Could anyone give me the hint to solve this problem ?

Tools of linear algebra work with any field ( https://secure.wikimedia.org/wikipedia/en/wiki/Field_(mathematics) ), including finite ones like integers mod 3. The problem is effectively Gaussian Elimination.

But we can't do transformations with antiequations such as with equations. For example, let we have a system:
x1    != 0 (mod 3)
   x2 != 2 (mod 3).
It has a solution (2; 0)
If we add the antiequations, we'll get this one:
x1+x2 != 2 (mod 3), but it's wrong for solution (2; 0) of the system!
Am I right?

Edited by author 29.03.2011 00:53

Uhm, sorry. I was wrong.

I think that your advice is absolutely right!
Let for example det(Aij)<>0.
Then Answer is 2^k,because (A*X)i in {0,1,2}\{bi}
If det(Aij)==0
we can use gauss method to make standard worm of the matrix A.