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Thread: Unique Solution

  1. #1
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    Unique Solution

    $\displaystyle
    2x_{1} - 6ax_{2} = 3
    $
    $\displaystyle
    3ax_{1} - x_{2} = 3/2
    $

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

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

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


    This is a matrix equation of the form

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

    Using some matrix algebra...

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

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

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



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

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


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


    Now simplify and solve for $\displaystyle x_1$ and $\displaystyle x_2$.
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