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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}