Why is the volume of a parallelogram $P$ formed from the unit square $I^2$ with an integer matrix $A$ equal to the number of integer points in $P$? I am aware of this existing post: Number of fixed points of a hyperbolic toral automorphism and its accepted answer: https://math.stackexchange.com/a/3371094/820472. Unfortunately the last two proposed methods do not seem obvious to me and seem a bit disconnected (note: what is and isn't connected is very subjective; here by connected I am referring to something that a first year undergraduate student would get after basic linear algebra). And then, while I like the method (Pick's formula) behind the first proposed solution it seems to assume a priori a relationship between the determinant of $A - I$ and the number of integer lattice points in a parallelogram.
Question: My question is: Given an integer matrix $A$ such that $A - I$ is invertible, how do you justify that the number of integer points in a parallelogram spanned by $(A - I)(0, 1)$ and $(A - I)(1, 0)$ is equal to $\mathrm{det}(A - I)$. Bonus question: Do we need that $A$ is an integer matrix? If it is the case, what do we know about a general real matrix $A$? Like a lower/upper bound for such integer points?
Thanks for the help!
 A: Let us consider a parallelogram $P$ in the plane. Since the problem is invariant under rotations and translations$^*$ of $P$, we may assume that one side of $P$ is aligned with the $x$-axis. To make things a bit easier, we may translate $P$ a tiny bit further so that its boundary intersects no integer points (since there are only countably many integer points). Imagine the lower-left corner to be placed at the coordinate $(-\epsilon,\epsilon)$, say. Denote the length of the bottom side by $l$ and the height by $h$. Let us verify that the number of integer points inside $P$ is equal to $\lceil l \rceil \lceil h \rceil$. In particular, it equals the area of $P$ if $l$ and $h$ are integers.
The intersection of the $x$-axis and $P$ contains $\lceil l \rceil$ many integer points ($\epsilon$ is tiny). If $P$ was a rectangle, then the intersection of the $y$-axis and $P$ contained $\lceil h \rceil$ many integer points, yielding a total of $\lceil l \rceil \lceil h \rceil$ for the whole of $P$. If $P$ is sheared, then the number of integer points that are in the rectangle but not in $P$ on the left side is, by symmetry, exactly equal to the number of integer points that are in $P$ but not in the rectangle on the right side. So, the number of integer points for a sheared $P$ is the same as for a rectangular $P$.
EDIT:$^*$ for a non-integer parallelogram, the argument shows that the maximum number of integer points is $\lceil l \rceil \lceil h \rceil$, but $P$ could be shifted so as to contain less. A lower bound would be $\lfloor l \rfloor \lfloor h \rfloor$ by an analogous argument.
