y = cd - af.

    if bd - ae isn't 0 (it might be), we can divide by it, and get:

    y = (cd - af)/(bd - ae), and then find x by substitution (it turns out that x = (ce - bf)/(ae - bd)).

    the point being, none of the things i did with the equations above depended on any special properties of x or y (we made no assumptions about what they might be), so if x and y satisfy the original equations:

    ax + by = c
    dx + ey = f

    then they still satisfy the rearranged equations:

    x = (ce - bf)/(ae - bd)
    y = (cd - af)/(bd - ae) <---provided the denominator isn't 0 (if it is, we have a "degenerate" system of equations. it turns out that if we express this in matrix terms, we have a non-invertible matrix).

    you are quite correct, that all row operations do is mimic (or as mathematicians like to say: formalize) the process you learned way back when, of "eliminating variables".
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  3. #3
    Junior Member Tclack's Avatar
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    Re: prove elementary row operations

    that makes sense. I shouldn't have given up so easily, I probably could have thought this up if I put more effort into it.

    that's the 3rd thing you've helped me with today Deveno!!!
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