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System of Linear Equations

  1. Jun 13, 2010 #1
    1. The problem statement, all variables and given/known data

    This is from a solved problem:

    So, how did they get this solution? I tried solving this and I just don't get it!

    3. The attempt at a solution

    The matrix corresponds to the set of equations:

    x=0
    -x=0
    3x-5y=0

    If I take "y" to be the free variable, I have x=5/3y therefore the solution to the system will be

    [tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {5/3 \\ 1\\0 \end{bmatrix} y + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

    But if I take "x" to be the free variable I get y=3/5x, so I get the following solution:

    [tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {1 \\ 3/5\\0 \end{bmatrix} x + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

    How do I know which one (x or y) should be taken as the free variable?

    Could anyone please explain to me how to get the right answer? :frown: I just can't see how to deal with this type of problems.
     
  2. jcsd
  3. Jun 13, 2010 #2

    Mark44

    Staff: Mentor

    From these equations, x is obviously 0. With x = 0, the 3rd equation implies that y = 0. This means that z is the free variable.
     
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