TITLE: Proof that $\gcd(ax+by,cx+dy)=\gcd(x,y)$ if $ad-bc= \pm 1$
QUESTION [9 upvotes]: I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory.

Given $x$ and $y$, let $m=ax+by$, $n=cx+dy$, where $ad-bc= \pm 1$. Prove that $(m,n)=(x,y)$.

I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice!
My proof:

We observe that since $ad-bc= \pm 1$, $ad=bc \pm 1$, and $(ad,bc)=1$.
  Now, $(a,b) \mid m$ and $(c,d) \mid n$ but
$$(a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.$$
Similarly, we determine $(c,d)=1$. So, $1=(a,b) \mid m$ and $1=(c,d)
> \mid n$. But $(x,y)$ also divide $m$ and $n$. Since $(x,y) \geq (a,b)=(c,d)=1$, this implies that $(x,y)=m,n$. Hence
  $(m,n)=((x,y),(x,y))=(x,y)$.

REPLY [2 votes]: Hint $ $ Homogeneous reduce to the case $\:(x,y) = 1\:$ by cancelling any gcd. Note  Bezout's GCD identity is equivalent to:  $\ (\color{#c00}{x,y}) = 1\iff \color{#c00}{\begin{smallmatrix}x\\ y\end{smallmatrix}}\,$ is a column in a matrix of determinant $\,1,\,$ thus
$\ \ \ \ \ \ (\color{#c00}{x,y}) = 1 \ \Rightarrow\  1 =  \begin{vmatrix} a & b \\ c & d\end{vmatrix}\ \begin{vmatrix} \color{#c00}x & u \\ \color{#c00}y & v \end{vmatrix} = \begin{vmatrix} ax+by & s \\ cx+dy & t \end{vmatrix}\,$ $ \Rightarrow\ (ax+by,\ cx+dy) = 1$
The converse follows the same way using the inverse transformation (or by easy arithmetic).