Linear Algebra Problem Set

1.已知$AX=B$,其中$A=\left[\begin{array}{ll}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right], B=\left[\begin{array}{ccc}{2} & {5} & {-1} \\ {4} & {10} & {-2} \\ {7} & {9} & {3}\end{array}\right]$,求$X$.


法一.由于\[\left[\begin{array}{cc}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right]\left[\begin{array}{lll}{x_{11}} & {x_{12}} & {x_{13}} \\ {x_{21}} & {x_{22}} & {x_{23}}\end{array}\right]=\left[\begin{array}{ccc}{2} & {5} & {-1} \\ {4} & {10} & {-2} \\ {7} & {9} & {3}\end{array}\right],\]

则|\[\left[\begin{array}{ll}{1} & {2} \\ {3} & {5}\end{array}\right]\left[\begin{array}{lll}{x_{11}} & {x_{12}} & {x_{13}} \\ {x_{21}} & {x_{22}} & {x_{23}}\end{array}\right]=\left[\begin{array}{ccc}{2} & {5} & {-1} \\ {7} & {9} & {3}\end{array}\right],\]

于是\[\left[\begin{array}{ccc}{x_{11}} & {x_{12}} & {x_{13}} \\ {x_{21}} & {x_{22}} & {x_{23}}\end{array}\right]=\left[\begin{array}{cc}{1} & {2} \\ {3} & {5}\end{array}\right]^{-1}\left[\begin{array}{ccc}{2} & {5} & {-1} \\ {7} & {9} & {3}\end{array}\right]=\left[\begin{array}{ccc}{4} & {-7} & {11} \\ {-1} & {6} & {-6}\end{array}\right].\]

法二.注意到\[\left[\begin{array}{cc}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right]\left[\begin{array}{l}{x_{11}} \\ {x_{21}}\end{array}\right]=\left[\begin{array}{l}{2} \\ {4} \\ {7}\end{array}\right],\left[\begin{array}{ll}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right]\left[\begin{array}{l}{x_{12}} \\ {x_{22}}\end{array}\right]=\left[\begin{array}{c}{5} \\ {10} \\ {9}\end{array}\right],\left[\begin{array}{cc}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right]\left[\begin{array}{c}{x_{13}} \\ {x_{23}}\end{array}\right]=\left[\begin{array}{c}{-1} \\ {-2} \\ {3}\end{array}\right],\]

则\[\left[\begin{array}{l}{x_{11}} \\ {x_{21}}\end{array}\right]=\left[\begin{array}{l}{4} \\ {-1}\end{array}\right],\quad\left[\begin{array}{l}{x_{12}} \\ {x_{22}}\end{array}\right]=\left[\begin{array}{c}{-7} \\ {6}\end{array}\right],\quad\left[\begin{array}{l}{x_{13}} \\ {x_{23}}\end{array}\right]=\left[\begin{array}{c}{11} \\ {-6}\end{array}\right].\]因此\[X=\left[\begin{array}{ccc}{4} & {-7} & {11} \\ {-1} & {6} & {-6}\end{array}\right].\]


Common Series \ [(1 + x) ^ {1 / x} = e- \ frac {ex} {2} + \ frac {11 ex ^ {2}} {24} - \ frac {7 ex ^ {3 }} {16} + \ frac {2447 ex ^ {4}} {5760} + O \ left (x ^ {5} \ right), \ quad x \ to 0 \]

\[\int_{0}^{1} \sqrt{\frac{x}{1-x}} d x=\frac{\pi}{2}.\]

 

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Origin www.cnblogs.com/Eufisky/p/10987674.html