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Basic Augmented Matrix Problem

  1. Nov 1, 2005 #1
    I have an augmented matrix:

    1 2 3 4
    0 3 4 5
    3 12 1 2

    Now this matrix simplifies to:

    1 2 3 4
    0 3 4 5
    0 0 0 0

    So there are infinite solutions, however I have to write all the solutions to the equation. I wanted to do this with vectors.

    So I first solved for x and for y and got:

    y = (4/3)z - 5/3

    x = -2((4/3)z - 5/3) -3z + 4

    So now what do I do, can I simply write the answer as:

    <-2((4/3)z - 5/3) -3z + 4, (4/3)z -5/3, z>

    which then goes to

    z<-2(4/3) - 3, 4/3, 1> + <-2(-5/3) +4, 5/3, 0>

    ???

    This doesn't seem correct to me though. Did I solve the equation correctly, and if so, is what I wrote the correct answer for finding all solutions to the equation?

    Thanks for the help.
     
  2. jcsd
  3. Nov 2, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    You did the reduction wrong. The bottom row does not become 0.
     
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