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Math Help - Matrix Row Reduction

  1. #1
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    Matrix Row Reduction

    I’m having serious problems in comprehending matrix row reduction. I know how to turn an equation into a matrix and just to clarify I think these are the rules:

    1. I can switch any two rows at any given point.
    2. I can multiply or divide a row by any number of my choosing.
    3. I can add or subtract any two rows with each other. The resulting answer is then replaced into the row that should change (ex: row1+row2 means row 2 is replaced. If row1-row2 then row1 is replaced).
    4. I should be concentrating on an outcome of:
    1 0 0 x
    0 1 0 y
    0 0 1 z
    5. If I'm able to get x, y or z separately by itself I can use regular algebra to solve for the rest. (Ex: I have 1 0 0 x but the rows of y and z still don't look as they should. I can simply plug in what x is with the linear equations given above to solve for y and z.)

    By this point, I don't understand the logical steps that should be taken when solving a problem. In other words, when I see examples I do not get why someone is choosing to multiply a row or subtract a row. Could someone step me through how I would solve:

    2x-y+4z=-3
    x-4z=5
    6x-y+2z=10

    Thanks.
    Last edited by riker; March 6th 2008 at 10:34 PM.
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  2. #2
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    Westwood, Los Angeles, CA
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    I can try - the best way to do these is to just practice - you'll start to get a more intuitive sense of what works and what doesn't. You start with:

    2 -1 4 -3
    1 0 -4 5
    6 -1 2 10

    If you can get two rows with 0s in the same column, it's often helpful. Since I have a 0 in R2, C2, I'm going to get a 0 in R3, C2. New R3 = R3 - R1. Note that the ONLY row I'm actually changing is R3, so R1 and R2 get copied exactly as they are.

    2 -1 4 -3
    1 0 -4 5
    4 0 -2 13

    Now, I can get another zero in R3 using a combination of R2. I'll get rid of the 4 in R3. New R3 = R3 - 4R2. Again, R1 and R2 don't change.

    2 -1 4 -3
    1 0 -4 5
    0 0 14 -7

    Let's finish R3 by dividing by 14.

    2 -1 4 -3
    1 0 -4 5
    0 0 1 -.5

    Now, using R3, I can easily eliminate anything I want in column 3 without messing anything else up. Start by cleaning up R2. New R2 = R2 + 4R3.

    2 -1 4 -3
    1 0 0 3
    0 0 1 -.5

    Now, let's clean up R1. New R1 = R1 - 4R3.

    2 -1 0 -1
    1 0 0 3
    0 0 1 -.5

    Clean up the first position in R1. New R1 = R1 - 2R2.

    0 -1 0 -7
    1 0 0 3
    0 0 1 -.5

    Rearrange and multiply the current R1 by -1.

    1 0 0 3
    0 1 0 7
    0 0 1 -.5

    So x = 3, y = 7, z = -.5

    Let's check:

    2x - y + 4z = -3

    2(3) - 7 - 2 = 3. Check.

    x - 4z = 5

    3 + 2 = 5 Check.

    6x - y + 2z = 10

    6(3) - 7 - 1 = 10 Check.

    Important things to remember:

    Getting two zeros in the same column is incredibly useful because you can combine those rows to get a zero in an additional column without messing up what you've done. Once you've got a row with two zeros in it, you can pretty quickly clean up the rest. So, aim to get a zero in the same position in two different rows.

    Also, as I learned while doing this, copy the problem down correctly.

    Hope this helps!
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