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\begin_body
\begin_layout Section
(Gauss) 消元法
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
简介
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
在第一章里我们已经研究过线性方程组的一种特殊情形, 即
\series bold
线性方程组所含方程的个数等于未知量的个数
\series default
, 且方程组的系数行列式不等于零的情形.
\end_layout
\begin_layout Standard
求解线性方程组是线性代数最主要的任务之一, 此类问题在科学技术与经济管理领域有着相当广泛的应用, 因而有必要从更普遍的角度来讨论线性方程组的一般理论.
\end_layout
\begin_layout Standard
本章主要讨论
\series bold
一般线性方程组的
\color brown
解法
\series default
\color inherit
,
\series bold
线性方程组解的
\color brown
存在性
\series default
\color inherit
和
\series bold
线性方程组
\color brown
解的结构
\series default
\color inherit
等内容.
\end_layout
\end_deeper
\begin_layout Subsection
线性方程组的初等变换
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
消元法的一个例子
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
用消元法求解下列线性方程组:
\begin_inset Formula $\systeme[x_{1},x_{2},x_{3}]{2x_{1}+2x_{2}-x_{3}=6,x_{1}-2x_{2}+4x_{3}=3,5x_{1}+7x_{2}+x_{3}=28}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align}
\systeme[x_{1},x_{2},x_{3}]{2x_{1}+2x_{2}-x_{3}=6,x_{1}-2x_{2}+4x_{3}=3,5x_{1}+7x_{2}+x_{3}=28} & \Longrightarrow\systeme[x_{1},x_{2},x_{3}]{x_{1}-2x_{2}+4x_{3}=3,2x_{1}+2x_{2}-x_{3}=6,5x_{1}+7x_{2}+x_{3}=28}\longleftrightarrow\begin{bmatrix}1 & -2 & 4 & 3\\
2 & 2 & -1 & 6\\
5 & 7 & 1 & 28
\end{bmatrix}\label{eq:3.1-1}\\
& \hspace{-5em}\Longrightarrow\systeme[x_{1},x_{2},x_{3}]{x_{1}-2x_{2}+4x_{3}=3,6x_{2}-9x_{3}=0,17x_{2}-19x_{3}=13}\longleftrightarrow\begin{bmatrix}1 & -2 & 4 & 3\\
0 & 6 & -9 & 0\\
0 & 17 & -19 & 13
\end{bmatrix}\label{eq:3.1-2}\\
& \hspace{-5em}\Longrightarrow\systeme[x_{1},x_{2},x_{3}]{x_{1}-2x_{2}+4x_{3}=3,6x_{2}-9x_{3}=0,\frac{13}{2}x_{3}=-\frac{221}{6}}\longleftrightarrow\begin{bmatrix}1 & -2 & 4 & 3\\
0 & 6 & -9 & 0\\
0 & 0 & \frac{13}{2} & -\frac{221}{6}
\end{bmatrix}\label{eq:3.1-3}\\
& \hspace{-5em}\Longrightarrow\systeme[x_{1},x_{2},x_{3}]{x_{1}-2x_{2}+4x_{3}=3,x_{2}-\frac{3}{2}x_{3}=0,x_{3}=-\frac{17}{3}}\longleftrightarrow\begin{bmatrix}1 & -2 & 4 & 3\\
0 & 1 & -\frac{3}{2} & 0\\
0 & 0 & 1 & -\frac{17}{3}
\end{bmatrix},\label{eq:3.1-4}
\end{align}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align}
\systeme[x_{1},x_{2},x_{3}]{2x_{1}+2x_{2}-x_{3}=6,x_{1}-2x_{2}+4x_{3}=3,5x_{1}+7x_{2}+x_{3}=28} & \Longrightarrow\systeme[x_{1},x_{2},x_{3}]{x_{1}-2x_{2}+4x_{3}=3,x_{2}-\frac{3}{2}x_{3}=0,x_{3}=-\frac{17}{3}}\longleftrightarrow\begin{bmatrix}1 & -2 & 4 & 3\\
0 & 1 & -\frac{3}{2} & 0\\
0 & 0 & 1 & -\frac{17}{3}
\end{bmatrix}\label{eq:3.1-5}\\
& \hspace{-5em}\Longrightarrow\systeme[x_{1},x_{2},x_{3}]{x_{1}=3-4x_{3}+2x_{2}=\frac{26}{3},x_{2}=\frac{3}{2}x_{3}=-\frac{17}{2},x_{3}=-\frac{17}{3}}\longleftrightarrow\begin{bmatrix}1 & 0 & 0 & \frac{26}{3}\\
0 & 1 & 0 & -\frac{17}{2}\\
0 & 0 & 1 & -\frac{17}{3}
\end{bmatrix}.\label{eq:3.1-6}
\end{align}
\end_inset
\end_layout
\begin_layout Standard
通常把过程 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
)-(
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-4"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 称为
\series bold
消元过程
\series default
, 矩阵 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-4"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 就是行阶梯形矩阵, 与之对应的方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-4"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 则称为
\series bold
行阶梯方程组
\series default
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求解线性方程与矩阵的初等变换
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
从上述解题过程可以看出, 用消元法求解线性方程组的具体作法就是对方程组反复实施以下三种变换:
\end_layout
\begin_layout Standard
(1).
交换某两个方程的位置;
\end_layout
\begin_layout Standard
(2).
用一个非零数乘某一个方程的两边;
\end_layout
\begin_layout Standard
(3).
将一个方程的倍数加到另一个方程上去.
\end_layout
\begin_layout Standard
以上这三种变换称为
\series bold
线性方程组的初等变换
\series default
.
而消元法的目的就是利用方程组的初等变换将原方程组
\series bold
化为阶梯形方程组
\series default
, 显然这个阶梯形方程组与原线性方程组同解, 解这个阶梯形方程组得到原方程组的解.
\end_layout
\begin_layout Standard
如果用矩阵表示其系数及常数项, 则将原方程组化为行阶梯形方程组的过程就是将对应矩阵化为行阶梯形矩阵的过程.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
回代过程与增广矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
将一个方程组化为行阶梯形方程组的步骤并不是唯一的, 所以,
\series bold
同一个方程组的行阶梯形方程组也不是唯一的
\series default
.
\end_layout
\begin_layout Standard
特别地, 我们还可以将一个一般的行阶梯形方程组化为
\series bold
行最简形方程组
\series default
, 从而使我们能直接 “读” 出该线性方程组的解.
\end_layout
\begin_layout Standard
通常把过程 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-5"
plural "false"
caps "false"
noprefix "false"
\end_inset
)-(
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-6"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 称为
\series bold
回代过程
\series default
.
\end_layout
\begin_layout Standard
从引例我们可得到如下启示: 用消元法解三元线性方程组的过程, 相当于对该方程组的
\series bold
增广矩阵
\series default
作初等行变换.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
线性方程组有解的判别定理
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
线性方程组解的一般理论
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
对一般线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-gnrl"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 是否有同样的结论? 答案是肯定的, 以下就一般线性方程组求解的问题进行讨论.
\end_layout
\begin_layout Standard
设有线性方程组
\begin_inset Formula
\begin{equation}
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=b_{1}\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=b_{2}\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}=b_{m}
\end{cases}\label{eq:3.1-gnrl}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
newpage
\end_layout
\end_inset
\end_layout
\begin_layout Standard
其矩阵形式为
\begin_inset Formula $Ax=b$
\end_inset
, 其中
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
A=\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{bmatrix},\quad x=\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix},\quad b=\begin{bmatrix}b_{1}\\
b_{2}\\
\vdots\\
b_{m}
\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Standard
称矩阵
\begin_inset Formula $\begin{bmatrix}A & b\end{bmatrix}$
\end_inset
(有时记为
\begin_inset Formula $\widetilde{A}$
\end_inset
) 为
\series bold
线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-gnrl"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的增广矩阵
\series default
.
\end_layout
\begin_layout Standard
当
\begin_inset Formula $b_{i}=0$
\end_inset
,
\begin_inset Formula $i=1,2,\cdots,m$
\end_inset
时, 线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-gnrl"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 称为
\series bold
齐次的
\series default
; 否则称为
\series bold
非齐次的
\series default
.
显然, 齐次线性方程组的矩阵形式为
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
Ax=0.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
线性方程组有解的判别定理
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
设
\begin_inset Formula $A=\left(a_{ij}\right)_{m\times n}$
\end_inset
,
\begin_inset Formula $n$
\end_inset
元齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
有非零解的充要条件是系数矩阵的秩
\begin_inset Formula $r(A)<n$
\end_inset
.
如:
\end_layout
\begin_layout Theorem
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&*&0
\backslash
\backslash
0&0&1&*&0
\backslash
\backslash
0&0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray},
\backslash
begin{bNiceArray}{ccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&0
\backslash
\backslash
0&1&*&0
\backslash
\backslash
0&0&0&0
\backslash
\backslash
0&0&0&0
\backslash
\backslash
0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray},
\backslash
xcancel{
\backslash
begin{bNiceArray}{ccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&0
\backslash
\backslash
0&1&*&0
\backslash
\backslash
0&0&
\backslash
boxed{1}&0
\backslash
\backslash
0&0&0&0
\backslash
\backslash
0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}}
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem
设
\begin_inset Formula $A=\left(a_{ij}\right)_{m\times n}$
\end_inset
,
\begin_inset Formula $n$
\end_inset
元非齐次线性方程组
\begin_inset Formula $Ax=b$
\end_inset
有解的充要条件是系数矩阵
\begin_inset Formula $A$
\end_inset
的秩等于增广矩阵
\begin_inset Formula $\widetilde{A}=\begin{bmatrix}A & b\end{bmatrix}$
\end_inset
的秩, 即
\begin_inset Formula
\[
r(A)=r(\widetilde{A}).
\]
\end_inset
如:
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&*&*
\backslash
\backslash
0&1&*&*&*
\backslash
\backslash
0&0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray},
\backslash
begin{bNiceArray}{ccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&*
\backslash
\backslash
0&1&*&*
\backslash
\backslash
0&0&1&*
\backslash
\backslash
0&0&0&0
\backslash
\backslash
0&0&0&0
\backslash
\backslash
0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray},
\backslash
xcancel{
\backslash
begin{bNiceArray}{ccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&*
\backslash
\backslash
0&1&*&*
\backslash
\backslash
0&0&1&*
\backslash
\backslash
0&0&0&
\backslash
boxed{1}
\backslash
\backslash
0&0&0&
\backslash
boxed{2}
\backslash
\backslash
0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}}
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
线性方程组有解的判别定理
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
记
\begin_inset Formula $\begin{bmatrix}A & b\end{bmatrix}=\widetilde{A}$
\end_inset
, 则上述定理的结果, 可简要总结如下:
\end_layout
\begin_layout ColumnsCenterAligned
\end_layout
\begin_deeper
\begin_layout Column
8cm
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
%
\backslash
begin{center}
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[ color=black, draw, fill=red!30,] (r1) { $r(A)=r(
\backslash
widetilde{A})=n
\backslash
Leftrightarrow Ax=b$ 有唯一解; };
\backslash
\backslash
\backslash
vspace{4mm}
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[ color=black, draw, fill=green!30,] (r2) { $r(A)=r(
\backslash
widetilde{A})<n
\backslash
Leftrightarrow Ax=b$ 有无穷多解; };
\backslash
\backslash
\backslash
vspace{4mm}
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[ color=black, draw, fill=blue!30,] (r3) { $r(A)
\backslash
neq r(
\backslash
widetilde{A})
\backslash
Leftrightarrow Ax=b$ 无解; };
\backslash
\backslash
\backslash
vspace{4mm}
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[ color=black, draw, fill=red!30,] (r4) { $r(A)=n
\backslash
Leftrightarrow Ax=0$ 只有零解; };
\backslash
\backslash
\backslash
vspace{4mm}
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[ color=black, draw, fill=green!30,] (r5) { $r(A)<n
\backslash
Leftrightarrow Ax=0$ 有非零解.
};
\end_layout
\begin_layout Plain Layout
%
\backslash
end{center}
\end_layout
\end_inset
\end_layout
\begin_layout Column
4cm
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
begin{center}
\end_layout
\begin_layout Plain Layout
\backslash
scriptsize
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[fill=red!30] (n1) {$
\backslash
begin{bNiceArray}{ccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&*
\backslash
\backslash
0&1&*&*
\backslash
\backslash
0&0&1&*
\backslash
\backslash
0&0&0&0
\backslash
\backslash
0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}$};
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[fill=red!30] (n2) {$
\backslash
begin{bNiceArray}{ccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&*
\backslash
\backslash
0&1&*&*
\backslash
\backslash
0&0&1&*
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}$};
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[fill=green!30] (n3) {$
\backslash
begin{bNiceArray}{ccccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&*&*&*
\backslash
\backslash
0&1&*&*&*&*
\backslash
\backslash
0&0&1&*&*&*
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}$};
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
node[fill=blue!30] (n4) {$
\backslash
begin{bNiceArray}{ccccc|c}
\end_layout
\begin_layout Plain Layout
1&*&*&*&*&*
\backslash
\backslash
0&1&*&*&*&*
\backslash
\backslash
0&0&1&*&*&*
\backslash
\backslash
0&0&0&0&0&*
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}$};
\end_layout
\begin_layout Plain Layout
\backslash
end{center}
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Block
\begin_inset Argument 1
status open
\begin_layout Plain Layout
1
\end_layout
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
draw[overlay,->,very thick,red,opacity=.5] (r1.east) -- (n1.west);
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
draw[overlay,->,very thick,red,opacity=.5] (r1.east) -- (n2.west);
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Block
\begin_inset Argument 1
status open
\begin_layout Plain Layout
2
\end_layout
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
draw[overlay,->,very thick,green,opacity=.5] (r2.east) -- (n3.west);
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Block
\begin_inset Argument 1
status open
\begin_layout Plain Layout
3
\end_layout
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
draw[overlay,->,very thick,blue,opacity=.5] (r3.east) -- (n4.west);
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Block
\begin_inset Argument 1
status open
\begin_layout Plain Layout
4
\end_layout
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
draw[overlay,->,very thick,purple,opacity=.5] (r4.east) -- (n1.west);
\end_layout
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
draw[overlay,->,very thick,purple,opacity=.5] (r4.east) -- (n2.west);
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Block
\begin_inset Argument 1
status open
\begin_layout Plain Layout
5
\end_layout
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
tikz[remember picture]
\backslash
draw[overlay,->,very thick,black,opacity=.5] (r5.east) -- (n3.west);
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
线性方程组有解的判别定理总结
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
而定理的证明实际上给出了求解线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.1-gnrl"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的方法:
\end_layout
\begin_layout Itemize
对非齐次线性方程组
\end_layout
\begin_layout Itemize
将增广矩阵
\begin_inset Formula $\widetilde{A}$
\end_inset
化为行阶梯形矩阵, 便可直接判断其是否有解;
\end_layout
\begin_deeper
\begin_layout Itemize
若有解, 化为行最简形矩阵, 便可直接写出其全部解.
\end_layout
\begin_layout Itemize
其中要注意, 当
\begin_inset Formula $r(A)=r(\widetilde{A})=r<n$
\end_inset
时,
\begin_inset Formula $\widetilde{A}$
\end_inset
的行阶梯形矩阵中含有
\begin_inset Formula $r$
\end_inset
个非零行, 把这
\begin_inset Formula $r$
\end_inset
行的第一个非零元所对应的未知量作为非自由量, 其余
\begin_inset Formula $n-r$
\end_inset
个作为自由未知量.
\end_layout
\end_deeper
\begin_layout Itemize
对齐次线性方程组, 将其系数矩阵化为行最简形矩阵, 便可直接写出其全部解.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
例题选讲
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
例题选讲
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
判断下列方程组是否有解? 如有解, 是否有唯一的一组解?
\begin_inset Formula
\[
\begin{cases}
x_{1}+2x_{2}-3x_{3}+x_{4}=1\\
x_{1}+x_{2}+x_{3}+x_{4}=0
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
方程组的系数矩阵
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 2 & -3 & 1\\
1 & 1 & 1 & 1
\end{bmatrix},
\]
\end_inset
显然
\begin_inset Formula $A$
\end_inset
有一个
\begin_inset Formula $2$
\end_inset
阶子式
\begin_inset Formula $\begin{vmatrix}1 & 2\\
1 & 1
\end{vmatrix}=-1\neq0$
\end_inset
, 因此
\begin_inset Formula $r(A)=2$
\end_inset
.
增广矩阵
\begin_inset Formula $\widetilde{A}=\begin{bmatrix}1 & 2 & -3 & 1 & 1\\
1 & 1 & 1 & 1 & 0
\end{bmatrix}$
\end_inset
, 显然
\begin_inset Formula $r(\widetilde{A})=2$
\end_inset
, 因此该方程组有解.
但方程组的未知数个数为
\begin_inset Formula $4$
\end_inset
, 因此应有无穷多组解.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
例题选讲
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
判断方程组是否有解?
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\end_layout
\begin_layout Example
\begin_inset Formula
\[
\begin{cases}
-3x_{1}+x_{2}+4x_{3}=-1\\
x_{1}+x_{2}+x_{3}=0\\
-2x_{1}+x_{3}=-1\\
x_{1}+x_{2}-2x_{3}=0
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
利用初等变换法求增广矩阵
\begin_inset Formula $\widetilde{A}$
\end_inset
的秩.
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\begin{aligned}\begin{bmatrix}-3 & 1 & 4 & -1\\
1 & 1 & 1 & 0\\
-2 & 0 & 1 & -1\\
1 & 1 & -2 & 0
\end{bmatrix} & \xrightarrow{r_{1}\leftrightarrow r_{2}}\begin{bmatrix}1 & 1 & 1 & 0\\
-3 & 1 & 4 & -1\\
-2 & 0 & 1 & -1\\
1 & 1 & -2 & 0
\end{bmatrix}\xrightarrow[r_{4}-r_{1}]{{r_{2}+3r_{1}\atop r_{3+2r_{1}}}}\begin{bmatrix}1 & 1 & 1 & 0\\
0 & 4 & 7 & -1\\
0 & 2 & 3 & -1\\
0 & 0 & -3 & 0
\end{bmatrix}\\
& \hspace{-5em}\xrightarrow{r_{2}\leftrightarrow r_{3}}\begin{bmatrix}1 & 1 & 1 & 0\\
0 & 2 & 3 & -1\\
0 & 4 & 7 & -1\\
0 & 0 & -3 & 0
\end{bmatrix}\xrightarrow{r_{3}-2r_{2}}\begin{bmatrix}1 & 1 & 1 & 0\\
0 & 2 & 3 & -1\\
0 & 0 & 1 & 1\\
0 & 0 & -3 & 0
\end{bmatrix}\xrightarrow{r_{4}+3r_{3}}\begin{bmatrix}1 & 1 & 1 & 0\\
0 & 2 & 3 & -1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 3
\end{bmatrix}.
\end{aligned}
\]
\end_inset
因此
\begin_inset Formula $r(A)=3$
\end_inset
,
\begin_inset Formula $r(\widetilde{A})=4$
\end_inset
.
由于
\begin_inset Formula $r(A)\neq r(\widetilde{A})$
\end_inset
, 故原方程组无解.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
例题选讲
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:ref-from-sec.6"
\end_inset
\begin_inset Argument 1
status open
\begin_layout Plain Layout
E01
\end_layout
\end_inset
求解齐次线性方程组
\begin_inset Formula $\begin{cases}
x_{1}+2x_{2}+2x_{3}+x_{4}=0\\
2x_{1}+x_{2}-2x_{3}-2x_{4}=0\\
x_{1}-x_{2}-4x_{3}-3x_{4}=0
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Solution*
对系数矩阵
\begin_inset Formula $A$
\end_inset
施行初等行变换.
\begin_inset Formula
\[
\begin{aligned}A & =\begin{bmatrix}1 & 2 & 2 & 1\\
2 & 1 & -2 & -2\\
1 & -1 & -4 & -3
\end{bmatrix}\xrightarrow[r_{3}-r_{1}]{r_{2}-2r_{1}}\begin{bmatrix}1 & 2 & 2 & 1\\
0 & -3 & -6 & -4\\
0 & -3 & -6 & -4
\end{bmatrix}\\
& \xrightarrow[r_{2}\div(-3)]{r_{3}-r_{2}}\begin{bmatrix}1 & 2 & 2 & 1\\
0 & 1 & 2 & 4/3\\
0 & 0 & 0 & 0
\end{bmatrix}\xrightarrow{r_{1}-2r_{2}}\begin{bmatrix}1 & 0 & -2 & -5/3\\
0 & 1 & 2 & 4/3\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{aligned}
\]
\end_inset
即得与原方程同解的方程组
\begin_inset Formula
\[
\begin{cases}
x_{1}=2x_{3}+\frac{5}{3}x_{4}\\
x_{2}=-2x_{3}-\frac{4}{3}x_{4}
\end{cases}\left(x_{3},x_{4}\text{ 可取任意实数. }\right)
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
令
\begin_inset Formula $x_{3}=c_{1}$
\end_inset
,
\begin_inset Formula $x_{4}=c_{2}$
\end_inset
, 把它写成向量形式为
\begin_inset Formula
\[
\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}=c_{1}\begin{bmatrix}2\\
-2\\
1\\
0
\end{bmatrix}+c_{2}\begin{bmatrix}\frac{5}{3}\\
-\frac{4}{3}\\
0\\
1
\end{bmatrix}.
\]
\end_inset
它表达了方程组的全部解.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
例题选讲
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
E02
\end_layout
\end_inset
解线性方程组
\begin_inset Formula $\begin{cases}
x_{1}+5x_{2}-x_{3}-x_{4}=-1\\
x_{1}-2x_{2}+x_{3}+3x_{4}=3\\
3x_{1}+8x_{2}-x_{3}+x_{4}=1\\
x_{1}-9x_{2}+3x_{3}+7x_{4}=7
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Solution*
对增广矩阵
\begin_inset Formula $\begin{bmatrix}A & b\end{bmatrix}$
\end_inset
施以初等变换, 化为阶梯形矩阵:
\begin_inset Formula
\[
\begin{aligned}\begin{bmatrix}A & b\end{bmatrix} & =\begin{bmatrix}1 & 5 & -1 & -1 & -1\\
1 & -2 & 1 & 3 & 3\\
3 & 8 & -1 & 1 & 1\\
1 & -9 & 3 & 7 & 7
\end{bmatrix}\rightarrow\begin{bmatrix}1 & 5 & -1 & -1 & -1\\
0 & -7 & 2 & 4 & 4\\
0 & -7 & 2 & 4 & 4\\
0 & -14 & 4 & 8 & 8
\end{bmatrix}\\
& \rightarrow\begin{bmatrix}1 & 5 & -1 & -1 & -1\\
0 & -7 & 2 & 4 & 4\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 5 & -1 & -1 & -1\\
0 & 1 & -\frac{2}{7} & -\frac{4}{7} & -\frac{4}{7}\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{bmatrix},
\end{aligned}
\]
\end_inset
所以
\begin_inset Formula $r(\begin{bmatrix}A & b\end{bmatrix})=r(A)=2<4$
\end_inset
, 故方程组有无穷多解.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
利用上式回代
\begin_inset Formula
\[
\xrightarrow{\text{ 回代 }}\begin{bmatrix}1 & 0 & 3/7 & 13/7 & 13/7\\
0 & 1 & -2/7 & -4/7 & -4/7\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}\Longrightarrow\begin{cases}
x_{1}=\frac{13}{7}-\frac{3}{7}x_{3}-\frac{13}{7}x_{4}\\
x_{2}=-\frac{4}{7}+\frac{2}{7}x_{3}+\frac{4}{7}x_{4}
\end{cases}
\]
\end_inset
取
\begin_inset Formula $x_{3}=c_{1}$
\end_inset
,
\begin_inset Formula $x_{4}=c_{2}$
\end_inset
, (
\begin_inset Formula $c_{1},c_{2}$
\end_inset
为任意常数), 由此方程组的全部解为
\begin_inset Formula
\[
\begin{cases}
x_{1}=\frac{13}{7}-\frac{3}{7}c_{1}-\frac{13}{7}c_{2}\\
x_{2}=-\frac{4}{7}+\frac{2}{7}c_{1}+\frac{4}{7}c_{2}\\
x_{3}=c_{1}\\
x_{4}=c_{2}
\end{cases}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
例题选讲
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
解线性方程组
\begin_inset Formula $\begin{cases}
x_{1}+x_{2}+2x_{3}+3x_{4}=1\\
x_{2}+x_{3}-4x_{4}=1\\
x_{1}+2x_{2}+3x_{3}-x_{4}=4\\
2x_{1}+3x_{2}-x_{3}-x_{4}=-6
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A & b\end{bmatrix} & =\begin{bmatrix}1 & 1 & 2 & 3 & 1\\
0 & 1 & 1 & -4 & 1\\
1 & 2 & 3 & -1 & 4\\
2 & 3 & -1 & -1 & -6
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 1 & 2 & 3 & 1\\
0 & 1 & 1 & -4 & 1\\
0 & 1 & 1 & -4 & 3\\
0 & 1 & -5 & -7 & -8
\end{bmatrix}\\
& \longrightarrow\begin{bmatrix}1 & 1 & 2 & 3 & 1\\
0 & 1 & 1 & -4 & 1\\
0 & 0 & 0 & 0 & 2\\
0 & 0 & -6 & -3 & -9
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 1 & 2 & 3 & 1\\
0 & 1 & 1 & -4 & 1\\
0 & 0 & 6 & 3 & 9\\
0 & 0 & 0 & 0 & 2
\end{bmatrix}
\end{align*}
\end_inset
因为
\begin_inset Formula $r(A)=3$
\end_inset
,
\begin_inset Formula $r\left(\begin{bmatrix}A & b\end{bmatrix}\right)=4$
\end_inset
,
\begin_inset Formula $r\left(\begin{bmatrix}A & b\end{bmatrix}\right)\neq r(A)$
\end_inset
, 所以原方程组无解.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
例题选讲
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
证明方程组
\begin_inset Formula $\begin{cases}
x_{1}-x_{2}=a_{1}\\
x_{2}-x_{3}=a_{2}\\
x_{3}-x_{4}=a_{3}\\
x_{4}-x_{5}=a_{4}\\
x_{5}-x_{1}=a_{5}
\end{cases}$
\end_inset
有解的充要条件是
\begin_inset Formula $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0$
\end_inset
.
在有解的情况下, 求出它的全部解.
\end_layout
\begin_layout Proof
对增广矩阵
\begin_inset Formula $\widetilde{A}$
\end_inset
进行初等变换:
\begin_inset Formula
\[
\widetilde{A}=\begin{bmatrix}1 & -1 & 0 & 0 & 0 & a_{1}\\
0 & 1 & -1 & 0 & 0 & a_{2}\\
0 & 0 & 1 & -1 & 0 & a_{3}\\
0 & 0 & 0 & 1 & -1 & a_{4}\\
-1 & 0 & 0 & 0 & 1 & a_{5}
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & -1 & 0 & 0 & 0 & a_{1}\\
0 & 1 & -1 & 0 & 0 & a_{2}\\
0 & 0 & 1 & -1 & 0 & a_{3}\\
0 & 0 & 0 & 1 & -1 & a_{4}\\
0 & 0 & 0 & 0 & 0 & \sum\limits _{i=1}^{5}a_{i}
\end{bmatrix}
\]
\end_inset
所以方程组有解当且仅当
\begin_inset Formula $r(A)=r(\widetilde{A})=4\Longleftrightarrow\sum\limits _{i=1}^{5}a_{i}=0$
\end_inset
, 即方程组有解的充要条件是
\begin_inset Formula $\sum\limits _{i=1}^{5}a_{i}=0$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proof
在有解的情况下, 原方程组等价于方程组
\begin_inset Formula $\begin{cases}
x_{1}-x_{2}=a_{1}\\
x_{2}-x_{3}=a_{2}\\
x_{3}-x_{4}=a_{3}\\
x_{4}-x_{5}=a_{4}
\end{cases}$
\end_inset
, 故所求全部解
\begin_inset Formula
\[
\begin{cases}
x_{1}=a_{1}+a_{2}+a_{3}+a_{4}+x_{5}\\
x_{2}=a_{2}+a_{3}+a_{4}+x_{5}\\
x_{3}=a_{3}+a_{4}+x_{5}\\
x_{4}=a_{4}+x_{5}
\end{cases},
\]
\end_inset
其中
\begin_inset Formula $x_{5}$
\end_inset
为任意实数.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
例题选讲
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
讨论线性方程组
\begin_inset Formula $\begin{cases}
x_{1}+x_{2}+2x_{3}+3x_{4}=1,\\
x_{1}+3x_{2}+6x_{3}+x_{4}=3,\\
3x_{1}-x_{2}-px_{3}+15x_{4}=3,\\
x_{1}-5x_{2}-10x_{3}+12x_{4}=t,
\end{cases}$
\end_inset
, 当
\begin_inset Formula $p,t$
\end_inset
取何值时, 方程组无解? 有唯一解? 有无穷多解? 在方程组有无穷多解的情况下, 求出全部解.
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
begin{align*}
\end_layout
\begin_layout Plain Layout
B & =
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
1&3&6&1&3
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
3&-1&-p&15&3
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
1&-5&-10&12&t
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
rightarrow
\end_layout
\begin_layout Plain Layout
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&2&4&-2&2
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&-4&-p-6&6&0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&-6&-12&9&t-1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
rightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&1&2&-1&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&-p+2&2&4
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&3&t+5
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\end_layout
\begin_layout Plain Layout
\backslash
end{align*}
\end_layout
\end_inset
(1) 当
\begin_inset Formula $p\neq2$
\end_inset
时,
\begin_inset Formula $r(A)=r(B)=4$
\end_inset
, 方程组有唯一解;
\end_layout
\begin_layout Solution*
(2) 当
\begin_inset Formula $p=2$
\end_inset
时, 有
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
B
\backslash
longrightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&1&2&-1&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&2&4
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&3&t+5
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
longrightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&1&2&-1&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&1&2
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&0&t-1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
{
\backslash
color{gray}{B
\backslash
longrightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&1&2&-1&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&2&4
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&3&t+5
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
longrightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&1&2&-1&1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&1&2
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&0&0&0&t-1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\end_layout
\begin_layout Plain Layout
}}
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
当
\begin_inset Formula $t\neq1$
\end_inset
时,
\begin_inset Formula $r(A)=3<r(B)=4$
\end_inset
, 方程组无解;
\end_layout
\begin_layout Solution*
当
\begin_inset Formula $t=1$
\end_inset
时,
\begin_inset Formula $r(A)=r(B)=3$
\end_inset
, 方程组有无穷多解.
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
B
\backslash
longrightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
0&1&2&-1&1
\backslash
\backslash
0&0&0&1&2
\backslash
\backslash
0&0&0&0&t-1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
longrightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&1&2&3&1
\backslash
\backslash
0&1&2&-1&1
\backslash
\backslash
0&0&0&1&2
\backslash
\backslash
0&0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
longrightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&0&0&0&-8
\backslash
\backslash
0&1&2&0&3
\backslash
\backslash
0&0&0&1&2
\backslash
\backslash
0&0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray},
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
B
\backslash
longrightarrow
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1&0&0&0&-8
\backslash
\backslash
0&1&2&0&3
\backslash
\backslash
0&0&0&1&2
\backslash
\backslash
0&0&0&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray},
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
即
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
begin{center}
\end_layout
\begin_layout Plain Layout
\backslash
systeme[x_1,x_2,x_3,x_4]{x_1=-8,x_2+2x_3=3,x_4=2},
\end_layout
\begin_layout Plain Layout
\backslash
end{center}
\end_layout
\end_inset
\end_layout
\begin_layout Solution*
故原方程组的全部解为
\begin_inset Formula
\[
\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}=k\begin{bmatrix}0\\
-2\\
1\\
0
\end{bmatrix}+\begin{bmatrix}-8\\
3\\
0\\
2
\end{bmatrix},\quad(k\in\RR).
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Subsection
作业
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
作业
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Problem
求解非齐次方程组
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
systeme{x_{1}-2x_{2}+3x_{3}-x_{4}=1,3x_{1}-x_{2}+5x_{3}-3x_{4}=2,2x_{1}+x_{2}+2x
_{3}-2x_{4}=3}.
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
求解非齐次方程组
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
systeme{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}=-1/2}.
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
\begin_inset Formula $a$
\end_inset
取何值时, 方程组
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
systeme[x_1,x_2,x_3]{x_{1}+x_{2}+x_{3}=a,ax_{1}+x_{2}+x_{3}=1,x_{1}+x_{2}+ax_{3}
=1}
\end_layout
\end_inset
有解, 并求其解.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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