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Reduced a big matrix, now the parametric form is not right, :\

  1. Sep 21, 2005 #1
    Hello everyone I did the following problem:
    Click http://img220.imageshack.us/img220/8486/untitled1copy4oq.jpg to view the problem and my answer.

    The row reduced form is:
    1 5 0 0 -7 6 -7
    0 0 1 0 -1 1 -1
    0 0 0 1 - 2 -4 8

    Any help would be great
     
  2. jcsd
  3. Sep 22, 2005 #2

    TD

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    Your matrix seems to be correct, and is:

    [tex]\left( {\begin{array}{*{20}c}
    1 & 5 & 0 & 0 & { - 7} & 6 &\vline & { - 7} \\
    0 & 0 & 1 & 0 & { - 1} & 1 &\vline & { - 1} \\
    0 & 0 & 0 & 1 & { - 2} & { - 4} &\vline & 8 \\

    \end{array} } \right)[/tex]

    Now we have 3 lineair indepedant equations with 6 variables which means we can 'choose' 3 variables. We keep the variables [tex]x_1[/tex], [tex]x_3[/tex] and [tex]x_4[/tex]. Let [tex]x_2 = s[/tex], [tex]x_5 = t[/tex] and [tex]x_6 = u[/tex].

    This gives the solution of the system:

    [tex]\left\{ \begin{gathered}
    x_1 + 5s - 7t + 6u = - 7 \hfill \\
    x_3 - t + u = - 1 \hfill \\
    x_4 - 2t - 4u = 8 \hfill \\
    \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
    x_1 = - 7 - 5s + 7t - 6u \hfill \\
    x_3 = - 1 + t - u \hfill \\
    x_4 = 8 + 2t + 4u \hfill \\
    \end{gathered} \right[/tex]

    Can you put it in vectorial notation now?
     
  4. Sep 22, 2005 #3
    hm...i don't see how 'im still missing this:
    i entered in:
    -7 -5 7 6
    0 1 0 0
    -1 0 1 -1
    8 0 0 4
    0 0 1 0
    0 0 0 1

    still wrong, thanks for the reply!
     
  5. Sep 23, 2005 #4

    TD

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    I think you for got the 2 in column 3 for [itex]x_4[/itex] since that has "2t) in its solution.
     
  6. Sep 23, 2005 #5
    I'm stilling screwing somthing up...I fixed what you said, but still not the right answer, and I know its row reduced correctly because I checked it with my calculator! here is the picture of what I wrote and submitted:
    Picture Thanks for the help!
     
  7. Sep 23, 2005 #6

    TD

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    I see you entered "6" in the last column for [itex]x_1[/itex] and I think that has to be "-6" (check my last system).
     
  8. Sep 23, 2005 #7
    ah ha! my bad! thanks for the help! :biggrin:
     
  9. Sep 23, 2005 #8

    TD

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    It was correct now? Glad I could help :smile:
     
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