例题 6.2

例题 6.2.

解非齐次线性方程组

$$\{\begin{array}{ll}{x}_1-x_2-x_3+x_4& =0,\\ x_1-x_2+x_3-3x_4& =1,\\ x_1-x_2-2x_3+3x_4& =-\frac{1}{2}.\end{array}$$

解 对增广矩阵 $\tilde{\mathbf{A}}$ 作初等行变换,即

$$\begin{array}{rl}\tilde{A}& =\left(\begin{array}{rrrrr}1& -1& -1& 1& 0\\ 1& -1& 1& -3& 1\\ 1& -1& -2& 3& -\frac{1}{2}\end{array}\right)\\ & \to \left(\begin{array}{rrrrr}1& -1& -1& 1& 0\\ 0& 0& 2& -4& 1\\ 0& 0& -1& 2& -\frac{1}{2}\end{array}\right)\\ & \to \left(\begin{array}{rrrrr}1& -1& -1& 1& 0\\ 0& 0& -1& 2& -\frac{1}{2}\\ 0& 0& 2& -4& 1\end{array}\right)\\ & \to \left(\begin{array}{rrrrr}1& -1& -1& 1& 0\\ 0& 0& -1& 2& -\frac{1}{2}\\ 0& 0& 0& 0& 0\end{array}\right).\end{array}$$

可知 $r\left(\tilde{\mathbf{A}}\right)=r\left(\mathbf{A}\right)=2$ , 故方程组有解,我们有

$$\{\begin{array}{ll}{x}_1-x_2-x_3+x_4& =0,\\ -x_3+2x_4& =-\frac{1}{2}.\end{array}$$

由于 $n-r\left(\mathbf{A}\right)=2$ , 所以取自由未知量 $x_2,x_4$ , 得到

$$\{\begin{array}{l}{x}_1=x_2+x_4+\frac{1}{2}\\ x_3=2x_4+\frac{1}{2}\end{array}$$

取 $x_2=x_4=0$ , 得 $x_1=x_3=\frac{1}{2}$ , 即得原方程组的一个特解 ${\gamma }_0=\left(\frac{1}{2},0,\frac{1}{2},0\right)$ .

对应的齐次方程组为

$$\{\begin{array}{l}{x}_1=x_2+x_4\\ x_3=2x_4\end{array}$$

分别取 $\left(x_2,x_4\right)=\left(1,0\right),\left(0,1\right)$ , 则有 $\left(x_1,x_3\right)=\left(1,0\right),\left(1,2\right)$ , 即得对应的齐次方程组的基础解系为

$${\eta }_1=\left(1,1,0,0\right),{\eta }_2=\left(1,0,2,1\right).$$

于是所求的非齐次方程组的通解为

$$\begin{array}{rl}& (x_1,x_2,x_3,x_4)\\ & ={\gamma }_0+k_1{\eta }_1+k_2{\eta }_2\\ & =\left(\frac{1}{2},0,\frac{1}{2},0\right)+k_1\left(1,1,0,0\right)+k_2\left(1,0,2,1\right).\end{array}$$

这里 $k_1,k_2$ 为任意常数.