线性方程解法

痴心男http://im.cx/n posted @ 2008年11月21日 00:58 in 未分类 , 1421 阅读

$\[ \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 3 & 1 & { - 1} \\ 5 & { - 2} & 3 \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right] \]$

设最左边的矩阵等于Ａ，等式两边都乘以Ａ的逆矩阵。那么：

$\left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 3 & 1 & { - 1} \\ 5 & { - 2} & 3 \\ \end{array}} \right]^{ - 1} \left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right]$

$\left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} { - 0.0417} & {0.2083} & {0.0833} \\ {0.5833} & {0.0833} & { - 0.1667} \\ {0.4583} & { - 0.2917} & {0.0833} \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right]$