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Solution space to homogenous equations.

  1. Dec 2, 2008 #1
    Ok so my question was:

    The directions state:

    Find the solution space of the following systems of linear homogeneous equations:


    Is the same as finding the solutions, because I did that and got (-z, -3z, 3z, z). So i said
    the solution space is actually the subspace you get. For instance, this problem yields a one dimensional subspace.
    And someone responded:

    I don't think that solution set is right.

    The vector -1,-3,3,1 does not give you 0 when used as weights in the original equation. It is a system of 4 unknowns in 3 rows there has to be a free variable somewhere. That means when you write your free variable in terms of the other variables, there should be a 0 in that solution set.

    The solution (-1,-3,3,1) does not satisfy ax=0.


    Am i right or am i wrong here?
  2. jcsd
  3. Dec 2, 2008 #2


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    Homework Helper

    I think you made an error somewhere. Substituting your answer into the first equation doesn't give 0. How did you get (-z, 3z, 3z, z) ?
  4. Dec 3, 2008 #3


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    Staff Emeritus
    Science Advisor

    I suggest you try that again. That is not at all what I get.

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