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Math Help - Find all values of K for which the system of linear equations..

  1. #1
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    Find all values of K for which the system of linear equations..

    Three planes have the following equations...

    x+2y+z = 3
    3x+4y+5z=k
    2x-y+kz=1

    \[<br />
\emph{Augumented Matrix} =<br />
\left( {\begin{array}{cccc}<br />
 1 & 2 & 1 & 3  \\<br />
 3 & 4 & 5 & k \\<br />
 2 & -1 & k & 1<br />
 \end{array} } \right)<br />
\]

    \[<br />
\emph{Reduced - Row echelon form} =<br />
\left( {\begin{array}{cccc}<br />
 1 & 0 & 0 & k+\frac{3}{2}  \\<br />
 0 & 1 & 0 & \frac{-k}{2}+2 \\<br />
 0 & 0 & 1 & \frac{-5}{2}<br />
 \end{array} } \right)<br />
\]

    \emph{Find all values of k for which the system of linear equations}

    \emph{(i) has infinitely many solutions}
    \emph{(ii) has a unique solution}
    \emph{(iii) has no solutions}

    Not sure on how to approach this question..
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  2. #2
    A Plied Mathematician
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    I would work with determinants. Check this thread out. That might give you a few ideas. Make sure to read all the way to the end, if you're interested in a rigorous proof, as you can see that my first idea was invalid.
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  3. #3
    MHF Contributor

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    Quote Originally Posted by Oiler View Post
    Three planes have the following equations...

    x+2y+z = 3
    3x+4y+5z=k
    2x-y+kz=1

    \[<br />
\emph{Augumented Matrix} =<br />
\left( {\begin{array}{cccc}<br />
 1 & 2 & 1 & 3  \\<br />
 3 & 4 & 5 & k \\<br />
 2 & -1 & k & 1<br />
 \end{array} } \right)<br />
\]
    This can't be right. In reducing the first two columns, you will get a term involving k in the third column of the third row (I get "k- 6"), then have to divide by that term. You should have a "k" in the denominator.
    \[<br />
\emph{Reduced - Row echelon form} =<br />
\left( {\begin{array}{cccc}<br />
 1 & 0 & 0 & k+\frac{3}{2}  \\<br />
 0 & 1 & 0 & \frac{-k}{2}+2 \\<br />
 0 & 0 & 1 & \frac{-5}{2}<br />
 \end{array} } \right)<br />
\]

    \emph{Find all values of k for which the system of linear equations}

    \emph{(i) has infinitely many solutions}
    \emph{(ii) has a unique solution}
    \emph{(iii) has no solutions}

    Not sure on how to approach this question..
    Last edited by HallsofIvy; February 27th 2011 at 11:14 AM.
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  4. #4
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    I solved this using Maxima and its answer agrees with Oiler's reduced echelon form.

    Just use that form to get expressions for your variables, Oiler. You can read them right off. Looking at the last row, what's z? Then find x and y similarly (these depend on k). Is there any k that won't give an answer, or will give more than one answer?
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  5. #5
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  6. #6
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    Oops, I answered without thinking this through. It looks like those programs cancel factors in the denominators. If it's reduced manually then there are ways to get more than one solution.
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