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线性代数:求非齐次线性方程组的通解的详细过程和易错点

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)

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