张宇考研数学书籍LaTeX模板

定制一个新版张宇考研数学书籍的LaTeX格式

最近可能是吃饱了没事做，看着一本张宇张宇线性代数9讲，萌生了一种想法，我给他重排一份？算了，战线可能无限期拉长。不过，折腾了一番，总算有所收获，大体上定制出了一份LaTeX的格式，当然，有部分格式并不与原书一致。在这里，实现了同一章节的习题与答案之间的交叉引用，带超链接的那种，不过并没有用我博客/知乎上的方法，感觉直接这么搞起来更容易一点。然后呢，绘制了一个封面，还是比较人模狗样的。至于目录的话，大家可以选择直接用\tableofcontents输出，也可以用我放在正文中的tcolorbox定制的环境输出。为了实现一些方正书版的输出格式，我也是费了不少力气。把封面以及第二章的代码分享给大家，算是自己做的第一个模板吧

\documentclass[openany]{ctexbook}
\usepackage[centering,
           top=2.5cm,bottom=2.5cm,right=2.4cm,left=2.4cm,
           headsep=25pt,headheight=20pt]{geometry}


\usepackage{newtxtext,newtxmath}
  \DeclareSymbolFont{CMlargesymbols}{OMX}{cmex}{m}{n}
  \let\sumop\relax\let\prodop\relax
  \DeclareMathSymbol{\sumop}{\mathop}{CMlargesymbols}{"50}
  \DeclareMathSymbol{\prodop}{\mathop}{CMlargesymbols}{"51}
\usepackage{amsmath}
\allowdisplaybreaks

\usepackage{mathdots}  %   提供副对角线\iddots命令
\usepackage{extarrows} %  提供长等于号
\usepackage{array,arydshln}
\newcommand\xline{\mathord{
  \begin{tikzpicture}[baseline = (a.base)]
      \node [inner sep=0pt] (a) {\phantom{N}};
      \draw[dash pattern = on 1pt off 1pt](a.south)--(a.north);
  \end{tikzpicture}
  }
}

%--------------------------------------------



%----------------------------------------------------------定义选择题
\def\CA{A}
\def\CB{B}
\def\CC{C}
\def\CD{D}
%%------------------这么定义的目的是为了方便修改.

\newcommand{\kh}{(\rule[-2pt]{0.6cm}{0pt})}
\usepackage{ifthen}
\newlength{\la}
\newlength{\lb}
\newlength{\lc}
\newlength{\ld}
\newlength{\lhalf}
\newlength{\lquarter}
\newlength{\lmax}
\newcommand{\xx}[4]{\hfill\kh\\[.5pt]%
\settowidth{\la}{\CA.#1~~~}
\settowidth{\lb}{\CB.#2~~~}
\settowidth{\lc}{\CC.#3~~~}
\settowidth{\ld}{\CD.#4~~~}
\ifthenelse{\lengthtest{\la>\lb}}{\setlength{\lmax}{\la}}{\setlength{\lmax}{\lb}}
\ifthenelse{\lengthtest{\lc>\lmax}}{\setlength{\lmax}{\lc}}{}
\ifthenelse{\lengthtest{\ld>\lmax}}{\setlength{\lmax}{\ld}}{}
\setlength{\lhalf}{\dimexpr 0.5\linewidth-\parindent/2}
\setlength{\lquarter}{\dimexpr 0.25\linewidth-\parindent/4}
\ifthenelse{\lengthtest{\lmax>\lhalf}}{\indent \CA.#1 \\\indent \CB.#2 \\\indent \CC.#3 \\\indent \CD.#4}{
\ifthenelse{\lengthtest{\lmax>\lquarter}}{\hspace*{2em}\makebox[\lhalf][l]{\CA.#1~~~}%
\makebox[\lhalf][l]{\CB.#2~~~}\\

\indent\makebox[\lhalf][l]{\CC.#3~~~}%
\makebox[\lhalf][l]{\CD.#4~~~}}%
{\hspace*{2em}\makebox[\lquarter][l]{\CA.#1~~~}%
\makebox[\lquarter][l]{\CB.#2~~~}%
\makebox[\lquarter][l]{\CC.#3~~~}%
\makebox[\lquarter][l]{\CD.#4~~~}}}}
%--------------------------------------------------------------------------



%-----------------------------------------字体选项
\setCJKmainfont[BoldFont={方正黑体_GBK},ItalicFont={方正楷体_GBK},Mapping=fullwidth-stop]{方正书宋_GBK}

\defaultfontfeatures{Mapping=tex-text}
\XeTeXlinebreaklocale ”zh”
\XeTeXlinebreakskip = 0pt plus 1pt
%\setmainfont{Times New Roman}
%------------------------------------------------

\usepackage[perpage]{footmisc}
\usepackage[colorlinks=true,linkcolor=blue,linktoc=all,pdfauthor={向禹}]{hyperref}


%---------------------------------------定义常用符号
\let\bd\boldsymbol
\def\ft#1{\fcolorbox{cyan}{white}{\text{#1}}}
\def\VA{\bd A}
\def\VB{\bd B}
\def\VC{\bd C}
\def\VD{\bd D}
\def\VE{\bd E}
\def\VO{\bd O}
\def\VQ{\bd Q}
\def\VP{\bd P}
\def\ba{\bd \alpha}
\def\bb{\bd \beta}
\def\bg{\bd \gamma}
\def\vx{\bd x}
\def\vy{\bd y}
\def\TT{^{\mathrm T}}

\def\ba{\bd \alpha}
\def\jian#1{{\color{cyan}#1}}
\let\le\leqslant
\let\ge\geqslant
\usepackage[Symbol]{upgreek}
\renewcommand{\pi}{\uppi}%%%直立pi
\newcommand\kong[1][2]{\underline{\hspace{#1 cm}}}
\newcommand\bline{\textcolor{cyan}{\,\rule[0.5ex]{3mm}{0.6pt}}\,}


%--------------------------------------



\usepackage{pgfornament-han}
%------------------------------------------章节格式设置
\usepackage[many]{tcolorbox}
\usetikzlibrary{shapes.symbols,shapes.geometric}
\def \mybox#1{
    \begin{tikzpicture}[baseline=(a.base)]
          \node[inner ysep=0pt, inner xsep=2pt, cloud ,draw=cyan, aspect=4, cloud puffs=40,
           fill=cyan!20]  (a) {#1};
    \end{tikzpicture}
  }

\robustify{\mybox}
\newcommand\exercise{ \begin{center} \mybox{习题} \end{center}}
\newcommand\answer{ \begin{center} \mybox{解析} \end{center}}


\newcommand*\circled[2][circle]{\tikz[baseline=(char.base)]{
            \node[draw,inner sep=0.1pt,minimum height=1em,#1] (char) {#2};}}
\newcommand*\scircled[2][circle]{\tikz[baseline=(char.base)]{
            \node[draw,inner sep=0.1pt,minimum size=0.5em,#1] (char) {$\scriptstyle #2$};}}

\newcommand*\Circled[1]{\tikz[baseline=(char.base)]{
            \node[draw,inner ysep=2pt,inner xsep=6pt, fill=cyan!20,rounded corners] (char) {#1};}}

\def\mytitle#1{
\begin{tikzpicture}[baseline=(a.base),remember picture]
   \node(a){\color{blue}#1};
   \fill[rounded corners,cyan!20] (a.north west)--(a.north east)--(a.south east)[sharp corners]--(a.south west)[bend right=40]to (a.north west)--cycle;
   \draw[rounded corners,cyan] (a.north west)--(a.north east)--(a.south east)--(a.south west);
   \node(a){\color{blue}#1};
 \end{tikzpicture}
}
%------------------关于section编号的过度定制，只是为了和张宇的书一致，大家用的话不要用这个
\def\myshift#1{%
\ifcase\value{#1}%
0mm \or0.76mm \or0.05mm \or-0.05mm \or0.05mm \or0mm \or0mm \or-0.05mm
\fi}
\def\shift{\myshift{section}}

\def\mynumber#1{
  \begin{tikzpicture}[baseline=(name.base),remember picture]
    \node (name) {\color{blue} #1};
    \node at ([yshift=\shift]name.center) {\pgfornamenthan[scale=0.1,color=cyan!40]{51}};
  \end{tikzpicture}
}
\renewcommand\thechapter{\arabic{chapter}}
\renewcommand\thesubsection{\arabic{subsection}.}
\renewcommand\thesubsubsection{(\arabic{subsubsection})}
\setcounter{secnumdepth}{3}
\setcounter{tocdepth}{1}
\usepackage{tocloft}
\renewcommand\cftsecaftersnum{、}  %% 目录section编号后面加一个顿号
\renewcommand\contentsname{\centerline{ \Huge\CJKfontspec{华文行楷}目 录 }}


\ctexset{
  chapter={
    name   = {第,讲},
    number = \arabic{chapter}
  },
  section = {
    titleformat+  = \raggedright\mytitle,
    nameformat    = \kern-0.4cm\protect\mynumber,
    aftername     = \hskip-1.13em\relax,
    number        = \chinese{section}
  },
  subsection={
    format     = \Large\kern1.2em\color{cyan}\CJKfontspec{汉仪大宋简},
    nameformat = \bfseries,
    aftername  = \;
  },
  subsubsection={
    format     = \bfseries\large\kern1.8em,
    aftername  = \,,
    aftertitle = .
  },
  punct=kaiming
}

\usepackage{fancyhdr}
\usetikzlibrary{fadings}
\pagestyle{fancy}
\fancyhf{}

\fancyfoot[RO]{\tikz
    {\node[draw,rounded corners,inner xsep=1em,scope fading = west,fill=cyan] {\thepage};
    \node[rounded corners,inner xsep=1em] {\thepage};
    }}

\fancyfoot[LE]{\tikz
    {\node[draw,rounded corners,inner xsep=1em,scope fading = east,fill=cyan] {\thepage};
    \node[rounded corners,inner xsep=1em] {\thepage};
    }}
\fancyhead[RO]{\leftmark}
\fancyhead[LE]{\rightmark}
\fancyhead[RE,LO]{\href{yuxtech.github.io}{我的博客:yuxtech.github.io}}
\renewcommand\sectionmark[1]{%
\markright{\CTEXifname{\CTEXthesection、}{}#1}}

%-------------------------------------常用环境
\newcounter{prop}
\counterwithin{prop}{chapter}
\renewcommand\theprop{\arabic{prop}}
\newcommand{\propname}{性质}
\newenvironment{prop}{\par%
  \refstepcounter{prop}%
  \textbf{\propname\theprop}\label{prop\thechapter-\theprop}\quad
}{\par}

\newcounter{thm}
\counterwithin{thm}{chapter}
\renewcommand\thethm{\arabic{thm}}
\newcommand{\thmname}{定理}
\newenvironment{thm}{\par%
  \refstepcounter{thm}%
  \textbf{\thmname\thethm}\label{thm\thechapter-\thethm}\quad
}{\par}

\newcounter{example}
\counterwithin{example}{chapter}
\renewcommand\theexample{\thechapter.\arabic{example}}

\newcommand{\examplename}{例}
\newenvironment{example}{\par%
  \refstepcounter{example}%
  \Circled{\textbf{\examplename\theexample}\label{ex\thechapter-\arabic{example}}}
}{\par}

\newcounter{method}
\counterwithin{method}{example}
\renewcommand\themethod{方法\chinese{method}}
\def\method{\par\refstepcounter{method}\textbf{\themethod}
    \label{ex\thechapter-\arabic{example}-\arabic{method}}\quad}




\usepackage{manfnt}
\newenvironment{note}{\parskip=0.5ex\par\noindent\begin{tcolorbox}[sharp corners,boxrule=0.4pt,colframe=cyan,left=0.65cm,breakable,enhanced]
     【\textbf{注}】}
     {\end{tcolorbox}\par}

\newenvironment{solution}
               {\textcolor{cyan}{【\textbf{解}】}}
               {\par}
\newenvironment{proof}
               {\textcolor{cyan}{【\textbf{证}】}}
               {\par}
\newenvironment{analysis}
               {\textcolor{cyan}{【\textbf{分析}】}}
               {\par}

%--------------------------------------------------------------

%---------------------------------------------定义习题环境与答案环境，建立习题与答案交叉引用，但是这里仅限于题目和答案在同一章，如果需要最后统一输出答案，输出第X章答案前，需要\setcounter{chapter}{X}。更好的方式是用我的博客上的方法 https://yuxtech.github.io/2020/04/15/QandA/

\newcounter{xiti}
\counterwithin{xiti}{chapter}
\renewcommand\thexiti{\thechapter.\arabic{xiti}}

\newenvironment{xiti}{\par%
  \refstepcounter{xiti}%
  \Circled{\hyperlink{ans\thechapter-\arabic{xiti}}{\textbf{\thexiti}}}
  \hypertarget{xiti\thechapter-\arabic{xiti}}{}
}{\par}

\newcounter{ans}
\counterwithin{ans}{chapter}
\renewcommand\theans{\thechapter.\arabic{ans}}

\newenvironment{ans}{\par%
  \refstepcounter{ans}%
  \Circled{\hyperlink{xiti\thechapter-\arabic{ans}}{\textbf{\theans}}}
  \hypertarget{ans\thechapter-\arabic{ans}}{}
}{\par}
\newcounter{amethod}
\counterwithin{amethod}{ans}
\renewcommand\theamethod{方法\chinese{amethod}}
\def\amethod{\refstepcounter{amethod}\textbf{\theamethod}
    \label{ex\thechapter-\arabic{ans}-\arabic{amethod}}\quad}


\usepackage[inline]{enumitem}
\setlist{nosep}

%------------------------------修改公式编号
\usepackage{mathtools}

\newtagform{circle}{}{}
\usetagform{circle}
\renewcommand\theequation{\circled{\arabic{equation}}}
\DeclareMathOperator\tr{tr}

%------------------------------英文分号和冒号转化为中文分号
\catcode`\;=13
\newcommand{;}{\text{；}}
\catcode`\:=13
\newcommand{:}{\text{：}}

\begin{document}


\begin{titlepage}\definecolor{zy}{RGB}{78,81,125}
\begin{tikzpicture}[remember picture,overlay]
  \fill[color=zy](current page.south east)rectangle (current page.north west);
  \node[white,scale=0.3,anchor=north] (a) at ([shift={(2,-1.7)}]current page.north west){
\begin{tikzpicture}[line width=4pt,line cap=round]
  \draw[rounded corners=0.1cm] (0,0) rectangle (2.2,3);
  \draw[rounded corners=0.05cm] (0.4,1.6) -- (0.4, 0.5) -- (1.8,0.5) -- (1.8,1.6)
     -- (1,1.6) -- (1,1);
  \draw (0.4,2.1) -- (1.8,2.1) (0.4,2.5) -- (1.8,2.5);
\end{tikzpicture}
};
\node(b)[right=2mm,align=left,white,scale=1.5]at (a) {\scalebox{1}[1.3]{云图}\\[-3.4mm]\scalebox{0.5}{YUN\,TU}};
\node[right=6.5mm,scale=1.7,draw,rounded corners,fill=white]at(b){\phantom{2021版}};
\node[right=8mm,scale=1.5,draw,rounded corners,fill=white]at(b){2021版};

\node[text width=1em,anchor=north,white,scale=2,align=center,font=\CJKfontspec{汉仪大宋简}]
at([shift={(-1.5,-1.5)}]current page.north east)
{张\\[-2mm]宇\\[-2mm]数\\[-2mm]学\\[-2mm]教\\[-2mm]
育\\[-2mm]系\\[-2mm]列\\[-2mm]丛\\[-2mm]书\\[-2mm]\textbullet\\[-2mm] 二};

\node[anchor=north east] at ([shift={(-3,-1.5)}]current page.north east)
{\begin{tikzpicture}
  \node(c)[draw=zy,align=center,font=\CJKfontspec{汉仪大宋简},scale=5,fill=white,inner xsep=2mm,
  inner ysep=3mm]
    {张\\[-2mm]宇\\[-2mm]线\\[-2mm]性\\[-2mm]代\\[-2mm]数\\[-2mm]
    9\\[-2mm] 讲};
  \node(b)[xscale=0.9,yscale=0.98]{\tikz\draw[color=zy,line width=3pt](c.south west)rectangle (c.north east);};
  \node[xscale=0.85,yscale=0.97,anchor=center]at(b){\tikz\draw[color=zy,line width=1pt](c.south west)rectangle (c.north east);};
  \node[left=2mm,anchor=south east,text=white,align=center,scale=2] at(c.south west)
    {\tikz\draw[line width=1pt] circle(0.4\ccwd);\\主\\[-2mm]编\\张\\[-2mm]宇};
\end{tikzpicture}
};

\node[align=center,anchor=south west,text=white,font=\CJKfontspec{华文行楷},scale=3]
at ([shift={(2,1.7)}]current page.south west)
{高\\[-2mm]等\\[-2mm]教\\[-2mm]育\\[-2mm]出\\[-2mm]版\\[-2mm]社};

\end{tikzpicture}
\end{titlepage} 


%\tableofcontents   %  不想用下面的盒子风格输出目录的话就用原来的目录命令

\begin{tcolorbox}[enhanced,title=目\quad 录,sharp corners,breakable,
  colframe=blue!50!black,colback=blue!10!white,colbacktitle=blue!5!yellow!10!white,
  fonttitle=\bfseries,coltitle=black,attach boxed title to top center=
    {yshift=-0.25mm-\tcboxedtitleheight/2,yshifttext=2mm-\tcboxedtitleheight/2},
  boxed title style={boxrule=0.5mm,
    frame code={ \path[tcb fill frame] ([xshift=-4mm]frame.west)
    -- (frame.north west) -- (frame.north east) -- ([xshift=4mm]frame.east)
    -- (frame.south east) -- (frame.south west) -- cycle; },
    interior code={ \path[tcb fill interior] ([xshift=-2mm] interior.west)
    -- (interior.north west) -- (interior.north east)
    --([xshift=2mm] interior.east) -- (interior.south east) -- (interior.south west)
    --cycle;} }]
  \makeatletter
  \@starttoc{toc}
  \makeatother

\end{tcolorbox}

\setcounter{chapter}{1}
\chapter{余子式和代数余子式的计算\label{chap2}}
\mybox{知识结构}
\vskip\baselineskip
  \[
     \begin{cases}
       \ft{行列式}\bline k_1A_{i1}+k_2A_{i2}+\cdots+k_nA_{in}=\smash[t]{\begin{vmatrix}
         \ast\\
         k_1\quad k_2\quad \cdots \quad k_n\\
         \ast
       \end{vmatrix}} \\
       \ft{用矩阵}\bline |\VA|\ne0时, \VA^\ast=|\VA|\VA^{-1}\\
       \ft{用特征值}\bline \begin{aligned}[t]
           &\text{设$\VA$为3阶矩阵，当$\VA$可逆时，记其特征值为$\lambda_1,\lambda_2,\lambda_3$，}\\
           &\quad \text{则}A_{11}+A_{22}+A_{33}=\lambda_2\lambda_3+\lambda_1\lambda_3+\lambda_1\lambda_2
       \end{aligned}\\
       \ft{求余子式的问题}\bline M_{ij}=(-1)^{i+j}A_{ij}
     \end{cases}
  \]
\section{用行列式}
  \begin{flalign}
    &\qquad\text{由}& a_{i1}A_{i1} + a_{i2}A_{i2}+\cdots+a_{in}A_{in} =
       \begin{vmatrix}
         \ast\\
         a_{i1}\quad a_{i2} \quad \cdots \quad a_{in}\\
         \ast
       \end{vmatrix},&&\label{eq2.1}\\
    \text{则}&& k_1A_{i1}+k_2A_{i2}+\cdots+k_nA_{in}=\begin{vmatrix}
         \ast\\
         k_1\quad k_2\quad \cdots \quad k_n\\
         \ast
       \end{vmatrix},&&\label{eq2.2}
  \end{flalign}
 其中$\ast$处表示元素不变， \ref{eq2.1}， \ref{eq2.2} 的区别仅仅在于第$i$行的元素$a_{i1},a_{i2},\cdots,a_{in}$换成了$k_1,k_2,\cdots,k_n$，这样，给出不同的系数$k_1,k_2,\cdots,k_n$，即得到了不同的行列式.而若要求$k_1M_{i1}+k_2M_{i2}+\cdots+k_nM_{in}$，只需用$M_{ij}=(-1)^{i+j}A_{ij}$化为关于$A_{ij}$的线性组合即可.

  \jian{见例 \ref{ex2-1}，例 \ref{ex2-2}.}

\section{用矩阵}
    \textcolor{cyan}{(与第 \ref{chap3} 讲综合，考生需学习相关知识后再研读此点)}

    当$|\VA|\ne0$时， $\VA^\ast=|\VA|\VA^{-1}$.

    由于$\VA^\ast$由$A_{ij}$组成，求出$\VA^\ast$，即得到所有的$A_{ij}$，但要注意，此方法要求$|\VA|\ne0$，这是前提，也是一种限制.

    \jian{见例 \ref{ex2-3}，例 \ref{ex2-4}.}

\section{用特征值}
    \textcolor{cyan}{(与第 \ref{chap7} 讲综合，考生需学习相关知识后再研读此点)}

    设$\VA$为3阶矩阵，当$\VA$为可逆矩阵时，记其特征值为$\lambda_1,\lambda_2,\lambda_3$.则$\VA^{-1}$的特征值为$\lambda_1^{-1},\lambda_2^{-1},\lambda_3^{-1}$，且由$\VA^\ast=|\VA|\VA^{-1}=\lambda_1\lambda_2\lambda_3\VA^{-1}$，可知$\VA^\ast$的特征值为
      \[
        \lambda_1^\ast=\lambda_1\lambda_2\lambda_3 \cdot\lambda_1^{-1}=\lambda_2\lambda_3,
        \lambda_2^\ast=\lambda_1\lambda_2\lambda_3 \cdot\lambda_2^{-1}=\lambda_1\lambda_3,
        \lambda_3^\ast=\lambda_1\lambda_2\lambda_3 \cdot\lambda_3^{-1}=\lambda_1\lambda_2,
      \]
    \begin{flalign*}
      \text{故由} & & \VA^\ast = \begin{bmatrix}
        A_{11} & A_{21} & A_{31}\\
        A_{12} & A_{22} & A_{32}\\
        A_{13} & A_{23} & A_{33}
      \end{bmatrix}&&
    \end{flalign*}
   知$A_{11}+A_{22}+A_{33}=\tr(\VA^\ast)=\lambda_1^\ast+\lambda_2^\ast+\lambda_3^\ast=\lambda_2\lambda_3
   +\lambda_1\lambda_3+\lambda_1\lambda_2$.

   这公式易记，好用，考生应熟知.

   \jian{见例 \ref{ex2-5}，例 \ref{ex2-6}.}

\section{求余子式的问题}
    由于$M_{ij}=(-1)^{i+j}A_{ij}$，故先求出$A_{ij}$，乘以$(-1)^{i+j}$即可.
\begin{example}
  设
     \[
       D_n = \begin{vmatrix}
         1 & 2 & 3 & \cdots & n\\
         1 & 2 & 0 & \cdots & 0\\
         1 & 0 & 3 & \cdots & 0\\
         \vdots & \vdots & \vdots & & \vdots\\
         1 & 0 & 0 & \cdots & n
       \end{vmatrix},
     \]
  求$A_{11}+A_{12}+\cdots +A_{1n}$.
\end{example}
\begin{solution}
    \begin{align*}
      A_{11}+A_{12}+\cdots+A_{1n} & =\begin{vmatrix}
        1 & 1 & 1 & \cdots & 1\\
        1 & 2 & 0 & \cdots & 0\\
        1 & 0 & 3 & \cdots & 0\\
        \vdots & \vdots & \vdots & & \vdots \\
        1 & 0 & 0 & \cdots &n
      \end{vmatrix}\\
      & = \begin{vmatrix}
        1-\sum_{i=2}^n\frac1i & 1 & 1 & \cdots & 1\\
        0 & 2 & 0 & \cdots & 0\\
        0 & 0 & 3 & \cdots & 0\\
        \vdots & \vdots & \vdots & & \vdots \\
        0 & 0 & 0 & \cdots &n
      \end{vmatrix}\\
      & = \Big( 1-\sum_{i=2}^n\frac1i \Big)\cdot n!.
    \end{align*}
\end{solution}

\begin{example}
  已知5阶行列式 $D_5=\begin{vmatrix}
    1 & 2 & 3 & 4 & 5\\
    2 & 2 & 2 & 1 & 1\\
    3 & 1 & 2 & 4 & 5\\
    1 & 1 & 1 & 2 & 2\\
    4 & 3 & 1 & 5 & 0
  \end{vmatrix} = 27$. 求$A_{41}+A_{42}+A_{43}$及$A_{44}+A_{45}$.
\end{example}
\begin{solution}
    \hfil $(A_{41}+A_{42}+A_{43}) + 2(A_{44}+A_{45}) = D_5 =27$，  \hfill
    \begin{flalign*}
        \text{又}&& 2(A_{41}+A_{42}+A_{43}) + (A_{44}+A_{45}) =
                   \begin{vmatrix}
                      1 & 2 & 3 & 4 & 5\\
                      2 & 2 & 2 & 1 & 1\\
                      3 & 1 & 2 & 4 & 5\\
                      2 & 2 & 2 & 1 & 1\\
                      4 & 3 & 1 & 5 & 0
                   \end{vmatrix} = 0,  &&
     \end{flalign*}
    联立上式解得
      \[ A_{41}+A_{42}+A_{43}=-9, A_{44}+A_{45}=18. \]
\end{solution}

\begin{example}
    设 $n$ 阶行列式
       \[
          |\VA| = \begin{vmatrix}
            2 & 0 & 0 & \cdots & 0\\
            2 & 2 & 0 & \cdots & 0\\
            2 & 2 & 2 & \cdots & 0\\
            2 & 2 & 2 & \cdots & 0\\
            2 & 2 & 2 & \cdots & 2
          \end{vmatrix}_n.
       \]
    求\begin{enumerate*}[label=(\arabic*),itemjoin=\\]
       \item $|\VA|$中所有元素$a_{ij}$的代数余子式$A_{ij}$之和$\sum_{i=1}^n\sum_{j=1}^nA_{ij}$；
       \item $A_{11}+A_{22}+\cdots+A_{nn}$;
       \item $A_{k1}+A_{k2}+\cdots+A_{kn}(k=1,2,\cdots,n)$.
    \end{enumerate*}

\end{example}
\begin{solution}
     \[
       \VA = \begin{bmatrix}
            2 & 0 & 0 & \cdots & 0\\
            2 & 2 & 0 & \cdots & 0\\
            2 & 2 & 2 & \cdots & 0\\
            2 & 2 & 2 & \cdots & 0\\
            2 & 2 & 2 & \cdots & 2
          \end{bmatrix},
     \]
  $|\VA|=2^n\ne0$，故$\VA$可逆，且
      \[ \VA^\ast=|\VA|\VA^{-1}=2^n\VA^{-1}. \]

  由 $[\VA \xline\VE]\xlongrightarrow{\text{初等行变换}}[\VE \xline\VA^{-1}]$，有
    \begin{align*}
      [\VA \xline \VE] & = \left[
         \begin{array}{@{}ccccc:ccccc@{}}
            2 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\
            2 & 2 & 0 & \cdots & 0 & 0 & 1 & 0 & \cdots & 0 \\
            2 & 2 & 2 & \cdots & 0 & 0 & 0 & 1 & \cdots & 0 \\
            \vdots & \vdots & \vdots & & \cdots  & \vdots & \vdots & \vdots & & \cdots \\
            2 & 2 & 2 & \cdots & 2 & 0 & 0 & 0 & \cdots & 1
         \end{array}\right]\\
         &\to \left[
         \begin{array}{@{}ccccc:ccccc@{}}
            1 & 0 & 0 & \cdots & 0 & \frac12 & 0& 0 & \cdots & 0         \\
            0 & 1 & 0 & \cdots & 0 & -\frac12 & \frac12 & 0 & \cdots & 0 \\
            0 & 0 & 1 & \cdots & 0 & 0 & -\frac12 & \frac12 & \cdots & 0 \\
            \vdots & \vdots & \vdots & & \cdots  & \vdots & \vdots & \vdots & & \cdots \\
            0 & 0 & 0 & \cdots & 1 & 0 & 0 & 0 & \cdots & \frac12
         \end{array}
      \right]
    \end{align*}
  即
    \begin{align*}
      \VA^\ast & = \begin{bmatrix}
        A_{11} & A_{21} & \cdots & A_{n1}\\
        A_{12} & A_{22} & \cdots & A_{n2}\\
        \vdots & \vdots & & \vdots\\
        A_{1n} & A_{2n} & \cdots & A_{nn}
      \end{bmatrix}\\
      & = 2^n\VA^{-1} = 2^n
       \begin{bmatrix}
         \frac12 & 0& 0 & \cdots & 0 & 0 \\
         -\frac12 & \frac12 & 0 & \cdots & 0 & 0 \\
         0 & -\frac12 & \frac12 & \cdots & 0 & 0\\
         \vdots & \vdots & \vdots & & \cdots & \vdots \\
         0 & 0 & 0 & \cdots & \frac12 & 0 \\
         0 & 0 & 0 & \cdots & -\frac12 & \frac12
       \end{bmatrix}.
    \end{align*}
  故有
  \begin{enumerate*}[label=(\arabic*),itemjoin=\\]
      \item $\sum_{i=1}^n\sum_{j=1}^nA_{ij}=2^n\big[n\cdot\frac12 + (n-1)\cdot
         \big( -\frac12 \big)\big]=2^{n-1}$.
      \item $\sum_{i=1}^nA_{ii}=2^n\cdot n\cdot\frac12 = 2^{n-1}\cdot n$.
      \item 当$k=1,2,\cdots,n-1$时，
  \end{enumerate*}
        \[ A_{k1}+A_{k2}+\cdots+A_{kn} = 2^n\Big( \frac12-\frac12 \Big) = 0.  \]
   当$k=n$时， $A_{n1}+A_{n2}+\cdots+A_{nn}=2^n\cdot\frac12=2^{n-1}$.
\end{solution}

\begin{example}
  已知3阶行列式$|\VA|$的元素$a_{ij}$均为实数，且$a_{ij}$不全为0.若
      \[ a_{ij} = -A_{ij}(i,j=1,2,3), \]
  其中$A_{ij}$是$a_{ij}$的代数余子式. 则$|\VA|=$\kong.
\end{example}

\begin{solution}
  应填$-1$.

  由$a_{ij}=-A_{ij}$有$\VA^\ast=-\VA\TT$. 于是
      \[  \VA\VA^\ast = -\VA\VA\TT. \]

  又$\VA\VA^\ast=|\VA|\VE$，故
      \[   \VA\VA\TT = -|\VA|\VE. \]
  两边取行列式得
      \[ |\VA||\VA\TT| = -|\VA|^3,\,\text{即}\,|\VA|^2=-|\VA|^3, \]
  也即
   \begin{equation}\label{ex2.1}
     |\VA|^2 (|\VA|+1) = 0.  \tag{($\ast$)}
   \end{equation}

   由$a_{ij}$不全为0知，存在$a_{kj}\ne0$，将行列式$|\VA|$ 按第$k$行展开得
     \begin{align*}
       |\VA| & = a_{k1}A_{k1} + a_{k2}A_{k2} + a_{k3}A_{k3}\\
             & = -a_{k1}^2 -a_{k2}^2-a_{k3}^2 < 0.
     \end{align*}

   故由 \eqref{ex2.1}， $|\VA|=-1$.
\end{solution}

\begin{example}
  设$\VA=\begin{bmatrix}
    a_{11} & 1 & a_{13}\\
    a_{21} & 1 & a_{23}\\
    a_{31} & 1 & a_{33}
  \end{bmatrix}$，已知$\VA$的特征值是$2,1$，求$A_{11}A_{23}-A_{21}A_{13}$.
\end{example}
\begin{solution}
  由题设
    \begin{gather*}
      |\VA| = 2\times 1\times (-1) = -2,\\
        |\VA^\ast| = |\VA|^2 = 4.
    \end{gather*}
  由$\VA = \begin{bmatrix}
    a_{11} & 1 & a_{13}\\
    a_{21} & 1 & a_{23}\\
    a_{31} & 1 & a_{33}
  \end{bmatrix}$，有
  \begin{align*}
    & A_{11}+A_{21}+A_{31} = \begin{vmatrix}
      1 & 1 & a_{13} \\ 1 & 1 & a_{23}\\ 1 & 1 & a_{33}
    \end{vmatrix} = 0,\\
    & A_{12}+A_{22}+A_{32} = \begin{vmatrix}
      a_{11} & 1 & a_{13}\\ a_{21} & 1 & a_{23}\\ a_{31} & 1 & a_{33}
    \end{vmatrix} = -2,\\
    &A_{13}+A_{23}+A_{33} = \begin{vmatrix}
      a_{11} & 1 & 1\\ a_{21} & 1 & 1 \\ a_{31} & 1 & 1
    \end{vmatrix} = 0.
  \end{align*}
  故
    \begin{align*}
        |\VA^\ast| & = \begin{vmatrix}
          A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}
        \end{vmatrix}
                     = \begin{vmatrix}
                       A_{11} & A_{21} & A_{11}+A_{21}+A_{31}\\
                       A_{12} & A_{22} & A_{12}+A_{22}+A_{32}\\
                       A_{13} & A_{23} & A_{13}+A_{23}+A_{33}
                     \end{vmatrix}\\
                   & = \begin{vmatrix}
                     A_{11} & A_{21} & 0 \\ A_{12} & A_{22} & -2 \\ A_{13} & A_{23} & 0
                   \end{vmatrix}
                     = - (-2) \begin{vmatrix}
                       A_{11} & A_{21} \\ A_{13} & A_{23}
                     \end{vmatrix}.
    \end{align*}
    于是
        \[ A_{11}A_{23} - A_{21}A_{13} = 2. \]
\end{solution}

\begin{example}
  已知3阶方阵$\VA$的特征值为$-1,2,3$. 求$\VA$的行列式$|\VA|$中元素$a_{11},a_{22},a_{33}$的代数余子式的和$A_{11}+A_{22}+A_{33}$.
\end{example}
\begin{solution}
  \hfil $|\VA|=-1\times2\times3=-6\ne0$, \hfill

  \noindent 故$\VA$可逆，且$\VA^\ast=|\VA|\VA^{-1}=-6\VA^{-1}.\VA^{-1}$的特征值为$-1,\frac12,\frac13,\VA^\ast=-\VA^{-1}$的3个特征值为
    \begin{gather*}
      \lambda_1^\ast = (-6)\times(-1)=6,\lambda_2^\ast=(-6)\times\frac12=-3,\\
      \lambda_3^\ast = (-6)\times\frac13=-2.
    \end{gather*}
  \begin{flalign*}
    \text{故} && A_{11}+A_{22}+A_{33} & =\tr(\VA^\ast) =\lambda_1^\ast +\lambda_2^\ast +\lambda_3^\ast\\
              &&                      & = 6 - 3 - 2 =1. &&
  \end{flalign*}
\end{solution}

\exercise
\begin{xiti}
  设
     \[
        |\VA| = \begin{vmatrix}
          2 & 1 & -3 & 5\\ 1 & 1 & 1 & 2 \\ 4 & 2 & 3 & 1 \\ 2 & 5 & 1 & 3
        \end{vmatrix},
     \]
  计算$A_{41}+A_{42}+A_{43}+A_{44}$，其中$A_{ij}$是元素$a_{ij}(j=1,2,3,4)$ 的代数余子式.
\end{xiti}

\begin{xiti}
  设
    \[
      |\VA| = \begin{vmatrix}
        2 & 2 & 2 & 2 \\
        a_{11} & a_{22} & a_{23} & a_{24}\\
        a_{31} & a_{32} & a_{33} & a_{34}\\
        a_{41} & a_{42} & a_{43} & a_{44}
      \end{vmatrix} = a,
    \]
  计算$\sum_{i=1}^4\sum_{j=1}^4A_{ij}$.
\end{xiti}

\begin{xiti}
  设$n$阶行列式
    \[
      D_n = \begin{vmatrix}
        x & a & \cdots & a\\
        a & x & \cdots & a\\
        \vdots & \vdots &  & \vdots\\
        a & a & \cdots & x
      \end{vmatrix},
    \]
  $A_{ij}$是$D_n$中元素$a_{ij}(i,j=1,2,\cdots,n)$的代数余子式，求$D_n$的全部代数余子式之和.
\end{xiti}

\begin{xiti}
  证明：若行列式的某行元素全为$k(k\ne0)$，则这个行列式的全部代数余子式之和为该行列式值的$\frac1k$倍，即
  $\sum_{i=1}^n\sum_{j=1}^nA_{ij}=\frac1k|\VA|$.
\end{xiti}

\begin{xiti}
  设$\VA$是$n(n>2)$阶非零实矩阵， $A_{ij}$是$|\VA|$中元素$a_{ij}$的代数余子式，且
     \[  a_{ij}=A_{ij}(i,j=1,2,\cdots,n). \]
  证明$|\VA|=1$.
\end{xiti}

\answer

\begin{ans}
  \begin{analysis}
    $A_{4j}$是第4行元素$a_{4j}(j=1,2,3,4)$的代数余子式，其值只与第$1,2,3$行元素有关，与第4行元素无关.
  \end{analysis}
  \begin{solution}
    将$|\VA|$中第4行元素依次换为$1,1,1,1$，并不改变$A_{4j}(j=1,2,3,4)$的大小，有
        \begin{align*}
          A_{41}+A_{42}+A_{43}+A_{44} & =1\cdot A_{41}+1\cdot A_{42}+1\cdot A_{43}+1\cdot A_{44}\\
          & = \begin{vmatrix}
            2 & 1 & -3 & 5\\ 1 & 1 & 1 & 2\\ 4 & 2 & 3 & 1\\ 1 & 1 & 1 & 1
          \end{vmatrix}
          \xlongequal{\scircled{2}-\scircled{4}}
          \begin{vmatrix}
            2 & 1 & -3 & 5\\ 0 & 0 & 0 & 1\\ 4 & 2 & 3 & 1\\ 1 & 1 & 1 & 1
          \end{vmatrix}\\
          & = (-1)^{2+4}\begin{vmatrix}
            2 & 1 & -3 \\4 & 2 & 3\\ 1 & 1 & 1
          \end{vmatrix}
          \xlongequal[{[3]-[1]}]{[2]-[1]}
          \begin{vmatrix}
            2 & -1 & -5\\4 & -2 & -1\\1 & 0 & 0
          \end{vmatrix}=1-10=-9.
        \end{align*}
  \end{solution}
\end{ans}

\begin{ans}
  \begin{solution}
      \begin{align*}
        \sum_{i=1}^4\sum_{j=1}^4A_{ij} = {}& A_{11}+A_{12}+A_{13}+A_{14}+A_{22}+A_{23}+A_{24}\\
                     & A_{31}+A_{32}+A_{33}+A_{34}+A_{41}+A_{42}+A_{43}+A_{44}+\\
                     ={}&\frac12(2A_{11}+2A_{12}+2A_{13}+2A_{14})
                     +\frac12(A_{21}+2A_{22}+2A_{23}+2A_{24})+\\
                      & \frac12(2A_{31}+2A_{32}+2A_{33}+2A_{34})
                      +\frac12(2A_{41}+2A_{42}+2A_{43}+2A_{44})\\
                     ={}&\frac a2+0+0+0=\frac a2.
      \end{align*}
  \begin{note}
    本题若命制成“设$|\VA|$的某行元素全为2，且$|\VA|=a$，计算$\sum_{i=1}^4\sum_{j=1}^4A_{ij}$”估计会增加一些难度，但解法和过程与本题完全一样，只不过要求读者多写一句“不妨设$|\VA|$的第一行元素全为2”即可.应对考试，读者一定要多从命题人的角度出发，想想看如何“变着花样”出题(例如习题 \hyperlink{xiti2-4}{2.4}，知己知彼，才是科学的复习策略.
  \end{note}
  \end{solution}
\end{ans}

\begin{ans}
  $\sum_{j=1}^nA_{1j} = \begin{vmatrix}
        1 & 1 & \cdots & 1\\
        a & x & \cdots & a\\
        \vdots & \vdots &  & \vdots\\
        a & a & \cdots & x
      \end{vmatrix}=
      \begin{vmatrix}
        1 & 1 & \cdots & 1\\
        0 & x-a & \cdots & 0\\
        \vdots & \vdots &  & \vdots\\
        0 & 0 & \cdots & x-a
      \end{vmatrix}=(x-a)^{n-1},
      $
      \[
         \sum_{j=1}^nA_{2j} = \begin{vmatrix}
        x & a & a & \cdots & a\\
        1 & 1 & 1 & \cdots & 1\\
        a & a & x & \cdots & a\\
        \vdots & \vdots & \vdots & & \vdots\\
        a & a & a & \cdots & x
      \end{vmatrix} =
      \begin{vmatrix}
        x-a & 0 & 0 & \cdots & 0\\
        1 & 1 & 1 & \cdots & 1\\
        0 & 0 & x-a & \cdots & 0\\
        \vdots & \vdots & \vdots & & \vdots\\
        0 & 0 & 0 & \cdots & x-a
      \end{vmatrix} = (x-a)^{n-1}.
      \]
  同理， $\sum_{j=1}^nA_{ij}=(x-a)^{n-1},i=3,4,\cdots,n$. 故$\sum_{i=1}^n\sum_{j=1}^nA_{ij}=n(x-a)^{n-1}$.
\end{ans}

\begin{ans}
  不失一般性，设
    \[
       |\VA|=\begin{vmatrix}
         k & k & \cdots &k\\
         a_{11} & a_{22} & \cdots &a_{2n}\\
         \vdots & \vdots & &\vdots\\
         a_{n1} & a_{n2} & \cdots &a_{nn}
       \end{vmatrix} = k
       \begin{vmatrix}
         1 & 1 & \cdots & 1\\
         a_{21} & a_{22} & \cdots & a_{2n}\\
         \vdots & \vdots & &\vdots\\
         a_{n1} & a_{n2} & \cdots &a_{nn}
       \end{vmatrix} = k\sum_{j=1}^nA_{1j},
    \]
  且$\sum_{j=1}^nA_{ij}=0,i=2,3,\cdots,n$. 故$\sum_{i=1}^n\sum_{j=1}^nA_{ij}=\frac1k|\VA|$.
\end{ans}

\begin{ans}
  \begin{proof}
    由于$\VA\ne\VO$，不妨设$a_{11}\ne0$，则
     \begin{gather*}
       |\VA| = \sum_{j=1}^na_{1j}A_{1j} = \sum_{j=1}^na_{1j}^2 >0 (A_{ij}=a_{ij},i,j=1,2,\cdots,n).\\
       \begin{aligned}
         \begin{bmatrix}
           |\VA| & 0 & \cdots & 0\\
           0 & |\VA| & \cdots & 0\\
           \vdots & \vdots & & \vdots\\
           0 & 0 & \cdots & |\VA|
         \end{bmatrix} & =
         \begin{bmatrix}
         a_{11} & a_{12} & \cdots &a_{1n}\\
         a_{11} & a_{22} & \cdots &a_{2n}\\
         \vdots & \vdots & &\vdots\\
         a_{n1} & a_{n2} & \cdots &a_{nn}
       \end{bmatrix}
       \begin{bmatrix}
         A_{11} & A_{21} & \cdots &A_{n1}\\
         A_{12} & A_{22} & \cdots &A_{n2}\\
         \vdots & \vdots & &\vdots\\
         A_{1n} & A_{2n} & \cdots &A_{nn}
       \end{bmatrix}\\
       & = \begin{bmatrix}
         a_{11} & a_{12} & \cdots &a_{1n}\\
         a_{11} & a_{22} & \cdots &a_{2n}\\
         \vdots & \vdots & &\vdots\\
         a_{n1} & a_{n2} & \cdots &a_{nn}
       \end{bmatrix}
       \begin{bmatrix}
         a_{11} & a_{21} & \cdots &a_{n1}\\
         a_{12} & a_{22} & \cdots &a_{n2}\\
         \vdots & \vdots & &\vdots\\
         a_{1n} & a_{2n} & \cdots &a_{nn}
       \end{bmatrix} = \VA\VA\TT,
       \end{aligned}
     \end{gather*}
  两边取行列式，得
    \[
       \begin{vmatrix}
           |\VA| & 0 & \cdots & 0\\
           0 & |\VA| & \cdots & 0\\
           \vdots & \vdots & & \vdots\\
           0 & 0 & \cdots & |\VA|
       \end{vmatrix} = |\VA|^n = |\VA\VA\TT|=|\VA|^2,
    \]
  故$|\VA|^2(|\VA|^{n-2}-1)=0$，又$|\VA|>0$，则$|\VA|^{n-2}=1$，得$|\VA|=1$ ($\VA$是实矩阵， $|\VA|$ 是实数，且$|\VA|>0$，舍去$-1$).
  \end{proof}
\end{ans}



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