对于高等代数一下比较基础的概念的认识

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:
m∑j=1ajxj=bi(i=1,2,...,n)\sum\limits^m_{j=1}a_jx_j=b_i(i=1,2,...,n)
其中将aija_{ij}称为系数,将xjx_j称为未知量,bib_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:
⟮a11a12⋯a1mb1a21a22⋯a2mb2⋮⋮⋱⋮⋮an1an2⋯anmbn⟯\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:
⟮a′11a′12a′13⋯a′1mb′10a′22a′23⋯a′2mb′200a′33⋯a′3mb′3⋮⋮⋮⋱⋮⋮000⋯a′nmb′n⟯\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:
xn=b′na′nm,xn−1=b′n−1−an−1,mxna′n−1,m−1,...,x1=b′n−1−m∑j=2a1,jxja′11x_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Gauss-Jordan\text{Gauss-Jordan}算法.

观察能够获得x1,x2,…,xnx_1,x_2,\dots,x_n的条件,也就是Gauss-Jordan\text{Gauss-Jordan}算法的求解过程.
1.若经过消元以后a′i1,a′i2,…,a′ina'_{i1},a'_{i2},\dots,a'_{in}为零,而b′ib'_i不为零,则重新写为线性方程组我们得到0=b′i0=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×nn\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:
{a11x1+a12x2+a13x3+a14x4=b1a21x1+a22x2+a23x3+a24x4=b2a31x1+a32x2+a33x3+a34x4=b3a41x1+a42x2+a43x3+a44x4=b4\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.
⟮a11a12a13a14b1a21a22a23a24b2a31a32a33a34b3a41a42a43a44b4⟯\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
⟮1a12a11a13a11a14a11b1a111a22a21a23a21a24a21b2a211a32a31a33a31a34a31b3a311a42a41a43a41a44a41b4a41⟯\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} 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\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\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\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\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 \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\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\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})}}\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})}\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}} \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| \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}\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_4p_1p_2p_3p_4是{1,2,3,4}的一个全排列,该式一共有4!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| \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 }\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}\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)\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.\left\{\begin{aligned}a_{11}x_1=b_{1}\end{aligned}\right.
此时x_1=\frac{a_{11}}{b_1}.x_1=\frac{a_{11}}{b_1}.
定义|a|=a|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.\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}}\\ 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} 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\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\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\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\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)D'_n(2;2)乘以之前除去的a_1,a_2a_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|\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}D_n=\sum_{i=1}^{n}a_{ni}A_{ni}
其中A_{ni}A_{ni}为代数余子式,为D_nD_n的矩阵挖去nn行ii列剩余的矩阵的行列式.显见:D_nD_n有n!n!项,其符号由\tau(P)\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AA行列交换的矩阵为A^TA^T,称为转置矩阵,则|A^T|=|A||A^T|=|A|.
2.若D_n(i;j)=kD'_n(i,j)(1\leqslant j\leqslant n)D_n(i;j)=kD'_n(i,j)(1\leqslant j\leqslant n),其余相同,则D_n=kD'_nD_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|\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)D_n(i)=D'_n(j),D_n(j)=D'_n(i),其余相同,则D_n=-D'_nD_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|\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)D_n(i;j)=D'_n(i;j)+D''_n(i;j),其余相同,则D_n=D'_n+D''_nD_n=D'_n+D''_n
5.若D_n(i;j)=D'_n(i;j)+D_n(k;j)D_n(i;j)=D'_n(i;j)+D_n(k;j),其余相同,则D_n=D'_nD_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)\displaystyle D_n=\sum_{j=1}^na_{ij}A_{ij}(1\leqslant i\leqslant n).
这个式子叫做D_nD_n的第i行展开式.
那么思考一下,能否将D_nD_n按多行展开呢？
\text{Laplace}\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}\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II一共有\displaystyle\binom{n}{i}\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)!\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}\text{Laplace}定理.

我们定义矩阵乘法:
\left\lgroup \begin{matrix} a_{ij}\\ \end{matrix}\right\rgroup_{n\times s} \left\lgroup \begin{matrix} b_{kl}\\ \end{matrix}\right\rgroup_{s\times m} :=\left\lgroup \begin{matrix} \sum\limits_{r=1}^sa_{xr}b_{ry}\\ \end{matrix}\right\rgroup_{n\times m}\left\lgroup \begin{matrix} a_{ij}\\ \end{matrix}\right\rgroup_{n\times s} \left\lgroup \begin{matrix} b_{kl}\\ \end{matrix}\right\rgroup_{s\times m} :=\left\lgroup \begin{matrix} \sum\limits_{r=1}^sa_{xr}b_{ry}\\ \end{matrix}\right\rgroup_{n\times m}
用以下的例子感受矩阵的书写:
\begin{multline} \shoveleft{\sum\limits^m_{j=1}a_jx_j=b_i(i=1,2,...,n)\Leftrightarrow \sum\limits_{j=1}^m\left\lgroup\begin{matrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\\\end{matrix}\right\rgroup x_j=\left\lgroup\begin{matrix}b_{1}\\b_{2}\\\vdots\\b_{n}\\\end{matrix}\right\rgroup} \\ \shoveleft{令{\bf\alpha_i}=\left\lgroup\begin{matrix}a_{1i}\\a_{2i}\\\vdots\\a_{ni}\\\end{matrix}\right\rgroup,B=\left\lgroup\begin{matrix}b_{1}\\b_{2}\\\vdots\\b_{n}\\\end{matrix}\right\rgroup,有:} \\ \shoveleft{ \sum\limits_{j=1}^m \left\lgroup \begin{matrix} a_{1j}\\a_{2j}\\\vdots\\a_{nj}\\ \end{matrix} \right\rgroup x_j= \left\lgroup \begin{matrix} b_{1}\\b_{2}\\\vdots\\b_{n}\\ \end{matrix} \right\rgroup \Leftrightarrow\sum\limits_{j=1}^{m}{\bf\alpha_j}x_j={\bf\beta}\Leftrightarrow\left\lgroup\begin{matrix}\alpha_{1}&\alpha_{2}&\dots &\alpha_{m}\end{matrix}\right\rgroup\left\lgroup\begin{matrix}x_{1}\\x_{2}\\\vdots\\x_{n}\\\end{matrix}\right\rgroup=B}\\ \shoveleft{ 记\left\lgroup\begin{matrix}\alpha_{1}&\alpha_{2}&\dots &\alpha_{m}\end{matrix}\right\rgroup作A,\left\lgroup\begin{matrix}x_{1}\\x_{2}\\\vdots\\x_{n}\\\end{matrix}\right\rgroup作X,有该式可写作AX=B}\\ \end{multline}\\ \begin{multline} \shoveleft{\sum\limits^m_{j=1}a_jx_j=b_i(i=1,2,...,n)\Leftrightarrow \sum\limits_{j=1}^m\left\lgroup\begin{matrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\\\end{matrix}\right\rgroup x_j=\left\lgroup\begin{matrix}b_{1}\\b_{2}\\\vdots\\b_{n}\\\end{matrix}\right\rgroup} \\ \shoveleft{令{\bf\alpha_i}=\left\lgroup\begin{matrix}a_{1i}\\a_{2i}\\\vdots\\a_{ni}\\\end{matrix}\right\rgroup,B=\left\lgroup\begin{matrix}b_{1}\\b_{2}\\\vdots\\b_{n}\\\end{matrix}\right\rgroup,有:} \\ \shoveleft{ \sum\limits_{j=1}^m \left\lgroup \begin{matrix} a_{1j}\\a_{2j}\\\vdots\\a_{nj}\\ \end{matrix} \right\rgroup x_j= \left\lgroup \begin{matrix} b_{1}\\b_{2}\\\vdots\\b_{n}\\ \end{matrix} \right\rgroup \Leftrightarrow\sum\limits_{j=1}^{m}{\bf\alpha_j}x_j={\bf\beta}\Leftrightarrow\left\lgroup\begin{matrix}\alpha_{1}&\alpha_{2}&\dots &\alpha_{m}\end{matrix}\right\rgroup\left\lgroup\begin{matrix}x_{1}\\x_{2}\\\vdots\\x_{n}\\\end{matrix}\right\rgroup=B}\\ \shoveleft{ 记\left\lgroup\begin{matrix}\alpha_{1}&\alpha_{2}&\dots &\alpha_{m}\end{matrix}\right\rgroup作A,\left\lgroup\begin{matrix}x_{1}\\x_{2}\\\vdots\\x_{n}\\\end{matrix}\right\rgroup作X,有该式可写作AX=B}\\ \end{multline}\\ 我们把n\times 1n\times 1的矩阵习惯性称作n维列向量.于是从该方程形式,我们很容易看出XX作为一个“解”的向量出现,若解向量存在,解向量的每一个维数对应x_ix_i的一个可行解.

并且当我们熟练以后,矩阵形式能够使得我们减少大量符号的书写,比不引入矩阵记号要方便得多.

从矩阵记号的形式来看,n<script type="math/tex">n</script>元线性方程组和一元一次线性方程没有什么不同.这引人思考,于是对这种形式的未知量进行研究.

\text{Definition }\text{Definition }取定一个数域KK,作它的笛卡尔(Cartesian)乘积得到:
K^n=K\times K^{n-1}=\{(a_1,a_2,...,a_n)|\forall a_i\in K\}K^n=K\times K^{n-1}=\{(a_1,a_2,...,a_n)|\forall a_i\in K\}.
称其为nn维向量空间,其中元素自然就称为nn维(行)向量,这使人联想到刚才所写线性方程组中的\alpha,\beta\alpha,\beta,那种元素也称为n<script type="math/tex">n</script>维(列)向量.于是在K^nK^n中定义以下运算:
加法\alpha+\beta:=(a_1+b_1,...,a_n+b_n)\alpha+\beta:=(a_1+b_1,...,a_n+b_n).
数乘运算k\alpha:=\alpha+(k-1)\alpha且0\alpha=(0,...,0)=:0k\alpha:=\alpha+(k-1)\alpha且0\alpha=(0,...,0)=:0.
容易证明加法具有交换律,结合律,数乘具有结合律,加法和数乘之间有分配律.用抽象代数的说法,这两种运算使得nn维向量空间的向量运算有交换环的性质.

回想高中所学的知识.取定\vec{i}=(1,0,0),\vec{j}=(0,1,0),\vec{k}=(0,0,1)\vec{i}=(1,0,0),\vec{j}=(0,1,0),\vec{k}=(0,0,1)(直角坐标系),他们被称为三维直角坐标的基向量.取这种名字的原因是,任意空间点三元组p(a,b,c)=a\vec{i}+b\vec{j}+c\vec{k}p(a,b,c)=a\vec{i}+b\vec{j}+c\vec{k}可以由基向量表示.自然地,我们可以拓展这种定义:

\text{Definition }\text{Definition },\displaystyle \sum_{i=1}^s k_i\alpha_i=k_1\alpha_1+k_2\alpha_2+\dots+k_s\alpha_s\displaystyle \sum_{i=1}^s k_i\alpha_i=k_1\alpha_1+k_2\alpha_2+\dots+k_s\alpha_s称为向量组\alpha_1,\alpha_2,\dots,\alpha_s\alpha_1,\alpha_2,\dots,\alpha_s的一个线性组合,若\beta=\sum\limits_i k_i\alpha_i\beta=\sum\limits_i k_i\alpha_i,称\beta\beta由该向量组线性表出.
于是,由这种概念给出关于n<script type="math/tex">n</script>元线性方程组解的问题就转化为了\beta\beta是否由向量组A=(\alpha_1\ \alpha_2\ \dots\ \alpha_n)A=(\alpha_1\ \alpha_2\ \dots\ \alpha_n)线性表出的问题.
先暂且不看\beta\beta是否由AA线性表出,我们跳出这个问题,去思考AA能表出哪些向量.

对于具体的例子,在三维空间中,直角坐标系\vec{i},\vec{j},\vec{k}\vec{i},\vec{j},\vec{k},能够表示所有三维空间的点,我们又知道从原点出发,任意点pp唯一确定一个向量\beta=op=(a,b,c)\beta=op=(a,b,c),于是三维空间的任意向量\beta\beta可以由向量组\vec{i},\vec{j},\vec{k}\vec{i},\vec{j},\vec{k}线性表出,这提示我们,如果能够掌握这个向量组的所有信息,仿佛就能够囊括三维空间的所有向量,如果你有足够的想象力,你应该能够从下面一句描述感受到\vec{i},\vec{j},\vec{k}\vec{i},\vec{j},\vec{k}的强大力量:三维向量空间K^3K^3由向量组\vec{i},\vec{j},\vec{k}\vec{i},\vec{j},\vec{k}生成.这就是为什么我们把这三个向量称为三维空间的基向量的原因.

但是如果我们去掉三个向量的其中一个,很容易知道它们表现的力量就被大大削弱了,因为他们三个互有分工,完成三维空间三个维度的刻画.缺少一个,这个向量组生成的空间就被拍平成平面,缺少两个,这个向量组所生成的空间就只剩一条空间直线了.

但奇异却不奇怪的是,平面和空间直线仍然有K^3K^3下的封闭性质,如同空间中仍然嵌套了一个空间,所以我们把它们称为三维空间的子空间,它们由它们对应剩下的向量组生成.

\text{Definition }\text{Definition },K^nK^n中的s<script type="math/tex">s</script>个向量的所有线性组合构成K^nK^n的子空间,记为\left<\alpha_1,\alpha_2,\dots\alpha_s\right>\left.

于是\beta\beta能由AA线性表出的问题就转化为了\beta\beta是否属于K^nK^n子空间\left<\alpha_1,\alpha_2,\dots\alpha_s\right>\left的问题.

继续观察\vec{i},\vec{j},\vec{k}\vec{i},\vec{j},\vec{k},如果我们把\vec{k}=(0,0,1)\vec{k}=(0,0,1)换为了(1,1,0)(1,1,0),\vec{i},\vec{j},\vec{k}\vec{i},\vec{j},\vec{k}是否仍然能够完成张成K^3K^3的任务呢？答案是不能,稍加思考就能知道\vec{k}=\vec{i}+\vec{j}\vec{k}=\vec{i}+\vec{j},有你没你其实是一样的,这本质和只有两个向量没有什么区别.原因就是s<script type="math/tex">s</script>个向量组中,有的向量没有能够拥有这个团队中独特的能力,换一句话说就是该向量(的任务)由向量组中其他向量(成员)线性表出(分工完成)了.

\text{Definition }\text{Definition }说向量组是线性相关的,就是存在\alpha_i\in A\alpha_i\in A能够由向量组中其他向量线性表出,即使得下式成立可以有非零k_ik_i存在:
\sum_{i}k_i\alpha_i=0.\sum_{i}k_i\alpha_i=0.
否则称向量组是线性无关的.
至于为什么这个式子能够完成线性相关的定义,有兴趣的人可以下课研究研究_(:3」∠)_.

关于向量和解的关系的研究是平凡的,所以我在把这些概念引出以后就不再想讲得更多,因为书中所讲就已经能够使得学生完全理解个中理论的灵魂所在(其实就是懒…如果有人有需要,我还是可以讲一讲的,反正没人看QAQ).

考虑刚才引入的矩阵概念.
简单地证明就可以发现,设KK上s\times ns\times n矩阵集合为M_{s\times n}[K]M_{s\times n}[K],在该集合上矩阵加法和数乘运算满足前面介绍向量运算的所有性质.
特殊地,令s=ns=n,即考虑n级方阵集合M_{n\times n}[K]M_{n\times n}[K]中的矩阵乘法的性质.
那么矩阵乘法满足结合律:
(AB)C=A(BC)(AB)C=A(BC).
且单位元为I=[\delta_{ij}]_{n\times n},\delta_{ij}为\text{Kronnecker}函数,\delta_{ij}=[i=j]I=[\delta_{ij}]_{n\times n},\delta_{ij}为\text{Kronnecker}函数,\delta_{ij}=[i=j].
但是,ABAB的逆元A^{-1}A^{-1}使得AA^{-1}=IAA^{-1}=I并不能总是找到,于是M_{n\times n}[K]M_{n\times n}[K]不是群.
如果考虑AA有右逆元A^{-1}A^{-1},那么AA^{-1}=I.AA^{-1}=I.
\displaystyle AA^{-1}=I\Leftrightarrow [\sum_{k=1}^{n}a_{ik}a'_{kj}=\delta_{ij}]_{n\times n}\Leftrightarrow [\sum_{k=1}^{n}a'_{ki}a_{jk}=\delta_{ij}]_{n\times n}\Leftrightarrow A^{-1}A=I\displaystyle AA^{-1}=I\Leftrightarrow [\sum_{k=1}^{n}a_{ik}a'_{kj}=\delta_{ij}]_{n\times n}\Leftrightarrow [\sum_{k=1}^{n}a'_{ki}a_{jk}=\delta_{ij}]_{n\times n}\Leftrightarrow A^{-1}A=I
这意味着A^{-1}A^{-1}的右逆元是AA.
我们知道逆元是唯一的,并且由上证,左右逆元是相等的,于是AA的逆元总是A^{-1}A^{-1}.
从而所有K上n级可逆矩阵组成的集合GL_n(K)GL_n(K)成为一个群,称它为一般线性群.
考虑一个\delta_{ij}\delta_{ij}记号,如果AA^{-1}=IAA^{-1}=I,由第一节知识\displaystyle \sum_{k=1}^{n}a_{ik}A_{kj}=|A|\delta_{ij}\displaystyle \sum_{k=1}^{n}a_{ik}A_{kj}=|A|\delta_{ij},于是\displaystyle A^{-1}=\frac{1}{|A|}[A_{ij}]^{T}_{n\times n}\displaystyle A^{-1}=\frac{1}{|A|}[A_{ij}]^{T}_{n\times n},其中TT运算使某矩阵行列对应互换,称为矩阵的转置.记A^{*}=[A_{ji}]_{n\times n}A^{*}=[A_{ji}]_{n\times n},称为AA的伴随矩阵.

现在再倒回来看,如何判断AA是否可逆.
若A<script type="math/tex">A</script>可逆,则A^{-1}A^{-1}也可逆,总之若AB=IAB=I,则|AB|=|A||B|=|I|=1(|AB|=|A||B|).所以|A|\neq 0,|B|\neq 0.|AB|=|A||B|=|I|=1(|AB|=|A||B|).所以|A|\neq 0,|B|\neq 0.

对于一个初等行变换,总能够找到一个对应的矩阵PP使得PA=A'PA=A',其中A'A'是初等行变换之后的矩阵.
先引入一类矩阵名为基本矩阵,E_{i'j'}=[[i=i'][j=j']]_{n\times n}E_{i'j'}=[[i=i'][j=j']]_{n\times n},
即所有行列中,只有i'行,j'列i'行,j'列上的元素为1,其他为零.
对于:
1.交换矩阵的i,ji,j行,那么PP=
\left\lgroup\begin{matrix} 1&&&&&&\\ &\ddots&&&&&\\ &&&&1&&\\ &&&\ddots&&&\\ &&1&&&&\\ &&&&&&1\\ \end{matrix}\right\rgroup\left\lgroup\begin{matrix} 1&&&&&&\\ &\ddots&&&&&\\ &&&&1&&\\ &&&\ddots&&&\\ &&1&&&&\\ &&&&&&1\\ \end{matrix}\right\rgroup
即P=I-E_{ii}-E_{jj}+E_{ij}+E_{ji}P=I-E_{ii}-E_{jj}+E_{ij}+E_{ji}.
现在可以稍微体会一下EE矩阵的左乘作用,它的意思是把ii行的元素拿出来加到第jj行,那么PP的意思是,先取出一个A<script type="math/tex">A</script>矩阵,然后消去i,ji,j行的元素,然后把jj行元素放到ii行,然后把ii行元素放到jj行.(对应右乘如何?可以思考一下.)
2.将第ii行元素增大kk倍:P=I-E_{ii}+kE_{ii}=I+(k-1)E_{ii}:P=I-E_{ii}+kE_{ii}=I+(k-1)E_{ii}.
3.将第ii行元素kk倍加到第jj行:P=I+kE_{ji}P=I+kE_{ji}.
根据EE矩阵的意义,很容易求出PP矩阵的逆,所以所有PP可逆.

\text{Theorem }\text{Theorem }若A,BA,B可逆,那么ABAB可逆.
证明:
设A^{-1},B^{-1}A^{-1},B^{-1}是两者的逆矩阵.那么AB(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1}=IAB(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1}=I,所以ABAB可逆.

我们假设AA经过一系列初等行变换\prod P\prod P得到JJ矩阵,J:J:
\left\lgroup\begin{matrix}1&&&&&\\&1&&&&\\&&\ddots&&&\\&&&1&&\\&&&&0&\\&&&&&\ddots\\\end{matrix}\right\rgroup\left\lgroup\begin{matrix}1&&&&&\\&1&&&&\\&&\ddots&&&\\&&&1&&\\&&&&0&\\&&&&&\ddots\\\end{matrix}\right\rgroup
这个矩阵一定是可逆矩阵,那么J=IJ=I(一定能得到JJ吗,答案是肯定的).

从这个层面看AA,我们发现如果他不可逆,那么J\neq IJ\neq I,我们给JJ的11的个数取一个名字,叫做秩.

\text{Theorem }\text{Theorem }如果\mathrm{rank}(A)=n\mathrm{rank}(A)=n,那么A<script type="math/tex">A</script>可逆,否则不可逆.

自此,我觉得一些高等代数比较重要的基础概念如何被引入,就讲完了…以后可能会补充一些比较重要的知识吧..

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