How to find an inverse of this type of functions? Let $F(x,y)$ and $G(x,y)$ Be functions from where $x,y$ are whole numbers. ($Z^2 \to Z^2$)
$F(x,y) = (x+3y,x+5y)$
$G(x,y) = (2x+3y,3x+5y)$
The question:
One of these functions has an inverse, Prove it and find its inverse functions.
My question:
I don't really know how to decide which one is inversible. I tried to show that $f(x,y)$ is not surjective but with no success.
Also, I don't know how to find the inverse of a function like this.
Can someone hint me to the solution? (I want to solve it myself). Maybe through an example or something.
Thank you.
 A: If the function
$(ax+by, cx+dy)$
is invertable,
then,
given $(u, v)$
you can solve
$ax+by = u\\
cx+dy = v
$
for $x$ and $y$.
This means that the determinant 
$\bigg|
\begin{array}\\
a & b\\
c & d 
\end{array}
\bigg|
$
is non-zero.
So,
what are the determinants of
the two systems?
(added a bit later)
I realized that you want integers.
So the determinant
has to be
$\pm 1$.
Check that, too.
A: For the first function for example, you have $$F(x,y)=(u,v)=(x+3y,x+5y)$$
This can be written as 
$u=x+3y$ and $v=x+5y$
So the inverse $F^{-1}$ of $F$ must have the property $F^{-1}(u,v)=(x,y)$ i.e. to get $F^{-1}$ you have to find $x$ and $y$ in terms of $u$ and $v$.
A: An example : consider the function $h : \mathbb{Z}^2 \to \mathbb{Z}^2$ given by $h(x,y) = (3x+y,5x+2y)$. It is easy to see that $h$ is injective. Now, we claim that $h$ is surjective. To see this, given $(u,v) \in \mathbb{Z}^2$ we need to find $(x,y) \in \mathbb{Z}^2$ such that $(u,v) = h(x,y)$. 
Observe that the latter reduces to solve a system of linear equations:
$$\begin{cases}
3x+y = u \\ 5x+2y = v
\end{cases}$$
Solving this, we obtain that $(x,y) = (2u-v,-5u+3v)$, which means that
$$h(2u-v,-5u+3v) = (u,v).$$
Defining $g : \mathbb{Z}^2 \to \mathbb{Z}^2$ by $g(u,v) = (2u-v,-5u+3v)$ the latter can be rewritten as $h(g(u,v)) = (u,v)$, but this means precisely that the inverse of $h$ is $g$.
A: Hint:  The inverses of $\pmatrix{1&&3\\1&&5}$ and $\pmatrix{2&&3\\3&&5}$ are 
$\pmatrix{\frac52&&-\frac32\\-\frac12&&\frac12}$ and $\pmatrix{5&&-3\\-3&&2},$ respectively.
