2

线性方程组与向量组其实是一回事

我们来看一般的非齐次线性方程组，aija_{ij}就是系数

\begin{align}\left\{\begin{array}{c} a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 n} x_{n}=b_{1}, \\ a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 n} x_{n}=b_{2}, \\ \cdots \cdots \\ a_{m 1} x_{1}+a_{m 2} x_{2}+\cdots+a_{m n} x_{n}=b_{m} \end{array}\right.\end{align}

下面就是该方程组的系数矩阵，Am×nA_{m \times n}其中mm表示所给方程的个数而nn表示未知数的个数

\begin{align}\boldsymbol{A}=\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} \end{array}\right]\end{align}

下面方程由若干个列向量拼成的，且其增广矩阵

\begin{align}\left[\begin{array}{cccc:c} a_{11} & a_{12} & \cdots & a_{1 n} & b_{1} \\ a_{21} & a_{22} & \cdots & a_{2 n} & b_{2} \\ \vdots & \vdots & & \vdots & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} & b_{m} \end{array}\right]\end{align}

而如果我们把最开始的非齐次线性方程组整理成向量组，就可以得到

\begin{align}x_{1}\left(\begin{array}{c} a_{11} \\ a_{21} \\ \vdots \\ a_{m 1} \end{array}\right)+x_{2}\left(\begin{array}{c} a_{12} \\ a_{22} \\ \vdots \\ a_{m 2} \end{array}\right)+\cdots+x_{n}\left(\begin{array}{c} a_{1 n} \\ a_{2 n} \\ \vdots \\ a_{m n} \end{array}\right)=\left(\begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \end{array}\right)\end{align}

而把向量组(a11a21⋮am1)\left(\begin{array}{c} a_{11} \\ a_{21} \\ \vdots \\ a_{m 1} \end{array}\right)看成$\alpha $，那么就可以化简就可以得到

\begin{align}x_{1}\alpha_{1}+x_{2}\alpha_{2}+\cdots +x_{n}\alpha_{n}=\beta\end{align}

齐次方程

方程组，后面用（I）来表示，称为mm个方程，nn个未知量的齐次线性方程组

\begin{align}\left\{\begin{array}{c} a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 n} x_{n}=0, \\ a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 n} x_{n}=0, \\ \cdots \cdots \\ a_{m 1} x_{1}+a_{m 2} x_{2}+\cdots+a_{m n} x_{n}=0 \end{array}\right.\end{align}

其向量形式为

\begin{align}x_{1} \boldsymbol{\alpha}_{1}+x_{2} \boldsymbol{\alpha}_{2}+\cdots+x_{n} \boldsymbol{\alpha}_{n}=\mathbf{0}\end{align}

其中

\begin{align}\boldsymbol{\alpha}_{j}=\left[\begin{array}{c} a_{1 j} \\ a_{2 j} \\ \vdots \\ a_{m j} \end{array}\right], j=1,2, \cdots, n\end{align}

其矩阵形式为

\begin{align}\boldsymbol{A}_{m \times n} \boldsymbol{x}=\mathbf{0}\end{align}

其中

\begin{align}\boldsymbol{A}_{m \times n}=\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} \end{array}\right], \quad \boldsymbol{x}=\left[\begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array}\right]\end{align}

有解的条件

当r(A)=nr(\boldsymbol{A})=n时（α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}线性无关），方程组（I）有唯一零解

当r(A)=r<nr(\boldsymbol{A})=r<n时（α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}线性相关），方程组（I）有非零解，且有n−rn-r个线性无关解

解的性质

若Aξ1=0,Aξ2=0\boldsymbol{A} \boldsymbol{\xi}_{1}=\mathbf{0}, \boldsymbol{A} \boldsymbol{\xi}_{2}=\mathbf{0}，则A(k1ξ1+k2ξ2)=0\boldsymbol{A}\left(k_{1} \boldsymbol{\xi}_{1}+k_{2} \boldsymbol{\xi}_{2}\right)=\mathbf{0}，其中k1,k2k_{1}, k_{2}是任意常数

基础解系和解的结构

基础解系

设ξ1,ξ2,⋯ ,ξn−r\boldsymbol{\xi}_{1}, \boldsymbol{\xi}_{2}, \cdots, \boldsymbol{\xi}_{n-r}满足

  • 是方程组Ax=0\boldsymbol{A x}=\mathbf{0}的解
  • 线性无关
  • 方程组Ax=0\boldsymbol{A x}=\mathbf{0}的任一解均可由ξ1,ξ2,⋯ ,ξn−r\boldsymbol{\xi}_{1}, \boldsymbol{\xi}_{2}, \cdots, \boldsymbol{\xi}_{n-r}线性表出，则称ξ1,ξ2,⋯ ,ξn−r\boldsymbol{\xi}_{1}, \boldsymbol{\xi}_{2}, \cdots, \boldsymbol{\xi}_{n-r}为Ax=0\boldsymbol{A x}=\mathbf{0}的基础解系

通解

设ξ1,ξ2,⋯ ,ξn−r\boldsymbol{\xi}_{1}, \boldsymbol{\xi}_{2}, \cdots, \boldsymbol{\xi}_{n-r}是Ax=0\boldsymbol{A x}=\mathbf{0}的基础解系，则k1ξ1+k2ξ2+⋯+kn−rξn−rk_{1} \boldsymbol{\xi}_{1}+k_{2} \boldsymbol{\xi}_{2}+\cdots+k_{n-r} \boldsymbol{\xi}_{n-r}是方程组Ax=0\boldsymbol{A} \boldsymbol{x}=\mathbf{0}的通解，其中k1,k2,⋯ ,kn−rk_{1}, k_{2}, \cdots, k_{n-r}是任意常数

求解方法与步骤

注意只能进行行变换

  • 将系数矩阵A\boldsymbol{A}作初等行变换化成阶梯形矩阵B\boldsymbol{B}（或最简阶梯形矩阵B\boldsymbol{B}），初等行变换将方程组化为同解方程组，故Ax=0\boldsymbol{A} \boldsymbol{x}=\mathbf{0}和Bx=0\boldsymbol{B} \boldsymbol{x}=\mathbf{0}同解，只需解Bx=0\boldsymbol{B} \boldsymbol{x}=\mathbf{0}即可。设r(A)=rr(\boldsymbol{A})=r

    \begin{align}\boldsymbol{A} \xrightarrow{\text { 初等行变换 }} \boldsymbol{B}=\left[\begin{array}{cccccc} c_{11} & c_{12} & \cdots & c_{1 r} & \cdots & c_{1 n} \\ 0 & c_{22} & \cdots & c_{2 r} & \cdots & c_{2 n} \\ \vdots & \vdots & & \vdots & & \vdots \\ 0 & 0 & \cdots & c_{r r} & \cdots & c_{r n} \\ 0 & 0 & \cdots & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots & & \vdots \\ 0 & 0 & \cdots & 0 & \cdots & 0 \end{array}\right]_{m \times n}\end{align}

    其中，mm是原方程组中方程个数，nn是末知量个数

  • 按列找出一个秩为rr的子矩阵，剩余列位置的末知数设为自由变量

  • 按基础解系定义求出ξ1,ξ2,⋯ ,ξn−r\boldsymbol{\xi}_{1}, \boldsymbol{\xi}_{2}, \cdots, \boldsymbol{\xi}_{n-r}，并写出通解

例题

求齐次线性方程组的通解

\begin{align}\left\{\begin{array}{l} x_{1}+x_{2}-3 x_{4}-x_{5}=0, \\ x_{1}-x_{2}+2 x_{3}-x_{4}=0, \\ 4 x_{1}-2 x_{2}+6 x_{3}+3 x_{4}-4 x_{5}=0, \\ 2 x_{1}+4 x_{2}-2 x_{3}+4 x_{4}-7 x_{5}=0 \end{array}\right.\end{align}

解：将系数矩阵作初等行变换，化成阶梯形矩阵

()括号里面的运算表示行变换

\begin{align}\begin{array}{l} \boldsymbol{A}=\left[\begin{array}{ccccc} 1 & 1 & 0 & -3 & -1 \\ 1 & -1 & 2 & -1 & 0 \\ 4 & -2 & 6 & 3 & -4 \\ 2 & 4 & -2 & 4 & -7 \end{array}\right] \\ \xrightarrow[\text { (4) }-2(1)]{\substack{(2)-(1) \\ (3)-4(1)}}\left[\begin{array}{ccccc} 1 & 1 & 0 & -3 & -1 \\ 0 & -2 & 2 & 2 & 1 \\ 0 & -6 & 6 & 15 & 0 \\ 0 & 2 & -2 & 10 & -5 \end{array}\right] \\ \xrightarrow[\text { (4) }+ \text { (2) }]{\text { (3) }-3(2)}\left[\begin{array}{ccccc} 1 & 1 & 0 & -3 & -1 \\ 0 & -2 & 2 & 2 & 1 \\ 0 & 0 & 0 & 9 & -3 \\ 0 & 0 & 0 & 12 & -4 \end{array}\right] \\ \xrightarrow[\frac{1}{3}(3)]{\text { (4) }-\frac{4}{3}(3)}\left[\begin{array}{ccccc} 1 & 1 & 0 & -3 & -1 \\ 0 & -2 & 2 & 2 & 1 \\ 0 & 0 & 0 & 3 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]=\boldsymbol{B}\\ \end{array}\end{align}

行阶梯型进阶

B\boldsymbol{B}为行阶梯型

  • 若有0行，全在下方
  • 从行上看，自左边起出现连续0的个数自上而下严格单增

然后这边还能化成行最简阶梯型

\begin{align}\begin{aligned} \boldsymbol{B} & =\left[\begin{array}{rrrrr} 1 & 1 & 0 & -3 & -1 \\ 0 & -2 & 2 & 2 & 1 \\ 0 & 0 & 0 & 3 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow[\frac{1}{3}(3)]{-\frac{1}{2}(2)}\left[\begin{array}{rrrrrr} 1 & 1 & 0 & -3 & -1 \\ 0 & 1 & -1 & -1 & -\frac{1}{2} \\ 0 & 0 & 0 & 1 & -\frac{1}{3} \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \\ & \xrightarrow{(1)-(2)}\left[\begin{array}{rrrrr} 1 & 0 & 1 & -2 & -\frac{1}{2} \\ 0 & 1 & -1 & -1 & -\frac{1}{2} \\ 0 & 0 & 0 & 1 & -\frac{1}{3} \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{(1)+2(3)}\left[\begin{array}{lllll} 1 & 0 & 1 & 0 & -\frac{7}{6} \\ 0 & 1 & -1 & 0 & -\frac{5}{6} \\ 0 & 0 & 0 & 1 & -\frac{1}{3} \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]=\boldsymbol{C} \end{aligned}\end{align}

行最简阶梯型（包含行阶梯型的所有要求）

  • 台脚（以C\boldsymbol{C}来看，第二行第三行都是台脚）位置元素都是1
  • 台脚正上方元素都为0

则Ax=0\boldsymbol{A} \boldsymbol{x}=\mathbf{0}和Bx=0\boldsymbol{B} \boldsymbol{x}=\mathbf{0}是同解方程组，且r(A)=r(B)=3r(\boldsymbol{A})=r(\boldsymbol{B})=3

按列找出一个秩为3的子矩阵，可取第一、二、四列，则剩余第三、五列位置的元素x3,x5x_{3}, x_{5}即设为自由末知量，我们求通解

方法一：

取自由未知量x3=k1,x5=3k2x_{3}=k_{1}, x_{5}=3 k_{2}，代人方程得

\begin{align}\begin{array}{l} x_{4}=k_{2}, \\ x_{2}=x_{3}+x_{4}+\frac{1}{2} x_{5}=k_{1}+\frac{5}{2} k_{2}, \\ x_{1}=-x_{2}+3 x_{4}+x_{5}=-k_{1}+\frac{7}{2} k_{2} . \end{array}\end{align}

由此得通解

\begin{align}\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right]=\left[\begin{array}{c} -k_{1}+\frac{7}{2} k_{2} \\ k_{1}+\frac{5}{2} k_{2} \\ k_{1}+0 \\ 0+k_{2} \\ 0+3 k_{2} \end{array}\right]=k_{1}\left[\begin{array}{c} -1 \\ 1 \\ 1 \\ 0 \\ 0 \end{array}\right]+k_{2}\left[\begin{array}{c} \frac{7}{2} \\ \frac{5}{2} \\ 0 \\ 1 \\ 3 \end{array}\right]\end{align}

其中k1,k2k_{1}, k_{2}是任意常数

方法二：

\begin{align}\boldsymbol{\xi}_{1}=(h_1,h_2,h_3,h_4,h_5)^\mathrm{T}\\\boldsymbol{\xi}_{2}=(l_1,l_2,l_3,l_4,l_5)^\mathrm{T}\end{align}

中ξ1\boldsymbol{\xi}_{1}和ξ2\boldsymbol{\xi}_{2}要线性无关，那么我们可以先把剩余第三、五列位置的元素x3,x5x_{3}, x_{5}设为1、0和0、1，然后把B\boldsymbol{B}的每一行（至下而上）与ξ1\boldsymbol{\xi}_{1}（之后而前）相乘使其等于0，然后在把B\boldsymbol{B}的每一行（至下而上）与ξ2\boldsymbol{\xi}_{2}（之后而前）相乘使其等于0，我们就可以得到

\begin{align}\boldsymbol{\xi}_{1}=(1,1,1,0,0)^\mathrm{T}\\\boldsymbol{\xi}_{2}=(\frac{7 }{2},\frac{5 }{2},0,1,3)^\mathrm{T}\end{align}

所以k1ξ1+k2ξ2k_1\boldsymbol{\xi}_{1}+k_2\boldsymbol{\xi}_{2}是通解

非齐次线性方程组

方程组，后面用（II）来表示，称为mm个方程，nn个未知量的非齐次线性方程组

\begin{align}\left\{\begin{array}{c} a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 n} x_{n}=b_{1}, \\ a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 n} x_{n}=b_{2}, \\ \cdots \cdots \\ a_{m 1} x_{1}+a_{m 2} x_{2}+\cdots+a_{m n} x_{n}=b_{m} \end{array}\right.\end{align}

其向量形式为x1α1+x2α2+⋯+xnαn=bx_{1} \boldsymbol{\alpha}_{1}+x_{2} \boldsymbol{\alpha}_{2}+\cdots+x_{n} \boldsymbol{\alpha}_{n}=\boldsymbol{b}，其中

\begin{align}\boldsymbol{\alpha}_{j}=\left[\begin{array}{c} a_{1 j} \\ a_{2 j} \\ \vdots \\ a_{m j} \end{array}\right], j=1,2, \cdots, n, \quad \boldsymbol{b}=\left[\begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \end{array}\right]\end{align}

其矩阵形式为Ax=b\boldsymbol{A x}=\boldsymbol{b}，其中

\begin{align}\boldsymbol{A}=\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} \end{array}\right], \quad \boldsymbol{x}=\left[\begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array}\right]\end{align}

矩阵[a11a12⋯a1nb1a21a22⋯a2nb2⋮⋮⋮⋮am1am2⋯amnbm]\left[\begin{array}{cccc:c}a_{11} & a_{12} & \cdots & a_{1 n} & b_{1} \\ a_{21} & a_{22} & \cdots & a_{2 n} & b_{2} \\ \vdots & \vdots & & \vdots & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} & b_{m}\end{array}\right]称为矩阵A\boldsymbol{A}的增广矩阵，简记成[Ab][\begin{array}{c:c}\boldsymbol{A}&\boldsymbol{b}\end{array}]

有解的条件

若r(A)≠r([A,b])r(\boldsymbol{A}) \neq r([\boldsymbol{A}, \boldsymbol{b}])（b\boldsymbol{b}不能由α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}线性表出），则方程组（II）无解

若r(A)=r([A,b])=nr(\boldsymbol{A})=r([\boldsymbol{A}, \boldsymbol{b}])=n（即α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}线性无关，α1,α2,⋯ ,αn,b\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n}, \boldsymbol{b}线性相关），则方程组（II）有唯一解

若r(A)=r([A,b])=r<nr(\boldsymbol{A})=r([\boldsymbol{A}, \boldsymbol{b}])=r<n，则方程组（II）有无穷多解

解的性质

设η1,η2,η\boldsymbol{\eta}_{1}, \boldsymbol{\eta}_{2}, \boldsymbol{\eta}是非齐次线性方程组Ax=b\boldsymbol{A x}=\boldsymbol{b}的解，ξ\boldsymbol{\xi}是对应齐次线性方程组Ax=0\boldsymbol{A x}=\mathbf{0}的解，则 :

  • η1−η2\boldsymbol{\eta}_{1}-\boldsymbol{\eta}_{2}是Ax=0\boldsymbol{A x}=\mathbf{0}的解
  • kξ+ηk \boldsymbol{\xi}+\boldsymbol{\eta}是Ax=b\boldsymbol{A x}=\boldsymbol{b}的解

求解方法与步骤

注意只能进行行变换

将增广矩阵作初等行变换化成阶梯形（或最简阶梯形）矩阵，求出对应齐次线性方程组的通解，再加上一个非齐次线性方程组的特解即是非齐次线性方程组的通解

  1. 写出Ax=b\boldsymbol{A x}=\boldsymbol{b}的导出方程组Ax=0\boldsymbol{A x}=\mathbf{0}，并求Ax=0\boldsymbol{A x}=\mathbf{0}的通解k1ξ1+k2ξ2+⋯+kn−ξn−rk_{1} \boldsymbol{\xi}_{1}+k_{2} \boldsymbol{\xi}_{2}+\cdots+k_{n-} \boldsymbol{\xi}_{n-r}
  2. 求出Ax=b\boldsymbol{A x}=\boldsymbol{b}的一个特解η\boldsymbol{\eta}
  3. 则Ax=b\boldsymbol{A x}=\boldsymbol{b}的通解为k1ξ1+k2ξ2+⋯+kn−rξn−r+ηk_{1} \boldsymbol{\xi}_{1}+k_{2} \boldsymbol{\xi}_{2}+\cdots+k_{n-r} \boldsymbol{\xi}_{n-r}+\boldsymbol{\eta}，其中k1,k2,⋯ ,kn−rk_{1}, k_{2}, \cdots, k_{n-r}为任意常数

例题

求解非齐次线性方程组，并用对应的齐次线性方程组的基础解系表示通解

\begin{align}\left\{\begin{array}{l} x_{1}+5 x_{2}-x_{3}-x_{4}=-1, \\ x_{1}-2 x_{2}+x_{3}+3 x_{4}=3, \\ 3 x_{1}+8 x_{2}-x_{3}+x_{4}=1, \\ x_{1}-9 x_{2}+3 x_{3}+7 x_{4}=7 \end{array}\right.\end{align}

()括号里面的运算表示行变换

解：对增广矩阵作初等行变换化成阶梯形矩阵

\begin{align}\begin{array}{l}\left[\begin{array}{l:l} \boldsymbol{A} & \boldsymbol{b} \end{array}\right]=\left[\begin{array}{cccc:c} 1 & 5 & -1 & -1 & -1 \\ 1 & -2 & 1 & 3 & 3 \\ 3 & 8 & -1 & 1 & 1 \\ 1 & -9 & 3 & 7 & 7 \end{array}\right] \xrightarrow[(4)-(1)]{\substack{(2)-(1) \\ (3)-3(1)}}\left[\begin{array}{cccc:c} 1 & 5 & -1 & -1 & -1 \\ 0 & -7 & 2 & 4 & 4 \\ 0 & -7 & 2 & 4 & 4 \\ 0 & -14 & 4 & 8 & 8 \end{array}\right]\\ \xrightarrow[(4)-2(2)]{(3)-(2)}\left[\begin{array}{cccc:c} 1 & 5 & -1 & -1 & -1 \\ 0 & -7 & 2 & 4 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\\ \end{array}\end{align}

解的条件：

  • 对增广矩阵进行初等变换成阶梯形矩阵

  • 判断阶梯型矩阵的秩

    • 当r(A)≠(Ab)r(\boldsymbol{A})\ne (\begin{array}{l:l}\boldsymbol{A} & \boldsymbol{b}\end{array})，无解

    • 当r(A)=(Ab)=nr(\boldsymbol{A})= (\begin{array}{l:l}\boldsymbol{A} & \boldsymbol{b}\end{array})=n，有唯一解，其中nn的个数为未知数的个数，就是题目中的x1,x2,x3,x4x_1,x_2,x_3,x_4

    • 当r(A)=(Ab)<nr(\boldsymbol{A})= (\begin{array}{l:l}\boldsymbol{A} & \boldsymbol{b}\end{array})<n，有无穷多解，nn的个数同上

  • 求出齐次方程的通解（增广矩阵中(Ab)(\begin{array}{l:l}\boldsymbol{A} & \boldsymbol{b}\end{array})，求通解只求A\boldsymbol{A}，不需要关里面的b\boldsymbol{b}）加上一个非齐次方程的特解

方法一：

令自由未知量x3=k1,x4=k2x_{3}=k_{1}, x_{4}=k_{2}，代人得

\begin{align}\begin{array}{c} x_{2}=-\frac{1}{7}\left(4-2 k_{1}-4 k_{2}\right)=-\frac{4}{7}+\frac{2}{7} k_{1}+\frac{4}{7} k_{2}, \\ x_{1}=-1+k_{1}+k_{2}-5\left(-\frac{4}{7}+\frac{2}{7} k_{1}+\frac{4}{7} k_{2}\right)=\frac{13}{7}-\frac{3}{7} k_{1}-\frac{13}{7} k_{2} \end{array}\end{align}

得通解为

\begin{align}\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right]=\left[\begin{array}{c} \frac{13}{7}-\frac{3}{7} k_{1}-\frac{13}{7} k_{2} \\ -\frac{4}{7}+\frac{2}{7} k_{1}+\frac{4}{7} k_{2} \\ k_{1} \\ k_{2} \end{array}\right]=\left[\begin{array}{c} \frac{13}{7} \\ -\frac{4}{7} \\ 0 \\ 0 \end{array}\right]+k_{1}\left[\begin{array}{c} -\frac{3}{7} \\ \frac{2}{7} \\ 1 \\ 0 \end{array}\right]+k_{2}\left[\begin{array}{c} -\frac{13}{7} \\ \frac{4}{7} \\ 0 \\ 1 \end{array}\right]\end{align}

其中k1,k2k_{1}, k_{2}是任意常数，[−37,27,1,0]T,[−137,47,0,1]T\left[-\frac{3}{7}, \frac{2}{7}, 1,0\right]^{\mathrm{T}},\left[-\frac{13}{7}, \frac{4}{7}, 0,1\right]^{\mathrm{T}}为对应的齐次线性方程组的基础解系

方法二：

得出通解为：

\begin{align}\boldsymbol{\xi}_{1}=(-3,2,7,0)^\mathrm{T}\\\boldsymbol{\xi}_{2}=(-13,4,0,7)^\mathrm{T}\end{align}

接下来求齐次方程的特解，凡是自由项的位置统统为0，然后需要带进去求解：

(0∗4)+(0∗2)+(−7∗u1)=4(0*4)+(0*2)+(-7*u_1)=4求的u1=−47u_1=-\frac{4}{7}

(0∗−1)+(0∗−1)+(5∗u1)+(1∗u2)=−1(0*-1)+(0*-1)+(5*u_1)+(1*u_2)=-1，我们直接把u1=−47u_1=-\frac{4}{7}代入方程，可以求得u2=137u_2=\frac{13}{7}，就可以得出非齐次方程的特解为：

\begin{align}\eta =(\frac{13}{7},-\frac{4}{7},0 ,0)^\mathrm{T}\end{align}

所以k1ξ1+k2ξ2+ηk_1\boldsymbol{\xi}_{1}+k_2\boldsymbol{\xi}_{2}+\eta是通解

已知线性方程组

\begin{align}\left\{\begin{array}{l} x_{1}+x_{2}+x_{3}+x_{4}=1, \\ 3 x_{1}+2 x_{2}+x_{3}+x_{4}=a, \\ x_{2}+2 x_{3}+2 x_{4}=3, \\ 5 x_{1}+4 x_{2}+3 x_{3}+4 x_{4}=b \end{array}\right.\end{align}

则a,ba, b为何值时，方程组无解？a,ba, b为何值时，方程组有解？方程组有解时，求其全部解

解：对方程组的增广矩阵[Ab][\begin{array}{l:l}\boldsymbol{A} & \boldsymbol{b}\end{array}]作初等行变换

\begin{align}\begin{aligned} {\left[\begin{array}{l:l} \boldsymbol{A} & \boldsymbol{b} \end{array}\right] } & =\left[\begin{array}{llll:l} 1 & 1 & 1 & 1 & 1 \\ 3 & 2 & 1 & 1 & a \\ 0 & 1 & 2 & 2 & 3 \\ 5 & 4 & 3 & 4 & b \end{array}\right] \rightarrow\left[\begin{array}{cccc:c} 1 & 1 & 1 & 1 & 1 \\ 0 & -1 & -2 & -2 & a-3 \\ 0 & 1 & 2 & 2 & 3 \\ 0 & -1 & -2 & -1 & b-5 \end{array}\right] \\ & \longrightarrow\left[\begin{array}{llll:c} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 2 & 3 \\ 0 & 0 & 0 & 0 & a \\ 0 & 0 & 0 & 1 & b-2 \end{array}\right] \longrightarrow\left[\begin{array}{cccc:c} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 2 & 3 \\ 0 & 0 & 0 & 1 & b-2 \\ 0 & 0 & 0 & 0 & a \end{array}\right] \\ & \longrightarrow\left[\begin{array}{cccc:c} 1 & 0 & -1 & -1 & -2 \\ 0 & 1 & 2 & 2 & 3 \\ 0 & 0 & 0 & 1 & b-2 \\ 0 & 0 & 0 & 0 & a \end{array}\right] . \end{aligned}\end{align}

当a≠0,ba \neq 0, b任意时，r(A)=3≠r([A,b])=4r(\boldsymbol{A})=3 \neq r([\boldsymbol{A}, \boldsymbol{b}])=4，方程组无解

当a=0,ba=0, b任意时，r(A)=3=r([A,b])<4r(\boldsymbol{A})=3=r([\boldsymbol{A}, \boldsymbol{b}])<4，方程组有无穷多解

后面通解懒得写了，解法一样的