"Milos Milenkovic" <firstname.lastname@example.org> wrote in message <email@example.com>... > ....... > This is ok, but now I have to make this multiplication with respect to sign of these (ai,bi,ci) numbers, so if (ai,bi,ci) and (di,ei,fi) are >0 (they are >0, if bi and ei >0) then the upper solution is ok, > but when bi or ei is <0 than I have to apply next rule: > (ai,bi,ci)*(di,ei,fi) = (ei*ai+bi*(fi-ei), bi*ei , ei*ci+bi(di-ei)) > Or, if (ai,bi,ci) and (di,ei,fi) are both <0 (they are <0, if bi and ei <0) > (ai,bi,ci)*(di,ei,fi) = (ei*ci+bi*(fi-ei), bi*ei , ei*ai+bi(di-ei)) > ....... - - - - - - - - - - - Hello again Milos. The rules I gave you earlier can easily be revised to the following for the two new cases:
However, neither of these two new definitions are associative. In other words, if P, Q, and R are matrices of this kind, it is no longer true that P*(Q*R) is identically equal to (P*Q)*R where '*' represents either of your new multiplication rules. Your original special multiplication was in fact associative as I mentioned earlier. This makes me suspicious of your two new kinds of multiplication operators. Any "multiplication" which is not even associative is in my opinion a very ugly kind of multiplication and hardly worthy of the name. What in the world are you using it for?