# Probability that a determinant is equal to zero about 2*2 or 3*3 Integer matrix.

In fact, it is an expansion of this problem. But I restricted the elements of the matrix to be integers only. Obviously this probability is related to the range of random $$\text{item}\in[0,n]$$.

For the 2*2 matrix, I solved some probability values related to $$n$$ by math software violencely:

But I don't know what the general formula is. For the 3*3 matrix, I only found 5 terms:

But how to calculate the analytical solution of this problem?

• Why do you believe that this problem has an "analytical solution"? Commented Jan 15, 2022 at 22:45
• @BenGrossmann I'm not sure, but it might be worth trying to find out. I've been thinking about this off and on for years, but I'm so bad at math that I'm asking here for help.
– mayi
Commented Jan 15, 2022 at 22:48
• It is astronomically unlikely that there is a "nice" answer to this question. Even the answer for the case of $n=1$ (for different matrix sizes) has no known analytic form. Commented Jan 15, 2022 at 22:53
• @BenGrossmann In my post I just don't limit the range of random values, which I think is much easier than not limiting the size of the matrix. As you can imagine, if the random values are continuous, unbounded size means that there are an infinite number of integral variables, but here I only have 4
– mayi
Commented Jan 16, 2022 at 6:24
• I suppose you have a point. At the very least, I suspect that there might be a nice general approach to the $2 \times 2$ case, if not an analytic formula Commented Jan 16, 2022 at 6:31

Well, the analytical solution about $$\small 2*2$$ case is:
$$\text{probability}(n)=\frac{2 \sum\limits _{r=1}^n \phi (r) \lfloor\frac{n}{r}\rfloor^2+(n+1) (3 n+1)}{(n+1)^4}$$