求非齐次线性方程组的通解

$\left{\begin{array}{c}{x_{1}+x_{4}+2 x_{5}=1} \ {x_{1}+2 x_{3}+3 x_{4}+2 x_{5}=3} \ {4 x_{1}+5 x_{2}+3 x_{3}+2 x_{4}+3 x_{5}=2} \ {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1}\end{array}\right.$

解:

行阶梯化:

$\left[\begin{matrix}{1} & {0} & {0} & {1} & {2} & {1} \ {1} & {0} & {2} & {3} & {2} & {3} \ {4} & {5} & {3} & {2} & {3} & {2} \ {1} & {1} & {1} & {1} & {1} & {1}\end{matrix}\right] \Rightarrow\left[\begin{array}{cccccc}{1} & {0} & {0} & {1} & {2} & {1} \ {0} & {0} & {2} & {2} & {0} & {2} \ {0} & {5} & {3} & {-2} & {-5} & {-2} \ {0} & {1} & {1} & {0} & {-1} & {0}\end{array}\right]$

$\Rightarrow\left[\begin{array}{cccccc}{1} & {0} & {0} & {2} & {2} & {1} \ {0} & {0} & {1} & {1} & {0} & {1} \ {0} & {0} & {-2} & {-2} & {0} & {-2} \ {0} & {1} & {1} & {0} & {-1} & {0}\end{array}\right] \Rightarrow\left[\begin{array}{cccccc}{1} & {0} & {0} & {1} & {2} & {1} \ {0} & {1} & {1} & {0} & {-1} & {0} \ {0} & {0} & {1} & {1} & {0} & {1} \ {0} & {0} & {0} & {0} & {0} & {0}\end{array}\right]$

错误 1: 没有把等式右方的主元用自由未知量表示:

$\left(\begin{array}{l}{x_{1}} \ {x_{2}} \ {x_{1}} \ {x_{4}} \ {x_{5}}\end{array}\right)$ $=\left[\begin{array}{l}{1} \ {0} \ {1} \ {0} \ {0}\end{array}\right]$ $+x_{3}\left[\begin{array}{c}{0} \ {-1} \ {0} \ {0} \ {0}\end{array}\right]+x_{4}\left[\begin{array}{c}{-1} \ {0} \ {-1} \ {1} \ {0}\end{array}\right]+x_{5}\left[\begin{array}{c}{-2} \ {1} \ {0} \ {0}\{1}\end{array}\right]$

取 $\eta\_0=\left [\begin {array}{l}{1} \ {0} \ {1} \ {0} \ {0}\end {array}\right]$ 为一个特戒,$\xi\_1 = \left [\begin {array}{c}{0} \ {-1} \ {0} \ {0} \ {0}\end {array}\right]$ ,$\xi\_2 =\left [\begin {array}{c}{-1} \ {0} \ {-1} \ {1} \ {0}\end {array}\right]$, $\xi\_3 =\left [\begin {array}{c}{-2} \ {1} \ {0} \ {0} \ {1}\end {array}\right]$ 为导出组的一个基础解系.

从而方程组的通解为: $\eta=\eta_{0}+k_{1} \xi_{1}+k_{2} \xi_{2}+k_{3} \xi\_{3}$ ($k\_1, k\_2, k\_3$ 为任意常数.)

再解:

$\left[\begin{array}{cccccc}{1} & {0} & {0} & {1} & {2} & {1} \ {0} & {1} & {1} & {0} & {-1} & {0} \ {0} & {0} & {1} & {1} & {0} & {1} \ {0} & {0} & {0} & {0} & {0} & {0}\end{array}\right]$

等价表示为:

$\left{\begin{array}{l}{x_{1}+x_{4}+2 x_{5}=1} \ {x_{2}+x_{3}-x_{5}=0} \ {x_{3}+x_{4}=1}\end{array}\right.$ $\Rightarrow\left{\begin{array}{l}{x_{1}=-x_{4}-2 x_{5}+1} \ {x_{2}=-x_{3}+x_{5}} \ {x_{3}=1-x_{4}}\end{array}\right.$

$x_{2}=-\left (1-x_{4}\right)+x_{5}=-1+x_{4}+x_{5}$ (注意这里:把等式右方出现的主元 $x_2$ 用自由未知量表示)

则有: (PS: 下面左边的 x 就是主元,右边的 x 就是自由未知量)

$\left{\begin{array}{l}{x_{1}=-x_{4}-2 x_{5}+1} \ {x_{2}=x_{4}+x_{5}-1} \ {x_{3}=1-x_{4}}\end{array}\right.$

错误三:通解没有去掉常数项,没有解导出组的基础解系.

分别令

$\left[\begin{array}{l}{x_{4}} \ {x_{5}}\end{array}\right]=\left[\begin{array}{l}{0} \ {0}\end{array}\right]\left[\begin{array}{l}{1} \ {0}\end{array}\right]\left[\begin{array}{l}{0} \ {1}\end{array}\right]$

分别得

$\left[\begin{array}{c}{x_{1}} \ {x_{2}} \ {x_{3}} \ {x_{4}} \ {x_{5}}\end{array}\right]=\left[\begin{array}{c}{1} \ {-1} \ {1} \ {0} \ {0}\end{array}\right]\left[\begin{array}{c}{0} \ {0} \ {0} \ {1} \ {0}\end{array}\right] \quad\left[\begin{array}{c}{-1} \ {0} \ {1} \ {0} \ {1}\end{array}\right]$

再解:

对于

$\left{\begin{array}{l}{x_{1}=-x_{4}-2 x_{5}+1} \ {x_{2}=x_{4}+x_{5}-1} \ {x_{3}=1-x_{4}}\end{array}\right.$

令 $\left [\begin {array}{l}{x_{4}} \ {x_{5}}\end {array}\right]=\left [\begin {array}{l}{0} \ {0}\end {array}\right]$ 得到特解一个: $\left [\begin {array}{c}{1} \ {-1} \ {1} \ {0} \ {0}\end {array}\right]$

而导出组 (去除常数的方程组) 为

$\left{\begin{array}{l}{x_{1}=-x_{4}-2 x_{5}} \ {x_{2}=x_{4}+x_{5}} \ {x_{3}=-x_{4}}\end{array}\right.$

分别令

$\left[\begin{array}{l}{x_{4}} \ {x_{5}}\end{array}\right]=\left[\begin{array}{l}{1} \ {0}\end{array}\right]\left[\begin{array}{l}{0} \ {1}\end{array}\right]$

分别得到

$\left[\begin{array}{c}{x_{1}} \ {x_{2}} \ {x_{4}} \ {x_{4}} \ {x_{5}}\end{array}\right]=\left[\begin{array}{c}{-1} \ {1} \ {-1} \ {1} \ {0}\end{array}\right]\left[\begin{array}{c}{-2} \ {1} \ {0} \ {0} \ {1}\end{array}\right]$

因此得到非其次线性方程组的通解为:

$x=k_{1}(-1,1,-1,1,0)^{T}+k_{2}(-2,1,0,0,1)^{T}+(1,-1,1,0,0)^{T}$ ($k\_1, k\_2$ 为任意常数.)

(此为正解)