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Math Help - Complex Numbers?(Find value of determinant)

  1. #1
    Super Member fardeen_gen's Avatar
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    Complex Numbers?(Find value of determinant)

    Find the value of the determinant:
    \left|\begin{array}{ccc}e^{2iA}&e^{-iC}&e^{-iB}\\e^{-iC}&e^{2iB}&e^{-iA}\\e^{-iB}&e^{-iA}&e^{2iC}\end{array}\right|
    where A, B, C are the angles of a triangle.

    EDIT: I corrected the typo
    Last edited by fardeen_gen; May 25th 2009 at 11:08 AM. Reason: TYPO!
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  2. #2
    MHF Contributor red_dog's Avatar
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    If d is the determinant, then

    d=e^{2i(A+B+C)}+e^{-i(A+B+C)}+e^{-i(A+B+C)}-1-1-1=

    =e^{4\pi i}+2e^{-\pi i}-3=1-2-3=-4
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  3. #3
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    \left|\begin{array}{ccc}e^{2iA}&e^{-iC}&e^{-iB}\\e^{{\color{red}-}i{\color{red}C}}&e^{2iB}&e^{-iA}\\e^{-iB}&e^{-iA}&e^{2iC}\end{array}\right|<br />
    I think there's a typo in red. (which red_dog considered...)

    Sarrus' rule gives the value :

    e^{2i(A+B+C)}+e^{i(-C-B-A)}+e^{-i(A+B+C)} \quad - \quad e^{i(2B-B-B)}-e^{i(2C-C-C)}-e^{i(2A-A-A)}

    A+B+C=\pi and recall these values : e^{i\pi}=e^{-i\pi}=-1, e^{2i\pi}=1

    --> =1-1-1-1-1-1=-4


    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Another approach.

    Multiply the first row by e^{i(B+C)}, the second row by e^{i(A+C)}, the third row by e^{i(A+B)}.

    This gives (implicitly using e^{i(A+B+C)}=e^{i\pi}=-1) :

    D=e^{-i(B+C)-i(A+C)-i(A+B)} \begin{vmatrix} -e^{iA} & e^{iB} & e^{iC} \\ e^{iA} & -e^{iB} & e^{iC} \\ e^{iA} & e^{iB} & -e^{iC} \end{vmatrix}=e^{-i(A+B+C)}\begin{vmatrix} -1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}
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