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\end_header

\begin_body

\begin_layout Section
克莱姆法则
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
非齐次线性方程组
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
含有 
\begin_inset Formula $n$
\end_inset

 个未知数 
\begin_inset Formula $x_{1},x_{2},\cdots,x_{n}$
\end_inset

 的线性方程组
\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\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}=b_{n},
\end{cases}\label{eq:5-1}
\end{equation}

\end_inset

称为 
\series bold

\begin_inset Formula $n$
\end_inset

 元线性方程组
\series default
.
 当其右端的常数项 
\begin_inset Formula $b_{1},b_{2},\cdots,b_{n}$
\end_inset

 不全为零时, 线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 称为
\series bold
非齐次线性方程组
\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
系数行列式
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 的系数 
\begin_inset Formula $a_{ij}$
\end_inset

 构成的行列式称为该方程组的
\series bold
系数行列式
\series default
 
\begin_inset Formula $D$
\end_inset

, 即
\begin_inset Formula 
\[
D=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Subsection
克莱姆法则
\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
克莱姆法则
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Theorem
\begin_inset Argument 1
status open

\begin_layout Plain Layout
克莱姆法则
\end_layout

\end_inset

 若线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

)
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=b_{1},\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=b_{2},\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}=b_{n},
\end{cases}$
\end_inset


\end_layout

\end_inset

 的系数行列式 
\begin_inset Formula $D\neq0$
\end_inset

, 则线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 有唯一解, 其解为
\begin_inset Formula 
\begin{equation}
x_{j}=\frac{D_{j}}{D},\qquad(j=1,2,\cdots,n),\label{eq:5-3}
\end{equation}

\end_inset

其中 
\begin_inset Formula $D_{j}$
\end_inset

, (
\begin_inset Formula $j=1,2,\cdots,n$
\end_inset

), 是把 
\begin_inset Formula $D$
\end_inset

 中第 
\begin_inset Formula $j$
\end_inset

 列元素 
\begin_inset Formula $a_{1j},a_{2j},\cdots,a_{nj}$
\end_inset

 对应地换成常数项 
\begin_inset Formula $b_{1},b_{2},\cdots,b_{n}$
\end_inset

, 而其余各列保持不变所得到的行列式.
\end_layout

\begin_layout Standard
定理中的 
\begin_inset Formula $D_{j}$
\end_inset

 具有形式
\begin_inset Formula 
\[
D_{j}=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1,j-1} & b_{1} & a_{1,j+1} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2,j-1} & b_{2} & a_{2,j+1} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{n,j-1} & b_{n} & a_{2,j+1} & \cdots & a_{nn}
\end{vmatrix}.
\]

\end_inset


\end_layout

\begin_layout Standard
一般来说, 用克莱姆法则求线性方程组的解时, 计算量较大.
 
\end_layout

\begin_layout Standard

\xout on
对具体的线性方程组, 当未知数较多时往往用计算机来求解.
 用计算机求解线性方程组目前已经有了一整套成熟的方法.
\end_layout

\begin_layout Standard
克莱姆法则在一定条件下给出了
\series bold
线性方程组解的存在性、唯一性
\series default
, 其在计算方面的价值更多地体现在理论层面.
 
\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
克莱姆法则的证明
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Proof
\begin_inset Argument 1
status open

\begin_layout Plain Layout
证明: (存在性)
\end_layout

\end_inset

 因 
\begin_inset Formula $D\ne0$
\end_inset

, 故而只需验证 
\begin_inset Formula $x_{i}=\frac{D_{i}}{D}$
\end_inset

 满足线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

)
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=b_{1},\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=b_{2},\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}=b_{n},
\end{cases}$
\end_inset


\end_layout

\end_inset

 即可说明存在性.
 
\end_layout

\begin_layout Proof
将 
\begin_inset Formula $x_{1}=\frac{D_{1}}{D}$
\end_inset

, 
\begin_inset Formula $x_{2}=\frac{D_{2}}{D}$
\end_inset

, ..., 
\begin_inset Formula $x_{n}=\frac{D_{n}}{D}$
\end_inset

 代入方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 中的第 
\begin_inset Formula $i$
\end_inset

 个方程, 有
\begin_inset Formula 
\begin{align*}
a_{i1}x_{1}+a_{i2}x_{2}+\cdots+a_{in}x_{n} & =a_{i1}\frac{D_{1}}{D}+a_{i2}\frac{D_{2}}{D}+\cdots+a_{in}\frac{D_{n}}{D}.\\
 & =\frac{1}{D}\left(a_{i1}D_{1}+a_{i2}D_{2}+\cdots+a_{in}D_{n}\right)\\
 & =\frac{1}{D}\sum_{k=1}^{n}a_{ik}D_{k}
\end{align*}

\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
克莱姆法则的证明
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Proof
\begin_inset Argument 1
status open

\begin_layout Plain Layout
证明: (存在性)
\end_layout

\end_inset

注意 
\begin_inset Formula $D_{j}$
\end_inset

 (
\begin_inset Formula $j=1,2,\cdots,n$
\end_inset

) 的定义
\begin_inset Formula 
\[
D_{j}=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1,j-1} & b_{1} & a_{1,j+1} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2,j-1} & b_{2} & a_{2,j+1} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{n,j-1} & b_{n} & a_{2,j+1} & \cdots & a_{nn}
\end{vmatrix}={\color{magenta}\sum_{k=1}^{n}b_{k}A_{kj}},
\]

\end_inset

代入 
\begin_inset Formula $a_{i1}x_{1}+a_{i2}x_{2}+\cdots+a_{in}x_{n}=\frac{1}{D}\sum\limits _{k=1}^{n}a_{ik}{\color{magenta}D_{k}}$
\end_inset

, 有
\begin_inset Formula 
\begin{align*}
a_{i1}x_{1}+a_{i2}x_{2}+\cdots+a_{in}x_{n} & =\frac{1}{D}\sum\limits _{j=1}^{n}a_{ij}\left({\color{magenta}\sum_{k=1}^{n}b_{k}A_{kj}}\right)=\frac{1}{D}\sum_{k=1}^{n}\sum_{j=1}^{n}b_{k}a_{ij}A_{kj}\\
 & \hspace{-8em}=\frac{1}{D}\sum_{k=1}^{n}b_{k}\cdot\left({\color{teal}\sum_{j=1}^{n}a_{ij}A_{kj}}\right)=\frac{1}{D}\sum_{k=1}^{n}b_{k}\cdot{\color{teal}D\delta_{ik}}=\sum_{k=1}^{n}b_{k}\delta_{ik}=b_{i}.
\end{align*}

\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
克莱姆法则的证明
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Proof
\begin_inset Argument 1
status open

\begin_layout Plain Layout
证明: (唯一性)
\end_layout

\end_inset

 只需证明线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 任何解都等于 
\begin_inset Formula $x_{j}=\frac{D_{j}}{D}$
\end_inset

 即可.
 假设这个任何解是 
\begin_inset Formula $x'=\left(x_{1}',x_{2}',\cdots,x_{n}'\right)$
\end_inset

, 于是有
\begin_inset Formula 
\[
\begin{cases}
b_{1}=a_{11}x_{1}'+a_{12}x_{2}'+\cdots+a_{1n}x_{n}',\\
b_{2}=a_{21}x_{1}'+a_{22}x_{2}'+\cdots+a_{2n}x_{n}',\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
b_{n}=a_{n1}x_{1}'+a_{n2}x_{2}'+\cdots+a_{nn}x_{n}',
\end{cases}
\]

\end_inset

所以
\begin_inset Formula 
\begin{align*}
x_{j} & =\frac{1}{D}\cdot D_{j}=\frac{1}{D}\cdot\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1,j-1} & b_{1} & a_{1,j+1} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2,j-1} & b_{2} & a_{2,j+1} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{n,j-1} & b_{n} & a_{2,j+1} & \cdots & a_{nn}
\end{vmatrix}
\end{align*}

\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
克莱姆法则的证明
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Proof
\begin_inset Argument 1
status open

\begin_layout Plain Layout
证明: (唯一性)
\end_layout

\end_inset


\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\begin{align*}
x_{j} & =\frac{1}{D}\cdot\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1,j-1} & a_{11}x_{1}'+a_{12}x_{2}'+\cdots+a_{1n}x_{n}' & a_{1,j+1} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2,j-1} & a_{21}x_{1}'+a_{22}x_{2}'+\cdots+a_{2n}x_{n}' & a_{2,j+1} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{n,j-1} & a_{n1}x_{1}'+a_{n2}x_{2}'+\cdots+a_{nn}x_{n}' & a_{2,j+1} & \cdots & a_{nn}
\end{vmatrix}\\
 & \xlongequal[{i=1,2,\cdots,n\atop i\ne j}]{c_{j}-x_{i}'\times c_{i}}\frac{1}{D}\cdot\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1,j-1} & a_{1,j}x_{j}' & a_{1,j+1} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2,j-1} & a_{2,j}x_{j}' & a_{2,j+1} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{n,j-1} & a_{n,j}x_{j}' & a_{2,j+1} & \cdots & a_{nn}
\end{vmatrix}\\
 & \xlongequal{c_{j}/x_{j}'}\frac{x_{j}'}{D}\cdot\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1,j-1} & a_{1,j} & a_{1,j+1} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2,j-1} & a_{2,j} & a_{2,j+1} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{n,j-1} & a_{n,j} & a_{2,j+1} & \cdots & a_{nn}
\end{vmatrix}=x_{j}'.
\end{align*}

\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
克莱姆法则的理论价值
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
在求解公式 
\begin_inset Formula 
\begin{equation}
x_{j}=\frac{D_{j}}{D},\qquad(j=1,2,\cdots,n)\label{eq:5-5-sol}
\end{equation}

\end_inset

时, 克莱姆法则可叙述为下面的定理.
\end_layout

\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:5-2"

\end_inset

 如果线性方程组 
\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\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}=b_{n},
\end{cases}\label{eq:5-5}
\end{equation}

\end_inset

的系数行列式 
\begin_inset Formula $D\neq0$
\end_inset

, 则 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-5"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 一定有解, 且解是唯一的.
 
\end_layout

\begin_layout Standard
因为这个解就是公式 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-5-sol"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 所给出的.
\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
克莱姆法则的理论价值
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
在解题或证明中, 常用到定理 
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:5-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

 的逆否定理:
\end_layout

\begin_layout Theorem*
\begin_inset Argument 1
status open

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
ref{thm:5-2}'
\end_layout

\end_inset


\end_layout

\end_inset

 如果线性方程组 
\begin_inset Formula 
\[
\ensuremath{\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=b_{1},\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=b_{2},\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}=b_{n},
\end{cases}}
\]

\end_inset

无解或有两个不同的解, 则它的系数行列式必为零.
\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
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
当 
\begin_inset Formula $b_{1},b_{2},\cdots,b_{n}$
\end_inset

 全为零时, 线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 称为
\series bold
齐次线性方程组
\series default
, 即
\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\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}=0,
\end{cases}\label{eq:5-2}
\end{equation}

\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
齐次线性方程组的情况
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
对齐次线性方程组 
\begin_inset Formula 
\[
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=0,\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=0,\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}=0,
\end{cases}
\]

\end_inset

易见 
\begin_inset Formula 
\[
x_{1}=x_{2}=\cdots=x_{n}=0
\]

\end_inset

一定为该方程组的一组解, 称其为
\series bold
齐次线性方程组的零解
\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
齐次线性方程组的情况
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
零解和非零解
\end_layout

\begin_layout Itemize
每个未知量的值都等于零的解称为
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
fcolorbox{gray}{yellow}{零解}
\end_layout

\end_inset

;
\end_layout

\begin_layout Itemize
至少有一个未知量不等于零的解称为
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
fcolorbox{gray}{yellow}{非零解}
\end_layout

\end_inset

.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
问题: 齐次线性方程组在什么情况下才有非零解?
\end_layout

\end_deeper
\begin_layout Frame

\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
齐次线性方程组的解
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
把克莱姆法则 (定理 
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:5-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 应用于齐次线性方程组.
\end_layout

\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:5-3"

\end_inset

 如果齐次线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 的系数行列式 
\begin_inset Formula $D\neq0$
\end_inset

, 则齐次线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 只有零解.
\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Theorem*
\begin_inset Argument 1
status open

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
ref{thm:5-3}'
\end_layout

\end_inset


\end_layout

\end_inset

 如果齐次方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 有非零解, 则它的系数行列式 
\begin_inset Formula $D=0$
\end_inset

.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Remark*
在第四章中还将进一步证明, 如果齐次线性方程组的系数行列式 
\begin_inset Formula $D=0$
\end_inset

, 则齐次线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 有非零解.
\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
用克莱姆法则求线性方程组
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
用克莱姆法则求解线性方程组:
\begin_inset Formula 
\[
\begin{cases}
2x_{1}+3x_{2}+5x_{3}=2\\
x_{1}+2x_{2}=5\\
3x_{2}+5x_{3}=4
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula 
\begin{align*}
D & =\begin{vmatrix}2 & 3 & 5\\
1 & 2 & 0\\
0 & 3 & 5
\end{vmatrix}\xlongequal{r_{1}-r_{3}}\begin{vmatrix}2 & 0 & 0\\
1 & 2 & 0\\
0 & 3 & 5
\end{vmatrix}=2\begin{vmatrix}2 & 0\\
3 & 5
\end{vmatrix}=2\times2\times5=20,\\
D_{1} & =\begin{vmatrix}2 & 3 & 5\\
5 & 2 & 0\\
4 & 3 & 5
\end{vmatrix}\xlongequal{r_{1}-r_{3}}\begin{vmatrix}-2 & 0 & 0\\
5 & 2 & 0\\
4 & 3 & 5
\end{vmatrix}=(-2)\times2\times5=-20,\\
D_{2} & =\begin{vmatrix}2 & 2 & 5\\
1 & 5 & 0\\
0 & 4 & 5
\end{vmatrix}\xlongequal{r_{1}-2r_{2}}\begin{vmatrix}0 & -8 & 5\\
1 & 5 & 0\\
0 & 4 & 5
\end{vmatrix}\xlongequal{r_{1}\leftrightarrow r_{2}}-\begin{vmatrix}1 & 5 & 0\\
0 & -8 & 5\\
0 & 4 & 5
\end{vmatrix}=-\begin{vmatrix}-8 & 5\\
4 & 5
\end{vmatrix}=60,\\
D_{3} & =\begin{vmatrix}2 & 3 & 2\\
1 & 2 & 5\\
0 & 3 & 4
\end{vmatrix}\xlongequal{r_{1}-2r_{2}}\begin{vmatrix}0 & -1 & -8\\
1 & 2 & 5\\
0 & 3 & 4
\end{vmatrix}\xlongequal{r_{1}\leftrightarrow r_{2}}-\begin{vmatrix}1 & 2 & 5\\
0 & -1 & -8\\
0 & 3 & 4
\end{vmatrix}=-\begin{vmatrix}-1 & -8\\
3 & 4
\end{vmatrix}=-20.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
由克莱姆法则,
\begin_inset Formula 
\[
x_{1}=\frac{D_{1}}{D}=-1,\quad x_{2}=\frac{D_{2}}{D}=3,\quad x_{3}=\frac{D_{3}}{D}=-1.
\]

\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
用克莱姆法则求线性方程组
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
用克莱姆法则解方程组 
\begin_inset Formula $\begin{cases}
2x_{1}+x_{2}-5x_{3}+x_{4}=8,\\
x_{1}-3x_{2}-6x_{4}=9,\\
2x_{2}-x_{3}+2x_{4}=-5,\\
x_{1}+4x_{2}-7x_{3}+6x_{4}=0.
\end{cases}$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula 
\begin{align*}
D & =\begin{vmatrix}2 & 1 & -5 & 1\\
1 & -3 & 0 & -6\\
0 & 2 & -1 & 2\\
1 & 4 & -7 & 6
\end{vmatrix}\xlongequal[r_{4}-r_{2}]{r_{1}-2r_{2}}\begin{vmatrix}0 & 7 & -5 & 13\\
1 & -3 & 0 & -6\\
0 & 2 & -1 & 2\\
0 & 7 & -7 & 12
\end{vmatrix}=-\begin{vmatrix}7 & -5 & 13\\
2 & -1 & 2\\
7 & -7 & 12
\end{vmatrix}\\
 & \xlongequal[c_{3}+2c_{2}]{c_{1}+2c_{2}}-\begin{vmatrix}-3 & -5 & 3\\
0 & -1 & 0\\
-7 & -7 & -2
\end{vmatrix}=\begin{vmatrix}-3 & 3\\
-7 & -2
\end{vmatrix}=27.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
同理计算
\begin_inset Formula 
\[
\begin{aligned}D_{1}=\begin{vmatrix}8 & 1 & -5 & 1\\
9 & -3 & 0 & -6\\
-5 & 2 & -1 & 2\\
0 & 4 & -7 & 6
\end{vmatrix}=81, & \qquad D_{2}=\begin{vmatrix}2 & 8 & -5 & 1\\
1 & 9 & 0 & -6\\
0 & -5 & -1 & 2\\
1 & 0 & -7 & 6
\end{vmatrix}=-108,\\
D_{3}=\begin{vmatrix}2 & 1 & 8 & 1\\
1 & -3 & 9 & -6\\
0 & 2 & -5 & 2\\
1 & 4 & 0 & 6
\end{vmatrix}=-27, & \qquad D_{4}=\begin{vmatrix}2 & 1 & -5 & 8\\
1 & -3 & 0 & 9\\
0 & 2 & -1 & -5\\
1 & 4 & -7 & 0
\end{vmatrix}=27,
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
所以
\begin_inset Formula 
\begin{align*}
x_{1}=\frac{D_{1}}{D}=\frac{81}{27}=3, & \qquad x_{2}=\frac{D_{2}}{D}=\frac{-108}{27}=-4,\\
x_{3}=\frac{D_{3}}{D}=\frac{-27}{27}=-1, & \qquad x_{4}=\frac{D_{4}}{D}=\frac{27}{27}=1.
\end{align*}

\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
求曲线方程 (
\strikeout on
插值公式
\strikeout default
)
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设曲线 
\begin_inset Formula $y=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}$
\end_inset

 通过四点 
\begin_inset Formula $(1,3)$
\end_inset

, 
\begin_inset Formula $(2,4)$
\end_inset

, 
\begin_inset Formula $(3,3)$
\end_inset

, 
\begin_inset Formula $(4,-3)$
\end_inset

, 求系数 
\begin_inset Formula $a_{0},a_{1},a_{2},a_{3}$
\end_inset

.
\end_layout

\begin_layout Solution*
把四个点的坐标代入曲线方程, 得线性方程组
\begin_inset Formula 
\[
\begin{cases}
a_{0}+a_{1}+a_{2}+a_{3}=3\\
a_{0}+2a_{1}+4a_{2}+8a_{3}=4\\
a_{0}+3a_{1}+9a_{2}+27a_{3}=3\\
a_{0}+4a_{1}+16a_{2}+64a_{3}=-3,
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
其系数行列式为范德蒙行列式 
\begin_inset Formula $D=\begin{vmatrix}1 & 1 & 1 & 1\\
1 & 2 & 4 & 8\\
1 & 3 & 9 & 27\\
1 & 4 & 18 & 64
\end{vmatrix}=1\cdot2\cdot3\cdot1\cdot2\cdot1=12$
\end_inset

, 而 
\begin_inset Formula 
\begin{align*}
D_{1} & =\begin{vmatrix}3 & 1 & 1 & 1\\
4 & 2 & 4 & 8\\
3 & 3 & 9 & 27\\
-3 & 4 & 16 & 64
\end{vmatrix}\xlongequal[c_{1}-3c_{2}]{\substack{c_{4}-c_{3}\\
c_{3}-c_{2}
}
}\begin{vmatrix}0 & \boxed{1} & 0 & 0\\
-2 & 2 & 2 & 4\\
-6 & 3 & 6 & 18\\
-15 & 4 & 12 & 48
\end{vmatrix}=(-1)^{1+2}\begin{vmatrix}-2 & 2 & 4\\
-6 & 6 & 18\\
-15 & 12 & 48
\end{vmatrix}\\
 & \xlongequal{c_{1}+c_{2}}-\begin{vmatrix}0 & 2 & 4\\
0 & 6 & 18\\
-3 & 12 & 48
\end{vmatrix}=-(-3)\begin{vmatrix}2 & 4\\
6 & 18
\end{vmatrix}=36.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
类似地, 计算得:
\begin_inset Formula 
\[
\begin{aligned}D_{2} & =\begin{vmatrix}1 & 3 & 1 & 1\\
1 & 4 & 4 & 8\\
1 & 3 & 9 & 27\\
1 & -3 & 16 & 64
\end{vmatrix}=-18;\ D_{3}=\begin{vmatrix}1 & 1 & 3 & 1\\
1 & 2 & 4 & 8\\
1 & 3 & 3 & 27\\
1 & 4 & -3 & 64
\end{vmatrix}=24;\ D_{4}=\begin{vmatrix}1 & 1 & 1 & 3\\
1 & 2 & 4 & 4\\
1 & 3 & 9 & 3\\
1 & 4 & 16 & -3
\end{vmatrix}=-6;\end{aligned}
\]

\end_inset

故由克莱姆法则, 得唯一解
\begin_inset Formula 
\[
a_{0}=3,\ a_{1}=-\frac{3}{2},\ a_{2}=2,\ a_{3}=-\frac{1}{2},
\]

\end_inset

即曲线方程为
\begin_inset Formula 
\[
y=3-\frac{3}{2}x+2x^{2}-\frac{1}{2}x^{3}.
\]

\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
含参线性方程组
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
问 
\begin_inset Formula $\lambda$
\end_inset

 为何值时, 齐次方程组 
\begin_inset Formula $\begin{cases}
(1-\lambda)x_{1}-2x_{2}+4x_{3}=0\\
2x_{1}+(3-\lambda)x_{2}+x_{3}=0\\
x_{1}+x_{2}+(1-\lambda)x_{3}=0
\end{cases}$
\end_inset

 有非零解?
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\[
\begin{aligned}D & =\begin{vmatrix}1-\lambda & -2 & 4\\
2 & 3-\lambda & 1\\
1 & 1 & 1-\lambda
\end{vmatrix}\xlongequal{c_{2}+c_{1}}\begin{vmatrix}1-\lambda & -3+\lambda & 4\\
2 & 1-\lambda & 1\\
1 & 0 & 1-\lambda
\end{vmatrix}\\
 & =(1-\lambda)^{3}+(\lambda-3)-4(1-\lambda)-2(1-\lambda)(-3+\lambda)\\
 & =(1-\lambda)^{3}+2(1-\lambda)^{2}+\lambda-3=\lambda(\lambda-2)(3-\lambda),
\end{aligned}
\]

\end_inset

由于齐次线性方程组有非零解, 则 
\begin_inset Formula $D=0$
\end_inset

, 解得 
\begin_inset Formula $\lambda=0$
\end_inset

, 
\begin_inset Formula $\lambda=2$
\end_inset

 或 
\begin_inset Formula $\lambda=3$
\end_inset

 时齐次线性方程组有非零解.
\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
含参线性方程组
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设方程组 
\begin_inset Formula $\begin{cases}
x+y+z=a+b+c\\
ax+by+cz=a^{2}+b^{2}+c^{2}\\
bcx+cay+abz=3abc
\end{cases}$
\end_inset

, 试问 
\begin_inset Formula $a,b,c$
\end_inset

 满足什么条件时, 方程组有惟一解, 并求出惟一解.
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\begin{align*}
D & =\begin{vmatrix}1 & 1 & 1\\
a & b & c\\
bc & ca & ab
\end{vmatrix}\xlongequal{\substack{c_{1}-c_{2}\\
c_{2}-c_{3}
}
}\begin{vmatrix}0 & 0 & 1\\
a-b & b-c & c\\
c(b-a) & a(c-b) & ab
\end{vmatrix}\\
 & \xlongequal{\substack{c_{1}\div(a-b)\\
c_{2}\div(b-c)
}
}(a-b)(b-c)\begin{vmatrix}0 & 0 & 1\\
1 & 1 & c\\
-c & -a & ab
\end{vmatrix}=(a-b)(b-c)\begin{vmatrix}1 & 1\\
-c & -a
\end{vmatrix}\\
 & =(a-b)(b-c)(c-a).
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
显然, 当 
\begin_inset Formula $a,b,c$
\end_inset

 互不相等时, 
\begin_inset Formula $D\neq0$
\end_inset

, 该方程组有唯一解.
 又
\begin_inset Formula 
\begin{align*}
D_{1} & =\begin{vmatrix}a+b+c & 1 & 1\\
a^{2}+b^{2}+c^{2} & b & c\\
3abc & ca & ab
\end{vmatrix}\xlongequal{c_{1}-bc_{2}-cc_{3}}\begin{vmatrix}a & 1 & 1\\
a^{2} & b & c\\
abc & ca & ab
\end{vmatrix}\\
 & \xlongequal{c_{1}\div a}a\begin{vmatrix}1 & 1 & 1\\
a & b & c\\
bc & ca & ab
\end{vmatrix}=aD.
\end{align*}

\end_inset


\end_layout

\begin_layout Solution*
同理可得 
\begin_inset Formula $D_{2}=bD$
\end_inset

, 
\begin_inset Formula $D_{3}=cD$
\end_inset

, 于是
\begin_inset Formula 
\[
x=\frac{D_{1}}{D}=a,\quad y=\frac{D_{2}}{D}=b,\quad z=\frac{D_{3}}{D}=c.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Subsection
结式*
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
什么是结式?
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
结式 (resultant) 是代数学术语, 指由两个多项式的系数所构成的一种行列式, 或称 Sylvester 行列式, 结式可判断两个多项式是否有公根、是否
互素, 以及判断多项式是否有重根.
 
\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
结式的定义
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
scriptsize
\end_layout

\end_inset

设 
\begin_inset Formula 
\[
\begin{aligned}f(x) & =a_{0}x^{n}+a_{1}x^{n-1}+\ldots+a_{n},\\
g(x) & =b_{0}x^{m}+b_{1}x^{m-1}+\ldots+b_{m}.
\end{aligned}
\]

\end_inset

定义下列 
\begin_inset Formula $m+n$
\end_inset

 阶行列式 
\begin_inset Formula 
\[
\mathrm{Res}(f,g)=\begin{vmatrix}a_{0} & a_{1} & a_{2} & \cdots & \cdots & a_{n} & 0\\
0 & a_{0} & a_{1} & \cdots & \cdots & a_{n-1} & a_{n} & \ddots\\
 & 0 & a_{0} & \cdots & \cdots & a_{n-2} & a_{n-1} & \ddots & 0\\
 &  & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\
 &  &  & 0 & a_{0} & \cdots & \cdots & \cdots & a_{n}\\
b_{0} & b_{1} & b_{2} & \cdots & b_{m-1} & b_{m} & 0\\
0 & b_{0} & b_{1} & \cdots & \cdots & b_{m-1} & b_{m} & \ddots\\
 & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & 0\\
 &  & 0 & b_{0} & b_{1} & \cdots & \cdots & b_{m-1} & b_{m}
\end{vmatrix}
\]

\end_inset

为 
\begin_inset Formula $f(x)$
\end_inset

 与 
\begin_inset Formula $g(x)$
\end_inset

 的结式或称 Sylvester 行列式.
 
\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
多项式的公共根
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:sylvester"

\end_inset

 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
scriptsize
\end_layout

\end_inset

多项式 
\begin_inset Formula $f(x)$
\end_inset

 与 
\begin_inset Formula $g(x)$
\end_inset

 有公共根 (在复数域中) 的充分必要条件是它们的结式 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\mathrm{Res}(f,g)=\begin{vmatrix}a_{0} & a_{1} & a_{2} & \cdots & \cdots & a_{n} & 0\\
0 & a_{0} & a_{1} & \cdots & \cdots & a_{n-1} & a_{n} & \ddots\\
 & 0 & a_{0} & \cdots & \cdots & a_{n-2} & a_{n-1} & \ddots & 0\\
 &  & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\
 &  &  & 0 & a_{0} & \cdots & \cdots & \cdots & a_{n}\\
b_{0} & b_{1} & b_{2} & \cdots & b_{m-1} & b_{m} & 0\\
0 & b_{0} & b_{1} & \cdots & \cdots & b_{m-1} & b_{m} & \ddots\\
 & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & 0\\
 &  & 0 & b_{0} & b_{1} & \cdots & \cdots & b_{m-1} & b_{m}
\end{vmatrix}=0.
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Proof
由条件设多项式 
\begin_inset Formula $f(x)$
\end_inset

 与 
\begin_inset Formula $g(x)$
\end_inset

 有公共根 
\begin_inset Formula $x=t_{0}$
\end_inset

, 不妨设 
\begin_inset Formula $a_{n}$
\end_inset

 与 
\begin_inset Formula $b_{m}$
\end_inset

 不同时零, (否则结式中最后一列为零, 结论显然), 因此 
\begin_inset Formula $x_{0}\ne0$
\end_inset

.
 考虑到
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset

 
\begin_inset Formula 
\[
\begin{cases}
a_{0}t_{0}^{m+n-1}+a_{1}t_{0}^{m+n-2}+\ldots+a_{n}t_{0}^{m-1}=0\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{0}t_{0}^{n+1}+a_{1}t_{0}^{n}+\ldots+a_{n}t_{0}=0\\
a_{0}t_{0}^{n}+a_{1}t_{0}^{n-1}+\ldots+a_{n}=0
\end{cases}
\]

\end_inset


\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\begin{cases}
b_{0}t_{0}^{m+n-1}+b_{1}t_{0}^{m+n-2}+\ldots+b_{m}t_{0}^{n-1}=0\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
b_{0}t_{0}^{m+1}+b_{1}t_{0}^{m}+\ldots+b_{m}t_{0}=0\\
b_{0}t_{0}^{m}+b_{1}t_{0}^{m-1}+\ldots+b_{m}=0
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Proof
现在令 
\begin_inset Formula $x_{r}=t_{0}^{r}$
\end_inset

, (
\begin_inset Formula $r=0,1,2,\cdots,m+n-1$
\end_inset

), 则 
\begin_inset Formula $(m+n)$
\end_inset

 元一次方程组
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset

 
\begin_inset Formula 
\[
\begin{cases}
a_{0}x_{m+n-1}+a_{1}x_{m+n-2}+\ldots+a_{n}x_{m-1}=0,\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{0}x_{n}+a_{1}x_{n-1}+\ldots+a_{n}x_{0}=0\\
b_{0}x_{m+n-1}+b_{1}x_{m+n-2}+\ldots+b_{m}x_{n-1}=0\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
b_{0}x_{m}+b_{1}x_{m-1}+\ldots+b_{m}x_{0}=0
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Proof
有非零解 
\begin_inset Formula $(x_{0},x_{1},x_{2},\cdots,x_{m+n-1})=(1,t_{0},t_{0}^{2},\cdots,t_{0}^{m+n-1})$
\end_inset

, 由齐次线性方程组的克莱姆法则的逆定理, 上式对应的系数行列式为零.
 也即
\begin_inset ERT
status open

\begin_layout Plain Layout

{
\backslash
tiny{
\end_layout

\end_inset


\begin_inset Formula 
\[
\mathrm{Res}(f,g)=\begin{vmatrix}a_{0} & a_{1} & a_{2} & \cdots & \cdots & a_{n} & 0\\
0 & a_{0} & a_{1} & \cdots & \cdots & a_{n-1} & a_{n} & \ddots\\
 & 0 & a_{0} & \cdots & \cdots & a_{n-2} & a_{n-1} & \ddots & 0\\
 &  & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\
 &  &  & 0 & a_{0} & \cdots & \cdots & \cdots & a_{n}\\
b_{0} & b_{1} & b_{2} & \cdots & b_{m-1} & b_{m} & 0\\
0 & b_{0} & b_{1} & \cdots & \cdots & b_{m-1} & b_{m} & \ddots\\
 & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & 0\\
 &  & 0 & b_{0} & b_{1} & \cdots & \cdots & b_{m-1} & b_{m}
\end{vmatrix}=0.
\]

\end_inset


\begin_inset ERT
status open

\begin_layout Plain Layout

}}
\end_layout

\end_inset


\end_layout

\begin_layout Corollary
对 
\begin_inset Formula $f(x),g(x)\in\mathbb{C}[X]$
\end_inset

, 方程组 
\begin_inset Formula $\begin{cases}
f(x)=0\\
g(x)=0
\end{cases}$
\end_inset

 有复数解当且仅当结式 
\begin_inset Formula $\mathrm{Res}(f,g)=0$
\end_inset

.
 
\end_layout

\begin_layout Standard
这就把一元方程组是否有解归纳为一个常数 
\begin_inset Formula $\mathrm{Res}(f,g)$
\end_inset

 是否为 
\begin_inset Formula $0$
\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
多元高次方程组的消元法
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
现设 
\begin_inset Formula $f(x,y),g(x,y)\in F[X,Y]$
\end_inset

 为二元多项式, 现在解方程 
\begin_inset Formula 
\[
\begin{cases}
f(x,y)=0,\\
g(x,y)=0.
\end{cases}
\]

\end_inset

可以把 
\begin_inset Formula $f$
\end_inset

 和 
\begin_inset Formula $g$
\end_inset

 看作不定元 
\begin_inset Formula $x$
\end_inset

 的 (以 
\begin_inset Formula $y$
\end_inset

 的多项式为系数的) 多项式, 即 
\begin_inset Formula 
\[
\begin{cases}
f(x,y)=a_{0}(y)x^{n}+\cdots+a_{n}(y)\in F[y][x],\\
g(x,y)=b_{0}(y)x^{m}+\cdots+b_{m}(y)\in F[y][x].
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
于是由定理 
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:sylvester"
plural "false"
caps "false"
noprefix "false"

\end_inset

 (其中 
\begin_inset Formula $a_{k}(y)$
\end_inset

 与 
\begin_inset Formula $b_{k}(y)$
\end_inset

 为变量 
\begin_inset Formula $y$
\end_inset

 的多项式), 如果上述方程有解 
\begin_inset Formula $(x,y)=(x_{0},y_{0})$
\end_inset

 使得
\begin_inset Formula 
\[
\begin{cases}
f(x_{0},y_{0})=0,\\
g(x_{0},y_{0})=0.
\end{cases}
\]

\end_inset

则必有
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\mathrm{Res}(f,g,x)\coloneqq\begin{vmatrix}a_{0} & a_{1} & a_{2} & \cdots & \cdots & a_{n} & 0 & \cdots & 0\\
0 & a_{0} & a_{1} & \cdots & \cdots & a_{n-1} & a_{n} & \cdots & 0\\
0 & 0 & a_{0} & \cdots & \cdots & a_{n-2} & a_{n-1} & \cdots & 0\\
\vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\
0 & 0 & \cdots & 0 & a_{0} & \cdots & \cdots & \cdots & a_{n}\\
b_{0} & b_{1} & b_{2} & \cdots & \cdots & \cdots & b_{m} & \cdots & 0\\
0 & b_{0} & b_{1} & \cdots & \cdots & \cdots & b_{m-1} & b_{m} & \vdots\\
\vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\
0 & \cdots & 0 & b_{0} & b_{1} & \cdots & \cdots & \cdots & b_{m}
\end{vmatrix}(y)=0.
\]

\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
如果下列齐次线性方程组有非零解, 
\begin_inset Formula $k$
\end_inset

 应取何值?
\end_layout

\begin_layout Problem
\begin_inset Formula 
\[
\begin{cases}
\phantom{(+2)\ \ }kx_{1}\phantom{+2x_{2}+3x_{3}}+\phantom{k}x_{4}=0\\
\phantom{(k+2)}x_{1}+2x_{2}\phantom{+3x_{3}}-\phantom{k}x_{4}=0\\
(k+2)x_{1}-\phantom{2}x_{2}\phantom{+3x_{3}}+4x_{4}=0\\
\phantom{(k+)\ \ }2x_{1}+\phantom{\ }x_{2}+3x_{3}+kx_{4}=0
\end{cases}\text{. }
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Problem
判定齐次线性方程组 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
systeme[x_1,x_2,x_3,x_4]{x_1+x_2+2x_3+3x_4=0,x_1+2x_2+3x_3-x_4=0,3x_1-x_2-x_3-2x
_4=0,2x_1+3x_2-x_3-x_4=0}
\end_layout

\end_inset

 是否仅有零解.
\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
作业
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Problem
用克莱姆法则求解如下线性方程组:
\end_layout

\begin_layout Problem
(1).
 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
systeme{2x_{1}+2x_{2}-x_{3}-x_{4}=4,4x_{1}+3x_{2}-x_{3}+2x_{4}=6,8x_{1}+5x_{2}-3
x_{3}-4x_{4}=12,3x_{1}+3x_{2}-2x_{3}-2x_{4}=6};
\end_layout

\end_inset


\end_layout

\begin_layout Problem
(2).
 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
systeme{2x_{1}+3x_{2}+11x_{3}+5x_{4}=2,x_{1}+x_{2}+5x_{3}+2x_{4}=1,2x_{1}+x_{2}+
3x_{3}-2x_{4}=-3,x_{1}+x_{2}+3x_{3}+4x_{4}=-3};
\end_layout

\end_inset


\end_layout

\begin_layout Problem
(3).
 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
systeme{2x_{1}+5x_{2}+4x_{3}+x_{4}=20,x_{1}+3x_{2}+2x_{3}+x_{4}=11,2x_{1}+10x_{2
}+9x_{3}+7x_{4}=40,3x_{1}+8x_{2}+9x_{3}+2x_{4}=37}.
\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
作业
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Problem
\begin_inset CommandInset label
LatexCommand label
name "prob:1.5-18"

\end_inset

 
\begin_inset Argument 1
status open

\begin_layout Plain Layout
\begin_inset Formula $\star$
\end_inset


\end_layout

\end_inset

 有同样多个未知量的两个线性方程组 (但不一定有同样多个方程), 如果第一组的任何解满足第二组, 且第二组的任何解满足第一组, 则称它们是等价的
 (有同样多末知量的两个方程组, 如果每一个都没有解, 也认为是等价的).
\end_layout

\begin_layout Problem
证明: 线性方程组的下列任一变换, 将给定方程组变为等价的方程组: 
\end_layout

\begin_layout Problem
(a).
 对调两个方程; 
\end_layout

\begin_layout Problem
(b).
 用任一不为零的数乘一个方程的两端;
\end_layout

\begin_layout Problem
(c).
 将一个方程乘以任一数后加到另一个方程中去.
\end_layout

\begin_layout Problem
改变末知量的编号是否把给定方程组变为等价的方程组? 当解方程组时, 是否容许改变未知量的编号?
\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
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
tiny
\end_layout

\end_inset

证明: 任何线性方程组 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\sum_{j=1}^{n}a_{ij}x_{j}=b_{i},\quad i=1,2,\cdots,s
\]

\end_inset

利用问题 
\begin_inset CommandInset ref
LatexCommand ref
reference "prob:1.5-18"
plural "false"
caps "false"
noprefix "false"

\end_inset

 中 
\begin_inset Formula $(a),(b),(c)$
\end_inset

 的变换和改变未知量的编号, 可以把它化到如下形式 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\sum_{j=1}^{n}c_{ij}y_{j}=d_{i},\quad i=1,2,\cdots,s
\]

\end_inset

后者满足下列三组条件之一且仅满足一组: 
\end_layout

\begin_layout Problem
(a).
 
\begin_inset Formula $c_{ii}\neq0$
\end_inset

, 
\begin_inset Formula $i=1,2,\cdots,n$
\end_inset

; 
\begin_inset Formula $c_{ij}=0$
\end_inset

, 当 
\begin_inset Formula $i>j$
\end_inset

; (特别地, 第 
\begin_inset Formula $n$
\end_inset

 个以后的所有方程 (当 
\begin_inset Formula $s>n$
\end_inset

 时) 中的末知量的系数都是零), 
\begin_inset Formula $d_{i}=0$
\end_inset

, 当 
\begin_inset Formula $i=n+1,\cdots,s$
\end_inset

 (在这种情形, 我们说: 方程组被化为三角形形式);
\end_layout

\begin_layout Problem
(b).
 存在整数 
\begin_inset Formula $r$
\end_inset

, 
\begin_inset Formula $0\le r\le n-1$
\end_inset

, 使得 
\begin_inset Formula $c_{ii}\neq0$
\end_inset

, 
\begin_inset Formula $i=1,2,\cdots,r$
\end_inset

; 
\begin_inset Formula $c_{ij}=0$
\end_inset

, 当 
\begin_inset Formula $i>j$
\end_inset

; 
\begin_inset Formula $c_{ij}=0$
\end_inset

, 对 
\begin_inset Formula $i>r$
\end_inset

 和等于 
\begin_inset Formula $1,2,\cdots,n$
\end_inset

 的任何 
\begin_inset Formula $j$
\end_inset

; 
\begin_inset Formula $d_{i}=0$
\end_inset

, 对 
\begin_inset Formula $i=r+1,r+2,\cdots,s$
\end_inset

.
 
\end_layout

\begin_layout Problem
(c).
 存在整数 
\begin_inset Formula $r$
\end_inset

, 
\begin_inset Formula $0\le r\le n$
\end_inset

, 使得: 
\begin_inset Formula $c_{ii}\neq0$
\end_inset

, 当 
\begin_inset Formula $i=1,2,\cdots,r$
\end_inset

; 
\begin_inset Formula $;c_{ij}=0$
\end_inset

, 当 
\begin_inset Formula $i>j$
\end_inset

; 
\begin_inset Formula $c_{ij}=0$
\end_inset

, 对 
\begin_inset Formula $i>r$
\end_inset

 和任何 
\begin_inset Formula $j=1,2,\cdots,n$
\end_inset

.
 存在整数 
\begin_inset Formula $k$
\end_inset

, 
\begin_inset Formula $r+1\le k\le s$
\end_inset

, 使得 
\begin_inset Formula $d_{k}\neq0$
\end_inset

.
\end_layout

\begin_layout Problem
证明: 如果在组 (2) 中恢复未知量原来的编号, 则得到与原方程组 (1) 等价的方程组.
\end_layout

\begin_layout Problem
然后证明: 
\end_layout

\begin_layout Problem
在情形 (a), 组 (2) (从而, 组 (1)) 有唯一解; 
\end_layout

\begin_layout Problem
在情形 (b), 组 (2) 有无穷多解, 并且对末知量 
\begin_inset Formula $y_{r+1},\cdots,y_{n}$
\end_inset

 的任何值, 存在其余末知量 
\begin_inset Formula $y_{1},\cdots,y_{r}$
\end_inset

 的唯一一组值; 
\end_layout

\begin_layout Problem
在情形 (c), 组 (2) 没有解.
 这一定理给出了解线性方程组的消元法的根据.
\end_layout

\end_deeper
\begin_layout Frame

\end_layout

\end_body
\end_document