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Thread: 27.2 solve the system Y'={}y

  1. #1
    Super Member bigwave's Avatar
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    27.2 solve the system Y'={}y

    Solve the system $\displaystyle Y'=\begin{bmatrix}
    1 & 3 & -3 \\
    0 & 1 & 0 \\
    6 & 3 & -8
    \end{bmatrix}Y
    $

    not real sure but W|A returned this but no steps
    so assume first thing we so is Eigenvalues

    subtracting $\lambda$ from the diagonal entries of the given matrix and Find the determinant of the obtained matrix:
    $\displaystyle \left[
    \begin{array}{ccc} - \lambda+1&2&-3\\
    0&-\lambda+1&0\\
    6&3&-\lambda-8
    \end{array} \right]
    =-18\lambda
    +\left(-\lambda-8\right)
    \left(- \lambda + 1\right)^{2} + 18$

    27.2 solve the system Y'={}y-27.1.png
    Last edited by bigwave; Apr 5th 2019 at 02:00 PM.
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  2. #2
    MHF Contributor
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    Re: 27.2 solve the system Y'={}y

    it's solved just like the 2 dimensional case but now there are 3 terms instead of 2
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: 27.2 solve the system Y'={}y

    Quote Originally Posted by bigwave View Post
    Solve the system $\displaystyle Y'=\begin{bmatrix}
    1 & 3 & -3 \\
    0 & 1 & 0 \\
    6 & 3 & -8
    \end{bmatrix}Y
    $

    not real sure but W|A returned this but no steps
    so assume first thing we so is Eigenvalues

    subtracting $\lambda$ from the diagonal entries of the given matrix and Find the determinant of the obtained matrix:
    $\displaystyle \left[
    \begin{array}{ccc} - \lambda+1&2&-3\\
    0&-\lambda+1&0\\
    6&3&-\lambda-8
    \end{array} \right]
    =-18\lambda
    +\left(-\lambda-8\right)
    \left(- \lambda + 1\right)^{2} + 18$

    Click image for larger version. 

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    Okay, so what are the eigenvalues and eigenvectors?

    -Dan
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  4. #4
    MHF Contributor

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    Re: 27.2 solve the system Y'={}y

    Yes, you have the equation for the eigenvalues and it is fairly simple to solve it: the eigenvalues are 1, 1, and 8.
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