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
行列式的性质与计算
\end_layout
\begin_layout Subsection
行列式的计算
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
按照
\begin_inset Formula $n$
\end_inset
阶行列式的定义,
\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
人们很少通过上式直接计算行列式的值, 上述定义常用于理论证明, 也即仅有理论价值, 对于通常的计算大多采用行列式的性质给出.
\end_layout
\begin_layout Standard
在本节, 我们致力于学习和理解这些计算行列式时常用的性质.
\end_layout
\end_deeper
\begin_layout Subsection
行列式的性质与计算
\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
行列式的性质与计算1
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
将行列式
\begin_inset Formula $D$
\end_inset
的
\series bold
行与列互换
\series default
后得到的行列式, 称为
\begin_inset Formula $D$
\end_inset
的
\series bold
转置行列式
\series default
, 记为
\begin_inset Formula $D^{T}$
\end_inset
或
\begin_inset Formula $D^{\prime}$
\end_inset
, 即若
\begin_inset Formula
\[
D=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix},\quad\text{ 则 }\quad D^{T}=\begin{vmatrix}a_{11} & a_{21} & \cdots & a_{n1}\\
a_{12} & a_{22} & \cdots & a_{n2}\\
\vdots & \vdots & \ddots & \vdots\\
a_{1n} & a_{2n} & \cdots & a_{nn}
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Proposition
\begin_inset CommandInset label
LatexCommand label
name "prop:1"
\end_inset
行列式与它的转置行列式相等, 即
\begin_inset Formula $D=D^{T}$
\end_inset
.
\end_layout
\begin_layout Remark*
由以上命题知, 行列式中的行与列具有相同的地位.
行列式的行具有的性质, 它的列也同样具有.
\end_layout
\begin_layout Example
若
\begin_inset Formula $D=\begin{vmatrix}1 & 2 & 3\\
-1 & 0 & 1\\
0 & 1 & \sqrt{2}
\end{vmatrix}$
\end_inset
, 则
\begin_inset Formula $D^{T}=\begin{vmatrix}1 & -1 & 0\\
2 & 0 & 1\\
3 & 1 & \sqrt{2}
\end{vmatrix}=D$
\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
行列式的性质与计算2
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
\begin_inset ERT
status open
\begin_layout Plain Layout
<1->
\end_layout
\end_inset
\begin_inset CommandInset label
LatexCommand label
name "prop:2"
\end_inset
交换行列式的两行 (列), 行列式变号.
\end_layout
\begin_layout Corollary
\begin_inset ERT
status open
\begin_layout Plain Layout
<2>
\end_layout
\end_inset
\begin_inset CommandInset label
LatexCommand label
name "cor:2"
\end_inset
若行列式中有两行 (列) 的对应元素相同, 则此行列式为零.
\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
行列式的性质与计算2
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
(1)
\begin_inset Formula $\begin{vmatrix}1 & 2 & 1\\
0 & 1 & -1\\
2 & -1 & 0
\end{vmatrix}=-\begin{vmatrix}0 & 1 & -1\\
1 & 2 & 1\\
2 & -1 & 0
\end{vmatrix}$
\end_inset
, (第一、二行互换).
\end_layout
\begin_layout Example
(2)
\begin_inset Formula $\begin{vmatrix}1 & 2 & 1\\
0 & 1 & -1\\
2 & -1 & 0
\end{vmatrix}=-\begin{vmatrix}1 & 1 & 2\\
0 & -1 & 1\\
2 & 0 & -1
\end{vmatrix}$
\end_inset
, (第二、 三列互换).
\end_layout
\begin_layout Example
(3)
\begin_inset Formula $\begin{vmatrix}1 & 1 & 0\\
1 & 1 & 0\\
5 & \sqrt{2} & 7
\end{vmatrix}=0$
\end_inset
, (第一、二两行相等).
\end_layout
\begin_layout Example
(4)
\begin_inset Formula $\begin{vmatrix}-2 & 1 & 1\\
4 & 2 & 2\\
7 & -3 & -3
\end{vmatrix}=0$
\end_inset
, (第二、三列相等).
\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
行列式的性质与计算2 - 应用
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
求直线的方程
\end_layout
\end_inset
已知平面上两点的坐标
\begin_inset Formula $A(x_{A},y_{A})$
\end_inset
,
\begin_inset Formula $B(x_{B},y_{B})$
\end_inset
, 则由
\begin_inset Formula $A,B$
\end_inset
确定的直线方程满足
\begin_inset Formula
\[
\begin{vmatrix}x & y & 1\\
x_{A} & y_{A} & 1\\
x_{B} & y_{B} & 1
\end{vmatrix}=0.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
求平面的方程
\end_layout
\end_inset
已知空间中不共线三点
\begin_inset Formula $A,B,C$
\end_inset
的坐标分别为
\begin_inset Formula $\left(x_{A},y_{A},z_{A}\right)$
\end_inset
,
\begin_inset Formula $\left(x_{B},y_{B},z_{B}\right)$
\end_inset
,
\begin_inset Formula $\left(x_{C},y_{C},z_{C}\right)$
\end_inset
, 则由
\begin_inset Formula $A,B,C$
\end_inset
确定的平面方程为
\begin_inset Formula
\[
\begin{vmatrix}x & y & z & 1\\
x_{A} & y_{A} & z_{A} & 1\\
x_{B} & y_{B} & z_{B} & 1\\
x_{C} & y_{C} & z_{C} & 1
\end{vmatrix}=0.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
求圆的方程
\end_layout
\end_inset
已知平面上三点的坐标
\begin_inset Formula $A(x_{A},y_{A})$
\end_inset
,
\begin_inset Formula $B(x_{B},y_{B})$
\end_inset
,
\begin_inset Formula $C(x_{C},y_{C})$
\end_inset
, 则过
\begin_inset Formula $A,B,C$
\end_inset
三点的圆的方程为
\begin_inset Formula
\[
\begin{vmatrix}x^{2}+y^{2} & x & y & 1\\
x_{A}^{2}+y_{A}^{2} & x_{A} & y_{A} & 1\\
x_{B}^{2}+y_{B}^{2} & x_{B} & y_{B} & 1\\
x_{C}^{2}+y_{C}^{2} & x_{C} & y_{C} & 1
\end{vmatrix}=0.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem*
多少个点能够确定圆锥曲线的方程?
\end_layout
\begin_layout Itemize
使用行列式求曲线方程能否避免解方程.
(思考前面小节的作业
\begin_inset CommandInset ref
LatexCommand ref
reference "prob:inter"
plural "false"
caps "false"
noprefix "false"
\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
行列式的性质与计算3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Proposition
\begin_inset CommandInset label
LatexCommand label
name "prop:3"
\end_inset
用数
\begin_inset Formula $k$
\end_inset
乘行列式的某一行 (列), 等于用数
\begin_inset Formula $k$
\end_inset
乘此行列式, 即
\begin_inset Formula
\[
D_{1}=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
ka_{i1} & ka_{i2} & \cdots & ka_{in}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}=k\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{i1} & a_{i2} & \cdots & a_{in}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}=kD.
\]
\end_inset
\end_layout
\begin_layout Standard
第
\begin_inset Formula $i$
\end_inset
行 (列) 乘以
\begin_inset Formula $k$
\end_inset
, 记为
\begin_inset Formula $r_{i}\times k$
\end_inset
(或
\begin_inset Formula $c_{i}\times k$
\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
行列式的性质与计算3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Corollary
\begin_inset CommandInset label
LatexCommand label
name "cor:3-1"
\end_inset
行列式的某一行 (列) 中所有元素的公因子可以提到行列式符号的外面.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Corollary
如果行列式中某行 (列) 的元素全为零, 则此行列式为零.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Corollary
\begin_inset CommandInset label
LatexCommand label
name "cor:3-2"
\end_inset
行列式中若有两行 (列) 元素成比例, 则此行列式为零.
\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
行列式的性质与计算3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
(1)
\begin_inset Formula $\begin{vmatrix}1 & -1 & 2\\
0 & 1 & 5\\
\sqrt{2} & -\sqrt{2} & 2\sqrt{2}
\end{vmatrix}=0$
\end_inset
因为第三行是第一行的
\begin_inset Formula $\sqrt{2}$
\end_inset
倍.
\end_layout
\begin_layout Example
(2)
\begin_inset Formula $\begin{vmatrix}1 & 4 & 1 & 0\\
2 & 8 & 3 & 5\\
0 & 0 & 1 & 4\\
-1 & -4 & -5 & 7
\end{vmatrix}=0$
\end_inset
因为第一列与第二列成比例, 即第二列是第一列的
\begin_inset Formula $4$
\end_inset
倍.
\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
行列式的性质与计算3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
若
\begin_inset Formula $D=\begin{vmatrix}1 & 0 & 2\\
3 & -1 & 0\\
1 & 2 & -1
\end{vmatrix}$
\end_inset
, 则
\begin_inset Formula $\begin{vmatrix}-2 & 0 & -4\\
3 & -1 & 0\\
1 & 2 & -1
\end{vmatrix}=(-2)\begin{vmatrix}1 & 0 & 2\\
3 & -1 & 0\\
1 & 2 & -1
\end{vmatrix}=-2D$
\end_inset
.
\end_layout
\begin_layout Example
又
\begin_inset Formula
\[
\begin{vmatrix}4 & 0 & 2\\
12 & -1 & 0\\
4 & 2 & -1
\end{vmatrix}=4\begin{vmatrix}1 & 0 & 2\\
3 & -1 & 0\\
1 & 2 & -1
\end{vmatrix}=4D\text{. }
\]
\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
行列式的性质与计算3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $\begin{vmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{vmatrix}=1$
\end_inset
, 求
\begin_inset Formula $\begin{vmatrix}6a_{11} & -2a_{12} & -10a_{13}\\
-3a_{21} & a_{22} & 5a_{23}\\
-3a_{31} & a_{32} & 5a_{33}
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
<2>
\end_layout
\end_inset
利用行列式性质, 有
\begin_inset Formula
\[
\begin{aligned}\begin{vmatrix}6a_{11} & -2a_{12} & -10a_{13}\\
-3a_{21} & a_{22} & 5a_{23}\\
-3a_{31} & a_{32} & 5a_{33}
\end{vmatrix} & =-2\begin{vmatrix}-3a_{11} & a_{12} & 5a_{13}\\
-3a_{21} & a_{22} & 5a_{23}\\
-3a_{31} & a_{32} & 5a_{33}
\end{vmatrix}=-2\cdot(-3)\cdot5\begin{vmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{vmatrix}\\
& =-2\cdot(-3)\cdot5\cdot1=30.
\end{aligned}
\]
\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
行列式的性质与计算3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
证明奇数阶反对称行列式的值为零.
\end_layout
\begin_layout Remark*
这里, 反对称行列式是满足行列式转置后所有元素变为其相反数的行列式.
换句话说, 反对称行列式具有如下形式:
\begin_inset Formula
\[
D=\begin{vmatrix}0 & a_{12} & a_{13} & \cdots & a_{1n}\\
-a_{12} & 0 & a_{23} & \cdots & a_{2n}\\
-a_{13} & -a_{23} & 0 & \cdots & a_{3n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
-a_{1n} & -a_{2n} & -a_{3n} & \cdots & 0
\end{vmatrix}
\]
\end_inset
\end_layout
\begin_layout Proof
设反对称行列式
\begin_inset Formula
\[
D=\begin{vmatrix}0 & a_{12} & a_{13} & \cdots & a_{1n}\\
-a_{12} & 0 & a_{23} & \cdots & a_{2n}\\
-a_{13} & -a_{23} & 0 & \cdots & a_{3n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
-a_{1n} & -a_{2n} & -a_{3n} & \cdots & 0
\end{vmatrix}
\]
\end_inset
其中
\begin_inset Formula $a_{ij}=-a_{ji}$
\end_inset
(
\begin_inset Formula $i\neq j$
\end_inset
时),
\begin_inset Formula $a_{ij}=0$
\end_inset
(
\begin_inset Formula $i=j$
\end_inset
时).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proof
利用行列式性质
\begin_inset CommandInset ref
LatexCommand ref
reference "prop:1"
plural "false"
caps "false"
noprefix "false"
\end_inset
及性质
\begin_inset CommandInset ref
LatexCommand ref
reference "prop:3"
plural "false"
caps "false"
noprefix "false"
\end_inset
的推论
\begin_inset CommandInset ref
LatexCommand ref
reference "cor:3-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
, 有
\begin_inset Formula
\[
D=D^{T}=(-1)^{n}\begin{vmatrix}0 & a_{12} & a_{13} & \cdots & a_{1n}\\
-a_{12} & 0 & a_{23} & \cdots & a_{2n}\\
-a_{13} & -a_{23} & 0 & \cdots & a_{3n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
-a_{1n} & -a_{2n} & -a_{3n} & \cdots & 0
\end{vmatrix}=(-1)^{n}D,
\]
\end_inset
当
\begin_inset Formula $n$
\end_inset
为奇数时有
\begin_inset Formula $D=-D$
\end_inset
, 即
\begin_inset Formula $D=0$
\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
行列式的性质与计算4
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Proposition
\begin_inset Argument 1
status open
\begin_layout Plain Layout
行列式分解
\end_layout
\end_inset
\begin_inset CommandInset label
LatexCommand label
name "prop:4"
\end_inset
若行列式的某一行 (列) 的元素都是两数之和, 例如,
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
D=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
b_{i1}+c_{i1} & b_{i2}+c_{i2} & \cdots & b_{in}+c_{in}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}
\]
\end_inset
则
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
D=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
b_{i1} & b_{i2} & \cdots & b_{in}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}+\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
c_{i1} & c_{i2} & \cdots & c_{in}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}=D_{1}+D_{2}.
\]
\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
行列式的性质与计算4
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
(1)
\begin_inset Formula $\begin{vmatrix}2 & 3\\
1 & 1
\end{vmatrix}=\begin{vmatrix}1+1 & 3+0\\
1 & 1
\end{vmatrix}=\begin{vmatrix}1 & 3\\
1 & 1
\end{vmatrix}+\begin{vmatrix}1 & 0\\
1 & 1
\end{vmatrix}$
\end_inset
;
\end_layout
\begin_layout Example
(2)
\begin_inset Formula $\begin{vmatrix}1 & 1+\sqrt{2} & 5\\
0 & 3-2 & 7\\
2 & -1-\sqrt{2} & -1
\end{vmatrix}=\begin{vmatrix}1 & 1+(\sqrt{2}) & 5\\
0 & 3+(-2) & 7\\
2 & -1+(-\sqrt{2}) & -1
\end{vmatrix}=\begin{vmatrix}1 & 1 & 5\\
0 & 3 & 7\\
2 & -1 & -1
\end{vmatrix}+\begin{vmatrix}1 & \sqrt{2} & 5\\
0 & -2 & 7\\
2 & -\sqrt{2} & -1
\end{vmatrix}$
\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
行列式的性质与计算4
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
因为
\begin_inset Formula $\begin{vmatrix}3+1 & 2-2\\
-1+2 & 3+0
\end{vmatrix}=\begin{vmatrix}4 & 0\\
1 & 3
\end{vmatrix}=12$
\end_inset
, 而
\begin_inset Formula $\begin{vmatrix}3 & 2\\
-1 & 3
\end{vmatrix}+\begin{vmatrix}1 & -2\\
2 & 0
\end{vmatrix}=(9+2)+(0+4)=15$
\end_inset
.
\end_layout
\begin_layout Example
因此
\begin_inset Formula
\[
\begin{vmatrix}3+1 & 2-2\\
-1+2 & 3+0
\end{vmatrix}\neq\begin{vmatrix}3 & 2\\
-1 & 3
\end{vmatrix}+\begin{vmatrix}1 & -2\\
2 & 0
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Remark*
一般来说下式是不成立的
\begin_inset Formula
\[
\begin{vmatrix}a_{11}+b_{11} & a_{12}+b_{12}\\
a_{21}+b_{21} & a_{22}+b_{22}
\end{vmatrix}\neq\begin{vmatrix}a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}+\begin{vmatrix}b_{11} & b_{12}\\
b_{21} & b_{22}
\end{vmatrix}.
\]
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Proposition
\begin_inset CommandInset label
LatexCommand label
name "prop:5"
\end_inset
将行列式的某一行 (列) 的所有元素都乘以数
\begin_inset Formula $k$
\end_inset
后加到另一行 (列) 对应位置的元素上, 行列式不变.
\end_layout
\begin_layout Remark
以数
\begin_inset Formula $k$
\end_inset
乘第
\begin_inset Formula $j$
\end_inset
行加到第
\begin_inset Formula $i$
\end_inset
行上, 记作
\begin_inset Formula
\[
r_{i}+kr_{j};
\]
\end_inset
以数
\begin_inset Formula $k$
\end_inset
乘第
\begin_inset Formula $j$
\end_inset
列加到第
\begin_inset Formula $i$
\end_inset
列上, 记作
\begin_inset Formula
\[
c_{i}+kc_{j}.
\]
\end_inset
助记:
\begin_inset Formula $r$
\end_inset
表示单词 row 的首字母,
\begin_inset Formula $c$
\end_inset
表示 column 的首字母.
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
(1)
\begin_inset Formula $\begin{vmatrix}1 & 3 & -1\\
1 & 4 & -1\\
2 & 3 & 1
\end{vmatrix}\xlongequal{r_{2}-r_{1}}\begin{vmatrix}1 & 3 & -1\\
0 & 1 & 0\\
2 & 3 & 1
\end{vmatrix}$
\end_inset
, 上式表示第一行乘以
\begin_inset Formula $-1$
\end_inset
后加第二行上去, 其值不变.
\end_layout
\begin_layout Example
(2)
\begin_inset Formula $\begin{vmatrix}1 & 3 & -1\\
1 & 4 & -1\\
2 & 3 & 1
\end{vmatrix}\xlongequal{c_{3}+c_{1}}\begin{vmatrix}1 & 3 & 0\\
1 & 4 & 0\\
2 & 3 & 3
\end{vmatrix}$
\end_inset
, 上式表示第一列乘以
\begin_inset Formula $1$
\end_inset
后加到第三列上去, 其值不变.
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算行列式
\begin_inset Formula $D=\begin{vmatrix}3 & 6 & 12\\
2 & -3 & 0\\
5 & 1 & 2
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Overprint
\begin_inset Argument item:1
status open
\begin_layout Plain Layout
2
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Solution*
先将第一行的公因子
\begin_inset Formula $3$
\end_inset
提出来:
\begin_inset Formula
\[
\begin{vmatrix}3 & 6 & 12\\
2 & -3 & 0\\
5 & 1 & 2
\end{vmatrix}=3\begin{vmatrix}1 & 2 & 4\\
2 & -3 & 0\\
5 & 1 & 2
\end{vmatrix},
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Overprint
\begin_inset Argument item:1
status open
\begin_layout Plain Layout
3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Solution*
再计算
\begin_inset Formula
\begin{align*}
D & =3\begin{vmatrix}1 & 2 & 4\\
2 & -3 & 0\\
5 & 1 & 2
\end{vmatrix}=3\begin{vmatrix}1 & 2 & 4\\
0 & -7 & -8\\
0 & -9 & -18
\end{vmatrix}=27\begin{vmatrix}1 & 2 & 4\\
0 & 7 & 8\\
0 & 1 & 2
\end{vmatrix}=54\begin{vmatrix}1 & 2 & 2\\
0 & 7 & 4\\
0 & 1 & 1
\end{vmatrix}=54\begin{vmatrix}1 & 0 & 2\\
0 & 3 & 4\\
0 & 0 & 1
\end{vmatrix}\\
& =54\times3=162.
\end{align*}
\end_inset
\end_layout
\end_deeper
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算
\begin_inset Formula $D=\begin{vmatrix}3 & 1 & -1 & 2\\
-5 & 1 & 3 & -4\\
2 & 0 & 1 & -1\\
1 & -5 & 3 & -3
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Overprint
\begin_inset Argument item:1
status open
\begin_layout Plain Layout
2
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}D & \xlongequal{c_{1}\leftrightarrow c_{2}}-\begin{vmatrix}1 & 3 & -1 & 2\\
1 & -5 & 3 & -4\\
0 & 2 & 1 & -1\\
-5 & 1 & 3 & -3
\end{vmatrix}\xlongequal{\substack{r_{2}-r_{1}\\
r_{4}+5r_{1}
}
}-\begin{vmatrix}1 & 3 & -1 & 2\\
0 & -8 & 4 & -6\\
0 & 2 & 1 & -1\\
0 & 16 & -2 & 7
\end{vmatrix}\end{aligned}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Overprint
\begin_inset Argument item:1
status open
\begin_layout Plain Layout
3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}D & {\color{gray}\xlongequal{c_{1}\leftrightarrow c_{2}}-\begin{vmatrix}1 & 3 & -1 & 2\\
1 & -5 & 3 & -4\\
0 & 2 & 1 & -1\\
-5 & 1 & 3 & -3
\end{vmatrix}\xlongequal{\substack{r_{2}-r_{1}\\
r_{4}+5r_{1}
}
}-\begin{vmatrix}1 & 3 & -1 & 2\\
0 & -8 & 4 & -6\\
0 & 2 & 1 & -1\\
0 & 16 & -2 & 7
\end{vmatrix}}\\
& \xlongequal{r_{2}\leftrightarrow r_{3}}\begin{vmatrix}1 & 3 & -1 & 2\\
0 & 2 & 1 & -1\\
0 & -8 & 4 & -6\\
0 & 16 & -2 & 7
\end{vmatrix}\xlongequal[r_{4}-8r_{2}]{r_{3}+4r_{2}}\begin{vmatrix}1 & 3 & -1 & 2\\
0 & 2 & 1 & -1\\
0 & 0 & 8 & -10\\
0 & 0 & -10 & 15
\end{vmatrix}
\end{aligned}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Overprint
\begin_inset Argument item:1
status open
\begin_layout Plain Layout
4
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}D & {\color{gray}\xlongequal{r_{2}\leftrightarrow r_{3}}\begin{vmatrix}1 & 3 & -1 & 2\\
0 & 2 & 1 & -1\\
0 & -8 & 4 & -6\\
0 & 16 & -2 & 7
\end{vmatrix}\xlongequal[r_{4}-8r_{2}]{r_{3}+4r_{2}}\begin{vmatrix}1 & 3 & -1 & 2\\
0 & 2 & 1 & -1\\
0 & 0 & 8 & -10\\
0 & 0 & -10 & 15
\end{vmatrix}}\\
& \xlongequal{r_{4}+\frac{5}{4}r_{3}}\begin{vmatrix}1 & 3 & -1 & 2\\
0 & 2 & 1 & -1\\
0 & 0 & 8 & -10\\
0 & 0 & 0 & 5/2
\end{vmatrix}=40.
\end{aligned}
\]
\end_inset
\end_layout
\end_deeper
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算
\begin_inset Formula $D=\begin{vmatrix}3 & 1 & 1 & 1\\
1 & 3 & 1 & 1\\
1 & 1 & 3 & 1\\
1 & 1 & 1 & 3
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
<2>
\end_layout
\end_inset
\begin_inset Formula
\begin{align*}
D & \xlongequal{r_{1}+r_{2}+r_{3}+r_{4}}\left|\begin{array}{llll}
6 & 6 & 6 & 6\\
1 & 3 & 1 & 1\\
1 & 1 & 3 & 1\\
1 & 1 & 1 & 3
\end{array}\right|=6\left|\begin{array}{llll}
1 & 1 & 1 & 1\\
1 & 3 & 1 & 1\\
1 & 1 & 3 & 1\\
1 & 1 & 1 & 3
\end{array}\right|\\
& \xlongequal[r_{4}-r_{1}]{\substack{r_{2}-r_{1}\\
r_{3}-r_{1}
}
}6\left|\begin{array}{llll}
1 & 1 & 1 & 1\\
0 & 2 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 2
\end{array}\right|=48.
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Remark*
仿照上述方法可得到更一般的结果:
\begin_inset Formula
\[
\left|\begin{array}{rrrrr}
a & b & b & \cdots & b\\
b & a & b & \cdots & b\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
b & b & b & \cdots & a
\end{array}\right|=[a+(n-1)b](a-b)^{n-1}.
\]
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算
\begin_inset Formula $\begin{vmatrix}a_{1} & -a_{1} & 0 & 0\\
0 & a_{2} & -a_{2} & 0\\
0 & 0 & a_{3} & -a_{3}\\
1 & 1 & 1 & 1
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
根据行列式的特点, 可将第
\begin_inset Formula $1$
\end_inset
列加至第
\begin_inset Formula $2$
\end_inset
列, 然后将第
\begin_inset Formula $2$
\end_inset
列加至第
\begin_inset Formula $3$
\end_inset
列, 再将第
\begin_inset Formula $3$
\end_inset
列加至第
\begin_inset Formula $4$
\end_inset
列, 目的是使
\begin_inset Formula $D_{4}$
\end_inset
中的零元素增多.
\begin_inset Formula
\[
\begin{aligned}D_{4} & \xlongequal{c_{2}+c_{1}}\begin{vmatrix}a_{1} & 0 & 0 & 0\\
0 & a_{2} & -a_{2} & 0\\
0 & 0 & a_{3} & -a_{3}\\
1 & 2 & 1 & 1
\end{vmatrix}\xlongequal{c_{3}+c_{2}}\begin{vmatrix}a_{1} & 0 & 0 & 0\\
0 & a_{2} & 0 & 0\\
0 & 0 & a_{3} & -a_{3}\\
1 & 2 & 3 & 1
\end{vmatrix}\\
& \xlongequal{c_{4}+c_{3}}\begin{vmatrix}a_{1} & 0 & 0 & 0\\
0 & a_{2} & 0 & 0\\
0 & 0 & a_{3} & 0\\
1 & 2 & 3 & 4
\end{vmatrix}=4a_{1}a_{2}a_{3}.
\end{aligned}
\]
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算
\begin_inset Formula $D=\begin{vmatrix}a & b & c & d\\
a & a+b & a+b+c & a+b+c+d\\
a & 2a+b & 3a+2b+c & 4a+3b+2c+d\\
a & 3a+b & 6a+3b+c & 10a+6b+3c+d
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
<2>
\end_layout
\end_inset
从第
\begin_inset Formula $4$
\end_inset
行开始, 后一行减前一行:
\begin_inset Formula
\[
\begin{aligned}D & \xlongequal[r_{2}-r_{1}]{\substack{r_{4}-r_{3}\\
r_{3}-r_{2}
}
}\begin{vmatrix}a & b & c & d\\
0 & a & a+b & a+b+c\\
0 & a & 2a+b & 3a+2b+c\\
0 & a & 3a+b & 6a+3b+c
\end{vmatrix}\xlongequal[r_{3}-r_{2}]{r_{4}-r_{3}}\begin{vmatrix}a & b & c & d\\
0 & a & a+b & a+b+c\\
0 & 0 & a & 2a+b\\
0 & 0 & a & 3a+b
\end{vmatrix}.\\
& \xlongequal{r_{4}-r_{3}}\begin{vmatrix}a & b & c & d\\
0 & a & a+b & a+b+c\\
0 & 0 & a & 2a+b\\
0 & 0 & 0 & a
\end{vmatrix}=a^{4}.
\end{aligned}
\]
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
书本例 8 的其它证法
\end_layout
\end_inset
\begin_inset CommandInset label
LatexCommand label
name "exa:laplace"
\end_inset
设
\begin_inset Formula $D=\begin{vmatrix}a_{11} & \cdots & a_{1k} & 0 & \cdots & 0\\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\
a_{k1} & \cdots & a_{kk} & 0 & \cdots & 0\\
c_{11} & \cdots & c_{1k} & b_{11} & \cdots & b_{1n}\\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\
c_{n1} & \cdots & c_{nk} & b_{n1} & \cdots & b_{nn}
\end{vmatrix}$
\end_inset
,
\begin_inset Formula $D_{1}=\mathrm{det}(a_{ij})=\begin{vmatrix}a_{11} & \cdots & a_{1k}\\
\vdots & \ddots & \vdots\\
a_{k1} & \cdots & a_{kk}
\end{vmatrix}$
\end_inset
,
\begin_inset Formula $D_{2}=\mathrm{det}(b_{ij})=\begin{vmatrix}b_{11} & \cdots & b_{1n}\\
\vdots & \ddots & \vdots\\
b_{n1} & \cdots & b_{nn}
\end{vmatrix}$
\end_inset
, 证明:
\begin_inset Formula $D=D_{1}D_{2}$
\end_inset
.
\end_layout
\begin_layout Proof
对
\begin_inset Formula $D_{1}$
\end_inset
作运算
\begin_inset Formula $r_{i}+kr_{j}$
\end_inset
, 对
\begin_inset Formula $D_{2}$
\end_inset
作运算
\begin_inset Formula $c_{i}+kc_{j}$
\end_inset
, 可分别把
\begin_inset Formula $D_{1}$
\end_inset
和
\begin_inset Formula $D_{2}$
\end_inset
化为下三角形行列式.
\begin_inset Formula
\[
D_{1}=\begin{vmatrix}p_{11} & \cdots & 0\\
\vdots & \ddots & \vdots\\
p_{k1} & \cdots & p_{kk}
\end{vmatrix}=p_{11}\cdots p_{kk};\ D_{2}=\begin{vmatrix}q_{11} & \cdots & 0\\
\vdots & \ddots & \vdots\\
q_{n1} & \cdots & q_{nn}
\end{vmatrix}=q_{11}\cdots q_{nn}\cdot
\]
\end_inset
对
\begin_inset Formula $D$
\end_inset
的前
\begin_inset Formula $k$
\end_inset
行作与对
\begin_inset Formula $D_{1}$
\end_inset
相同的运算
\begin_inset Formula $r_{i}+kr_{j}$
\end_inset
, 再对后
\begin_inset Formula $n$
\end_inset
列作与对
\begin_inset Formula $D_{2}$
\end_inset
相同的运算
\begin_inset Formula $c_{i}+kc_{j}$
\end_inset
, 即把
\begin_inset Formula $D$
\end_inset
化为下三角形行列式, 且
\begin_inset Formula $D=p_{11}\cdots p_{kk}\cdot q_{11}\cdots q_{nn}=D_{1}D_{2}$
\end_inset
, 证毕.
\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
行列式的性质与计算5
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
解方程
\begin_inset Formula $\begin{vmatrix}a_{1} & a_{2} & a_{3} & \cdots & a_{n-1} & a_{n}\\
a_{1} & a_{1}+a_{2}-x & a_{3} & \cdots & a_{n-1} & a_{n}\\
a_{1} & a_{2} & a_{2}+a_{3}-x & \cdots & a_{n-1} & a_{n}\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
a_{1} & a_{2} & a_{3} & \cdots & a_{n-2}+a_{n-1}-x & a_{n}\\
a_{1} & a_{2} & a_{3} & \cdots & a_{n-1} & a_{n-1}+a_{n}-x
\end{vmatrix}=0$
\end_inset
.
\end_layout
\begin_layout Solution*
从第二行开始每一行都减去第一行得
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\begin{align*}
\begin{vmatrix}a_{1} & a_{2} & a_{3} & \cdots & a_{n-1} & a_{n}\\
0 & a_{1}-x & 0 & \cdots & 0 & 0\\
0 & 0 & a_{2}-x & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & a_{n-2}-x & 0\\
0 & 0 & 0 & \cdots & 0 & a_{n-1}-x
\end{vmatrix}\\
\qquad\qquad=a_{1}\left(a_{1}-x\right)\left(a_{2}-x\right)\cdots\left(a_{n-2}-x\right)\left(a_{n-1}-x\right),
\end{align*}
\end_inset
由
\begin_inset Formula $a_{1}\left(a_{1}-x\right)\left(a_{2}-x\right)\cdots\left(a_{n-2}-x\right)\left(a_{n-1}-x\right)=0$
\end_inset
, 解得方程的
\begin_inset Formula $n-1$
\end_inset
个根:
\begin_inset Formula
\[
x_{1}=a_{1},\ x_{2}=a_{2},\cdots,\ x_{n-2}=a_{n-2},\ x_{n-1}=a_{n-1}.
\]
\end_inset
\end_layout
\begin_layout Remark*
事实上, 根据
\begin_inset Formula $n$
\end_inset
阶行列式的定义, 方程左边的行列式是
\begin_inset Formula $x$
\end_inset
的多项式, 且次数不超过
\begin_inset Formula $x$
\end_inset
出现的行数 (或列数), 也即: 所解的方程最多是个
\begin_inset Formula $(n-1)$
\end_inset
次方程.
结合推论
\begin_inset CommandInset ref
LatexCommand ref
reference "cor:2"
plural "false"
caps "false"
noprefix "false"
\end_inset
并观察行列式的结构和规律可以猜到
\begin_inset Formula $x_{1}=a_{1}$
\end_inset
,
\begin_inset Formula $x_{2}=a_{2}$
\end_inset
, ...,
\begin_inset Formula $x_{n-2}=a_{n-2}$
\end_inset
,
\begin_inset Formula $x_{n-1}=a_{n-1}$
\end_inset
是方程的
\begin_inset Formula $(n-1)$
\end_inset
个解, 解数恰等于方程次数的上界
\begin_inset Formula $(n-1)$
\end_inset
, 所以方程的解已经解出.
\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
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
\series bold
在计算行列式时
\series default
, 常把它化为三角形行列式来计算.
以化为
\series bold
上三角形行列式
\series default
为例, 其步骤是:
\end_layout
\begin_layout Enumerate
如果第一列第一个元素为
\begin_inset Formula $0$
\end_inset
, 先将第一行与其它行交换使得第一列第一个元素不为
\begin_inset Formula $0$
\end_inset
;
\end_layout
\begin_layout Enumerate
然后把第一行分别乘以适当的数加到其它各行, 使得第一列除第一个元素外其余元素全为
\begin_inset Formula $0$
\end_inset
;
\end_layout
\begin_layout Enumerate
再用同样的方法处理除去第一行和第一列后余下的低一阶行列式, 如此继续下去, 直至使它成为
\series bold
上三角形行列式
\series default
, 这时主对角线上元素的乘积就是所求行列式的值.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\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 Exercise
计算下列行列式
\end_layout
\begin_layout Exercise
(1).
\begin_inset Formula $\begin{vmatrix}0 & -1 & -1 & 2\\
1 & -1 & 0 & 2\\
-1 & 2 & -1 & 0\\
2 & 1 & 1 & 0
\end{vmatrix}$
\end_inset
; (2).
\begin_inset Formula $\begin{vmatrix}1 & 1 & 1 & 1\\
1 & -1 & 1 & 1\\
1 & 1 & -1 & 1\\
1 & 1 & 1 & -1
\end{vmatrix}$
\end_inset
; (3).
\begin_inset Formula $\begin{vmatrix}0 & 1 & 1 & 1\\
1 & 1 & 1 & 1\\
1 & 1 & 0 & 1\\
1 & 1 & 1 & 0
\end{vmatrix}$
\end_inset
;
\end_layout
\begin_layout Exercise
(4).
\begin_inset Formula $\begin{vmatrix}2 & -5 & 1 & 2\\
-3 & 7 & -1 & 4\\
5 & -9 & 2 & 7\\
4 & -6 & 1 & 2
\end{vmatrix}$
\end_inset
; (5).
\begin_inset Formula $\begin{vmatrix}-3 & 9 & 3 & 6\\
-5 & 8 & 2 & 7\\
4 & -5 & -3 & -2\\
7 & -8 & -4 & -5
\end{vmatrix}$
\end_inset
; (6).
\begin_inset Formula $\begin{vmatrix}3 & -3 & -5 & 8\\
-3 & 2 & 4 & -6\\
2 & -5 & -7 & 5\\
-4 & 3 & 5 & -6
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
不展开而计算行列式:
\begin_inset Formula
\[
\begin{vmatrix}a & b & c & 1\\
b & c & a & 1\\
c & a & b & 1\\
\frac{b+c}{2} & \frac{c+a}{2} & \frac{a+b}{2} & 1
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
不展开行列式证明下列恒等式:
\begin_inset Formula
\[
\begin{vmatrix}0 & x & y & z\\
x & 0 & z & y\\
y & z & 0 & x\\
z & y & x & 0
\end{vmatrix}=\begin{vmatrix}0 & 1 & 1 & 1\\
1 & 0 & z^{2} & y^{2}\\
1 & z^{2} & 0 & x^{2}\\
1 & y^{2} & x^{2} & 0
\end{vmatrix}.
\]
\end_inset
(
\series bold
Hint:
\series default
分别用
\begin_inset Formula $yz$
\end_inset
,
\begin_inset Formula $xz$
\end_inset
和
\begin_inset Formula $xy$
\end_inset
乘等式左边行列式的第二, 第三和第四列.)
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
如果
\begin_inset Formula $\begin{vmatrix}x & y & z\\
3 & 0 & 2\\
1 & 1 & 1
\end{vmatrix}=1$
\end_inset
, 计算下面各行列式之值:
\end_layout
\begin_layout Exercise
(1).
\begin_inset Formula $\begin{vmatrix}2x & 2y & 2z\\
\frac{3}{2} & 0 & 1\\
1 & 1 & 1
\end{vmatrix}$
\end_inset
; (2).
\begin_inset Formula $\begin{vmatrix}x & y & z\\
3x+3 & 3y & 3z+2\\
x+1 & y+1 & z+1
\end{vmatrix}$
\end_inset
; (3).
\begin_inset Formula $\begin{vmatrix}x-1 & y-1 & z-1\\
4 & 1 & 3\\
1 & 1 & 1
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶方阵,
\begin_inset Formula $\alpha$
\end_inset
为
\begin_inset Formula $n\times1$
\end_inset
矩阵,
\begin_inset Formula $\beta$
\end_inset
为
\begin_inset Formula $1\times n$
\end_inset
矩阵, 且
\begin_inset Formula $\begin{vmatrix}A & \alpha\\
\beta & b
\end{vmatrix}=0$
\end_inset
.
求证:
\begin_inset Formula $\begin{vmatrix}A & \alpha\\
\beta & c
\end{vmatrix}=(c-b)\mathrm{det}A$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
证明如下行列式:
\end_layout
\begin_layout Exercise
(1).
\begin_inset Formula
\[
\begin{vmatrix}a & b & c & d\\
-b & a & d & -c\\
-c & -d & a & b\\
-d & c & -b & a
\end{vmatrix}=\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2};
\]
\end_inset
\end_layout
\begin_layout Exercise
(2).
\begin_inset Formula
\[
\begin{vmatrix}0 & 1 & 1 & a\\
1 & 0 & 1 & b\\
1 & 1 & 0 & c\\
a & b & c & d
\end{vmatrix}=a^{2}+b^{2}+c^{2}-2ab-2bc-2ac+2d.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
计算
\begin_inset Formula $n$
\end_inset
阶行列式
\begin_inset Formula $\begin{vmatrix}a & b & b & \cdots & b\\
b & a & b & \cdots & b\\
b & b & a & \cdots & b\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
b & b & b & \cdots & a
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
设
\begin_inset Formula $n$
\end_inset
阶行列式
\begin_inset Formula $\mathrm{det}A$
\end_inset
的元素
\begin_inset Formula $a_{ij}$
\end_inset
都是变数
\begin_inset Formula $t$
\end_inset
的可微函数.
试证明行列式的微分可如下计算:
\end_layout
\begin_layout Exercise
(1)
\begin_inset Formula $\frac{\mathrm{d}(\mathrm{det}A)}{\mathrm{d}t}=\mathrm{det}A_{1}+\cdots+\mathrm{det}A_{n}$
\end_inset
, 其中
\begin_inset Formula $A_{i}$
\end_inset
为对
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $i$
\end_inset
行微分 (其余行不变) 所得方阵
\begin_inset Formula $(i=1,\cdots,n)$
\end_inset
;
\end_layout
\begin_layout Exercise
(2)
\begin_inset Formula $\frac{\mathrm{d}(\mathrm{det}A)}{\mathrm{d}t}=\sum_{i,j=1}^{n}\frac{\mathrm{d}a_{ij}(t)}{\mathrm{d}t}A_{ij}(t)$
\end_inset
.
\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 )