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Math Help - MAtrix Problem using Guassian Method

  1. #1
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    MAtrix Problem using Guassian Method

    Hi

    Can someone tell me what should be the next step:
    I have tried many ways but cannot seem to get the right answer.
    This is what i have done:

    \begin{bmatrix}<br />
1 & 2 & 1 & | & 6 \\<br />
2 & -1 & -1 & |& -1\\<br />
3 & 2 & 1 & | & 8<br />
\end{bmatrix}

    2R1-R2 -> R2
    3R1-R3 -> R3

    \begin{bmatrix}<br />
1 & 2 & 1 & | & 6 \\<br />
0 & 6 & 3 & |& 13 \\<br />
0 & 4 & 2 & | & 10<br />
\end{bmatrix}

    P.S
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  2. #2
    A Plied Mathematician
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    Well, on a computer, you wouldn't multiply row 2 by 1/6, but when doing things by hand, that's what I would do next. It makes it slightly easier to contemplate the next step of getting a zero in the 3,2 position.
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  3. #3
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    These are linear algebra questions! Stop posting these in Advanced Applied Math!
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  4. #4
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    Quote Originally Posted by Ackbeet View Post
    These are linear algebra questions! Stop posting these in Advanced Applied Math!
    But shouldn't this be posted here, i was told by mod, that on one of my previous post on matrix it was in pre-algebra section, however in fact it should of been in Applied Math.
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  5. #5
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    The pre-algebra forum is in the Pre-University Math Help forum. I guarantee you that your problems would fit correctly in the University Math Help/Linear and Abstract Algebra subforum.
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  6. #6
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    After multiplying by 1/6 to row 2, i then done the following, however i get stuck again because i end up with:

    a+2b+c = 6
    b+\frac{1}{2}c = \frac{13}{6}
    a=1

    Do you have any suggestions of what i should do instead?

    \begin{bmatrix}<br />
1 & 2 & 1 & | & 6 \\<br />
0 & 1 & \frac{1}{2} & | & \frac{13}{6} \\<br />
0 & 4 & 2 & | & 10<br />
\end{bmatrix}

    R3-2R1 -> R3

    \begin{bmatrix}<br />
1 & 2 & 1 & | & 6 \\<br />
0 & 1 & \frac{1}{2} & | & \frac{13}{6} \\<br />
-2 & 0 & 0 & | & -2<br />
\end{bmatrix}
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  7. #7
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    Hmm. I double-checked your earlier work, and I think you made a mistake. Your original system is this:

    <br />
\left[\begin{matrix}1&2&1\\2&-1&-1\\3&2&1\end{matrix}\left|\begin{matrix}6\\-1\\8\end{matrix}\right].<br />

    After performing the following elementary row operations, as you suggested:

    2R1-R2 -> R2
    3R1-R3 -> R3,

    I get this:

    <br />
\left[\begin{matrix}1&2&1\\0&5&3\\0&4&2\end{matrix}\left  |\begin{matrix}6\\13\\10\end{matrix}\right].<br />

    The reason I thought there was an error, was because the original matrix definitely has a solution, but your simplified matrix after those two row operations was inconsistent. So, at this point, I would divide row 2 by 5. What do you get then?
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  8. #8
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    This is want i got when i divided by 5:

    \begin{bmatrix}<br />
1 & 2 & 1 & | & 6\\<br />
0 & 1 & \frac{3}{5} & | & \frac{13}{5}\\<br />
0 & 4 & 2 & | & 10<br />
\end{bmatrix}

    R3-4R2 -> R3

    \begin{bmatrix}<br />
1 & 2 & 1 & | & 6\\<br />
0 & 1 & \frac{3}{5} & | & \frac{13}{5}\\<br />
0 & 0 & \frac{-2}{5} & | & \frac{-2}{5}<br />
\end{bmatrix}

    a+2b+c=6
    a=1

    b+\frac{3}{5}
    c=\frac{13}{5}
    c=\frac{-2}{5}*\frac{-5}{21}
    c=1

    b=\frac{13}{5}-\frac{3}{5}
    b=2
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  9. #9
    Senior Member yeKciM's Avatar
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    yes it's correct u always when u get some results just put them in your starting system of equations and see is it solution that u get the right one

    edit: I apologize, I'm not gonna do that again
    Last edited by yeKciM; July 24th 2010 at 09:21 AM. Reason: Ackbeet post :D
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  10. #10
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    Concur with yeKciM. Incidentally, yeKciM: on this forum it's considered bad form to use anything but full and correct English. Please don't use chat-lingo or texting abbreviations! Thanks! (For reference, look at Rule #12 here.)
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