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Math Help - [SOLVED] Linear Equation

  1. #16
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    Quote Originally Posted by ronaldo_07 View Post
    \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]<br />
\left[\begin{array} {cc} x \\ y \end{array}\right]=<br />
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]

    \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]^{-1} \left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]= \left[\begin{array} {cc} x \\ y \end{array}\right]

    \frac{1}{17-i}\left[\begin{array} {cc} 3-2i&3i \\ -3i&2+i \end{array}\right]=\left[\begin{array} {cc} x \\ y \end{array}\right]

    After this it gets really messy and I am unable to get a solution.
    The determinant is not 17-i. You need to re-do that part of the calculation (or post your working so that the mistake can be located). Once you have the correct inverse you have to do the matrix multiplication. This is not dificult - it is simply tedious and it's your job to do it (however, I am willing to check your answer once you get it).
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  2. #17
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    I have got the determinant to be -1-i

    and for 1/(-1-i) =-1+i? im unsure if thats correct

    \left[\begin{array} {cc} -1+5i&-3+3i \\ 3+3i&-3+i \end{array}\right]<br />
\left[\begin{array} {cc} 3+i \\ 3-3i \end{array}\right]=<br />
\left[\begin{array} {cc} x \\ y \end{array}\right]

    How would I do matrix multiplication for this 2x2 x 1x1
    Last edited by mr fantastic; January 17th 2009 at 07:08 PM. Reason: Merged posts
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  3. #18
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    Quote Originally Posted by ronaldo_07 View Post
    I have got the determinant to be -1-i

    and for 1/(-1-i) =-1+i? im unsure if thats correct
    \frac{1}{-1-i} = \frac{1}{2} \, (-1 + i).

    It should be:
    Quote Originally Posted by ronaldo_07 View Post
    {\color{red}\frac{1}{2} } \left[\begin{array} {cc} -1+5i&-3 {\color{red}-} 3i \\ 3+3i&-3+i \end{array}\right]<br />
\left[\begin{array} {cc} 3+i \\ 3-3i \end{array}\right]=<br />
\left[\begin{array} {cc} x \\ y \end{array}\right]

    How would I do matrix multiplication for this 2x2 x 1x1
    Now multiply the two matrices in the usual way using the usual arithmetic of complex numbers.
    Last edited by mr fantastic; January 17th 2009 at 07:08 PM. Reason: Merged posts
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  4. #19
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    <br />
\frac{1}{2}\left[\begin{array} {cc} -1+5i&-3-3i \\ 3+3i&-3+i \end{array}\right]=\left[\begin{array} {cc} (\frac{-1+5i}{2})(3+i)&(\frac{-3-3i}{2})(3+i) \\ (\frac{3+3i}{2})(3-3i)&(\frac{-3+i}{2})(3-3i) \end{array}\right]=<br />
\left[\begin{array} {cc} x \\ y \end{array}\right]<br />

    Is this correct or am I multiplying it the wrong way because i still dont seem to be getting the correct answer when I sub it in.
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  5. #20
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    Quote Originally Posted by ronaldo_07 View Post
    <br />
\frac{1}{2}\left[\begin{array} {cc} -1+5i&-3-3i \\ 3+3i&-3+i \end{array}\right]=\left[\begin{array} {cc} (\frac{-1+5i}{2})(3+i)&(\frac{-3-3i}{2})(3+i) \\ (\frac{3+3i}{2})(3-3i)&(\frac{-3+i}{2})(3-3i) \end{array}\right]=<br />
\left[\begin{array} {cc} x \\ y \end{array}\right]<br />

    Is this correct or am I multiplying it the wrong way because i still dont seem to be getting the correct answer when I sub it in.
    This is not correct. Do you know how to multiply two matrices? You need to review that ... Read this: Scalar and Matrix Multiplication

    Then try doing the multiplication again.
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  6. #21
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    \frac{1}{2}\left[\begin{array} {cc} -1+5i \\ 3+3i \end{array}\right]=\left[\begin{array} {cc} (\frac{-1+5i}{2})&(\frac{-3-3i}{2}) \\ (\frac{3+3i}{2})&(\frac{-3+i}{2})\end{array}\right]

    I did the multiplication like this. after this i multiplied this matix to

    \left[\begin{array} {cc} 3+i \\ 3-3i \end{array}\right]

    That is how came to the previous answer.
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  7. #22
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    Quote Originally Posted by ronaldo_07 View Post
    \frac{1}{2}\left[\begin{array} {cc} -1+5i \\ 3+3i \end{array}\right]=\left[\begin{array} {cc} (\frac{-1+5i}{2})&(\frac{-3-3i}{2}) \\ (\frac{3+3i}{2})&(\frac{-3+i}{2})\end{array}\right]

    I did the multiplication like this. after this i multiplied this matix to

    \left[\begin{array} {cc} 3+i \\ 3-3i \end{array}\right]

    That is how came to the previous answer.
    The way you multiplied the two matrices is wrong. Review matrix multiplication like I said.
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  8. #23
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    Quote Originally Posted by ronaldo_07 View Post
    \frac{1}{2}\left[\begin{array} {cc} -1+5i \\ 3+3i \end{array}\right]=\left[\begin{array} {cc} (\frac{-1+5i}{2})&(\frac{-3-3i}{2}) \\ (\frac{3+3i}{2})&(\frac{-3+i}{2})\end{array}\right]
    So this part is correct? I will try the other part again
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  9. #24
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    Quote Originally Posted by ronaldo_07 View Post
    So this part is correct? I will try the other part again
    Go back to the correct matrix equation in post #18 and do the multiplication again - after you have reviewed matrix multiplication.
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  10. #25
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    Thanks I have managed to solve succesfully and I have checked my answer by susbtitutiong to the origional equation'
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  11. #26
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    Quote Originally Posted by ronaldo_07 View Post
    Thanks I have managed to solve succesfully and I have checked my answer by susbtitutiong to the origional equation'
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