Probability of determinants being coprime

This answer was obtained numerical, but it is nevertheless exact

There are $N=9^9$ possibilities to choose the matrix $m$ (each with equal probability). There are 1913 possible outcomes for $\det m$. The most likely is $\det m =0$. The probability for this event is $$P[\det m = 0] = \frac{218605}{14348907} \approx 1.5\%.$$ The rest is distributed over different integer number. The largest determinant is a matrix with $\det m = 1216$. There are 3 possibilities for this matrix (therefore the probability is $P[\det m = 1216] = 3/N$). There are also 3 matrices for which $\det m=-1216$. (in fact this is always true. If there are $n$ matrices with $\det m= x$ then there are exactly $n$ matrices with $\det m=-x$). To find out if the determinants are coprime the sign does not matter, therefore we only need to know the probabilities $$P[|\det m| = x].$$ Generically, the probability decreases for increasing $\det m$.

The absolute value of the determinant may assume 957 values. Therefore, there are $957*(957+1)/2 =458403$ pairs of matrices $m_1$ and $m_2$. Out of them there are 274487 pairs coprime. (The fraction of matrices which are coprime is $274487/458403 \approx 60\%$)

Counting each of the pair of matrices $m_1, m_2$ with the proper probability, we obtain the probability that two matrices are coprime $$P[ m_1 \text{ and } m_2 \text{ are coprime}]= \frac{7253958722902984}{16677181699666569} \approx 43\%.$$