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Hm. Seems there are no tests with small number of linear independent rows. I have two solutions with different answers to such tests, but both got OK. Admins, please, add some. 30 30 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 ... 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 30 30 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 x20 30 30 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 0 x10 if det(A) == 0 , what should I do ? If A - is not square matrix ( i.e. k != n ), how to calculate det(A) ? Could anyone give me the hint to solve this problem ? 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. |
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