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 :P so [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 goal mike8901: do that mike8901: 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 |