$\displaystyle \begin{pmatrix} k & -2\\1-k & k \end{pmatrix}$, 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:

$\displaystyle \begin{pmatrix} k & -2\\1-k & k \end{pmatrix}$$\displaystyle \begin{pmatrix} x\\2x \end{pmatrix}$=$\displaystyle \begin{pmatrix} x\\2x \end{pmatrix}$

implies $\displaystyle \begin{pmatrix} x(k-4) \\x(1+k) \end{pmatrix}$=$\displaystyle \begin{pmatrix} x\\2x \end{pmatrix}$

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