Let $A \in M\_n(\mathbb Z)$ and $\|A\| = \max |a\_{ij}|$.
Denote $$ S(r) = \sum\_{\substack{\|A\| \leq r \\\ \det{A} \neq 0}} \dfrac{1}{|\det{A}|} $$
- the sum over **all** matrices $A \in M\_n(\mathbb Z)$ with $\|A\| \leq r$ and $\det{A} \neq 0$.


I'm interested about the asymptotic behavior of the function $S(r)$:


* How does $S(r)$ grows?
* Is it true that $\lim\limits\_{r \to \infty} \dfrac{S(r)}{r^{n^2}} = 0 $ ?


Suggestions on special cases when $n = 2,3$ are also appreciated.



Sums of inverse determinants over matrices
RV above links to a paper of [Duke, Rudnick and Sarnak](http://www.ams.org/mathscinet-getitem?mr=1230289) where they establish the following (see Example 1.6 and Theorem 1.10):


The number of $n \times n$ matrices with $||A|| \leq r$ and $\det A = k$ is
$$c\_n r^{n^2-n} \sum\_{d\_1 d\_2 \cdots d\_n=k} \frac{1}{d\_2 d\_3^2 \cdots d\_n^{n-1}} + O(r^{n^2-n-1/n + \epsilon})$$
where
$$c\_n = \frac{\pi^{n^2-n}}{\Gamma\left(\frac{n}{2}\right) \Gamma\left(\frac{n^2-n+2}{2}\right) \zeta(2) \zeta(3) \cdots \zeta(n)}.$$


Taking $k=1$, we get the lower bound
$$S(r) > (c\_n-o(1)) r^{n^2-n}.$$
I didn't read the paper carefully enough to figure out how uniform the $O(\ )$ bound is. If we ignored the $O(\ )$ term altogether, we'd get
$$S(r) \approx c\_n r^{n^2-n} \sum\_k \sum\_{d\_1 d\_2 \cdots d\_n=k} \frac{1}{k d\_2 d\_3^2 \cdots d\_n^{n-1}} = c\_n r^{n^2-n} \sum\_{d\_1 d\_2 \cdots d\_n} \frac{1}{d\_1 d\_2^2 d\_3^3 \cdots d\_n^{n}}$$
$$ = c\_n r^{n^2-n} \sum\_{d\_1} \frac{1}{d\_1} \sum\_{d\_2} \frac{1}{d\_2^2} \sum\_{d\_3} \frac{1}{d\_3^3} \cdots \sum\_{d\_n} \frac{1}{d\_n^n} = c\_n r^{n^2-n} \left( \sum\_d \frac{1}{d} \right) \zeta(2) \zeta(3) \cdots \zeta(n)$$
Hmm, the sum diverges. That isn't good. 


Presumably, if we looked at the proof and the error bounds a little more carefully, we'd wind up cutting off the sum at some upper bound: $c\_n r^{n^2-n} \zeta(2) \zeta(3) \cdots \zeta(n) \sum\_{d \leq D} 1/d \approx c\_n r^{n^2-n} \zeta(2) \zeta(3) \cdots \zeta(n) \log D$. I'm not going to work hard enough to figure out what $D$ is, but surely $n! r^n$ is large enough, as there are no terms with determinant larger than that. On the other hand, I would guess that the leading term in the DRS bound dominates when $k \leq r^{1/(100 n)}$. Both of these are powers of $r$, so both of them have logarithm of the form $c \log r$. 


So I would guess that the asymptotics of $S(r)$ are like 
$$c r^{n^2-n} \log r$$
for some constant $c$ (dependent on $n$).


