Question:
$A= \begin{pmatrix} k & -2 \\ 1-k & k \end{pmatrix}\text{, where k is a constant}$
$\text {A transformation } T : \mathbb{R}^2 \rightarrow \mathbb{R}^2 \text{ is represented by the matrix A.}$
$\text {Find the value of k for which the line } y = 2x \text{ is mapped onto itself under T.}$
Working: $\begin{pmatrix} k & -2 \\ 1-k & k \end{pmatrix}\cdot\begin{pmatrix} x \\ 2x \end{pmatrix}=\begin{pmatrix} x \\ 2x \end{pmatrix}$ $\begin{pmatrix} x(k-4) \\ x(1+k) \end{pmatrix}=\begin{pmatrix} x \\ 2x \end{pmatrix}$ $x(k-4)=x$ $x(1+k)=2x$
Leaving me with $k=5$ and $k=1$, However the answer is $k = 9$ why?