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Math Help - Proving Identity Without Determinants

  1. #1
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    Proving Identity Without Determinants

    Prove the identity without evaluating the determinants

    [a1+b1t a2+b2t a3+b3t] [a1 a2 a3]
    [a1t + b1 a2t+b2 a3t+b3] = (1-t^2) [b1 b2 b3]
    [c1 c2 c3 ] [c1 c2 c3]

    I am not even sure where to begin with this as I do not know the method required to prove the identity.

    Any help would greatly appreciated.

    Thanks

    I'm just editing this to include a picture that should make the problem clearer (also, it's t^2, not t2 in the scalar before the matrix on the RHS of the equation):

    Last edited by Adrian; October 29th 2007 at 04:19 PM.
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  2. #2
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    To prove that \begin{vmatrix}a_1+b_1t&a_2+b_2t&a_3+b_3t\\ a_1t+b_1&a_2t+b_2&a_3t+b_3\\c_1&c_2&c_3\end{vmatri  x} = (1-t^2)\begin{vmatrix}a_1&a_2&a_3\\ b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix} without actually evaluating the determinants, you have to use the properties of elementary row operations applied to determinants. One of these says that you can add a multiple of one row to another row without changing the determinant. (So for example, starting with the matrix on the left side of that equation, you could subtract t times the middle row from the top row.) Another of these properties says that you can take a common factor out of one row ... .
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