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Math Help - Matrix Question

  1. #1
    Senior Member Stroodle's Avatar
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    Matrix Question

    The point of intersection of the lines 2x+3y=a and x-2y=5 is (2,1). Use a matrix method to find a.

    Thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Stroodle View Post
    The point of intersection of the lines 2x+3y=a and x-2y=5 is (2,1). Use a matrix method to find a.

    Thanks
    You need to apply row reduction methods to this matrix:

    \left[\begin{array}{cc|c}2&3&a\\1&-2&5\end{array}\right]\xrightarrow{R_1\leftrightarrow R_2}{}\left[\begin{array}{cc|c}1&-2&5\\2&3&a\end{array}\right]\xrightarrow{-2R_1+R_2\rightarrow R_2}{}\left[\begin{array}{cc|c}1&-2&5\\0&7&a-10\end{array}\right] \xrightarrow{\tfrac{1}{7}R_2\rightarrow R_2}\left[\begin{array}{cc|c}1&-2&5\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]\xrightarrow{2R_2+R_1\rightarrow R_1}{}\left[\begin{array}{cc|c}1&0&5+\tfrac{2}{7}\left(a-10\right)\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]

    Since the intersection point is (2,1), its matrix representation would be \left[\begin{array}{cc|c}1&0&2\\0&1&1\end{array}\right], it follows that \left[\begin{array}{cc|c}1&0&5+\tfrac{2}{7}\left(a-10\right)\\0&1&\tfrac{1}{7}\left(a-10\right)\end{array}\right]=\left[\begin{array}{cc|c}1&0&2\\0&1&1\end{array}\right]\implies \tfrac{1}{7}\left(a-10\right)=1 and 5+\tfrac{2}{7}\left(a-10\right)=2

    However, if I solve the first equation for a, I get a=17, but when I solve the second equation for a, I get a=-\frac{1}{2}....

    Did you copy the intersection pt correctly?
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  3. #3
    Senior Member Stroodle's Avatar
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    Awesome. Thanks for that. The text says the intersection point is (2,1) and that the answer is a=17, but I also get a=-\frac{1}{2}...
    Last edited by Stroodle; July 1st 2009 at 01:31 AM.
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  4. #4
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    The problem is that the second line doesn't even pass through the point of intersection!
    f(x,y)=x-2y,\ \ f(2,1)=0,\ \ 0\neq5
    Basically I think the second equation should be x-2y=0 and then as far as I can see a=7 is the answer, which makes sense given that if you plug (2,1) into the first line equation you get ... 7. No real need for matrices here I fear...
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