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Math Help - Unique Solution

  1. #1
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    Unique Solution

    <br />
2x_{1} - 6ax_{2} = 3<br />
    <br />
3ax_{1} - x_{2} = 3/2<br />

    The book says the answer is x_{1} = 3/2(3a+1) and x_{2} = -3/2(3a+1) but when I zero out 3ax_{1}, I get x_{2} = (-9a-3)/2(9a^2-1).
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  2. #2
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    Quote Originally Posted by Makall View Post
    <br />
2x_{1} - 6ax_{2} = 3<br />
    <br />
3ax_{1} - x_{2} = 3/2<br />

    The book says the answer is x_{1} = 3/2(3a+1) and x_{2} = -3/2(3a+1) but when I zero out 3ax_{1}, I get x_{2} = (-9a-3)/2(9a^2-1).
    Write in matrix form...

    \left[\begin{matrix}2 & -6a\\ 3a & -1\end{matrix}\right] \left[\begin{matrix}x_1\\ x_2 \end{matrix}\right] = \left[\begin{matrix} 3 \\ \frac{3}{2}\end{matrix}\right]


    This is a matrix equation of the form

    A\mathbf{x} = \mathbf{b}.

    Using some matrix algebra...

    A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}

    I\mathbf{x} = A^{-1}\mathbf{b}

    \mathbf{x} = A^{-1}\mathbf{b}.



    A^{-1} = \frac{1}{2(-1) - (-6a)3a}\left[\begin{matrix} -1 & 6a \\ -3a & 2 \end{matrix}\right]

     = \frac{1}{18a^2 - 2}\left[\begin{matrix} -1 & 6a \\ -3a & 2 \end{matrix}\right]


    So \left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right] = \frac{1}{18a^2 - 2}\left[\begin{matrix} -1 & 6a \\ -3a & 2 \end{matrix}\right] \left[\begin{matrix} 3 \\ \frac{3}{2} \end{matrix}\right].


    Now simplify and solve for x_1 and x_2.
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