初等变换解矩阵方程值会变吗

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用初等变换解矩阵方程AX=B，其中A=left( begin{array}{ccc} 1 & 1 & -1 0 & 2 & -5 1 & 0 & 1 end{array} right),B=left( begin{array}{c} 1 2 3 end{array} right)

解：仅作初等行变换，得

begin{split} left( begin{array}{ccc} A & {vdots} & B \ end{array} right)=&left( begin{array}{cccc} 1 & 1 & -1 & 1 \ 0 & 2 & -5 & 2 \ 1 & 0 & 1 & 3 \ end{array} right)\ &xrightarrow{mbox{①$times$(-1)+③}} left( begin{array}{cccc} 1 & 1 & -1 & 1 \ 0 & 2 & -5 & 2 \ 0 & -1 & 2 & 2 \ end{array} right)\ &xrightarrow{mbox{②$times(frac{1}{2})$+③}} left( begin{array}{cccc} 1 & 1 & -1 & 1 \ 0 & 2 & -5 & 2 \ 0 & 0 & -frac{1}{2} & 3 \ end{array} right)\ &xrightarrow{mbox{③$times$(-2)}} left( begin{array}{cccc} 1 & 1 & -1 & 1 \ 0 & 2 & -5 & 2 \ 0 & 0 & 1 & -6 \ end{array} right)\ &xrightarrow[text{③}times1+text{②}]{mbox{③$times$(5)+②}} left( begin{array}{cccc} 1 & 1 & 0 & -5 \ 0 & 2 & 0 & -28 \ 0 & 0 & 1 & -6 \ end{array} right)\ &xrightarrow{mbox{②$timesfrac{1}{2}$}} left( begin{array}{cccc} 1 & 1 & 0 & -5 \ 0 & 1 & 0 & -14 \ 0 & 0 & 1 & -6 \ end{array} right)\ &xrightarrow{mbox{②$times$(-1)+①}} left( begin{array}{cccc} 1 & 0 & 0 & 9 \ 0 & 1 & 0 & -14 \ 0 & 0 & 1 & -6 \ end{array} right). end{split} \

因此方程的解为X=left( begin{array}{c} 9 \ -14 \ -6 \ end{array} right).\

注记1： 初等（行）变换法求矩阵方程AX=B之原理

如果A可逆，则X=A^{-1}B，且A^{-1}也可逆. 根据定理，存在初等矩阵G_1,G_2,cdots,G_k，使

A^{-1}=G_1G_2cdots G_k, \

因此

A^{-1}A=G_1G_2cdots G_kA, \

即

{I=G_1G_2cdots G_kA},~~~~~~~(1) \ {A^{-1}B=G_1G_2cdots G_kB}.(2) \

式(1)表示对A施以若干次初等行变换化为I；

式(2)表示对B施以同样的初等行变换化为A^{-1}B.

注记2： 初等（行）变换法求矩阵方程AX=B之方法

(1). 作一个ntimes 2n的分块矩阵left( begin{array}{ccc} A & {vdots} &B \ end{array} right);

(2). 然后对此矩阵施以仅限于行的初等变换;

(3). 当子块A化为I的同时，子块B化为A^{-1}B，即

left( begin{array}{ccc} A & {vdots} &B \ end{array} right)xrightarrow{mbox{初等行变换}} left( begin{array}{ccc} I & {vdots} &A^{-1}B \ end{array} right). \