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    Found the answer now.
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    (Original post by morgan8002)
    I'm working from the AQA online textbook, page 99.

    Find all invariant lines of the form y = mx, for the matrix transformations:
    a. \begin{bmatrix} 5 & 15 \\ -2 & -8\end{bmatrix}

    I'm not too sure on how to do this and I'm not finding the textbook much help, so this is what I've done:
    \begin{bmatrix}5&15\\-2&-8\end{bmatrix}\begin{bmatrix}x\\  mx\end{bmatrix} = \begin{bmatrix}x' \\ mx' \end{bmatrix}

    \therefore \begin{bmatrix} 5x + 18mx\\-2x - 8mx\end{bmatrix} = \begin{bmatrix}x'\\mx' \end{bmatrix}


    Bottom = m times the top
    m(5x + 18mx) = -2x - 8mx
    5mx + 18m^2 x = -2x - 8mx
    18m^2 x + 13mx + 2x=0
    x(18m^2 + 13m + 2)= 0
    x(2m+1)(9m+2) = 0
    \therefore x = 0(always invariant), m = -\frac{1}{2}, -\frac{2}{9}
    \therefore y = -\frac{1}{2}x, y = -\frac{2}{9}

    However, the answer in the book is y = -\frac{1}{5}x, y = -\frac{2}{3}x


    Can someone tell me if the book's answer is correct? If it is, can you tell me where I went wrong. I've done this 10 times and haven't gotten their answer.

    this is probably a lot quicker by eigenvectors (so you can check it ...)
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    (Original post by TeeEm)
    this is probably a lot quicker by eigenvectors (so you can check it ...)
    Sorry, was tired and I read the question wrong 10 times. It should be 15 not 18 in the top right of my matrix and it comes out. I though it was me not understanding the question.
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    (Original post by morgan8002)
    Sorry, was tired and I read the question wrong 10 times. It should be 15 not 18 in the top right of my matrix and it comes out. I though it was me not understanding the question.
    no worries
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    (Original post by TeeEm)
    no worries
    Thanks anyway.
 
 
 
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