Слияние кода завершено, страница обновится автоматически
#LyX 2.3 created this file. For more info see http://www.lyx.org/
\lyxformat 544
\begin_document
\begin_header
\save_transient_properties true
\origin unavailable
\textclass beamer
\begin_preamble
% 如果没有这一句命令，XeTeX会出错，原因参见
% http://bbs.ctex.org/viewthread.php?tid=60547
\DeclareRobustCommand\nobreakspace{\leavevmode\nobreak\ }
% \usepackage{tkz-euclide}
% \usetkzobj{all}
\usepackage{multicol}
\usetheme[lw]{uantwerpen}
\useoutertheme[height=0pt,width=2cm,left]{sidebar}
\usecolortheme{rose,sidebartab}
\useinnertheme{circles}
\usefonttheme[only large]{structurebold}
\setbeamercolor{sidebar left}{bg=black!15}
\setbeamercolor{structure}{fg=red}
\setbeamercolor{author}{parent=structure}
\setbeamerfont{title}{series=\normalfont,size=\LARGER}
\setbeamerfont{title in sidebar}{series=\bfseries}
\setbeamerfont{author in sidebar}{series=\bfseries}
\setbeamerfont*{item}{series=}
\setbeamerfont{frametitle}{size=}
\setbeamerfont{block title}{size=\scriptsize}
\setbeamerfont{block body example}{size=\small}
\setbeamerfont{block body}{size=\footnotesize}
\setbeamerfont{block body alerted}{size=\small}
\setbeamerfont{subtitle}{size=\normalsize,series=\normalfont}
\AtBeginDocument{
\renewcommand\logopos{111.png}
\renewcommand\logoneg{111.png}
\renewcommand\logomonowhite{111.png}
\renewcommand\iconfile{111.png}
}
\setbeamertemplate{theorems}[numbered]
\AtBeginSection[]
{
\begin{frame}{章节内容}
\transfade%淡入淡出效果
\begin{multicols}{2}
\tableofcontents[sectionstyle=show/shaded,subsectionstyle=show/shaded/hide]
\end{multicols}
\addtocounter{framenumber}{-1} %目录页不计算页码
\end{frame}
}
\usepackage{amsmath, amsfonts, amssymb, mathtools, yhmath, mathrsfs}
% http://ctan.org/pkg/extarrows
% long equal sign
\usepackage{extarrows}
\DeclareMathOperator{\sech}{sech}
\DeclareMathOperator{\curl}{curl}
%\everymath{\color{blue}\everymath{}}
%\everymath\expandafter{\color{blue}\displaystyle}
%\everydisplay\expandafter{\the\everydisplay \color{red}}
\def\degree{^\circ}
\def\bt{\begin{theorem}}
\def\et{\end{theorem}}
\def\bl{\begin{lemma}}
\def\el{\end{lemma}}
\def\bc{\begin{corrolary}}
\def\ec{\end{corrolary}}
\def\ba{\begin{proof}[解]}
\def\ea{\end{proof}}
\def\ue{\mathrm{e}}
\def\ud{\,\mathrm{d}}
\def\GF{\mathrm{GF}}
\def\ui{\mathrm{i}}
\def\Re{\mathrm{Re}}
\def\Im{\mathrm{Im}}
\def\uRes{\mathrm{Res}}
\def\diag{\,\mathrm{diag}\,}
\def\be{\begin{equation}}
\def\ee{\end{equation}}
\def\bee{\begin{equation*}}
\def\eee{\end{equation*}}
\def\sumcyc{\sum\limits_{cyc}}
\def\prodcyc{\prod\limits_{cyc}}
\def\i{\infty}
\def\a{\alpha}
\def\b{\beta}
\def\g{\gamma}
\def\d{\delta}
\def\l{\lambda}
\def\m{\mu}
\def\t{\theta}
\def\p{\partial}
\def\wc{\rightharpoonup}
\def\udiv{\mathrm{div}}
\def\diam{\mathrm{diam}}
\def\dist{\mathrm{dist}}
\def\uloc{\mathrm{loc}}
\def\uLip{\mathrm{Lip}}
\def\ucurl{\mathrm{curl}}
\def\usupp{\mathrm{supp}}
\def\uspt{\mathrm{spt}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\providecommand{\abs}[1]{\left\lvert#1\right\rvert}
\providecommand{\norm}[1]{\left\Vert#1\right\Vert}
\providecommand{\paren}[1]{\left(#1\right)}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\FF}{\mathbb{F}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\WW}{\mathbb{W}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\PP}{\mathbb{P}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\TT}{\mathbb{T}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\pNN}{\mathbb{N}_{+}}
\newcommand{\cZ}{\mathcal{Z}}
\newcommand{\cM}{\mathcal{M}}
\newcommand{\cS}{\mathcal{S}}
\newcommand{\cX}{\mathcal{X}}
\newcommand{\cW}{\mathcal{W}}
\newcommand{\eqdef}{\xlongequal{\text{def}}}%
\newcommand{\eqexdef}{\xlongequal[\text{存在}]{\text{记为}}}%
\end_preamble
\options aspectratio = 1610, 12pt, UTF8
\use_default_options true
\begin_modules
theorems-ams
theorems-sec
\end_modules
\maintain_unincluded_children false
\language chinese-simplified
\language_package default
\inputencoding utf8-cjk
\fontencoding global
\font_roman "default" "default"
\font_sans "default" "default"
\font_typewriter "default" "default"
\font_math "auto" "auto"
\font_default_family default
\use_non_tex_fonts false
\font_sc false
\font_osf false
\font_sf_scale 100 100
\font_tt_scale 100 100
\font_cjk gbsn
\use_microtype false
\use_dash_ligatures true
\graphics default
\default_output_format pdf2
\output_sync 0
\bibtex_command default
\index_command default
\float_placement H
\paperfontsize default
\spacing single
\use_hyperref true
\pdf_bookmarks true
\pdf_bookmarksnumbered false
\pdf_bookmarksopen false
\pdf_bookmarksopenlevel 1
\pdf_breaklinks true
\pdf_pdfborder true
\pdf_colorlinks true
\pdf_backref false
\pdf_pdfusetitle true
\papersize default
\use_geometry true
\use_package amsmath 2
\use_package amssymb 2
\use_package cancel 1
\use_package esint 2
\use_package mathdots 1
\use_package mathtools 2
\use_package mhchem 1
\use_package stackrel 1
\use_package stmaryrd 1
\use_package undertilde 1
\cite_engine basic
\cite_engine_type default
\biblio_style plain
\use_bibtopic false
\use_indices false
\paperorientation portrait
\suppress_date false
\justification true
\use_refstyle 1
\use_minted 0
\index Index
\shortcut idx
\color #008000
\end_index
\leftmargin 2cm
\topmargin 2cm
\rightmargin 2cm
\bottommargin 2cm
\secnumdepth 3
\tocdepth 2
\paragraph_separation indent
\paragraph_indentation default
\is_math_indent 0
\math_numbering_side default
\quotes_style english
\dynamic_quotes 0
\papercolumns 1
\papersides 1
\paperpagestyle default
\tracking_changes false
\output_changes false
\html_math_output 0
\html_css_as_file 0
\html_be_strict false
\end_header
\begin_body
\begin_layout Section
\begin_inset Formula $n$
\end_inset
阶行列式
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
引言
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
现在, 为了研究
\begin_inset Formula $n$
\end_inset
元线性方程组, 需要进一步讨论
\begin_inset Formula $n$
\end_inset
阶行列式.
\end_layout
\begin_layout Standard
但应当指出的是: 主.
副对角线法则不易于向一般
\begin_inset Formula $n$
\end_inset
阶行列式推广.
\end_layout
\begin_layout Standard
例如, 对
\begin_inset Formula $4$
\end_inset
阶行列式:
\begin_inset Formula
\[
\begin{vmatrix}\boxed{a_{11}} & a_{12} & a_{13} & a_{14}\\
a_{21} & \boxed{a_{22}} & a_{23} & a_{24}\\
a_{31} & a_{32} & a_{33} & \boxed{a_{34}}\\
a_{41} & a_{42} & \boxed{a_{43}} & a_{44}
\end{vmatrix}
\]
\end_inset
其中一项
\begin_inset Formula $a_{11}a_{22}a_{34}a_{43}$
\end_inset
的符号是什么呢? 这时用主,副对角线法则就不好确定了!
\end_layout
\end_deeper
\begin_layout Subsection
\series bold
\begin_inset Formula $n$
\end_inset
级排列与逆序数
\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
\series bold
\begin_inset Formula $n$
\end_inset
级排列与逆序数
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout FrameSubtitle
排列与逆序
\end_layout
\begin_layout Definition
由自然数
\begin_inset Formula $1,2,\cdots,n$
\end_inset
组成的不重复的每一种有确定次序的排列, 称为一个
\series bold
\begin_inset Formula $n$
\end_inset
级排列
\series default
(简称为
\series bold
排列
\series default
).
\end_layout
\begin_layout Standard
例如,
\begin_inset Formula $1234$
\end_inset
和
\begin_inset Formula $4312$
\end_inset
都是
\begin_inset Formula $4$
\end_inset
级排列, 而
\begin_inset Formula $24315$
\end_inset
是一个
\begin_inset Formula $5$
\end_inset
级排列.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition
在一个
\begin_inset Formula $n$
\end_inset
级排列
\begin_inset Formula $\left(i_{1}i_{2}\cdots i_{t}\cdots i_{s}\cdots i_{n}\right)$
\end_inset
中, 若数
\begin_inset Formula $i_{t}>i_{s}$
\end_inset
, 则称数
\begin_inset Formula $i_{t}$
\end_inset
与
\begin_inset Formula $i_{s}$
\end_inset
构成一个
\series bold
逆序
\series default
.
一个
\begin_inset Formula $n$
\end_inset
级排列中逆序的总数称为该
\series bold
排列的逆序数
\series default
, 记为
\begin_inset Formula $N\left(i_{1}i_{2}\cdots i_{n}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
根据上述定义, 可按如下方法计算排列的逆序数:
\end_layout
\begin_layout Standard
设在一个
\begin_inset Formula $n$
\end_inset
级排列
\begin_inset Formula $i_{1}i_{2}\cdots i_{n}$
\end_inset
中, 比
\begin_inset Formula $i_{t}$
\end_inset
(
\begin_inset Formula $t=1,2,\cdots,n$
\end_inset
) 大的且排在
\begin_inset Formula $i_{t}$
\end_inset
前面的数共有
\begin_inset Formula $t_{i}$
\end_inset
个, 则
\begin_inset Formula $i_{t}$
\end_inset
的逆序的个数为
\begin_inset Formula $t_{i}$
\end_inset
, 而该排列中所有
\series bold
元素的逆序数
\series default
之和就是这个
\series bold
排列的逆序数
\series default
.
即
\begin_inset Formula
\[
N\left(i_{1}i_{2}\cdots i_{n}\right)=t_{1}+t_{2}+\cdots+t_{n}=\sum_{i=1}^{n}t_{i}.
\]
\end_inset
\end_layout
\begin_layout Definition
逆序数为奇数的排列称为
\series bold
奇排列
\series default
, 逆序数为偶数的排列称为
\series bold
偶排列
\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
计算逆序数
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算排列
\begin_inset Formula $32514$
\end_inset
的逆序数.
\end_layout
\begin_layout Solution*
在排列
\begin_inset Formula $32514$
\end_inset
中,
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $3$
\end_inset
排在首位, 故其逆序数为
\begin_inset Formula $0$
\end_inset
;
\end_layout
\begin_layout Itemize
\begin_inset Formula $2$
\end_inset
的前面比
\begin_inset Formula $2$
\end_inset
大的数只有
\begin_inset Formula $1$
\end_inset
个
\begin_inset Formula $3$
\end_inset
, 故其逆序数为
\begin_inset Formula $1$
\end_inset
;
\end_layout
\begin_layout Itemize
\begin_inset Formula $5$
\end_inset
的前面没有比
\begin_inset Formula $5$
\end_inset
大的数, 故其逆序数为
\begin_inset Formula $0$
\end_inset
;
\end_layout
\begin_layout Itemize
\begin_inset Formula $1$
\end_inset
的前面比
\begin_inset Formula $1$
\end_inset
大的数有
\begin_inset Formula $3$
\end_inset
个, 故其逆序数为
\begin_inset Formula $3$
\end_inset
;
\end_layout
\begin_layout Itemize
\begin_inset Formula $4$
\end_inset
的前面比
\begin_inset Formula $4$
\end_inset
大的数有
\begin_inset Formula $1$
\end_inset
个, 故其逆序数为
\begin_inset Formula $1$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Float table
wide false
sideways false
status open
\begin_layout Plain Layout
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
centering
\end_layout
\end_inset
\begin_inset Tabular
<lyxtabular version="3" rows="3" columns="6">
<features booktabs="true" tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
排列
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
3
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
2
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
5
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
1
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
4
\end_layout
\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
逆序
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
0
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
1
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
0
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
3
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
1
\end_layout
\end_inset
</cell>
</row>
</lyxtabular>
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Solution*
于是排列
\begin_inset Formula $32514$
\end_inset
的逆序数为
\begin_inset Formula $N=0+1+0+3+1=5$
\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
计算逆序数
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算排列
\begin_inset Formula $217986354$
\end_inset
的逆序数, 并讨论其偶性.
\end_layout
\begin_layout Solution*
\begin_inset Float table
wide false
sideways false
status open
\begin_layout Plain Layout
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
centering
\end_layout
\end_inset
\begin_inset Tabular
<lyxtabular version="3" rows="3" columns="10">
<features booktabs="true" tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
排列
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
2
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
1
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
7
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
9
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
8
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
6
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
3
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
5
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
4
\end_layout
\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
逆序
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
0
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
1
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
0
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
0
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
1
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
3
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
4
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
4
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
5
\end_layout
\end_inset
</cell>
</row>
</lyxtabular>
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Solution*
于是题设排列的逆序数为
\begin_inset Formula
\[
N=5+4+4+3+1+0+0+1+0=18。
\]
\end_inset
该排列是偶排列.
\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
计算逆序数
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求排列
\begin_inset Formula $n(n-1)(n-2)\cdots321$
\end_inset
的逆序数, 并讨论其奇偶性.
\end_layout
\begin_layout Solution*
\begin_inset Float table
wide false
sideways false
status open
\begin_layout Plain Layout
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
centering
\end_layout
\end_inset
\begin_inset Tabular
<lyxtabular version="3" rows="3" columns="8">
<features booktabs="true" tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
排列
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $n$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $n-1$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $n-2$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\cdots$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $3$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $2$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $1$
\end_inset
\end_layout
\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\cdots$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\downarrow$
\end_inset
\end_layout
\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
逆序
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $0$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $1$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $2$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $\cdots$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $n-3$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $n-2$
\end_inset
\end_layout
\end_inset
</cell>
<cell alignment="center" valignment="top" bottomline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\begin_inset Formula $n-1$
\end_inset
\end_layout
\end_inset
</cell>
</row>
</lyxtabular>
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Solution*
于是题设排列的逆序数为
\begin_inset Formula
\[
N=(n-1)+(n-2)+(n-3)+\cdots+2+1+0=\frac{n(n-1)}{2}\text{. }
\]
\end_inset
易见
\end_layout
\begin_deeper
\begin_layout Itemize
当
\begin_inset Formula $n=4k,4k+1$
\end_inset
时, 题设排列是偶排列;
\end_layout
\begin_layout Itemize
当
\begin_inset Formula $n=4k+2,4k+3$
\end_inset
时, 题设排列是奇排列.
\end_layout
\end_deeper
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
\begin_inset Formula $n$
\end_inset
阶行列式的定义
\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
\begin_inset Formula $n$
\end_inset
阶行列式的定义
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:def-det"
\end_inset
由
\begin_inset Formula $n^{2}$
\end_inset
个元素
\begin_inset Formula $a_{ij}$
\end_inset
(
\begin_inset Formula $i,j=1,2,\cdots,n$
\end_inset
) 组成的记号
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}
\]
\end_inset
称为
\series bold
\begin_inset Formula $n$
\end_inset
阶行列式
\series default
, 其中横排称为
\series bold
行
\series default
, 竖排称为
\series bold
列
\series default
, 它表示所有取自不同行、不同列的
\begin_inset Formula $n$
\end_inset
个元素乘积
\begin_inset Formula $a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}}$
\end_inset
的代数和,
\color brown
各项的符号是:
\color inherit
当该项各元素的行标按自然顺序排列后:
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-2mm}
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Itemize
若对应的列标构成的排列是偶排列则取
\series bold
正号
\series default
;
\end_layout
\begin_layout Itemize
若对应的列标构成的排列是奇排列则取
\series bold
负号
\series default
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset CommandInset ref
LatexCommand ref
reference "def:def-det"
plural "false"
caps "false"
noprefix "false"
\end_inset
'
\end_layout
\end_inset
即:
\begin_inset Formula
\[
\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}=\sum_{j_{1}j_{2}\cdots j_{n}}(-1)^{N(j_{1}j_{2}\cdots j_{n})}a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}},
\]
\end_inset
其中
\begin_inset Formula $\sum_{j_{1}j_{2}\cdots j_{n}}$
\end_inset
表示对所有
\begin_inset Formula $n$
\end_inset
级排列
\begin_inset Formula $j_{1}j_{2}\cdots j_{n}$
\end_inset
求和.
行列式有时也简记为
\begin_inset Formula ${\color{brown}\mathrm{det}(a_{ij})}$
\end_inset
或
\begin_inset Formula $\left|a_{ij}\right|$
\end_inset
\begin_inset Foot
status open
\begin_layout Plain Layout
行列式的定义是怎么来的
\begin_inset CommandInset href
LatexCommand href
name "阅读材料"
target "https://zhuanlan.zhihu.com/p/76526424"
literal "false"
\end_inset
.
\end_layout
\end_inset
, 这里数
\begin_inset Formula $a_{ij}$
\end_inset
称为
\series bold
行列式的元素
\series default
, 称
\begin_inset Formula $(-1)^{N(j_{1}j_{2}\cdots j_{n})}a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}}$
\end_inset
为
\series bold
行列式的一般项
\series default
.
\end_layout
\begin_layout Remark*
(1)
\begin_inset Formula $a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}}$
\end_inset
的符号为
\begin_inset Formula $(-1)^{N(j_{1}j_{2}\cdots j_{n})}$
\end_inset
(不算元素本身所带的符号);
\end_layout
\begin_layout Remark*
(2) 一阶行列式
\begin_inset Formula $|a|=a$
\end_inset
, 不要与绝对值记号相混淆.
\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
\begin_inset Formula $n$
\end_inset
阶行列式的定义
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算行列式
\begin_inset Formula $D=\begin{vmatrix}0 & 0 & 0 & 1\\
0 & 0 & 2 & 0\\
0 & 3 & 0 & 0\\
4 & 0 & 0 & 0
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
四阶行列式
\begin_inset Formula $D$
\end_inset
的一般项为
\begin_inset Formula $(-1)^{N(j_{1}j_{2}j_{3}j_{4})}a_{1j_{1}}a_{2j2}a_{3j_{3}}a_{4j_{4}}$
\end_inset
,
\end_layout
\begin_layout Solution*
\begin_inset Formula $D$
\end_inset
中第
\begin_inset Formula $1$
\end_inset
行的非零元素只有
\begin_inset Formula $a_{14}$
\end_inset
, 因而
\begin_inset Formula $j_{1}$
\end_inset
只需取
\begin_inset Formula $4$
\end_inset
;
\end_layout
\begin_layout Solution*
同理
\end_layout
\begin_layout Solution*
由
\begin_inset Formula $D$
\end_inset
中第
\begin_inset Formula $2,3,4$
\end_inset
行知,
\begin_inset Formula $j_{2}=3$
\end_inset
,
\begin_inset Formula $j_{3}=2$
\end_inset
,
\begin_inset Formula $j_{4}=1$
\end_inset
, 即行列式
\begin_inset Formula $D$
\end_inset
中的非零项只有一项, 即
\begin_inset Formula
\[
D=(-1)^{N(4321)}a_{14}a_{23}a_{32}a_{41}=(-1)^{N(4321)}1\cdot2\cdot3\cdot4=24.
\]
\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
\begin_inset Formula $n$
\end_inset
阶行列式的定义
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算上三角形行列式
\begin_inset Formula $\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
0 & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & a_{nn}
\end{vmatrix}$
\end_inset
, (
\begin_inset Formula $a_{11}a_{22}\cdots a_{nn}\neq0$
\end_inset
).
\end_layout
\begin_layout Standard
行列式中从左上角到右下角的对角线称为
\series bold
主对角线
\series default
.
\end_layout
\begin_layout Solution*
行列式的一般项为
\begin_inset Formula
\[
(-1)^{N(j_{1}j_{2}\cdots j_{n})}a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}},\ j_{n}=n,\ j_{n-1}=n-1,\cdots,\ j_{2}=2,\ j_{1}=1,
\]
\end_inset
所以不为零的项只有
\begin_inset Formula $a_{11}a_{12}\cdots a_{nn}$
\end_inset
, 所以
\begin_inset Formula
\[
\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
0 & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & a_{nn}
\end{vmatrix}=(-1)^{N(12\cdots n)}a_{11}a_{12}\cdots a_{nn}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
同理, 下三角形行列式
\begin_inset Formula
\[
\begin{vmatrix}a_{11} & 0 & \cdots & 0\\
a_{21} & a_{22} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}=a_{11}a_{22}\cdots a_{nn}.
\]
\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
\begin_inset Formula $n$
\end_inset
阶行列式的定义
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $D_{1}=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}$
\end_inset
,
\begin_inset Formula $D_{2}=\begin{vmatrix}a_{11} & a_{12}b^{-1} & \cdots & a_{1n}b^{1-n}\\
a_{21}b & a_{22} & \cdots & a_{2n}b^{2-n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1}b^{n-1} & a_{n2}b^{n-2} & \cdots & a_{nn}
\end{vmatrix}$
\end_inset
, 证明:
\begin_inset Formula $D_{1}=D_{2}$
\end_inset
.
\end_layout
\begin_layout Proof
由定义知,
\begin_inset Formula $D_{1}=\sum\limits _{j_{1}j_{2}\cdots j_{n}}(-1)^{N(j_{1}j_{2}\cdots j_{n})}a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}}$
\end_inset
, 再注意到
\begin_inset Formula $D_{2}$
\end_inset
中第
\begin_inset Formula $i$
\end_inset
行第
\begin_inset Formula $j$
\end_inset
列的元素可表为
\begin_inset Formula $a_{ij}b^{i-j}$
\end_inset
, 故
\begin_inset Formula
\[
D_{2}=\sum\limits _{j_{1}j_{2}\cdots j_{n}}(-1)^{N\left(j_{1}j_{2}\cdots j_{n}\right)}a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}}b^{(1+2+\cdots+n)-\left(j_{1}+j_{2}+\cdots+j_{n}\right)},
\]
\end_inset
然而
\begin_inset Formula $j_{1}+j_{2}+\cdots+j_{n}=1+2+\cdots+n$
\end_inset
.
所以
\begin_inset Formula
\[
D_{2}=\sum\limits _{j_{1}j_{2}\cdots j_{n}}(-1)^{N\left(j_{1}j_{2}\cdots j_{n}\right)}a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}}\Longrightarrow D_{1}=D_{2}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
对换
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
对换
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
为进一步研究
\begin_inset Formula $n$
\end_inset
阶行列式的性质, 先要讨论对换的概念及其与排列奇偶性的关系.
\end_layout
\begin_layout Definition
在排列中, 将任意两个元素对调, 其余的元素不动, 这种作出新排列的方式称为
\series bold
对换
\series default
.
将两个相邻元素对换, 称为
\series bold
相邻对换
\series default
.
\end_layout
\begin_layout Theorem
任意一个排列经过一个对换后, 其奇偶性改变.
\end_layout
\begin_layout Corollary
奇排列变成自然顺序排列的对换次数为奇数, 偶排列变成自然顺序排列的对换次数为偶数.
\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
对换
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
\begin_inset Formula $n$
\end_inset
个自然数
\begin_inset Formula $(n>1)$
\end_inset
共有
\begin_inset Formula $n!$
\end_inset
个
\begin_inset Formula $n$
\end_inset
级排列, 其中奇偶排列各占一半.
\end_layout
\begin_layout Remark*
\begin_inset Formula $n$
\end_inset
阶行列式是
\begin_inset Formula $n!$
\end_inset
项的代数和, 且冠以正号的项和冠以负号的项 (不算元素本身所带的符号) 各占一半;
\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
对换
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
试判断
\begin_inset Formula $a_{14}a_{23}a_{31}a_{42}a_{56}a_{65}$
\end_inset
和
\begin_inset Formula $-a_{32}a_{43}a_{14}a_{51}a_{25}a_{66}$
\end_inset
是否都是六阶行列式中的项.
\end_layout
\begin_layout Solution*
\begin_inset Formula $a_{14}a_{23}a_{31}a_{42}a_{56}a_{65}$
\end_inset
下标的逆序数为
\begin_inset Formula $N(431265)=0+1+2+2+0+1=6$
\end_inset
.
\end_layout
\begin_layout Solution*
所以
\begin_inset Formula $a_{14}a_{23}a_{31}a_{42}a_{56}a_{65}$
\end_inset
是六阶行列式中的项.
\end_layout
\begin_layout Solution*
而
\begin_inset Formula $a_{32}a_{43}a_{14}a_{51}a_{25}a_{66}$
\end_inset
两下标排列的逆序数和为
\begin_inset Formula $N(341526)+N(234156)=5+3=8$
\end_inset
,
\end_layout
\begin_layout Solution*
所以
\begin_inset Formula $-a_{32}a_{43}a_{14}a_{51}a_{25}a_{66}$
\end_inset
不是六阶行列式中的的项.
\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
对换
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
在六阶行列式中, 下列两项各应带什么符号
\end_layout
\begin_layout Example
(1)
\begin_inset Formula $a_{23}a_{31}a_{42}a_{56}a_{14}a_{65}$
\end_inset
;
\end_layout
\begin_layout Example
(2)
\begin_inset Formula $a_{32}a_{43}a_{14}a_{51}a_{66}a_{25}$
\end_inset
.
\end_layout
\begin_layout Solution*
(1)
\begin_inset Formula $a_{23}a_{31}a_{42}a_{56}a_{14}a_{65}=a_{14}a_{23}a_{31}a_{42}a_{56}a_{65}$
\end_inset
,
\begin_inset Formula $431265$
\end_inset
的逆序数为
\begin_inset Formula
\[
N=0+1+2+2+0+1=6,
\]
\end_inset
所以
\begin_inset Formula $a_{23}a_{31}a_{42}a_{56}a_{14}a_{65}$
\end_inset
前边应带正号.
\end_layout
\begin_layout Solution*
(2)
\begin_inset Formula $a_{32}a_{43}a_{14}a_{51}a_{66}a_{25}$
\end_inset
行标排列
\begin_inset Formula $341562$
\end_inset
的逆序数为
\begin_inset Formula $N=0+0+2+0+0+4=6$
\end_inset
,
\end_layout
\begin_layout Solution*
列标排列
\begin_inset Formula $234165$
\end_inset
的逆序数为
\begin_inset Formula $N=0+0+0+3+0+1=4$
\end_inset
,
\end_layout
\begin_layout Solution*
所以
\begin_inset Formula $a_{32}a_{43}a_{14}a_{51}a_{66}a_{25}$
\end_inset
前边应带正号.
\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
对换
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
用行列式的定义计算
\begin_inset Formula $D_{n}=\begin{vmatrix}0 & 0 & \cdots & 0 & 1 & 0\\
0 & 0 & \cdots & 2 & 0 & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
n-1 & 0 & \cdots & 0 & 0 & 0\\
0 & 0 & \cdots & 0 & 0 & n
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
根据行列式的定义, 有
\begin_inset Formula
\[
D_{n}=(-1)^{N}a_{1,n-1}a_{2,n-2}\cdots a_{n-1,1}a_{nn}=(-1)^{N}1\cdot2\cdots(n-1)\cdot(n-2)\cdot n=(-1)^{N}n!,
\]
\end_inset
其中
\begin_inset Formula $N=N[(n-1)(n-2)\cdots21n]=0+1+2+\cdots+(n-2)+0=\frac{(n-1)(n-2)}{2}$
\end_inset
,
\end_layout
\begin_layout Solution*
所以
\begin_inset Formula $D_{n}=(-1)^{\frac{(n-1)(n-2)}{2}}n!$
\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
行列式的其它定义
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
\begin_inset Formula $n$
\end_inset
阶行列式也定义为
\begin_inset Formula
\[
D=\sum(-1)^{S}a_{i_{1}j_{1}}a_{i_{2}j_{2}}\cdots a_{i_{n}j_{n}},
\]
\end_inset
其中
\begin_inset Formula $S$
\end_inset
为行标与列标排列的逆序数之和.
即
\begin_inset Formula $S=N\left(i_{1}i_{2}\cdots i_{n}\right)+N\left(j_{1}j_{2}\cdots j_{n}\right)$
\end_inset
.
\end_layout
\begin_layout Corollary
\begin_inset Formula $n$
\end_inset
阶行列式也可定义为
\begin_inset Formula
\[
D=\sum(-1)^{N\left(i_{1}i_{2}\cdots i_{n}\right)}a_{i_{1}1}a_{i_{2}2}\cdots a_{i_{n}n}.
\]
\end_inset
\end_layout
\begin_layout Corollary
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
\begin_inset Formula $n$
\end_inset
阶行列式的其它定义方式*
\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
\begin_inset Formula $n$
\end_inset
阶行列式的其它定义方式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
事实上, 三阶行列式还可以按第一行展开的方法得到行列式的值.
即
\begin_inset Formula
\[
D=\begin{vmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{vmatrix}=a_{1}A_{11}+a_{12}A_{12}+a_{13}A_{13},
\]
\end_inset
其中
\begin_inset Formula $A_{11},A_{12},A_{13}$
\end_inset
分别是第一行元素
\begin_inset Formula $a_{11},a_{12},a_{13}$
\end_inset
的代数余子式:
\begin_inset Formula
\[
A_{11}=(-1)^{1+1}\cdot\begin{vmatrix}a_{22} & a_{23}\\
a_{32} & a_{33}
\end{vmatrix}=a_{22}a_{33}-a_{23}a_{32}.
\]
\end_inset
\end_layout
\begin_layout Standard
同理
\begin_inset Formula
\[
\begin{aligned}A_{12} & =(-1)^{1+2}\cdot\begin{vmatrix}a_{21} & a_{23}\\
a_{31} & a_{33}
\end{vmatrix}=-\left(a_{21}a_{33}-a_{23}a_{31}\right),\\
A_{13} & =(-1)^{1+3}\cdot\begin{vmatrix}a_{21} & a_{22}\\
a_{31} & a_{32}
\end{vmatrix}=a_{21}a_{32}-a_{22}a_{31}.
\end{aligned}
\]
\end_inset
注意到,
\begin_inset Formula
\[
\begin{aligned}D & =\begin{vmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{vmatrix}=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}\\
& =a_{11}\cdot\left(a_{22}a_{33}-a_{23}a_{32}\right)\\
& \quad-a_{12}\cdot\left(a_{21}a_{33}-a_{23}a_{31}\right)\\
& \quad+a_{13}\cdot\left(a_{21}a_{32}-a_{22}a_{31}\right)
\end{aligned}
\]
\end_inset
不难看到, 这与用主, 副对角线法则得到的结果是一致的.
\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
例子
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
例如, 对行列式
\begin_inset Formula
\[
D=\begin{vmatrix}1 & 0 & 1\\
-1 & 2 & 3\\
3 & 1 & 1
\end{vmatrix},\quad A_{11}=-1,\quad A_{12}=10,\quad A_{13}=-7,
\]
\end_inset
从而行列式的值
\begin_inset Formula
\[
\begin{aligned}D & =a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}\\
& =1\times(-1)+0\times10+1\times(-7)=-8.
\end{aligned}
\]
\end_inset
与对角线法结果相同。
\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
规律总结
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
这一展开的规律启示我们:
\end_layout
\begin_layout Enumerate
可以用低阶行列式的值去定义高阶行列式的值;
\begin_inset Newline newline
\end_inset
即, 从二, 三阶行列式出发去定义一般的
\begin_inset Formula $n$
\end_inset
阶行列式.
\end_layout
\begin_layout Enumerate
这样的定义方式应该具有某种内在的一致性.
即, 这样定义的各阶行列式应该有统一的性质.
见第
\begin_inset CommandInset ref
LatexCommand ref
reference "sec:4"
plural "false"
caps "false"
noprefix "false"
\end_inset
节.
\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
\begin_inset Formula $n$
\end_inset
阶行列式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:det"
\end_inset
由
\begin_inset Formula $n\times n$
\end_inset
个数
\begin_inset Formula $a_{ij}$
\end_inset
(
\begin_inset Formula $i,j=1,2,\ldots,n$
\end_inset
) 组成的具有
\begin_inset Formula $n$
\end_inset
行
\begin_inset Formula $n$
\end_inset
列的式子
\begin_inset Formula
\[
D=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}=\left|a_{ij}\right|_{n\times n}
\]
\end_inset
叫做
\begin_inset Formula $n$
\end_inset
阶行列式 (Determinant), 并且规定其值为:
\end_layout
\begin_layout Definition
1) 当
\begin_inset Formula $n=1$
\end_inset
时,
\begin_inset Formula $D=\left|a_{11}\right|=a_{11}$
\end_inset
;
\end_layout
\begin_layout Definition
2) 当
\begin_inset Formula $n\geq2$
\end_inset
时,
\begin_inset Formula
\begin{align*}
D & =a_{11}A_{11}+a_{12}A_{12}+\cdots+a_{1n}A_{1n}\\
& =\sum_{j=1}^{n}a_{1j}A_{1j}
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset CommandInset ref
LatexCommand ref
reference "def:det"
plural "false"
caps "false"
noprefix "false"
\end_inset
\end_layout
\end_inset
\begin_inset Formula
\begin{align*}
D & =a_{11}A_{11}+a_{12}A_{12}+\cdots+a_{1n}A_{1n}\\
& =\sum_{j=1}^{n}a_{1j}A_{1j}
\end{align*}
\end_inset
其中
\begin_inset Formula $A_{1j}=(-1)^{1+j}M_{1j}$
\end_inset
,
\begin_inset Formula
\[
M_{1j}=\begin{vmatrix}a_{21} & \cdots & a_{2,j-1} & a_{2,j+1} & \cdots & a_{2n}\\
a_{31} & \cdots & a_{3,j-1} & a_{3,j+1} & \cdots & a_{3n}\\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\
a_{n1} & \cdots & a_{n,j-1} & a_{n,j+1} & \cdots & a_{nn}
\end{vmatrix},
\]
\end_inset
并称
\begin_inset Formula $M_{1j}$
\end_inset
为行列式
\begin_inset Formula $D$
\end_inset
的元素
\begin_inset Formula $a_{1j}$
\end_inset
的余子式,
\begin_inset Formula $A_{1j}$
\end_inset
为行列式
\begin_inset Formula $D$
\end_inset
的元素
\begin_inset Formula $a_{1j}$
\end_inset
的代数余子式.
\end_layout
\end_deeper
\begin_layout Subsection
作业
\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
作业
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Problem
求以下排列的逆序数
\end_layout
\begin_layout Problem
(1).
\begin_inset Formula $2,3,5,4,1$
\end_inset
; (2).
\begin_inset Formula $6,3,1,2,5,4$
\end_inset
; (3).
\begin_inset Formula $1,3,5,7,\cdots,2n-1,2,4,6,8,\cdots,2n$
\end_inset
; (4).
\begin_inset Formula $2,4,6,\cdots,2n,1,3,5,\cdots,2n-1$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
写出数
\begin_inset Formula $1,2,\cdots,n$
\end_inset
的一个排列, 其逆序数最大, 并求其逆序数.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
判断下列乘积在相应阶数的行列式中带有怎样的符号:
\end_layout
\begin_layout Problem
(1).
\begin_inset Formula $a_{43}a_{21}a_{35}a_{12}a_{54}$
\end_inset
; (2).
\begin_inset Formula $a_{61}a_{23}a_{45}a_{36}a_{12}a_{54}$
\end_inset
; (3).
\begin_inset Formula $a_{27}a_{36}a_{51}a_{74}a_{25}a_{43}a_{62}$
\end_inset
; (4).
\begin_inset Formula $a_{33}a_{16}a_{72}a_{27}a_{55}a_{61}a_{44}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
若
\begin_inset Formula $(-1)^{N(i432k)+N(52j14)}a_{i5}a_{42}a_{3j}a_{21}a_{k4}$
\end_inset
是五阶行列式的一项, 则
\begin_inset Formula $i,j,k$
\end_inset
应为何值? 此时该项的符号是什么?
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
选取
\begin_inset Formula $i$
\end_inset
和
\begin_inset Formula $k$
\end_inset
的值, 使得乘积
\begin_inset Formula $a_{62}a_{i5}a_{33}a_{k4}a_{46}a_{21}$
\end_inset
是相应
\begin_inset Formula $6$
\end_inset
阶行列式且带有负号.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
取
\begin_inset Formula $i$
\end_inset
和
\begin_inset Formula $k$
\end_inset
的值, 使得乘积
\begin_inset Formula $a_{47}a_{63}a_{1i}a_{55}a_{7k}a_{24}a_{31}$
\end_inset
是相应
\begin_inset Formula $7$
\end_inset
阶行列式中且带有正号.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
用行列式的定义计算下列行列式:
\begin_inset Formula $\begin{vmatrix}0 & 1 & 0 & 1\\
1 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 1
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
仅利用行列式的定义, 计算
\begin_inset Formula
\[
\begin{vmatrix}a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\
a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\
a_{31} & a_{32} & 0 & 0 & 0\\
a_{41} & a_{42} & 0 & 0 & 0\\
a_{51} & a_{52} & 0 & 0 & 0
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
已知
\begin_inset Formula $f(x)=\begin{vmatrix}x & 1 & 1 & 2\\
1 & x & 1 & -1\\
3 & 2 & x & 1\\
1 & 1 & 2x & 1
\end{vmatrix}$
\end_inset
, 求
\begin_inset Formula $x^{3}$
\end_inset
的系数.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
求出行列式
\begin_inset Formula
\[
\begin{vmatrix}5x & 1 & 2 & 3\\
x & x & 1 & 2\\
1 & 2 & x & 3\\
x & 1 & 2 & 2x
\end{vmatrix}
\]
\end_inset
包含
\begin_inset Formula $x^{4}$
\end_inset
和
\begin_inset Formula $x^{3}$
\end_inset
的项.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
Вы можете оставить комментарий после Вход в систему
Неприемлемый контент может быть отображен здесь и не будет показан на странице. Вы можете проверить и изменить его с помощью соответствующей функции редактирования.
Если вы подтверждаете, что содержание не содержит непристойной лексики/перенаправления на рекламу/насилия/вульгарной порнографии/нарушений/пиратства/ложного/незначительного или незаконного контента, связанного с национальными законами и предписаниями, вы можете нажать «Отправить» для подачи апелляции, и мы обработаем ее как можно скорее.
Опубликовать ( 0 )