线性方程组和矩阵记号的基本认识

至多是一次齐次方程的方程组称为线性方程组，每个方程组可以表示为如下形式：
$$\sum\limits^m_{j=1}a_jx_j=b_i(i=1,2,...,n)$$
其中将$a_{ij}$称为系数，将$x_j$称为未知量，$b_i$称为常数项。
把上面的n元线性方程组写成“表格”也就是矩阵形式：
$$\left\lgroup \begin{array}{cccc|c} a_{11}&a_{12}&\cdots&a_{1m}&b_1\\ a_{21}&a_{22}&\cdots&a_{2m}&b_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nm}&b_n\\ \end{array}\right\rgroup$$
竖线左边的部分是等式左边的系数，右边的部分则是等式右边的系数。
在高中的时候，我们是通过各个方程组相加减得到方程组的解的。
对比高中的朴素求解，我们得到了对矩阵的初等行变换。

1.把一行的倍数加到另一行上；
即把一个方程组两边加到另一个方程组两边。

2.互换两行的位置；
即把两个方程组的书写顺序调换。

3.用一个非零数乘以某一行。
即把某个方程组的两边同时乘以一个非零数。

于是我们可以在该矩阵中执行逐行消元操作，获得以下的矩阵：
$$\left\lgroup \begin{array}{ccccc|c} a_{11}'&a_{12}'&a_{13}'&\cdots&a_{1m}'&b_1'\\ 0&a_{22}'&a_{23}'&\cdots&a_{2m}'&b_2'\\ 0&0&a_{33}'&\cdots&a_{3m}'&b_3'\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ \\ 0&0&0&\cdots&a_{nm}'&b_n'\\ \end{array}\right\rgroup$$
若能化成该形式，并且以下各过程均能执行的话，有：
$$x_n=\frac{b_n'}{a_{nm}'},x_{n-1}=\frac{b_{n-1}'-a_{n-1,m}x_n}{a_{n-1,m-1}'},...,x_{1}=\frac{b_{n-1}'-\sum\limits_{j=2}^ma_{1,j}x_j}{a_{11}'}$$

事实上，仅仅通过初等行变换完成线性方程组的求解的方法，叫做$\text{Gauss-Jordan}$算法。

观察能够获得$x_1,x_2,\dots,x_n$的条件，也就是$\text{Gauss-Jordan}$算法的求解过程。
1.若经过消元以后$a'_{i1},a'_{i2},\dots,a'_{in}$为零，而$b'_i$不为零，则重新写为线性方程组我们得到$0=b'_i$，所以原方程组无解。
2.若1的情况不存在，且非零行的个数等于未知量的个数，那么我们能够进行上面的过程，因此得到唯一解。
3.若1的情况不存在，且非零行的个数少于未知量的个数，意味着有的未知量能够任意改变而同时能够通过其他元的配合使得线性方程组成立，于是该线性方程组有无穷多解。

行列式运算的基本认识

行列式是对方阵($n\times n$矩阵)的运算。
该定义源于线性方程组的书写。
考虑一个4元一次方程组:
$$\left\{ \begin{aligned} a_{11}x_1+a_{12}x_2+a_{13}x_3+a_{14}x_4=b_1\\ a_{21}x_1+a_{22}x_2+a_{23}x_3+a_{24}x_4=b_2\\ a_{31}x_1+a_{32}x_2+a_{33}x_3+a_{34}x_4=b_3\\ a_{41}x_1+a_{42}x_2+a_{43}x_3+a_{44}x_4=b_4\\\end{aligned}\right.$$
$$\left\lgroup\begin{array}{cccc|c} a_{11}&a_{12}&a_{13}&a_{14}&b_1\\ a_{21}&a_{22}&a_{23}&a_{24}&b_2\\ a_{31}&a_{32}&a_{33}&a_{34}&b_3\\ a_{41}&a_{42}&a_{43}&a_{44}&b_4\\ \end{array}\right\rgroup$$
$$\left\lgroup \begin{array}{cccc|c} 1&\frac{a_{12}}{a_{11}}&\frac{a_{13}}{a_{11}}&\frac{a_{14}}{a_{11}}&\frac{b_{1}}{a_{11}}\\ 1&\frac{a_{22}}{a_{21}}&\frac{a_{23}}{a_{21}}&\frac{a_{24}}{a_{21}}&\frac{b_{2}}{a_{21}}\\ 1&\frac{a_{32}}{a_{31}}&\frac{a_{33}}{a_{31}}&\frac{a_{34}}{a_{31}}&\frac{b_{3}}{a_{31}} \\1&\frac{a_{42}}{a_{41}}&\frac{a_{43}}{a_{41}}&\frac{a_{44}}{a_{41}}&\frac{b_{4}}{a_{41}}\\ \end{array}\right\rgroup$$
$$\left\lgroup \begin{array}{cccc|c} 1&\frac{a_{12}}{a_{11}}&\frac{a_{13}}{a_{11}}&\frac{a_{14}}{a_{11}}&\frac{b_{1}}{a_{11}}\\ 0&\frac{a_{22}}{a_{21}}-\frac{a_{12}}{a_{11}}&\frac{a_{23}}{a_{21}}-\frac{a_{13}}{a_{11}}&\frac{a_{24}}{a_{21}}-\frac{a_{14}}{a_{11}}&\frac{b_{2}}{a_{21}}-\frac{b_{1}}{a_{11}}\\ 0&\frac{a_{32}}{a_{31}}-\frac{a_{12}}{a_{11}}&\frac{a_{33}}{a_{31}}-\frac{a_{13}}{a_{11}}&\frac{a_{34}}{a_{31}}-\frac{a_{14}}{a_{11}}&\frac{b_{3}}{a_{31}}-\frac{b_{1}}{a_{11}} \\ 0&\frac{a_{42}}{a_{41}}-\frac{a_{12}}{a_{11}}&\frac{a_{43}}{a_{41}}-\frac{a_{13}}{a_{11}}&\frac{a_{44}}{a_{41}}-\frac{a_{14}}{a_{11}}&\frac{b_{4}}{a_{41}}-\frac{b_{1}}{a_{11}}\\ \end{array}\right\rgroup$$
$$\left\lgroup \begin{array}{cccc|c} a_{11}&a_{12}&a_{13}&a_{14}&b_1\\ 0&\frac{a_{22}a_{11}-a_{12}a_{21}}{a_{21}a_{11}}&\frac{a_{23}a_{11}-a_{13}a_{21}}{a_{21}a_{11}}&\frac{a_{24}a_{11}-a_{14}a_{21}}{a_{21}a_{11}}&\frac{b_{2}a_{11}-b_{1}a_{21}}{a_{21}a_{11}}\\ 0&\frac{a_{32}a_{11}-a_{12}a_{31}}{a_{31}a_{11}}&\frac{a_{33}a_{11}-a_{13}a_{31}}{a_{31}a_{11}}&\frac{a_{34}a_{11}-a_{14}a_{31}}{a_{31}a_{11}}&\frac{b_{3}a_{11}-b_{1}a_{31}}{a_{31}a_{11}}\\ 0&\frac{a_{42}a_{11}-a_{12}a_{41}}{a_{41}a_{11}}&\frac{a_{43}a_{11}-a_{13}a_{41}}{a_{41}a_{11}}&\frac{a_{44}a_{11}-a_{14}a_{41}}{a_{41}a_{11}}&\frac{b_{4}a_{11}-b_{1}a_{41}}{a_{41}a_{11}}\\ \end{array}\right\rgroup$$
暂时少写第一行第一列
$$\left\lgroup \begin{array}{ccc|c} 1&\frac{a_{23}a_{11}-a_{13}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}&\frac{a_{24}a_{11}-a_{14}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}&\frac{b_{2}a_{11}-b_{1}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}\\ 1&\frac{a_{33}a_{11}-a_{13}a_{31}}{a_{32}a_{11}-a_{12}a_{31}}&\frac{a_{34}a_{11}-a_{14}a_{31}}{a_{32}a_{11}-a_{12}a_{31}}&\frac{b_{3}a_{11}-b_{1}a_{31}}{a_{32}a_{11}-a_{12}a_{31}}\\ 1&\frac{a_{43}a_{11}-a_{13}a_{41}}{a_{42}a_{11}-a_{12}a_{41}}&\frac{a_{44}a_{11}-a_{14}a_{41}}{a_{42}a_{11}-a_{12}a_{41}}&\frac{b_{4}a_{11}-b_{1}a_{41}}{a_{42}a_{11}-a_{12}a_{41}}\\ \end{array}\right\rgroup$$

左半截
$$\left\lgroup \begin{array}{ccc|c} 1&\frac{a_{23}a_{11}-a_{13}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}&\frac{a_{24}a_{11}-a_{14}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}&\\ 0&\frac{a_{33}a_{11}-a_{13}a_{31}}{a_{32}a_{11}-a_{12}a_{31}}-\frac{a_{23}a_{11}-a_{13}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}&\frac{a_{34}a_{11}-a_{14}a_{31}}{a_{32}a_{11}-a_{12}a_{31}}-\frac{a_{24}a_{11}-a_{14}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}&\\ 0&\frac{a_{43}a_{11}-a_{13}a_{41}}{a_{42}a_{11}-a_{12}a_{41}}-\frac{a_{23}a_{11}-a_{13}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}&\frac{a_{44}a_{11}-a_{14}a_{41}}{a_{42}a_{11}-a_{12}a_{41}}-\frac{a_{24}a_{11}-a_{14}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}&\\ \end{array}\right\rgroup$$
右半截
$$\left\lgroup \begin{array}{|c} \frac{b_{2}a_{11}-b_{1}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}\\ \frac{b_{3}a_{11}-b_{1}a_{31}}{a_{32}a_{11}-a_{12}a_{31}}-\frac{b_{2}a_{11}-b_{1}a_{21}}{a_{22}a_{11}-a_{12}a_{21}}\\ \frac{b_{4}a_{11}-b_{1}a_{41}}{a_{42}a_{11}-a_{12}a_{41}}-\frac{b_{2}a_{11}-b_{1}a_{21}}{a_{22}a_{11}-a_{12}a_{21}} \end{array}\right\rgroup$$

$\displaystyle$
暂时省去第二行第二列
左半截
$$\left\lgroup \begin{array}{cc|} 1&\frac{(a_{34}a_{11}-a_{14}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{24}a_{11}-a_{14}a_{21})}{(a_{33}a_{11}-a_{13}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{23}a_{11}-a_{13}a_{21})}\\ 1&\frac{(a_{44}a_{11}-a_{14}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{24}a_{11}-a_{14}a_{21})}{(a_{43}a_{11}-a_{13}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{23}a_{11}-a_{13}a_{21})} \end{array}\right\rgroup$$
右半截
$$\left\lgroup \begin{array}{|c} \frac{(b_{3}a_{11}-b_{1}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(b_{2}a_{11}-b_{1}a_{21})}{(a_{33}a_{11}-a_{13}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{23}a_{11}-a_{13}a_{21})}\\ \frac{(b_{4}a_{11}-b_{1}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(b_{2}a_{11}-b_{1}a_{21})}{(a_{43}a_{11}-a_{13}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{23}a_{11}-a_{13}a_{21})} \end{array}\right\rgroup$$

第三行减去第四行，就大功告成了。
得到
$\displaystyle x_4= \frac{\frac{(b_{4}a_{11}-b_{1}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(b_{2}a_{11}-b_{1}a_{21})}{(a_{43}a_{11}-a_{13}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{23}a_{11}-a_{13}a_{21})}-\frac{(b_{3}a_{11}-b_{1}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(b_{2}a_{11}-b_{1}a_{21})}{(a_{33}a_{11}-a_{13}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{23}a_{11}-a_{13}a_{21})}} {\frac{(a_{44}a_{11}-a_{14}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{24}a_{11}-a_{14}a_{21})}{(a_{43}a_{11}-a_{13}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{23}a_{11}-a_{13}a_{21})}-\frac{(a_{34}a_{11}-a_{14}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{24}a_{11}-a_{14}a_{21})}{(a_{33}a_{11}-a_{13}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{23}a_{11}-a_{13}a_{21})}}$

考虑一个分母：
$\displaystyle\frac{(a_{44}a_{11}-a_{14}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{24}a_{11}-a_{14}a_{21})}{(a_{43}a_{11}-a_{13}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{23}a_{11}-a_{13}a_{21})}-\ \displaystyle\frac{(a_{34}a_{11}-a_{14}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{24}a_{11}-a_{14}a_{21})}{(a_{33}a_{11}-a_{13}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{23}a_{11}-a_{13}a_{21})}$
它等于
$\displaystyle\frac{ \begin{aligned}&[(a_{44}a_{11}-a_{14}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{24}a_{11}-a_{14}a_{21})]\\ \displaystyle&[(a_{33}a_{11}-a_{13}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{23}a_{11}-a_{13}a_{21})]\\ &-\\ \displaystyle&[(a_{43}a_{11}-a_{13}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{23}a_{11}-a_{13}a_{21})]\\ \displaystyle&[(a_{34}a_{11}-a_{14}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{24}a_{11}-a_{14}a_{21})] \end{aligned}} {\begin{aligned} \displaystyle&[(a_{43}a_{11}-a_{13}a_{41})(a_{22}a_{11}-a_{12}a_{21})-(a_{42}a_{11}-a_{12}a_{41})(a_{23}a_{11}-a_{13}a_{21})]\\ \displaystyle&[(a_{34}a_{11}-a_{14}a_{31})(a_{22}a_{11}-a_{12}a_{21})-(a_{32}a_{11}-a_{12}a_{31})(a_{24}a_{11}-a_{14}a_{21})] \end{aligned}}$

令其分子为
$\displaystyle a_{11}^2(a_{22}a_{11}-a_{12}a_{21})\left|\begin{array}{cccc} a_{11} &a_{12}&a_{13}&a_{14}\\ a_{21} &a_{22}&a_{23}&a_{24}\\ a_{31} &a_{32}&a_{33}&a_{34}\\ a_{41} &a_{42}&a_{43}&a_{44}\\ \end{array}\right|$
发现其为$\displaystyle a_{11}^2(a_{22}a_{11}-a_{12}a_{21})\sum_{p_1p_2p_3p_4}(-1)^{\tau(p_1p_2p_3p_4)}a_{1,p_1}a_{2,p_2}a_{3,p_3}a_{4,p_4}$，其中$p_1p_2p_3p_4$是{1,2,3,4}的一个全排列，该式一共有$4!$项。
发现这样一写，原式分子通分后的分子也可以表示为
$\displaystyle a_{11}^2(a_{22}a_{11}-a_{12}a_{21})\left|\begin{array}{cccc} a_{11} &a_{12}&a_{13}&b_{1}\\ a_{21} &a_{22}&a_{23}&b_{2}\\ a_{31} &a_{32}&a_{33}&b_{3}\\ a_{41} &a_{42}&a_{43}&b_{4}\\ \end{array}\right|$
于是对矩阵的一种运算行列式应运而生。
$\text{definition }$定义n阶行列式为
$\displaystyle \left| \begin{array}{cccc} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\ \end{array}\right|=\sum_{p_1p_2\dots p_n}(-1)^{\tau(p_1p_2\dots p_n)}a_{1,p_1}a_{2,p_2}\dots a_{n,p_n}$
其中$\tau(p_1p_2\dots p_n)$是与排列有关的函数，暂时不知道它的解析式。
我觉得刚刚的内容写丑了= =。
应该这样写：
保证以下所有方程组有唯一解。
对于n元1次方程组：
n=1时，
$\left\{\begin{aligned}a_{11}x_1=b_{1}\end{aligned}\right.$
此时$x_1=\frac{a_{11}}{b_1}.$
定义$|a|=a$.
n=2时，
$\left\{\begin{aligned}a_{11}x_1+a_{12}x_2=b_{1}\\a_{21}x_1+a_{22}x_2=b_{2}\end{aligned}\right.$
于是
$$\left\lgroup \begin{array}{cc|c} 1&\frac{a_{12}}{a_{11}}&\frac{b_1}{a_{11}}\\ 1&\frac{a_{22}}{a_{21}}&\frac{b_2}{a_{21}}\\ \end{array}\right\rgroup$$
$$\left\lgroup \begin{array}{cc|c} 1&\frac{a_{12}}{a_{11}}&\frac{b_1}{a_{11}}\\ 0&\frac{a_{22}}{a_{21}}-\frac{a_{12}}{a_{11}}&\frac{b_2}{a_{21}}-\frac{b_1}{a_{11}}\\ \end{array}\right\rgroup$$
$$\left\lgroup \begin{array}{cc|c} a_{11}&a_{12}&b_1\\ 0&a_{11}a_{22}-a_{12}a_{21}&a_{11}b_{2}-a_{21}b_{1}\\ \end{array}\right\rgroup$$
令 $$\left| \begin{array}{cc} a&b\\ c&d\\ \end{array}\right|=ad-bc$$
则 $$\left\lgroup \begin{array}{cc|c} \left| \begin{array}{cc} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{array}\right|&0&\left| \begin{array}{cc} b_{1}&a_{12}\\ b_{2}&a_{22}\\ \end{array}\right|\\ 0&\left| \begin{array}{cc} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{array}\right|&\left| \begin{array}{cc} a_{11}&b_{1}\\ a_{21}&b_{2}\\ \end{array}\right|\\ \end{array}\right\rgroup$$
n=3时，由n=2时的结论：
$$\left\lgroup \begin{array}{ccc|c} a_{11}&a_{12}&a_{13}&b_{1}\\ 0&\left| \begin{array}{cc} \frac{a_{22}}{a_{21}}-\frac{a_{12}}{a_{11}}&\frac{a_{23}}{a_{21}}-\frac{a_{13}}{a_{11}}\\ \frac{a_{32}}{a_{31}}-\frac{a_{12}}{a_{11}}&\frac{a_{33}}{a_{31}}-\frac{a_{13}}{a_{11}}\\ \end{array}\right|&0&\left| \begin{array}{cc} \frac{b_{2}}{a_{21}}-\frac{b_{1}}{a_{11}}&\frac{a_{23}}{a_{21}}-\frac{a_{13}}{a_{11}}\\ \frac{b_{3}}{a_{31}}-\frac{b_{1}}{a_{11}}&\frac{a_{33}}{a_{31}}-\frac{a_{13}}{a_{11}}\\ \end{array}\right|\\ 0&0&\left| \begin{array}{cc} \frac{a_{22}}{a_{21}}-\frac{a_{13}}{a_{11}}&\frac{a_{23}}{a_{21}}-\frac{a_{12}}{a_{11}}\\ \frac{a_{32}}{a_{31}}-\frac{a_{12}}{a_{11}}&\frac{a_{33}}{a_{31}}-\frac{a_{12}}{a_{11}}\\ \end{array}\right|&\left| \begin{array}{cc} \frac{a_{22}}{a_{21}}-\frac{a_{12}}{a_{11}}&\frac{b_{2}}{a_{21}}-\frac{b_{1}}{a_{11}}\\ \frac{a_{32}}{a_{31}}-\frac{a_{12}}{a_{11}}&\frac{b_{3}}{a_{31}}-\frac{b_{1}}{a_{11}}\\ \end{array}\right|\\ \end{array}\right\rgroup$$
将$D'_n(2;2)$乘以之前除去的$a_1,a_2$得 $$\left| \begin{array}{ccc} a&b&c\\ d&e&f\\ g&h&i\\ \end{array}\right|=g\left| \begin{array}{cc} b&c\\ e&f\\ \end{array}\right|-h\left| \begin{array}{cc} a&c\\ d&f\\ \end{array}\right|+i\left| \begin{array}{cc} a&b\\ d&e\\ \end{array}\right|$$
由此受到启发 $$D_n=\sum_{i=1}^{n}a_{ni}A_{ni}$$
其中$A_{ni}$为代数余子式，为$D_n$的矩阵挖去$n$行$i$列剩余的矩阵的行列式。显见：$D_n$有$n!$项，其符号由$\tau(P)$决定。
(学过的人，肯定知道这个函数是逆序函数，什么时候有兴趣写吧_(:3」∠)_毕竟我已经学完了，写这个东西挺枯燥的)
行列式有很多性质：
1.记矩阵$A$行列交换的矩阵为$A^T$，称为转置矩阵，则$|A^T|=|A|$。
2.若$D_n(i;j)=kD'_n(i,j)(1\leqslant j\leqslant n)$，其余相同，则$D_n=kD'_n$.
$$\left|\begin{array}{cccc}...\\ka_{i1}&ka_{i2}&\dots&ka_{in}\\...\end{array}\right|=k\left|\begin{array}{cccc}...\\a_{i1}&a_{i2}&\dots&a_{in}\\...\end{array}\right|$$
3.若$D_n(i)=D'_n(j),D_n(j)=D'_n(i)$，其余相同，则$D_n=-D'_n$
$$\left|\begin{array}{cccc}a_{i1}&a_{i2}&...&a_{in}\\...\\a_{j1}&a_{j2}&...&a_{jn}\end{array}\right|=\left|\begin{array}{cccc}a_{j1}&a_{j2}&...&a_{jn}\\...\\a_{i1}&a_{i2}&...&a_{in}\end{array}\right|$$
4.若$D_n(i;j)=D'_n(i;j)+D''_n(i;j)$，其余相同，则$D_n=D'_n+D''_n$
5.若$D_n(i;j)=D'_n(i;j)+D_n(k;j)$，其余相同，则$D_n=D'_n$
利用这个可以将刚才的内容推广
$\displaystyle D_n=\sum_{j=1}^na_{ij}A_{ij}(1\leqslant i\leqslant n)$.
这个式子叫做$D_n$的第i行展开式。
那么思考一下，能否将$D_n$按多行展开呢？
$\text{Laplace}$定理告诉我们，
$\displaystyle D_n=\sum_{I}(-1)^{\sum I_k+\sum J_k} A_{IJ}A_{U-I,U-J}$.
其中$U=\{1...n\},I是n-i组合,I_k是I组合的第k大数，J是n-i组合，J_k是J组合的第k大数$.
思考一下$I$一共有$\displaystyle\binom{n}{i}$种，n阶行列式有n!项，所以右侧有$\displaystyle\binom{n}{i}i!(n-i)!$项，而对于右侧的每一项，易证其是左侧的某一项。左侧每一项各不相同，右侧每一项各不相同，这就证明了$\text{Laplace}$定理.