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Suppose $Ax=b$ is a linear system and A is a $n \times n$ matrix and vector $b \neq 0$. Suppose all numbers $a_i$ in $A$ and $b_i$ in $b$ belong to $\mathbb{Z}$ and suppose they are in a fixed range such that $|a_i|

NOTE: For my $real$ purpose it suffices an answer when $A$ is a $3\times3$ matrix and c=10 FWIW.

Thanks.

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    I do not think that we can do any better than just counting all the cases. But this will be an enormous task. To estimate the probability, I would suggest a simulation. – 2017-02-26
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    FWIW some friend of mine and I tried to study the n=2 anc c=1 case looking at the rank of the matrixes and my friend estimated 729 total system and 504 with solution and 225 with no solution. But we aren't sure about that. Since my programming skills are very low can someone make a brief simulation as Peter suggested please? – 2017-02-26
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    Or you could try yourselves - consider monte Carlo simulations for starters – 2017-02-26
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    I created $10^7$ examples and $46515$ times there was no solution. So, we have a probability of about $0.00465$ or $1:215$ – 2017-02-26
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    The probability that $A$ has not full rank is only slightly higher, about $1:211$. This scenario includes the possibility that there are infinite many solutions. – 2017-02-26
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    In my simulations, I used $n=3$ and $c=11$ (entries from $-10$ to $10$) – 2017-02-26
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    Now, I realized that you want $c=10$ (entries from $-9$ to $9$). Here, we have the probabilities about $1:167.5$ that $A$ has not full rank and about $1:170$ that we have no solution. – 2017-02-26
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    Many thanks Peter, very useful. I add some calculations in the case $n=2$ for a given $c$. If we put $a_i, b_i$ in $\mathbb{Z}/c \mathbb{Z}$ we have $c^6$ systems and there are $c^3+c^2-c$ degenerate 2x2 matrix so the probability of a non-solution system is $\frac{c^2-1+(c^3+c^2-c-1)(c^2-c)}{c^6}$ which is asimptotic to $1/c$ so if we have c=10 the probability of non-solution system is 1/21 (we get out from $\mathbb{Z}/c \mathbb{Z}$) We look at the resut and if n=3 we find out 1:211 and we may conjecture that there is a 1/10 factor between dimensions – 2017-02-26

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