## #math - Fri 9 Mar 2007 between 17:39 and 20:02

### NY Lost Funds

 delta ihope, ok. a, A, d and D are arbitrary? ihope Yes, arbitrary real numbers. delta ihope, did you try something like p(X) = bX^2 + (a-d)X - c, etc? ihope What would that do? delta ihope, the discriminant of this trinom is the first term. Being < 0 meaning there's no real root.ihope, same for the second. So, the third one is the discriminant of p + q. ihope The third is the discriminant of p+q?(b+B)^2 - 4(a+A)(c+C) = b^2 - 4ac + B^2 - 4AC, isn't that? delta ? :)(p + q)(X) = (b+B)X^2 + (a - d + A - D)X - (c + C). ihope Oh yeah. delta So, if bB > 0, you're done.Investigate others cases maybe. ihope x^2 + (a+d)x + ad-bc is the polynomial I got the stuff from.At least, I think it is.Also, the existence of a nice definition of e^x is nice, no? mike8901 can someone guide me to the solution of this linear algebra question? The question is: If the vector (a,b) is a multiple of the vector(c,d) show that the vector(a,c) is a multiple of (b,d). The hint given(a very vague one) is that if A=[a,b;c,d] then AB=I and BA=I. However, I thought the matrix wouldn't be invertable.*wouldn't have an inverse seb-- mike8901: what do you mean 'multiple of the vector' ?mike8901: (a,b) = k (c,d) for some integer k? mike8901 (a,b)=s*(c,d) where s is a scalar multiplier FatalError mike8901, consider [a,b] = [c,d] and since [c,d] is a multiple of [a,b] we can rewrite it k[a,b] = [ka, kb] mike8901 right seb-- ok mike8901 FatalError: I got that far :Pso [ka,kb]=[c,d]then ka=c kb=d FatalError well then you have the vectors [a, ka] = j*[b,kb] -> a*[1,k] = j*b*[1,k] seb-- yes..i see it now too mike8901 hmm seb-- mike8901: so are you happy? mike8901 not really :/[a, ka] = j*[b,kb] how did you get there? seb-- mike8901: he worked backward from your goalmike8901: do thatmike8901: ASSUME what you are trying to prove and massage it into something obvious doyley|uk DCC SEND MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM analpear hi fruitbag Hey#not-not-math