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Matrix question

  1. Sep 20, 2005 #1
    Use the Gauss-Jordan algorithm to find all solutions of the following system of linear equations in C:

    x1 + x2 + x3 [] = 2
    2x1 [] + 2x3 + 2x4 = 2
    x1 + 2x2 + 2x3 [] = 1
    2x1 + 2x2 [] + x4 = 2

    [] signify a blank space in the equation. How do you even proceed to do this, I have never seen it. Any help would be appreciated. Thank you.
     
  2. jcsd
  3. Sep 20, 2005 #2

    TD

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    In matrix-form, it looks like this:

    [tex]\left( {\begin{array}{*{20}c}
    1 & 1 & 1 & 0 & 2 \\
    2 & 0 & 2 & 2 & 2 \\
    1 & 2 & 2 & 0 & 1 \\
    2 & 2 & 0 & 1 & 2 \\
    \end{array}} \right)[/tex]

    Do you know how Gaussian elimination works?
     
  4. Sep 20, 2005 #3

    saltydog

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    How about this:

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

    That's the augumented matrix Niteshadw.

    So, reduce it. Here, I'll do the first part. I'll multiply the top row by -2 and then add it to the second row yielding:

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

    Can you continue doing this until you get it to row-reduced form?
     
    Last edited: Sep 21, 2005
  5. Sep 20, 2005 #4

    HallsofIvy

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    Typo!! The first matrix in Saltydog's response should be
    [tex]\left(\begin{array}{cccc|c}1 & 1 & 1 & 0 & 2\\ 2 & 0 & 2 & 2 & 2 \\ 1 & 2 & 2 & 0 & 1 \\ 2 & 2 & 0 & 1 & 2 \end{array}\right)[/tex]

    Except for separating out the "augmenting" part, that's just what TD said.
     
  6. Sep 20, 2005 #5
    Hmm..looks simple enough, just did not know what to do with those blank spots...Ok, I got this,

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

    so um,
    x_4 = -2x_4
    x_3 = 0
    x_2 = -x_2
    x_1 = 3x_1

    are those the solutions?
    Thank you for the help thus far...
     
  7. Sep 21, 2005 #6

    TD

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    I think you row-reduced fine but the last column refers to the constants, not to an unknown. So the solution should be:

    [tex]\left\{ \begin{array}{l}
    x_1 = 3 \\
    x_2 = - 1 \\
    x_3 = 0 \\
    x_4 = - 2 \\
    \end{array} \right[/tex]
     
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