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\begin{document} \title[第二章]{行列式} \author[高等代数]{高等代数} % 显示作者 \institute[数学与计算机科学学院]{% \large 数学与计算机科学学院} \date{2023.04.19} % 显示日期
\begin{frame}
\thispagestyle{empty}%去掉封面的页眉页脚
\titlepage
\begin{figure}
\centering
\includegraphics[width=0.2\linewidth]{pics/jlu.jpg}
\end{figure}
\end{frame}
\begin{frame}{目录}
\tableofcontents % 显示目录,并超链接到对应的section
\end{frame}
\section{行列式的定义及性质}
\begin{frame}{行列式的定义}
\begin{block}{定义1}
$n$ 级行列式
\begin{equation}
\left|\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
a_{21} & a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right|
\end{equation}
等于所有取自不同行不同列的 $n$ 个元素的乘积
\begin{equation}\label{eq:pl}
a_{1 j_1} a_{2 j_2} \cdots a_{n j_n}
\end{equation}
的代数和, 这里 $j_1 j_2 \cdots j_n$ 是 $1,2, \cdots, n$ 的一个排列, 每一项\eqref {eq:pl} 都按 下列规则带有符号: 当 $j_1 j_2 \cdots j_n$ 是偶排列时, \eqref{eq:pl} 带有正 号, 当 $j_1 j_2 \cdots j_n$ 是奇排列时, \eqref{eq:pl} 带有负号. 这一定义可写成
\end{block}
\end{frame}
\begin{frame}{行列式的定义}
\begin{block}{定义1续}
\begin{equation}
\left|\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
a_{21} & a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right|=\sum_{j_1j_2\dots j_n}(-1)^{\tau\left(j_1j_2\dots j_n\right)} a_{1 j_1} a_{2 j_2} \cdots a_{n j_n},
\end{equation}
这里 $\displaystyle\sum_{j_1j_2\dots j_n}{ }$ 表示对所有 $n$ 级排列求和.
\end{block}
\end{frame}
\begin{frame}{行列式的性质}
\begin{block}{性质1:转置行列式}
行列互换,行列式不变,即
$$\left|\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right|=\left|\begin{array}{cccc}a_{11} & a_{21} & \cdots & a_{n 1} \\ a_{12} & a_{22} & \cdots & a_{n 2} \\ \vdots & \vdots & & \vdots \\ a_{1 n} & a_{2 n} & \cdots & a_{n n}\end{array}\right|$$
\end{block}
\end{frame}
\begin{frame}{行列式的性质}
\begin{block}{性质2:提公因子}
一行的公因子可以提出去,即
$$\left|\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ \vdots & \vdots & & \vdots \\ k a_{i 1} & k a_{i 2} & \cdots & k a_{i n} \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right|=k\left|\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ \vdots & \vdots & & \vdots \\ a_{i 1} & a_{i 2} & \cdots & a_{i n} \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right|$$
\end{block}
\end{frame}
\begin{frame}{行列式的性质}
\begin{block}{性质3:行(列)拆分,即}
$$\begin{aligned} &\left|\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ \vdots & \vdots & & \vdots \\ b_1+c_1 & b_2+c_2 & \cdots & b_n+c_n \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right| \\ &=\left|\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ \vdots & \vdots & & \vdots \\ b_1 & b_2 & \cdots & b_n \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right|+\left|\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ \vdots & \vdots & & \vdots \\ c_1 & c_2 & \cdots & c_n \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right|\end{aligned}$$
\end{block}
\end{frame}
\begin{frame}{行列式的性质}
\begin{block}{性质4:}
如果行列式中两行相同,那么行列式为零。
\end{block}
\begin{block}{性质5:}
如果行列式中两行成比例,那么行列式为零。
\end{block}
\begin{block}{性质6:}
把一行的非零倍数加到另一行,行列式不变。
\end{block}
\begin{block}{性质7:}
对换行列式中两行的位置,行列式反号。
\end{block}
\end{frame}
\section{相关例题}
\begin{frame}{方法论}
\begin{enumerate}
\item 行或列和相等
\item 大拆分
\item 小拆分
\item 大对角
\item 循环行列式
\end{enumerate}
\end{frame}
\begin{frame}{方法论}
\begin{block}{例1}
计算$n$阶行列式
$$d=\left|\begin{array}{ccccc}
a & b & b & \cdots & b \\
b & a & b & \cdots & b \\
b & b & a & \cdots & b \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & \cdots & a
\end{array}\right|$$
\end{block}
\end{frame}
\begin{frame}
\thispagestyle{empty}
\begin{align*}
d &=\left|\begin{array}{ccccc}
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{array}\right|
=\left|\begin{array}{ccccc}
(n-1) b+a & b & b & \cdots & b \\
(n-1) b+a & a & b & \cdots & b \\
(n-1) b+a & b & a & \cdots & b \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
(n-1) b+a & b & b & \cdots & a
\end{array}\right|\\
&=[a+(n-1)b]\left|\begin{array}{ccccc}
1 & b & b & \cdots & b \\
1 & a & b & \cdots & b \\
1 & b & a & \cdots & b \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & b & b & \cdots & a
\end{array}\right|
\end{align*}
\end{frame}
\begin{frame}
\thispagestyle{empty}
\begin{align*}
\quad &=[a+(n-1) b]\left|\begin{array}{ccccc}
1 & b & b & \cdots & b \\
0 & a-b & 0 & \cdots & 0 \\
0 & 0 & a-b & \cdots & 0 \\
\vdots & \vdots & \vdots & & \vdots \\
0 & 0 & 0 & \cdots & a-b
\end{array}\right|\\
& = [a+(n-1) b](a-b)^{n-1}
\end{align*}
\end{frame}
\begin{frame}{大拆分}
$d = \left|\begin{array}{ccccc}
b+(a-b) & b & b & \cdots & b \\
b & b+(a-b) & b & \cdots & b \\
b & b & b+(a-b) & \cdots & b \\
\vdots & \vdots & \vdots & & \vdots \\
b & b & b & \cdots & b+(a-b)
\end{array}\right|$
分为三种行列式
\begin{enumerate}
\item 有两列$b$
\item 有一列$b$
\item 全是$a$
\end{enumerate}
\end{frame}
\begin{frame}{小拆分}
\begin{align*}
D_n &=
\begin{vmatrix}
b+(a-b) & b & b & \cdots & b \\
b & a & b & \cdots & b \\
b & b & a & \cdots & b \\
\vdots & \vdots & \vdots & & \vdots \\
b & b & b & \cdots & a
\end{vmatrix}\\
&= (a-b)D_{n-1} +
\begin{vmatrix}
b & b & b & \cdots & b\\
b & a & b & \cdots & b\\
b & b & a & \cdots & b \\
\vdots & \vdots & \vdots & & \vdots \\
b & b & b & \cdots & a\\
\end{vmatrix}\\
&= (a-b)D_{n-1}+b(a-b)^{n-1}
\end{align*}
\end{frame}
\begin{frame}
\begin{block}{例2}
计算$n$阶行列式
$\begin{vmatrix}
1 & 2 & 2 & \cdots & 2\\
2 & 2 & 2 & \cdots & 2\\
2 & 2 & 3 & \cdots & 2\\
\vdots & \vdots & \vdots & & \vdots \\
2 & 2 & 2 & \cdots & n\\
\end{vmatrix}$
\end{block}
\end{frame}
\begin{frame}
\begin{block}{例3}
\begin{center}
$\begin{vmatrix}
a_1 & x & x & \cdots & x\\
x & a_2 & x & \cdots & x\\
x & x & a_3 & \cdots & x\\
\vdots & \vdots & \vdots & & \vdots \\
x & x & x & \cdots & a_n\\
\end{vmatrix}$
\end{center}
\end{block}
\end{frame}
\begin{frame}
\begin{block}{例4}
\begin{center}
$\left|\begin{array}{ccccc}
a_{1}+x & a_{2} & a_{3} & \cdots & a_{n}\\
a_{1} & a_{2}+x & a_{3} & \cdots & a_{n}\\
a_{1} & a_{2} & a_{3}+x & \cdots & a_{n}\\
\vdots & \vdots & \vdots & & \vdots\\
a_{1} & a_{2} & a_{3} & \cdots & a_{n}+x
\end{array}\right|$
\end{center}
\end{block}
\end{frame}
\begin{frame}
\begin{block}{例5}
\begin{center}
$
\begin{vmatrix}
a_{1}+b_{1}c_{1} & a_{2}+b_{1}c_{2} & \cdots & a_{n}+b_{1}c_{n}\\
a_{1}+b_{2}c_{1} & a_{2}+b_{2}c_{2} & \cdots & a_{n}+b_{2}c_{n}\\
\vdots & \vdots & & \vdots\\
a_{1}+b_{n}c_{1} & a_{2}+b_{n}c_{2} & \cdots & a_{n}+b_{n}c_{n}
\end{vmatrix}
$
\end{center}
\end{block}
\end{frame}
\begin{frame}
\begin{block}{例6}
\begin{center}
$
\begin{vmatrix}
x & a & a & \cdots & a\\
b & x & a & \cdots & a\\
b & b & x & \cdots & a\\
\vdots & \vdots & \vdots & & \vdots\\
b & b & b & \cdots & x
\end{vmatrix}
$
\end{center}
\end{block}
\end{frame}
\begin{frame}{经典题目}
\begin{block}{题目1}
\centering
$\begin{vmatrix}
3 & 1 & 1 & 1 \\
1 & 3 & 1 & 1 \\
1 & 1 & 3 & 1 \\
1 & 1 & 1 & 3
\end{vmatrix}$
\end{block}
\pause
这是一个4级行列式的计算,首先可以使用初等行(列)变换;如果效果不好,可以直接考虑行列式的定义。
\pause
\par\textbf{解:}原式$=
\begin{vmatrix}
3 & 1 & 1 & 1 \\
1 & 3 & 1 & 1 \\
1 & 1 & 3 & 1 \\
1 & 1 & 1 & 3
\end{vmatrix}$
\end{frame}
\begin{frame}{经典题目}
\begin{block}{题目2}
\centering
$\begin{vmatrix}
1 & 2 & 3 & \dots & n-1 & n \\
1 & -1 & 0 & \dots & 0 & 0 \\
0 & 2 & -2 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & n-1 & 1-n
\end{vmatrix}$
\end{block}
\end{frame}
\begin{frame}{经典题目}
\begin{block}{题目3}
\centering$
\begin{vmatrix}
1+a & 1 & 1 & \dots & 1 & 1 \\
2 & 2+a & 2 & \dots & 2 & 2 \\
3 & 3 & 3+a & \dots & 3 & 3 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n-1 & n-1 & n-1 & \dots & (n-1)+a & n-1 \\
n & n & n & \dots & n & n+a
\end{vmatrix}$
\end{block}
\end{frame}
\begin{frame}{经典题目}
\begin{block}{题目4}
\centering$
\begin{vmatrix}
a+b & a & 0 & \dots & 0 & 0 \\
b & a+b & a & \dots & 0 & 0 \\
0 & b & a+b & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & a+b & a \\
0 & 0 & 0 & \dots & b & a+b
\end{vmatrix}$
\end{block}
\end{frame}
\section{小结}
\begin{frame}{方法小结}
\begin{block}{总结}
\begin{itemize}
\item
\end{itemize}
\end{block}
\hspace*{\fill}
\begin{block}{未来工作}
\begin{itemize}
\item
\end{itemize}
\end{block}
\end{frame}
\begin{frame}
\thispagestyle{empty}
\begin{center}
\textcolor{blue}{\Huge{感谢聆听!}}
\end{center}
\end{frame}
\end{document}
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