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Homogeneous system of linear equation:

  1. Aug 30, 2005 #1
    I am having trouble finding the solution to the homogeneous system of linear equations:
    2x-2y+z=0
    -2x+y+z=0
     
  2. jcsd
  3. Aug 30, 2005 #2
    I guess I should have also put:

    How can I break this down?
     
  4. Aug 30, 2005 #3

    FredGarvin

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    Perhaps I have been out of the rigid math stuff for a while, but how are you going to solve a system with three unknowns and only two equations?
     
  5. Aug 30, 2005 #4

    TD

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    Choose one variable (e.g. z) as "t" and solve the system for x and y in function of t.
    You'll get an infinite number of solutions, for every t (so z), you have a couple (x,y).
     
  6. Aug 30, 2005 #5
    Hence my problem..
    The answers to choose from are:
    x=3/4t, y=-t,z=1/2t
    x=-3/4t,y=-t,z=1/2t
    z=-3/4t,y=t,z=1/2t
    z=3/4t,y=t,z=1/2t

    This is why I am stumped.
     
  7. Aug 30, 2005 #6

    TD

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    In your case, y was substituted for t. Then solve it as if t was a parameter for x and z.
     
  8. Aug 30, 2005 #7
    ? I am still confused.
     
  9. Aug 30, 2005 #8

    TD

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    You start with

    [tex]\left\{ \begin{gathered}
    2x - 2y + z = 0 \hfill \\
    - 2x + y + z = 0 \hfill \\
    \end{gathered} \right[/tex]

    Substitute y = t, t is now a parameter, and solve the following (2x2)-system for x and z

    [tex]\left\{ \begin{gathered}
    2x + z = 2t \hfill \\
    - 2x + z = - t \hfill \\
    \end{gathered} \right[/tex]
     
  10. Aug 30, 2005 #9

    HallsofIvy

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    You might also want to review your list of possible answers. "x" got changed to "z" in some of them!

    (Am I the only person who hates multiple choice questions in mathematics?)
     
  11. Aug 30, 2005 #10

    TD

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    (No :yuck:)
     
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