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\end_header
\begin_body
\begin_layout Section
线性方程组解的结构
\end_layout
\begin_layout Subsection
齐次线性方程组解的结构
\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
齐次线性方程组解的结构
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
设有齐次线性方程组
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\begin{equation}
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=0\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=0\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}=0
\end{cases}\label{eq:3.6-1}
\end{equation}
\end_inset
若记
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
A=\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{bmatrix},\quad x=\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix},
\]
\end_inset
则方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.6-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 可写为向量方程
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\begin{equation}
Ax=0,\label{eq:3.6-2}
\end{equation}
\end_inset
称方程 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.6-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解
\begin_inset Formula $x=\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix}$
\end_inset
为方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.6-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解向量.
\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
齐次线性方程组解的性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Proposition
若
\begin_inset Formula $\xi_{1},\xi_{2}$
\end_inset
为方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.6-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解, 则
\begin_inset Formula $\xi_{1}+\xi_{2}$
\end_inset
也是该方程组的解.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proposition
若
\begin_inset Formula $\xi_{1}$
\end_inset
为方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.6-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解,
\begin_inset Formula $k$
\end_inset
为实数, 则
\begin_inset Formula $k\xi_{1}$
\end_inset
也是 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.6-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
齐次线性方程组若有非零解, 则它就有无穷多个解.
\end_layout
\begin_layout Standard
由上节知: 线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的全体解向量所构成的集合对于加法和数乘是封闭的, 因此构成一个向量空间.
称
\series bold
此向量空间为齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的解空间
\series default
.
\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
齐次线性方程组的基础解系
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的有限个解
\begin_inset Formula $\eta_{1},\eta_{2},\cdots,\eta_{t}$
\end_inset
满足:
\end_layout
\begin_layout Definition
(1)
\begin_inset Formula $\eta_{1},\eta_{2},\cdots,\eta_{t}$
\end_inset
线性无关;
\end_layout
\begin_layout Definition
(2)
\begin_inset Formula $Ax=0$
\end_inset
的任意一个解均可由
\begin_inset Formula $\eta_{1},\eta_{2},\cdots,\eta_{t}$
\end_inset
线性表示.
则称
\series bold
\begin_inset Formula $\eta_{1},\eta_{2},\cdots,\eta_{t}$
\end_inset
是齐次线性方程组
\begin_inset Formula $AX=0$
\end_inset
的一个基础解系
\series default
.
\end_layout
\begin_layout Remark*
方程组
\begin_inset Formula $Ax=0$
\end_inset
的一个基础解系即为其解空间的一个基, 易见方程组
\begin_inset Formula $Ax=0$
\end_inset
基础解系不是唯一的, 其解空间是唯一的.
\end_layout
\begin_layout Remark*
按上述定义, 若
\begin_inset Formula $\eta_{1},\eta_{2},\cdots,\eta_{t}$
\end_inset
是齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的一个基础解系.
则
\begin_inset Formula $Ax=0$
\end_inset
的通解可表示为
\begin_inset Formula
\[
x=k_{1}\eta_{1}+k_{2}\eta_{2}+\cdots+k_{t}\eta_{t},
\]
\end_inset
其中
\begin_inset Formula $k_{1},k_{2},\cdots,k_{t}$
\end_inset
为任意常数.
\end_layout
\begin_layout Standard
当一个齐次线性方程组只有零解时, 该方程组没有基础解系; 而当一个齐次线性方程组有非零解时, 是否一定有基础解系呢? 如果有的话, 怎样去求它的基础解系?
下面的定理
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:1"
plural "false"
caps "false"
noprefix "false"
\end_inset
回答了这两个问题.
\end_layout
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:1"
\end_inset
对齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
, 若
\begin_inset Formula $r(A)=r<n$
\end_inset
, 则该方程组的基础解系一定存在, 且每个基础解系中所含解向量的个数均等于
\begin_inset Formula $n-r$
\end_inset
, 其中
\begin_inset Formula $n$
\end_inset
是方程组所含未知量的个数.
\end_layout
\begin_layout Remark*
定理
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:1"
plural "false"
caps "false"
noprefix "false"
\end_inset
的证明过程实际上已给出了求齐次线性方程组的基础解系的方法.
且
\end_layout
\begin_layout Remark*
若已知
\begin_inset Formula $\eta_{1},\eta_{2},\cdots,\eta_{n-r}$
\end_inset
是线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的一个基础解系, 则
\begin_inset Formula $Ax=0$
\end_inset
的全部解可表为
\begin_inset Formula
\begin{equation}
x=c_{1}\eta_{1}+c_{2}\eta_{2}+\cdots+c_{n-r}\eta_{n-r},\label{eq:3.6-4}
\end{equation}
\end_inset
其中
\begin_inset Formula $c_{1},c_{2},\cdots,c_{n-r}$
\end_inset
为任意实数.
称表达式 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:3.6-4"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的通解.
\end_layout
\end_deeper
\begin_layout Subsection
非齐次线性方程组解的结构
\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
非齐次线性方程组解的结构
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
设有非齐次线性方程组
\begin_inset Formula
\[
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=b_{1}\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=b_{2}\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}=b_{m}
\end{cases}
\]
\end_inset
它也可写作向量方程
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
Ax=b.
\]
\end_inset
\end_layout
\begin_layout Proposition
设
\begin_inset Formula $\eta_{1},\eta_{2}$
\end_inset
是非齐次线性方程组
\begin_inset Formula $Ax=b$
\end_inset
的解, 则
\begin_inset Formula $\eta_{1}-\eta_{2}$
\end_inset
是对应的齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的解.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proposition
设
\begin_inset Formula $\eta$
\end_inset
是非齐次线性方程组
\begin_inset Formula $Ax=b$
\end_inset
的解,
\begin_inset Formula $\xi$
\end_inset
为对应的齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的解, 则
\begin_inset Formula $\xi+\eta$
\end_inset
非齐次线性方程组
\begin_inset Formula $Ax=b$
\end_inset
的解.
\end_layout
\begin_layout Theorem
设
\begin_inset Formula $\eta^{*}$
\end_inset
是非齐次线性方程组
\begin_inset Formula $Ax=b$
\end_inset
的一个解,
\begin_inset Formula $\xi$
\end_inset
是对应齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
的通解, 则
\begin_inset Formula $x=\xi+\eta^{*}$
\end_inset
是非齐次线性方程组
\begin_inset Formula $Ax=b$
\end_inset
的通解.
\end_layout
\begin_layout Remark*
设有非齐次线性方程组
\begin_inset Formula $Ax=b$
\end_inset
, 而
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$
\end_inset
是系数矩阵
\begin_inset Formula $A$
\end_inset
的列向量组, 则下列四个命题等价:
\end_layout
\begin_layout Remark*
(1) 非齐次线性方程组
\begin_inset Formula $Ax=b$
\end_inset
有解;
\end_layout
\begin_layout Remark*
(2) 向量
\begin_inset Formula $b$
\end_inset
能由向量组
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$
\end_inset
线性表示;
\end_layout
\begin_layout Remark*
(3) 向量组
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$
\end_inset
与向量组
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{n},b$
\end_inset
等价;
\end_layout
\begin_layout Remark*
(4)
\begin_inset Formula $r(A)=r(\begin{bmatrix}A & b\end{bmatrix})$
\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
齐次线性方程组的基础解系
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求下列齐次线性方程组的一个基础解系:
\begin_inset Formula
\[
\begin{cases}
2x_{1}+x_{2}-2x_{3}+3x_{4}=0\\
3x_{1}+2x_{2}-x_{3}+2x_{4}=0\\
x_{1}+x_{2}+x_{3}-x_{4}=0
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
对此方程组的系数矩阵作如下初等行变换:
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\begin{align*}
A & =\begin{bmatrix}2 & 1 & -2 & 3\\
3 & 2 & -1 & 2\\
1 & 1 & 1 & -1
\end{bmatrix}\xrightarrow[r_{2}-3r_{3}]{r_{1}-2r_{3}}\begin{bmatrix}0 & -1 & -4 & 5\\
0 & -1 & -4 & 5\\
1 & 1 & 1 & -1
\end{bmatrix}\\
& \xrightarrow{r_{1}-r_{2}}\begin{bmatrix}0 & 0 & 0 & 0\\
0 & -1 & -4 & 5\\
1 & 1 & 1 & -1
\end{bmatrix}\xrightarrow{r_{1}\leftrightarrow r_{3}}\begin{bmatrix}1 & 1 & 1 & -1\\
0 & -1 & -4 & 5\\
0 & 0 & 0 & 0
\end{bmatrix}\\
& \xrightarrow{r_{1}+r_{2}}\begin{bmatrix}1 & 0 & -3 & 4\\
0 & -1 & -4 & 5\\
0 & 0 & 0 & 0
\end{bmatrix}\xrightarrow{(-1)r_{2}}\begin{bmatrix}1 & 0 & -3 & 4\\
0 & 1 & 4 & -5\\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
\end_inset
于是原方程组可同解地变为:
\begin_inset Formula
\[
\begin{cases}
x_{1}=3x_{3}-4x_{4}\\
x_{2}=-4x_{3}+5x_{4}
\end{cases}
\]
\end_inset
因此基础解系为
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\eta_{1}=(3,-4,1,0)^{T},\ \eta_{2}=(-4,5,0,1)^{T}.
\]
\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
齐次线性方程组的基础解系与通解
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:3.6-2"
\end_inset
\begin_inset Argument 1
status open
\begin_layout Plain Layout
例 12, p.
97
\end_layout
\end_inset
求齐次线性方程组
\begin_inset Formula $\begin{cases}
x_{1}+x_{2}-x_{3}-x_{4}=0,\\
2x_{1}-5x_{2}+3x_{3}+2x_{4}=0,\\
7x_{1}-7x_{2}+3x_{3}+x_{4}=0
\end{cases}$
\end_inset
的基础解系与通解.
\end_layout
\begin_layout Solution*
对系数矩阵
\begin_inset Formula $A$
\end_inset
作初等行变换, 化为行最简矩阵:
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\begin{align*}
A & =\begin{bmatrix}1 & 1 & -1 & -1\\
2 & -5 & 3 & 2\\
7 & -7 & 3 & 1
\end{bmatrix}\xrightarrow[r_{3}-7r_{1}]{r_{2}-2r_{1}}\begin{bmatrix}1 & 1 & -1 & -1\\
0 & -7 & 5 & 4\\
0 & -14 & 10 & 8
\end{bmatrix}\\
& \xrightarrow{r_{3}-2r_{2}}\begin{bmatrix}1 & 1 & -1 & -1\\
0 & -7 & 5 & 4\\
0 & 0 & 0 & 0
\end{bmatrix}\xrightarrow[r_{1}-r_{2}]{r_{2}\div(-7)}\begin{bmatrix}1 & 0 & -2/7 & -3/7\\
0 & 1 & -5/7 & -4/7\\
0 & 0 & 0 & 0
\end{bmatrix},
\end{align*}
\end_inset
得到原方程组的同解方程组
\begin_inset Formula
\begin{equation}
\begin{cases}
x_{1}=(2/7)x_{3}+(3/7)x_{4}\\
x_{2}=(5/7)x_{3}+(4/7)x_{4}
\end{cases}\label{eq:star}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
令
\begin_inset Formula $\begin{bmatrix}x_{3}\\
x_{4}
\end{bmatrix}=\begin{bmatrix}1\\
0
\end{bmatrix},\begin{bmatrix}0\\
1
\end{bmatrix}$
\end_inset
, 即得基础解系
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\eta_{1}=\begin{bmatrix}2/7\\
5/7\\
1\\
0
\end{bmatrix},\ \eta_{2}=\begin{bmatrix}3/7\\
4/7\\
0\\
1
\end{bmatrix}.
\]
\end_inset
并由此得到通解
\begin_inset Formula
\[
\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}=C_{1}\begin{bmatrix}2/7\\
5/7\\
1\\
0
\end{bmatrix}+C_{2}\begin{bmatrix}3/7\\
4/7\\
0\\
1
\end{bmatrix},\qquad\left(C_{1},C_{2}\in\RR\right).
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
在第一节中, 线性方程组的解法是从例
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:3.6-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
中的 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:star"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 式直接写出方程组的全部解 (通解)
\begin_inset Foot
status open
\begin_layout Plain Layout
\begin_inset Formula
\[
\begin{cases}
x_{1}=(2/7)x_{3}+(3/7)x_{4}\\
x_{2}=(5/7)x_{3}+(4/7)x_{4}
\end{cases}\left(x_{3},x_{4}\text{ 可取任意实数}.\right)
\]
\end_inset
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
\[
\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}=c_{1}\begin{bmatrix}2/7\\
5/7\\
1\\
0
\end{bmatrix}+c_{2}\begin{bmatrix}3/7\\
4/7\\
0\\
1
\end{bmatrix}.
\]
\end_inset
它表达了方程组的全部解.
\end_layout
\end_inset
.
实际上可从例
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:3.6-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
中的 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:star"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 式先取基础解系, 再写出通解, 两种解法其实没有多少区别.
\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
用基础解系求解线性方程组的通解
\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
用基础解系表示如下线性方程组的通解.
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}+x_{3}+4x_{4}-3x_{5}=0\\
x_{1}-x_{2}+3x_{3}-2x_{4}-x_{5}=0\\
2x_{1}+x_{2}+3x_{3}+5x_{4}-5x_{5}=0\\
3x_{1}+x_{2}+5x_{3}+6x_{4}-7x_{5}=0
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula $m=4$
\end_inset
,
\begin_inset Formula $n=5$
\end_inset
,
\begin_inset Formula $m<n$
\end_inset
, 因此所给方程组有无穷多个解.
对增广矩阵
\begin_inset Formula $A$
\end_inset
施以初等行变换:
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 1 & 1 & 4 & -3\\
1 & -1 & 3 & -2 & -1\\
2 & 1 & 3 & 5 & -5\\
3 & 1 & 5 & 6 & -7
\end{bmatrix}\rightarrow\begin{bmatrix}1 & 1 & 1 & 4 & -3\\
0 & -2 & 2 & -6 & 2\\
0 & -1 & 1 & -3 & 1\\
0 & -2 & 2 & -6 & 2
\end{bmatrix}\rightarrow\begin{bmatrix}1 & 0 & 2 & 1 & -2\\
0 & 0 & 0 & 0 & 0\\
0 & 1 & -1 & 3 & -1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}.
\]
\end_inset
即原方程组与下面方程组同解:
\begin_inset Formula
\[
\begin{cases}
x_{1}=-2x_{3}-x_{4}+2x_{5},\\
x_{2}=x_{3}-3x_{4}+x_{5},
\end{cases}
\]
\end_inset
其中
\begin_inset Formula $x_{3},x_{4},x_{5}$
\end_inset
为自由未知量.
\end_layout
\begin_layout Solution*
令自由未知量
\begin_inset Formula $\begin{bmatrix}x_{3}\\
x_{4}\\
x_{5}
\end{bmatrix}$
\end_inset
取值
\begin_inset Formula $\begin{bmatrix}1\\
0\\
0
\end{bmatrix},\begin{bmatrix}0\\
1\\
0
\end{bmatrix},\begin{bmatrix}0\\
0\\
1
\end{bmatrix}$
\end_inset
, 分别得方程组的解为
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\eta_{1}=(-2,1,1,0,0)^{T},\eta_{2}=(-1,-3,0,1,0)^{T},\eta_{1}=(2,1,0,0,1)^{T},
\]
\end_inset
\begin_inset Formula $\eta_{1},\eta_{2},\eta_{3}$
\end_inset
就是所给方程组的一个基础解系.
因此, 方程组的通解为
\begin_inset Formula $\eta=c_{1}\eta_{1}+c_{2}\eta_{2}+c_{3}\eta_{3}$
\end_inset
, (
\begin_inset Formula $c_{1},c_{2},c_{3}$
\end_inset
为任意常数).
\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
求解齐次线性方程组
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求解下列齐次线性方程组:
\end_layout
\begin_layout Example
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}-x_{3}+2x_{4}+x_{5}=0\\
x_{3}+3x_{4}-x_{5}=0\\
2x_{3}+x_{4}-2x_{5}=0
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
对方程组的系数矩阵作如下初等变换:
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\begin{aligned}A & =\begin{bmatrix}1 & 1 & -1 & 2 & 1\\
0 & 0 & 1 & 3 & -1\\
0 & 0 & 2 & 1 & -2
\end{bmatrix}\xrightarrow{r_{3}-2r_{2}}\begin{bmatrix}1 & 1 & -1 & 2 & 1\\
0 & 0 & 1 & 3 & -1\\
0 & 0 & 2 & 1 & -2
\end{bmatrix}\\
& \xrightarrow{\left(-\frac{1}{5}\right)r_{3}}\begin{bmatrix}1 & 1 & -1 & 2 & 1\\
0 & 0 & 1 & 3 & -1\\
0 & 0 & 2 & 1 & -2
\end{bmatrix}.
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
这个矩阵不符合要求, 因为它已经不可能仅用初等行变换变成所要求的左上角为单位块的形状了, 这是必须借助于列对调.
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
begin{align*}
\end_layout
\begin_layout Plain Layout
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1}&x_{2}&x_{3}&x_{4}&x_{5}
\backslash
\backslash
1&1&-1&2&1
\backslash
\backslash
0&0&1&3&-1
\backslash
\backslash
0&0&0&1&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\end_layout
\begin_layout Plain Layout
&
\backslash
xrightarrow{c_{2}
\backslash
leftrightarrow c_{3}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1}&x_{3}&x_{2}&x_{4}&x_{5}
\backslash
\backslash
1&-1&1&2&1
\backslash
\backslash
0&1&0&3&-1
\backslash
\backslash
0&0&0&1&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
xrightarrow{c_{3}
\backslash
leftrightarrow c_{4}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1}&x_{3}&x_{4}&x_{2}&x_{5}
\backslash
\backslash
1&-1&2&1&1
\backslash
\backslash
0&1&3&0&-1
\backslash
\backslash
0&0&1&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
xrightarrow{r_{1}+r_{2}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1}&x_{3}&x_{4}&x_{2}&x_{5}
\backslash
\backslash
1&0&5&1&0
\backslash
\backslash
0&1&3&0&-1
\backslash
\backslash
0&0&1&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
xrightarrow[r_{2}-3r_{3}]{r_{1}-5r_{3}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1}&x_{3}&x_{4}&x_{2}&x_{5}
\backslash
\backslash
1&0&0&1&1
\backslash
\backslash
0&1&0&0&-1
\backslash
\backslash
0&0&1&0&0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}.
\end_layout
\begin_layout Plain Layout
\backslash
end{align*}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
矩阵的秩等于
\begin_inset Formula $3$
\end_inset
, 未知数个数
\begin_inset Formula $n=5$
\end_inset
, 因此基础解系应含有
\begin_inset Formula $2$
\end_inset
个向量, 分别取自由变量
\begin_inset Formula $x_{2}=1$
\end_inset
,
\begin_inset Formula $x_{5}=0$
\end_inset
及
\begin_inset Formula $x_{2}=0$
\end_inset
,
\begin_inset Formula $x_{5}=1$
\end_inset
.
得到基础解系:
\begin_inset Formula
\[
\eta_{1}=(-1,1,0,0,0)^{T},\ \eta_{2}=(0,0,1,0,1)^{T},
\]
\end_inset
于是原方程组的解为
\begin_inset Formula
\[
c_{1}\eta_{1}+c_{2}\eta_{2},
\]
\end_inset
其中
\begin_inset Formula $c_{1},c_{2}$
\end_inset
为任意数.
\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
求解齐次线性方程组
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求解齐次线性方程组:
\begin_inset Formula
\[
\begin{cases}
x_{1}+2x_{2}+x_{4}-2x_{5}=0,\\
2x_{1}+4x_{2}+2x_{3}+2x_{4}+5x_{5}=0,\\
-x_{1}-2x_{2}+x_{3}+3x_{4}+8x_{5}=0,\\
3x_{1}+6x_{2}+x_{4}-2x_{5}=0.
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
对方程组的系数矩阵进行初等变换:
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
begin{eqnarray*}
\end_layout
\begin_layout Plain Layout
A &=&
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
1 & 2 & 0 & 1 & -2
\backslash
\backslash
2 & 4 & 2 & 2 & 5
\backslash
\backslash
-1 & -2 & 1 & 3 & 8
\backslash
\backslash
3 & 6 & 0 & 1 & -2
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix}
\backslash
xrightarrow[r_{4}-3r_{1}]{{r_{2}-2r_{1}
\backslash
atop r_{3}+r_{1}}}
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
1 & 2 & 0 & 1 & -2
\backslash
\backslash
0 & 0 & 2 & 0 & 9
\backslash
\backslash
0 & 0 & 1 & 4 & 6
\backslash
\backslash
0 & 0 & 0 & -2 & 4
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
xrightarrow{r_{2}-2r_{3}} &
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
1 & 2 & 0 & 1 & -2
\backslash
\backslash
0 & 0 & 0 & -8 & -3
\backslash
\backslash
0 & 0 & 1 & 4 & 6
\backslash
\backslash
0 & 0 & 0 & -2 & 4
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix}
\backslash
xrightarrow{r_{2}
\backslash
leftrightarrow r_{3}}
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
1 & 2 & 0 & 1 & -2
\backslash
\backslash
0 & 0 & 2 & 4 & 6
\backslash
\backslash
0 & 0 & 1 & -8 & -3
\backslash
\backslash
0 & 0 & 0 & -2 & 4
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
xrightarrow{c_{2}
\backslash
leftrightarrow c_{3}} &
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{2} & x_{4} & x_{5}
\backslash
\backslash
1 & 0 & 2 & 1 & -2
\backslash
\backslash
0 & 1 & 0 & 4 & 6
\backslash
\backslash
0 & 0 & 0 & -8 & -3
\backslash
\backslash
0 & 0 & 0 & -2 & 4
\backslash
end{bNiceMatrix}
\backslash
xrightarrow{
\backslash
left(-
\backslash
frac{1}{2}
\backslash
right)r_{4}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{2} & x_{4} & x_{5}
\backslash
\backslash
1 & 0 & 2 & 1 & -2
\backslash
\backslash
0 & 1 & 0 & 4 & 6
\backslash
\backslash
0 & 0 & 0 & -8 & -3
\backslash
\backslash
0 & 0 & 0 & 1 & -2
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\end_layout
\begin_layout Plain Layout
\backslash
end{eqnarray*}
\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
{
\backslash
footnotesize{
\backslash
begin{eqnarray*}
\end_layout
\begin_layout Plain Layout
A &
\backslash
longrightarrow &
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{2} & x_{4} & x_{5}
\backslash
\backslash
1 & 0 & 2 & 1 & -2
\backslash
\backslash
0 & 1 & 0 & 4 & 6
\backslash
\backslash
0 & 0 & 0 & -8 & -3
\backslash
\backslash
0 & 0 & 0 & 1 & -2
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
xrightarrow{r_{4}
\backslash
leftrightarrow r_{3}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{2} & x_{4} & x_{5}
\backslash
\backslash
1 & 0 & 2 & 1 & -2
\backslash
\backslash
0 & 1 & 0 & 4 & 6
\backslash
\backslash
0 & 0 & 0 & 1 & -2
\backslash
\backslash
0 & 0 & 0 & -8 & -3
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
xrightarrow{c_{4}
\backslash
leftrightarrow c_{3}} &
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{4} & x_{2} & x_{5}
\backslash
\backslash
1 & 0 & 1 & 2 & -2
\backslash
\backslash
0 & 1 & 4 & 0 & 6
\backslash
\backslash
0 & 0 & 1 & 0 & -2
\backslash
\backslash
0 & 0 & -8 & 0 & -3
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
xrightarrow{r_{4}+8r_{3}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{4} & x_{2} & x_{5}
\backslash
\backslash
0 & 0 & 1 & 2 & -2
\backslash
\backslash
0 & 1 & 4 & 0 & 6
\backslash
\backslash
0 & 0 & 1 & 0 & -2
\backslash
\backslash
0 & 0 & 0 & 0 & -19
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
xrightarrow{
\backslash
left(-
\backslash
frac{1}{19}
\backslash
right)r_{4}} &
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{4} & x_{2} & x_{5}
\backslash
\backslash
1 & 0 & 1 & 2 & -2
\backslash
\backslash
0 & 1 & 4 & 0 & 6
\backslash
\backslash
0 & 0 & 1 & 0 & -2
\backslash
\backslash
0 & 0 & 0 & 0 & 0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
xrightarrow[r_{3}+2r_{4}]{{r_{1}+2r_{4}
\backslash
atop r_{2}-6r_{4}}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{4} & x_{2} & x_{5}
\backslash
\backslash
1 & 0 & 1 & 2 & 0
\backslash
\backslash
0 & 1 & 4 & 0 & 0
\backslash
\backslash
0 & 0 & 1 & 0 & 0
\backslash
\backslash
0 & 0 & 0 & 0 & 1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
xrightarrow[r_{2}-4r_{3}]{r_{1}-r_{3}} &
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{4} & x_{2} & x_{5}
\backslash
\backslash
1 & 0 & 0 & 2 & 0
\backslash
\backslash
0 & 1 & 0 & 0 & 0
\backslash
\backslash
0 & 0 & 1 & 0 & 0
\backslash
\backslash
0 & 0 & 0 & 0 & 1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
xrightarrow{c_{4}
\backslash
leftrightarrow c_{5}}
\backslash
begin{bNiceMatrix}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{3} & x_{4} & x_{5} & x_{2}
\backslash
\backslash
1 & 0 & 0 & 0 & 2
\backslash
\backslash
0 & 1 & 0 & 0 & 0
\backslash
\backslash
0 & 0 & 1 & 0 & 0
\backslash
\backslash
0 & 0 & 0 & 1 & 0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}.
\end_layout
\begin_layout Plain Layout
\backslash
end{eqnarray*}
\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*
这个矩阵秩等于
\begin_inset Formula $4$
\end_inset
, 未知数个数
\begin_inset Formula $n=5$
\end_inset
, 因此基础解系只含有
\begin_inset Formula $1$
\end_inset
个向量, 取自由变量
\begin_inset Formula $x_{2}=1$
\end_inset
, 得
\begin_inset Formula
\[
\eta=(-2,1,0,0,0)^{T},
\]
\end_inset
原方程的通解为
\begin_inset Formula $c\eta$
\end_inset
, 其中
\begin_inset Formula $c$
\end_inset
可取任何数.
\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
关于秩的等式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
证明
\begin_inset Formula $r\left(A^{T}A\right)=r(A)$
\end_inset
.
\end_layout
\begin_layout Proof
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $m\times n$
\end_inset
矩阵,
\begin_inset Formula $x$
\end_inset
为
\begin_inset Formula $n$
\end_inset
维列向量.
\end_layout
\begin_layout Proof
若
\begin_inset Formula $x$
\end_inset
满足
\begin_inset Formula $Ax=0\longrightarrow A^{T}(Ax)=0$
\end_inset
, 即
\begin_inset Formula $\left(A^{T}A\right)x=0$
\end_inset
;
\end_layout
\begin_layout Proof
若
\begin_inset Formula $x$
\end_inset
满足
\begin_inset Formula $\left(A^{T}A\right)x=0\longmapsto x^{T}\left(A^{T}A\right)x=0$
\end_inset
, 即
\begin_inset Formula $(Ax)^{T}(Ax)=0\longmapsto Ax=0$
\end_inset
.
\end_layout
\begin_layout Proof
综上可知方程组
\begin_inset Formula $Ax=0$
\end_inset
与
\begin_inset Formula $\left(A^{T}A\right)x=0$
\end_inset
同解,
\end_layout
\begin_layout Proof
所以
\begin_inset Formula $r\left(A^{T}A\right)=r(A)$
\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
齐次线性方程组的基础解系
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求出一个齐次线性方程组, 使它的基础解系由下列向量组成
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\xi_{1}=\begin{bmatrix}1\\
2\\
3\\
4
\end{bmatrix},\quad\xi_{2}=\begin{bmatrix}4\\
3\\
2\\
1
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Solution*
设所求得齐次线性方程组为
\begin_inset Formula $Ax=0$
\end_inset
, 矩阵
\begin_inset Formula $A$
\end_inset
的行向量形如
\begin_inset Formula $\alpha^{T}=\left(a_{1},a_{2},a_{3},a_{4}\right)$
\end_inset
,
\end_layout
\begin_layout Solution*
根据题意, 有
\begin_inset Formula $\alpha^{T}\xi_{1}=0$
\end_inset
,
\begin_inset Formula $\alpha^{T}\xi_{2}=0$
\end_inset
, 即
\begin_inset Formula $\begin{cases}
a_{1}+2a_{2}+3a_{3}+4a_{4}=0\\
4a_{1}+3a_{2}+2a_{3}+a_{4}=0
\end{cases}$
\end_inset
\end_layout
\begin_layout Solution*
设这个方程组系数矩阵为
\begin_inset Formula $B$
\end_inset
, 对
\begin_inset Formula $B$
\end_inset
进行初等行变换, 得
\begin_inset Formula $B=\begin{bmatrix}1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 2 & 3 & 4\\
0 & -5 & -10 & -15
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & -1 & -2\\
0 & 1 & 2 & 3
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
这个方程组的同解方程组为
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{cases}
a_{1}-a_{3}-2a_{4}=0\\
a_{2}+2a_{3}+3a_{4}=0
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
其基础解系为
\begin_inset Formula $\begin{bmatrix}1\\
-2\\
1\\
0
\end{bmatrix},\begin{bmatrix}2\\
-3\\
0\\
1
\end{bmatrix}$
\end_inset
, 故可取矩阵
\begin_inset Formula $A$
\end_inset
的行向量为
\begin_inset Formula $\alpha_{1}^{T}=(1,-2,1,0)$
\end_inset
,
\begin_inset Formula $\alpha_{2}^{T}=(2,-3,0,1)$
\end_inset
,
\end_layout
\begin_layout Solution*
故所求齐次线性方程组的系数矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & -2 & 1 & 0\\
2 & -3 & 0 & 1
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
所求齐次线性方程组为
\begin_inset Formula $\begin{cases}
x_{1}-2x_{2}+x_{3}=0\\
2x_{1}-3x_{2}+x_{4}=0
\end{cases}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
非齐次线性方程组的通解
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
E03
\end_layout
\end_inset
求下列方程组的通解
\begin_inset Formula $\begin{cases}
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=7\\
3x_{1}+x_{2}+2x_{3}+x_{4}-3x_{5}=-2\text{. }\\
2x_{2}+x_{3}+2x_{4}+6x_{5}=23
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Solution*
对增广矩阵做初等变换:
\begin_inset Formula $\widetilde{A}=\begin{bmatrix}1 & 1 & 1 & 1 & 1 & 7\\
3 & 1 & 2 & 1 & -3 & -2\\
0 & 2 & 1 & 2 & 6 & 23
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 1/2 & 0 & -2 & -9/2\\
0 & 1 & 1/2 & 1 & 3 & 23/2\\
0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}$
\end_inset
,
\end_layout
\begin_layout Solution*
由
\begin_inset Formula $r(A)=r(\widetilde{A})$
\end_inset
, 知方程组有解.
\end_layout
\begin_layout Solution*
又
\begin_inset Formula $r(A)=2$
\end_inset
,
\begin_inset Formula $n-r=3$
\end_inset
, 所以方程组有无穷多解.
且原方程组等价于方程组
\begin_inset Formula
\[
\begin{cases}
x_{1}=-x_{3}/2+2x_{5}-9/2\\
x_{2}=-x_{3}/2-x_{4}-3x_{5}+23/2
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
令
\begin_inset Formula $\begin{bmatrix}x_{3}\\
x_{4}\\
x_{5}
\end{bmatrix}=\begin{bmatrix}1\\
0\\
0
\end{bmatrix},\begin{bmatrix}0\\
1\\
0
\end{bmatrix},\begin{bmatrix}0\\
0\\
1
\end{bmatrix}$
\end_inset
分别代入等价方程组对应的齐次方程组中求得基础解系
\begin_inset Formula
\[
\xi_{1}=\begin{bmatrix}-1/2\\
-1/2\\
1\\
0\\
0
\end{bmatrix},\ \xi_{2}=\begin{bmatrix}0\\
-1\\
0\\
1\\
0
\end{bmatrix},\ \xi_{3}=\begin{bmatrix}2\\
-3\\
0\\
0\\
1
\end{bmatrix}
\]
\end_inset
求特解: 令
\begin_inset Formula $x_{3}=x_{4}=x_{5}=0$
\end_inset
, 得
\begin_inset Formula $x_{1}=-9/2,x_{2}=23/2$
\end_inset
.
故所求通解为
\begin_inset Formula $x=C_{1}\begin{bmatrix}-1/2\\
-1/2\\
1\\
0\\
0
\end{bmatrix}+C_{2}\begin{bmatrix}0\\
-1\\
0\\
1\\
0
\end{bmatrix}+C_{3}\begin{bmatrix}2\\
-3\\
0\\
0\\
1
\end{bmatrix}+\begin{bmatrix}-9/2\\
23/2\\
0\\
0\\
0
\end{bmatrix}$
\end_inset
.
其中
\begin_inset Formula $C_{1},C_{2},C_{3}$
\end_inset
为任意常数.
\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
求解非齐次线性方程组
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求解下列非齐次线性方程组:
\end_layout
\begin_layout Example
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}-3x_{3}-x_{4}=1\\
3x_{1}-x_{2}-3x_{3}-4x_{4}=4\\
x_{1}+5x_{2}-9x_{3}-8x_{4}=0
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
对方程组的增广矩阵作如下初等变换:
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\begin_layout Plain Layout
\backslash
begin{eqnarray*}
\end_layout
\begin_layout Plain Layout
\backslash
widetilde{A}&=&
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
A & b
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix}=
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1 & 1 & -3 & -1 & 1
\backslash
\backslash
3 & -1 & -3 & 4 & 4
\backslash
\backslash
1 & 5 & -9 & 8 & 0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
xrightarrow[r_{3}-r_{1}]{r_{2}-3r_{1}}
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1 & 1 & -3 & -1 & 1
\backslash
\backslash
0 & -4 & 6 & 7 & 1
\backslash
\backslash
0 & 4 & -6 & -7 & -1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
\backslash
&
\backslash
xrightarrow{r_{3}+r_{2}}&
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1 & 1 & -3 & -1 & 1
\backslash
\backslash
0 & -4 & 6 & 7 & 1
\backslash
\backslash
0 & 0 & 0 & 0 & 0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
xrightarrow{-
\backslash
frac{1}{4}r_{2}}
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1 & 1 & -3 & -1 & 1
\backslash
\backslash
0 & 1 & -3/2 & -7/4 & -1/4
\backslash
\backslash
0 & 0 & 0 & 0 & 0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
\backslash
&
\backslash
xrightarrow{r_{1}-r_{2}}&
\backslash
begin{bNiceArray}{cccc|c}
\end_layout
\begin_layout Plain Layout
1 & 0 & -3/2 & 3/4 & 5/4
\backslash
\backslash
0 & 1 & -3/2 & -7/4 & -1/4
\backslash
\backslash
0 & 0 & 0 & 0 & 0
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}.
\end_layout
\begin_layout Plain Layout
\backslash
end{eqnarray*}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
在上面的初等变换中没有作过列对换, 因此可立即求出特解
\begin_inset Formula $\gamma$
\end_inset
和对应齐次线性方程组的基础解系:
\begin_inset Formula
\[
\gamma=\begin{bmatrix}5/4\\
-1/4\\
0\\
0
\end{bmatrix},\quad\eta_{1}=\begin{bmatrix}3/2\\
3/2\\
1\\
0
\end{bmatrix},\quad\eta_{2}=\begin{bmatrix}-3/4\\
7/4\\
0\\
1
\end{bmatrix}.
\]
\end_inset
原方程组的解为
\begin_inset Formula $x=\gamma+c_{1}\eta_{1}+c_{2}\eta_{2}$
\end_inset
, 其中
\begin_inset Formula $c_{1},c_{2}$
\end_inset
为任意数.
\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
求解非齐次线性方程组
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求解下列线性方程组:
\begin_inset Formula
\[
\begin{cases}
x_{1}+2x_{2}-x_{3}+3x_{4}+x_{5}=2\\
2x_{1}+4x_{2}-2x_{3}+6x_{4}+3x_{5}=6\\
-x_{1}-2x_{2}+x_{3}-x_{4}+3x_{5}=4
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
对方程组的增广矩阵作如下初等变换:
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\begin_layout Plain Layout
\backslash
begin{eqnarray*}
\end_layout
\begin_layout Plain Layout
\backslash
widetilde{A}&=&
\end_layout
\begin_layout Plain Layout
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
A & b
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix}=
\backslash
begin{bNiceArray}{ccccc|c}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{2} & x_{3} & x_{4} & x_{5}
\backslash
\backslash
1 & 2 & -1 & 3 & 1 & 2
\backslash
\backslash
2 & 4 & -2 & 6 & 3 & 6
\backslash
\backslash
-1 & -2 & 1 & -1 & 3 & 4
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
xrightarrow[r_{3}+r_{1}]{r_{2}-2r_{1}}
\backslash
begin{bNiceArray}{ccccc|c}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{2} & x_{3} & x_{4} & x_{5}
\backslash
\backslash
1 & 2 & -1 & 3 & 1 & 2
\backslash
\backslash
0 & 0 & 0 & 0 & 1 & 2
\backslash
\backslash
0 & 0 & 0 & 2 & 4 & 6
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
\backslash
&
\backslash
xrightarrow{
\backslash
frac{1}{2}r_{3}} &
\backslash
begin{bNiceArray}{ccccc|c}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{2} & x_{3} & x_{4} & x_{5}
\backslash
\backslash
1 & 2 & -1 & 3 & 1 & 2
\backslash
\backslash
0 & 0 & 0 & 0 & 1 & 2
\backslash
\backslash
0 & 0 & 0 & 1 & 2 & 3
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
xrightarrow[c_{3}
\backslash
leftrightarrow c_{5}]{c_{2}
\backslash
leftrightarrow c_{4}}
\backslash
begin{bNiceArray}{ccccc|c}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{4} & x_{5} & x_{2} & x_{3}
\backslash
\backslash
1 & 3 & 1 & 2 & -1 & 2
\backslash
\backslash
0 & 0 & 1 & 0 & 0 & 2
\backslash
\backslash
0 & 1 & 2 & 0 & 0 & 3
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
\backslash
&
\backslash
xrightarrow{r_{2}
\backslash
leftrightarrow r_{3}} &
\backslash
begin{bNiceArray}{ccccc|c}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{4} & x_{5} & x_{2} & x_{3}
\backslash
\backslash
1 & 3 & 1 & 2 & -1 & 2
\backslash
\backslash
0 & 1 & 2 & 0 & 0 & 3
\backslash
\backslash
0 & 0 & 1 & 0 & 0 & 2
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
xrightarrow{r_{1}-3r_{2}}
\backslash
begin{bNiceArray}{ccccc|c}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{4} & x_{5} & x_{2} & x_{3}
\backslash
\backslash
1 & 0 & -5 & 2 & -1 & -7
\backslash
\backslash
0 & 1 & 2 & 0 & 0 & 3
\backslash
\backslash
0 & 0 & 1 & 0 & 0 & 2
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
\backslash
&
\backslash
xrightarrow[r_{2}-2r_{3}]{r_{1}+5r_{3}} &
\backslash
begin{bNiceArray}{ccccc|c}[first-row]
\end_layout
\begin_layout Plain Layout
x_{1} & x_{4} & x_{5} & x_{2} & x_{3}
\backslash
\backslash
1 & 0 & 0 & 2 & -1 & 3
\backslash
\backslash
0 & 1 & 0 & 0 & 0 & -1
\backslash
\backslash
0 & 0 & 1 & 0 & 0 & 2
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}.
\end_layout
\begin_layout Plain Layout
\backslash
end{eqnarray*}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
在上面初等变换的整个过程中, 我们进行了两次列对换, 第一次是第
\begin_inset Formula $2$
\end_inset
列与第
\begin_inset Formula $4$
\end_inset
列对换, 第二次是第
\begin_inset Formula $3$
\end_inset
列与第
\begin_inset Formula $5$
\end_inset
列对换.
\begin_inset Formula
\[
\text{ 秩 }(\widetilde{A})=\text{ 秩 }(A)=3,
\]
\end_inset
未知数个数
\begin_inset Formula $n=5$
\end_inset
, 因此基础解系应含有
\begin_inset Formula $2$
\end_inset
个向量, 分别取自由变量
\begin_inset Formula
\[
x_{2}=0,\ x_{3}=0;\ x_{2}=1,\ x_{3}=0\text{ 及 }x_{2}=0,\ x_{3}=1\text{. }
\]
\end_inset
得特解
\begin_inset Formula $\gamma$
\end_inset
以及基础解系
\begin_inset Formula $\eta_{1},\eta_{2}$
\end_inset
:
\begin_inset Formula
\[
\gamma=(3,0,0,-1,2)^{T},\quad\eta_{1}=(-2,1,0,0,0)^{T},\quad\eta_{2}=(1,0,1,0,0)^{T}.
\]
\end_inset
\end_layout
\begin_layout Solution*
于是原线性方程组的通解为
\begin_inset Formula
\[
\gamma+c_{1}\eta_{1}+c_{2}\eta_{2}
\]
\end_inset
其中
\begin_inset Formula $c_{1},c_{2}$
\end_inset
为任意数.
\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
非齐次线性方程组的通解
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
E04
\end_layout
\end_inset
设四元非齐次线性方程组
\begin_inset Formula $AX=b$
\end_inset
的系数矩阵
\begin_inset Formula $A$
\end_inset
的秩为
\begin_inset Formula $3$
\end_inset
, 已经它的三个解向量为
\begin_inset Formula $\eta_{1},\eta_{2},\eta_{3}$
\end_inset
, 其中
\begin_inset Formula
\[
\eta_{1}=\begin{bmatrix}3\\
-4\\
1\\
2
\end{bmatrix},\quad\eta_{2}+\eta_{3}=\begin{bmatrix}4\\
6\\
8\\
0
\end{bmatrix},
\]
\end_inset
求该方程组的通解.
\end_layout
\begin_layout Solution*
依题意, 方程组
\begin_inset Formula $Ax=b$
\end_inset
的导出组的基础解系含
\begin_inset Formula $4-3=1$
\end_inset
个向量, 于是导出组的任何一个非零解都可作为其基础解系.
显然
\begin_inset Formula $\eta_{1}-\frac{1}{2}\left(\eta_{2}+\eta_{3}\right)=\begin{bmatrix}1\\
-7\\
-3\\
2
\end{bmatrix}\neq\boldsymbol{0}$
\end_inset
是导出组的非零解, 可作为其基础解系.
\end_layout
\begin_layout Solution*
故方程组
\begin_inset Formula $Ax=b$
\end_inset
的通解为
\begin_inset Formula
\[
x=\eta_{1}+c\left[\eta_{1}-\frac{1}{2}\left(\eta_{2}+\eta_{3}\right)\right]=\begin{bmatrix}3\\
-4\\
1\\
2
\end{bmatrix}+C\begin{bmatrix}1\\
-7\\
-3\\
2
\end{bmatrix},\qquad(C\text{ 为任意常数}).
\]
\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
作业
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Problem
求线性方程组
\begin_inset Formula
\[
\begin{cases}
x_{1}-x_{2}-x_{3}+x_{4}=0\\
x_{1}-x_{2}+x_{3}-3x_{4}=1\\
x_{1}-x_{2}-2x_{3}+3x_{4}=-1/2
\end{cases}
\]
\end_inset
的通解.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设矩阵
\begin_inset Formula $A=\left(a_{ij}\right)_{m\times n}$
\end_inset
,
\begin_inset Formula $B=\left(b_{ij}\right)_{n\times s}$
\end_inset
满足
\begin_inset Formula $AB=O$
\end_inset
并且
\begin_inset Formula $r(A)=r$
\end_inset
.
试证:
\begin_inset Formula $r(B)\leq n-r$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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