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Matrices - finding a general solution.

  1. Aug 13, 2006 #1

    Agg

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    Howdy,

    I have been asked to find the general solution of the following matrix (pic attached).

    The matrix does not have an inverse, so I am a bit confused guys. Cheers and thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Aug 13, 2006 #2

    HallsofIvy

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    Yes, it does not have an inverse- that's why you are asked for the general solution.

    One way to do this is "row reduction". Set up the augmented matrix
    [tex]\left [ \begin{array} {cccc}1 & 0& -1 & 0 \\0 & -2 & 2 & 0 \\-1 & 1 & 0 & 0 \end{array} \right ][/tex]
    and row-reduce. Because the matrix does not have an inverse, the final row will be all zero's but you could solve for, say x and y in terms of z.

    Or just treat it as a system of equations: x- z= 0, -2y- 2z= 0, -x+ y= 0.
    The first and third just say x= z and y= x= z. The second is then automatically solved. The general solution is (x, y, z)= (z, z, z)where z can be any number.
     
  4. Aug 13, 2006 #3

    mathwonk

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    yiou are given a linear map from 3 space to 3 space and are asked to find all vectors that map to 0. obviously it is the line defined by x=y=z.
     
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