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Math Help - linear algebra

  1. #1
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    Exclamation linear algebra

    [1 2 5 8
    2 3 0 5
    3 0 6 9
    0 4 7 11]

    ive tried 4 times to reduce this but my answers all look different and wrong. any suggestions?
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  2. #2
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    Quote Originally Posted by linearalgebra View Post
    [1 2 5 8
    2 3 0 5
    3 0 6 9
    0 4 7 11]

    ive tried 4 times to reduce this but my answers all look different and wrong. any suggestions?
    Why not show some work. I got this

    \left[ \begin{array}{cccc}<br />
1 & 0 & 0 &1 \\<br />
0 & 1 & 0 &1\\<br />
0 & 0 & 1 & 1\\<br />
0 & 0 & 0 & 0 \end{array}\right]
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  3. #3
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    thanks..im done some row operations and right now my matrix looks like this
    [1 -1 0 0
    -1 7 1 7
    1 0 2 3
    0 0 1 1 ]

    did you have that matrix at one point too, or should i just start over?
    also, I have used about 10 row operations to get to that- is that too much? am i not seeing something really simple that could solve this using fewer row operations?
    thanks.
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  4. #4
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    okay did more operations and now im at the folowing...
    [ 1 0 0 1
    0 1 0 1
    0 0 1 1
    0 0 0 -2]

    how can i change the last row to [0 0 0 0] without too many operations?
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  5. #5
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    would it be valid if i multiplied the entire last row by 0?
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  6. #6
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    Quote Originally Posted by linearalgebra View Post
    thanks..im done some row operations and right now my matrix looks like this
    [1 -1 0 0
    -1 7 1 7
    1 0 2 3
    0 0 1 1 ]

    did you have that matrix at one point too, or should i just start over?
    also, I have used about 10 row operations to get to that- is that too much? am i not seeing something really simple that could solve this using fewer row operations?
    thanks.
    Your row reducing seems rather random. I think it's best to have an order. Starting with

    <br />
\left[ \begin{array}{cccc}<br />
1 & 2 & 5 &8 \\<br />
2& 3 & 0 &5\\<br />
3 & 0 & 6 & 9\\<br />
0 & 4 & 7 & 11 \end{array}\right] target to get rid of the red numbers <br />
\left[ \begin{array}{cccc}<br />
1 & 2 & 5 &8 \\<br />
\color{red}{2}& 3 & 0 &5\\<br />
\color{red}{3} & 0 & 6 & 9\\<br />
0 & 4 & 7 & 11 \end{array}\right]

    <br />
\begin{array}{c} 2R_1 -R_2 \rightarrow R_2\\<br />
3R_1- R_3 \rightarrow R_3<br />
\end{array}<br />

    <br />
\left[ \begin{array}{cccc}<br />
1 & 2 & 5 &8 \\<br />
0 & 1 & 10 &11\\<br />
0 & 6 & 9 & 15\\<br />
0 & 4 & 7 & 11 \end{array}\right] then use the second row to eliminate <br />
\left[ \begin{array}{cccc}<br />
1 & 2 & 5 &8 \\<br />
0 & 1 & 10 &11\\<br />
0 & \color{red}{6} & 9 & 15\\<br />
0 & \color{red}{4} & 7 & 11 \end{array}\right] etc
    Last edited by Jester; March 12th 2009 at 06:15 AM. Reason: typo
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  7. #7
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    according to the way you did it, my next step looks liks this...
    [ 1 2 5 8
    0 1 10 11
    0 0 51 51
    0 0 33 33]

    am i on the right track?
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  8. #8
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    Quote Originally Posted by linearalgebra View Post
    according to the way you did it, my next step looks liks this...
    [ 1 2 5 8
    0 1 10 11
    0 0 51 51
    0 0 33 33]

    am i on the right track?
    Yes, excellent! Now dividing the last two rows by 51 an 33 respectively, we can then start working your way up.
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