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Math Help - y=2x mapped onto itself under transformation T

  1. #1
    Member ssadi's Avatar
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    y=2x mapped onto itself under transformation T

    \begin{pmatrix} k & -2\\1-k & k \end{pmatrix}, where k is constant
    A transformation T:R^2 arrow R^2 is represented by the matrix A.
    Find the value of k for which the line y=2x is mapped onto itself under T.


    What I did:
    \begin{pmatrix} k & -2\\1-k & k \end{pmatrix} \begin{pmatrix} x\\2x \end{pmatrix}= \begin{pmatrix} x\\2x \end{pmatrix}
    implies \begin{pmatrix} x(k-4) \\x(1+k) \end{pmatrix}= \begin{pmatrix} x\\2x \end{pmatrix}
    Therefore k=5 and k=1 !!
    What I should do:
    x(1+k)=2x(k-4)
    k=9
    Where is my principle wrong? Please point it out, because i fear its something fundamental and i may have to suffer direly for it later.
    edit:k=5, k=1
    Last edited by ssadi; September 1st 2009 at 11:33 AM.
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  2. #2
    Member
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    Hi,
    your principle is wrong because by = you're saying that a point (x,2x) must map to itself. This condition is too strong and as you showed it has no solution for k. What you have to consider is  \begin{pmatrix} k & -2\\1-k & k \end{pmatrix} \begin{pmatrix} x\\2x \end{pmatrix}=\begin{pmatrix} w\\2w \end{pmatrix} (because (x,2x) can be mapped to arbitrary point (w,2w))
    and after eliminating w from two simultaneous equations you get the right result (or you get necessary condition for k, and by linearity you see that it is also sufficient)
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