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Simple Gauss-Jordan Elimination

  1. Oct 1, 2009 #1
    I have done the following problem over and over and keep getting the same result. So either the book answer is wrong or I am making the same mistake over again:

    [tex]\left[\begin{array} {cccc}
    5&-2&6&0\\
    -2&1&3&1\\
    \end{array}\right] [/tex]

    [tex]\left[\begin{array} {cccc}
    1&-\frac25&\frac65&0\\
    -2&1&3&1\\
    \end{array}\right] [/tex]

    [tex]\left[\begin{array} {cccc}
    1&-\frac25&\frac65&0\\
    0&\frac15&\frac{17}{5}&1\\
    \end{array}\right] [/tex]

    [tex]\left[\begin{array} {cccc}
    1&0&8&2\\
    0&1&17&5\\
    \end{array}\right] [/tex]

    Am I screwing this up somewhere?
     
  2. jcsd
  3. Oct 1, 2009 #2

    Mark44

    Staff: Mentor

    I ended up with different numbers in the 3rd column.
    [tex]\left[\begin{array} {cccc}
    1&0&12&2\\
    0&1&27&5\\
    \end{array}\right] [/tex]

    In my first step, I added 2 times row 1 to -5 times row 2. Then I added 2 times row 2 to row 1. Finally, I replaced row 1 with 1/5 of itself. It is usually simpler to keep integer values for as long as possible.
     
  4. Oct 1, 2009 #3
    @Saladsamurai, [STRIKE]When you go to eliminate the -2, you want to add row 1 to row 2. You're subtracting it.[/STRIKE]

    3 + 12/5 = 15/5 + 12/5 = 27/5
     
  5. Oct 1, 2009 #4
    Yeah! I just caught that :/ Man.. I must have added 6/5+3=17/5
    like ten times! Stupid brain gets sucked into following
    a pattern

    thanks!! :smile:
     
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