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Am i setting up equations right? Matrices, finding nullspace

  1. Dec 11, 2005 #1
    Hello everyone i'm confused if i'm setting these equations up right:
    i have:
    1 0 1
    0 1 -2
    0 0 0

    so i said:
    x + z = 0;
    y - 2z = 0;
    z = ? because its a whole row of 0's, so u have no info about what z could be
    so i said let z = a;
    x + a = 0;
    y -2a = 0;
    z = a;

    x = a;
    y = 2a;
    z = a;

    so let a = 1;
    my nullspace is:

    am i doing that right?
  2. jcsd
  3. Dec 11, 2005 #2


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    Homework Helper

    You lost a minus sign, it should be x=-a.

    The nullspace is given by all solutions to this system, not a specific one. That is all vectors (-a,2a,a) make up the nullspace. You can look at a specific non-zero one if you're after a basis of the nullspace (since it has dimension one in this case).
  4. Dec 11, 2005 #3
    i c, thank u!! but the synatx i used is correct? -a, 2a, a correct? so if i factorted out an a i would get
    a[-1,21] because my professor got a different answer: he got:

    from this matrix:
    which is a not simplified version of the matrix i gave above:
    2 1 0
    0 -1 2
    0 0 0

    2x + y = 0;
    -y + 2z = 0;
    x = -a/2;
    z = a/2
    y = a;

    null(A) = [-1/2,1/2]
    and i thought he did it wrong.
  5. Dec 11, 2005 #4


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    Strictly as written, null(A)=[-1/2,1,1/2] is wrong, the nullspace is not one vector, but a set of vectors (even if it's just the zero vector).

    null(A)=span{[-1/2,1,1/2]} would be a way to write it.
  6. Dec 11, 2005 #5
    Yep that makes sense to me, thanks shmoe
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