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Math Help - Help solving linear equation system

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    52

    Help solving linear equation system

    Problem:
    Using only elementary row operations, solve the following system of linear equations:

    {3x - y + 5z + 4w = -11}
    {-2x + y -3z -3w = 8}

    My steps: I keep going in circles. I can't seem to get rid of enough variables to make a dent in this.

    R1 +R2 -> R1 {x + 2z + w = -3}
    {-2x + y -3 z - 3w = 8}

    R2 + 2R1 -> R2 {x +2z +w = -3}
    {y + z - w = 2}

    R1 - 2R2 -> R1 {x - 2y + 3w = -1}
    {y + z - w = 2}

    2R2 -> R2 {x - 2y + 3w = -1}
    {2y + 2z + 2w = 4}

    R1 + R2 -> R1 {x + 2z + w = 3}
    {2y + 2z - 2w = 4}

    R2 - R1 -> R2 {x + 2z + w = 3}
    {-x + 2y = 1}

    R1 + R2 -> R2 {x + 2z + w = 3}
    {2y + 2z + w = 2}

    Do you see how I'm going in circles? Am I even taking the correct first step?
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  2. #2
    Member
    Joined
    Aug 2008
    Posts
    80
    Hello,

    You should know what you mean by "solving" the system of linear equations.
    Since there are only 2 equations for 4 variables,
    the space of solutions is 2-dimensional.
    (If there were 2 equations for 2 variables, the solution space would often be 0-dimensional, yielding only one solution.)
    In this case, the solution space is parametrized by 2 variables.
    Assuming we take z and w for parameters,
    you are asked to express (x, y, z, w) in terms of z and w.

    R1 +R2 -> R1
    {x + 2z + w = -3}
    {-2x + y -3 z - 3w = 8}

    R2 + 2R1 -> R2
    {x +2z +w = -3}
    {y + z - w = 2}

    Thus,
    (x, y, z, w)=(-2z-w-3, -z+w+2, z, w)
    is the solution.
    The solution space is
    {(-3, 2, 0, 0) + z(-2, -1, 1, 0) + w(-1, 1, 0, 1)| z, w: arbitrary}.

    Bye.
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  3. #3
    Junior Member
    Joined
    Aug 2008
    Posts
    52
    Thank you so much. I didn't even put 2 and 2 together to realize that this required a parametric solution. Thanks again!
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