1.已知$AX=B$,其中$A=left[egin{array}{ll}{1} & {2} \ {2} & {4} \ {3} & {5}end{array} ight], B=left[egin{array}{ccc}{2} & {5} & {-1} \ {4} & {10} & {-2} \ {7} & {9} & {3}end{array} ight]$,求$X$.
法一.由于[left[egin{array}{cc}{1} & {2} \ {2} & {4} \ {3} & {5}end{array} ight]left[egin{array}{lll}{x_{11}} & {x_{12}} & {x_{13}} \ {x_{21}} & {x_{22}} & {x_{23}}end{array} ight]=left[egin{array}{ccc}{2} & {5} & {-1} \ {4} & {10} & {-2} \ {7} & {9} & {3}end{array} ight],]
则|[left[egin{array}{ll}{1} & {2} \ {3} & {5}end{array} ight]left[egin{array}{lll}{x_{11}} & {x_{12}} & {x_{13}} \ {x_{21}} & {x_{22}} & {x_{23}}end{array} ight]=left[egin{array}{ccc}{2} & {5} & {-1} \ {7} & {9} & {3}end{array} ight],]
于是[left[egin{array}{ccc}{x_{11}} & {x_{12}} & {x_{13}} \ {x_{21}} & {x_{22}} & {x_{23}}end{array} ight]=left[egin{array}{cc}{1} & {2} \ {3} & {5}end{array} ight]^{-1}left[egin{array}{ccc}{2} & {5} & {-1} \ {7} & {9} & {3}end{array} ight]=left[egin{array}{ccc}{4} & {-7} & {11} \ {-1} & {6} & {-6}end{array} ight].]
法二.注意到[left[egin{array}{cc}{1} & {2} \ {2} & {4} \ {3} & {5}end{array} ight]left[egin{array}{l}{x_{11}} \ {x_{21}}end{array} ight]=left[egin{array}{l}{2} \ {4} \ {7}end{array} ight],left[egin{array}{ll}{1} & {2} \ {2} & {4} \ {3} & {5}end{array} ight]left[egin{array}{l}{x_{12}} \ {x_{22}}end{array} ight]=left[egin{array}{c}{5} \ {10} \ {9}end{array} ight],left[egin{array}{cc}{1} & {2} \ {2} & {4} \ {3} & {5}end{array} ight]left[egin{array}{c}{x_{13}} \ {x_{23}}end{array} ight]=left[egin{array}{c}{-1} \ {-2} \ {3}end{array} ight],]
则[left[egin{array}{l}{x_{11}} \ {x_{21}}end{array} ight]=left[egin{array}{l}{4} \ {-1}end{array} ight],quadleft[egin{array}{l}{x_{12}} \ {x_{22}}end{array} ight]=left[egin{array}{c}{-7} \ {6}end{array} ight],quadleft[egin{array}{l}{x_{13}} \ {x_{23}}end{array} ight]=left[egin{array}{c}{11} \ {-6}end{array} ight].]因此[X=left[egin{array}{ccc}{4} & {-7} & {11} \ {-1} & {6} & {-6}end{array} ight].]
常用级数[(1+x)^{1/x}=e-frac{e x}{2}+frac{11 e x^{2}}{24}-frac{7 e x^{3}}{16}+frac{2447 e x^{4}}{5760}+Oleft(x^{5} ight),quad x o 0]
[int_{0}^{1} sqrt{frac{x}{1-x}} d x=frac{pi}{2}.]
[int_{0}^{1} int_{0}^{1} frac{log left(x-x^{2} ight)-log left(y-y^{2} ight)}{left(x-x^{2} ight)-left(y-y^{2} ight)} d x d y=7zeta(3).]