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Gauss-Jordan Elimination

  1. Dec 14, 2015 #1
    By using row operation, how can

    ##
    \left(
    \begin{array}{rrrr|r}
    2 & 0 & -1 & 0 & 0 \\ 6 & 0 & 0 & -2 & 0 \\ 0 & 2 & -2 & -1 & 0
    \end{array}
    \right)
    ##

    be

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & 0 & -\frac{1}{3} & 0 \\ 0 & 1 & 0 & -\frac{7}{6} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0
    \end{array}
    \right)
    ##

    ???

    I am trying to write my attempt on this problem but I don't know how to write matrix with row operation in Latex. Please show me example.
     
  2. jcsd
  3. Dec 14, 2015 #2

    BvU

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    Hi askor, :welcome:

    There's a link to the LateX Tutorial in the homework guidelines and an extensive one in general math.
    Is there a speecific question you need answered ?

    Your matrix looks rather straightforward to me:
    Subtract 1/3 x row 2 from row 1 to get 2 x new row 3.
    Add row 1 to new row 3 to get 2 x new row 1
    Add 2 x new row 3 to row 3 to get 2 x new row 2

    But I expect you already found this ? If so where do you get stuck ?

    --
     
  4. Dec 14, 2015 #3

    HallsofIvy

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    Mod note: Corrected what appeared to be a typo in the second vector, below. I'm almost certain that the corrected version was what was intended.
    It should be obvious, seeing [itex]\begin{pmatrix}0 \\ 0 \\ 2 \end{pmatrix}[/itex] in the middle column of the original matrix and [itex]\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}[/itex] in the result you want, that you will want to swap the second and third rows. If you do that as the first step, you get [tex]\left(\begin{array}{cccc|c} 2 & -1 & 0 & 1 & 0 \\ 0 & 2 & -2 & -1 & 0 \\ 6 & 0 & 0 & -2 & 0 \end{array}\right)[/tex].

    The rest should be easy.
     
    Last edited by a moderator: Dec 14, 2015
  5. Dec 14, 2015 #4
    Could you write it in matrix rather than words? I am confuse.

    My attempt is:

    ##
    \left(
    \begin{array}{rrrr|r}
    2 & 0 & -1 & 0 & 0 \\ 6 & 0 & 0 & -2 & 0 \\ 0 & 2 & -2 & -1 & 0
    \end{array}
    \right)
    ##

    R1 divided by 2, R2 swap to R3, then I get

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 2 & -2 & -1 & 0 \\ 6 & 0 & 0 & -2 & 0
    \end{array}
    \right)
    ##

    R2 divided by 2, then I get

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 6 & 0 & 0 & -2 & 0
    \end{array}
    \right)
    ##

    R3 - 6R1, then I get lower triangular matrix

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 3 & -2 & 0
    \end{array}
    \right)
    ##

    Why the result is different?
     
    Last edited: Dec 14, 2015
  6. Dec 14, 2015 #5

    BvU

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    Which one is confusing you ?

    Subtract 1/3 x row 2 from row 1 to get 2 x new row 3 is so clear to me that I don't know how to simplify it!
    Perhaps:

    Subtract 1/3 x row 2 from row 1 to and divide the resulting row by -2 to get new row 3 ​
    ?
     
  7. Dec 15, 2015 #6
    Could you write it in my style as above.
     
  8. Dec 15, 2015 #7

    BvU

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    I could, but if it is unclear the way I write it now, you still have a problem. What is it that makes it undecipherable for you ?
     
  9. Dec 15, 2015 #8
    You said "Subtract 1/3 x row 2 from row 1..."

    If written by R1 = row 1, R2 = row 2, which one of below is correct according your above statement?

    1/3*R2 - R1

    or

    R1 - 1/3*R2
     
  10. Dec 15, 2015 #9

    BvU

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    R1 - 1/3*R2

    Gives you ## (0,0,-1, {2\over 3}) ## which is minus the new row 3. (new meaning: as in the final answer)

    So not +2 or -2 x new row 3 as I stupidly typed - twice . I fixed the - sign in post #5 and now I fix the 2 that should have been a 1. o:)

    Sorry for the mistakes. No wonder you get confused.
     
  11. Dec 15, 2015 #10
    OK, R1 - 1/3*R2 give me

    ##
    \left(
    \begin{array}{rrrr|r}
    0 & 0 & -1 & \frac{2}{3} & 0 \\ 6 & 0 & 0 & -2 & 0 \\ 0 & 2 & -2 & -1 & 0
    \end{array}
    \right)
    ##

    Now, what is the next step?
     
  12. Dec 15, 2015 #11

    BvU

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    From what you post now (#10): line 3 minus 2 x line 1 is ## (0,2,0,-7/3)##, then divide by 2 to get line 2 in the answer in post #1.
     
  13. Dec 15, 2015 #12
    Why the result is different with the post #4?
     
  14. Dec 16, 2015 #13

    Svein

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    In that post you have not finished the computations!
     
  15. Dec 16, 2015 #14
    How to finish it? Please tell step-by-step.

    Here is my step:

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 3 & -2 & 0
    \end{array}
    \right)
    ##

    R3 / 3

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0
    \end{array}
    \right)
    ##

    The matrix is already in form of Lower triangular matrix, how to obtain the upper triangular matrix?

    What next?
     
  16. Dec 16, 2015 #15

    Mark44

    Staff: Mentor

    Use the bottom row (3rd column entry) to eliminate the entries directly above it.

    Some notation I have seen is this:
    ##R_2## ← ##R_2 + R_3##
    ##R_1## ← ##R_1 + \frac 1 2 R_3##
    The arrow, ←, means "is replaced by".
     
  17. Dec 17, 2015 #16
    OK, I rewrite it with matrix

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0
    \end{array}
    \right)
    ##

    ##R_2## ← ##R_2 + R_3##

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & -\frac{7}{6} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0
    \end{array}
    \right)
    ##

    ##R_1## ← ##R_1 + \frac 1 2 R_3##

    ##
    \left(
    \begin{array}{rrrr|r}
    1 & 0 & 0 & -\frac{2}{6} & 0 \\ 0 & 1 & 0 & -\frac{7}{6} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0
    \end{array}
    \right)
    ##
     
  18. Dec 17, 2015 #17

    BvU

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    Well done !
     
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