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Math Help - Determinant

  1. #1
    Member roshanhero's Avatar
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    Determinant

    Prove that
    \begin{vmatrix} cos\theta & 1 & 0 \\1 & 2cos\theta & 1 \\0 & 1 & 2cos\theta\end{vmatrix} =cos3\vartheta
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  2. #2
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    I'll write x for theta.
    If we expand along the third row, the determinant is
    -cos(x)+2cos(x)(2(cos(x))^2-1)=4(cos(x))^3-3cos(x)=cos(x)(4(cos(x))^2-3)=cos(x)(2(2(cos(x))^2-1)-1)=cos(x)(2cos(2x)-1)=2cos(2x)cos(x)-cos(x)=cos(2x+x)=cos(3x)
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  3. #3
    Member roshanhero's Avatar
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    I want to solve this problem by using the properties of determinants.
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  4. #4
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    Quote Originally Posted by roshanhero View Post
    I want to solve this problem by using the properties of determinants.
    The I suggest you do elementary row operations to get the determinant into upper triangular form and then apply the appropriate theorems to calculate the determinant.
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  5. #5
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    Quote Originally Posted by roshanhero View Post
    Prove that
    \begin{vmatrix} cos\theta & 1 & 0 \\1 & 2cos\theta & 1 \\0 & 1 & 2cos\theta\end{vmatrix} =cos3\vartheta
    You can do it by makin' R_1-R_2\cos\theta and then expand by minors the first column.

    Quote Originally Posted by roshanhero View Post
    I want to solve this problem by using the properties of determinants.
    If you want to solve it, do it, show some progress but you're gonna have to change the way on asking your questions.
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