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Thread: Imaginary roots factoring help

  1. #1
    Member diddledabble's Avatar
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    $\displaystyle r^{2}-6r+6=0$
    This might result in an $\displaystyle i$ root. I am not sure just stuck.

    This is the determinant of my matrix and I need to find the eigenvalues. No square roots or fractions allowed. Must be a whole number or something like $\displaystyle 2i$
    Last edited by mr fantastic; Jul 29th 2009 at 06:15 PM. Reason: Merged posts
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  2. #2
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    Quote Originally Posted by diddledabble View Post
    $\displaystyle r^{2}-6r+6=0$

    This might result in an $\displaystyle i$ root. I am not sure just stuck.
    $\displaystyle r^{2}-6r+6=0$

    I would complete the square here

    $\displaystyle (r^{2}-6r+9)+6-9=0$

    $\displaystyle (r^{2}-6r+9)-3=0$

    $\displaystyle (r-3)^{2}-3=0$

    $\displaystyle (r-3)^{2}=3$

    Now solving for r will determine the nature of the roots.

    $\displaystyle (r-3)^{2}=3$

    $\displaystyle r-3=\sqrt{3}$

    $\displaystyle r=3\pm\sqrt{3}$

    These look real to me.
    Last edited by mr fantastic; Jul 29th 2009 at 06:13 PM. Reason: Added quote
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by diddledabble View Post
    $\displaystyle r^{2}-6r+6=0$
    This might result in an $\displaystyle i$ root. I am not sure just stuck.
    [I take this to be the characteristic equation from a DE]

    To find r, use the quadratic formula. You should get $\displaystyle r=3\pm\sqrt{3}$.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by diddledabble View Post
    This is the determinant of my matrix and I need to find the eigenvalues. No square roots or fractions allowed. Must be a whole number or something like $\displaystyle 2i$
    What is the matrix you're working with?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by diddledabble
    $\displaystyle \mathbf{x}^{\prime}(t)=\begin{bmatrix}6&-3\\2&1\end{bmatrix}{\color{red}\mathbf{x}(t)}$ with initial value of $\displaystyle \mathbf{x}(0)=\begin{bmatrix}-10\\-6\end{bmatrix}$
    I think you left out an important part (in red)

    Let $\displaystyle \mathbf{A}=\begin{bmatrix}6&-3\\2&1\end{bmatrix}$.

    Then to find eigenvalues, we set $\displaystyle \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=0$.

    In this case, it turns out that $\displaystyle \mathbf{A}-\lambda \mathbf{I}=\begin{bmatrix}6-\lambda & -3\\2&1-\lambda\end{bmatrix}$.

    It now follows that $\displaystyle \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=\left(6-\lambda\right)\left(1-\lambda\right)-\left(2\right)\left(-3\right)=6-7\lambda+\lambda^2+6=\lambda^2-7\lambda+12$

    Thus, $\displaystyle \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=0\implies\lambda^2-7\lambda+12=0\implies\left(\lambda-4\right)\left(\lambda-3\right)=0\implies$ $\displaystyle \lambda=4$ or $\displaystyle \lambda=3$.
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