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How to solve a system with more unknowns than equations?

  1. May 18, 2014 #1
    Because if you reduce to Row-Echelon form you don't end up with a triangulated form. Here is what I ended up with (How do I use Latex Reference to construct a Matrix? Can someone help me with that by the way please?)

    1 -1 3 2 | 1
    0 1 1 3 | -1
    0 0 1 1 | -1
     
  2. jcsd
  3. May 18, 2014 #2

    micromass

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    Here is the LaTeX code:

    [tex]\left(\begin{array}{cccc|c}
    1 & -1 & 3 & 2 & 1\\
    0 & 1 & 1 & 3 & -1\\
    0 & 0 & 1 & 1 & -1
    \end{array}\right)[/tex]

    See our LaTeX tutorial: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

    Anyway, to the math. In cases like this, you will not get a unique solution, you will get infinitely many solutions which depend on a free parameter. Let's first transfrom the system back to the usual form:

    [tex]\left\{
    \begin{array}{l}
    x - y + 3z + 2t = 1\\
    y + z + 3t = -1\\
    z + t = -1
    \end{array}
    \right.[/tex]

    The idea is to write all variables in function of ##t## (or ##x## or whatever, it doesn't matter which one). We can certainly write ##z## in function of ##t## as follows:

    [tex]z = -1 - t[/tex]

    Now, we know ##y + z + 3t = -1##, we can substitute the above version of ##z## into this equation and get

    [tex]y + (-1 -t) + 3t = -1[/tex]

    Thus we can write ##y## in function of ##t##:

    [tex]y = -2t[/tex]

    Finally, we can do the same for ##x##, we have the equation ##x - y + 3z + 2t = 1##. Substituting ##y## and ##z##, we get

    [tex]x - (-2t) + 3(-1-t) + 2t = 1[/tex]

    Solving for ##x## gives us

    [tex]x = 4-t[/tex]

    So we get the following solutions for the system:

    [tex]x= 4-t,~y=-2t,~z=-1-t[/tex]

    This means that for every real number ##t## that we take, we get a solution for our system. For example, taking ##t=0##, we get ##(x,y,z,t) = (4,0,-1,0)##. On the other hand, taking ##t=1##, we get ##(x,y,z,t) = (3,-2,0,1)## as a solution. So we see that we have infinitely many solutions, one for each ##t##.
     
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