第一章，用行列式解线性方程组，02-二阶与三阶行列式

简介

这是《玩转线性代数》的学习笔记
少壮不努力，老大徒伤悲，学校里没学好，工作多年后又从头看，两行泪。。。

2.1 二阶行列式

2.1.1 定义

将符号∣a11a12a21a22∣\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}∣∣∣∣a11a21a12a22∣∣∣∣称为二阶行列式，它的值为
a11a22−a12a21a_{11}a_{22}-a_{12}a_{21}a11a22−a12a21，记为DDD，即
D=∣a11a12a21a22∣=a11a22−a12a21D=\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21} D=∣∣∣∣a11a21a12a22∣∣∣∣=a11a22−a12a21

2.1.2 对角线法则

主对角线
从左上角到右下角元素连线称为主对角线
副对角线
从左下角到右上角元素连线称为副对角线

用主对角线的乘积减去副对角线乘积的方式计算行列式的方法称为对角线法则

2.1.3 二元线性方程组解的行列式表示

解二元线性方程组：
{a11x1+a12x2=b1a21x1+a22x2=b2（2.1）\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. \quad（2.1） {a11x1+a12x2=b1a21x1+a22x2=b2（2.1）
解得解为
{x1=b1a22−b2a12a11a22−a12a21x2=b2a11−b1a21a11a22−a12a21（a11a22−a12a21≠0）\left\{ \begin{aligned} x_1 = \frac{b_{1}a_{22}-b_{2}a_{12}}{a_{11}a_{22}-a_{12}a_{21}}\\ x_2 = \frac{b_{2}a_{11}-b_{1}a_{21}}{a_{11}a_{22}-a_{12}a_{21}}\\ \end{aligned} \right. \quad（a_{11}a_{22}-a_{12}a_{21}\neq0） ⎩⎪⎪⎨⎪⎪⎧x1=a11a22−a12a21b1a22−b2a12x2=a11a22−a12a21b2a11−b1a21（a11a22−a12a21=0）
记
D1=∣b1a12b2a22∣=b1a22−b2a12D2=∣a11b1a21b2∣=a11b2−b1a21D_1 = \begin{vmatrix} b_{1} & a_{12} \\ b_{2} & a_{22}\end{vmatrix}=b_{1}a_{22}-b_{2}a_{12}\\ D_2 = \begin{vmatrix} a_{11} & b_{1} \\ a_{21} & b_{2}\end{vmatrix}=a_{11}b_{2}-b_{1}a_{21} D1=∣∣∣∣b1b2a12a22∣∣∣∣=b1a22−b2a12D2=∣∣∣∣a11a21b1b2∣∣∣∣=a11b2−b1a21
则当D=∣a11a12a21a22∣≠0D=\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}\neq0D=∣∣∣∣a11a21a12a22∣∣∣∣=0时，方程组的解惟一且可以表示为：
x1=D1D,x2=D2Dx_1=\frac{D_1}{D},x_2=\frac{D_2}{D} x1=DD1,x2=DD2

2.2 三阶行列式

2.2.1 定义

将符号∣a11a12a13a21a22a23a31a32a33∣\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}∣∣∣∣∣∣a11a21a31a12a22a32a13a23a33∣∣∣∣∣∣称为三阶行列式，它的值为
a11a22a33+a12a23a31+a13a21a32−a13a22a31−a11a23a32−a12a21a33a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31}+ a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}a11a22a33+a12a23a31+a13a21a32−a13a22a31−a11a23a32−a12a21a33

2.2.2 对角线法则

将元素循环移位，可以看到同样满足对角线法则
图来自原地址

2.2.3 三元线性方程组解的行列式表示

解三元线性方程组：
{a11x1+a12x2+a13x3=b1a21x1+a22x2+a23x3=b2a31x1+a32x2+a33x3=b3\left\{ \begin{aligned} a_{11}x_1+a_{12}x_2+a_{13}x_3 = b_1\\ a_{21}x_1+a_{22}x_2+a_{23}x_3 = b_2\\ a_{31}x_1+a_{32}x_2+a_{33}x_3 = b_3\\ \end{aligned} \right. \quad ⎩⎪⎨⎪⎧a11x1+a12x2+a13x3=b1a21x1+a22x2+a23x3=b2a31x1+a32x2+a33x3=b3
令D=∣a11a12a13a21a22a23a31a32a33∣≠0,D1=∣b1a12a13b2a22a23b3a32a33∣,D2=∣a11b1a13a21b2a23a31b3a33∣,D3=∣a11a12b1a21a22b2a31a32b3∣(D1,D2,D3的意义与上同)D=\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}\neq0, D_1=\begin{vmatrix} b_{1} & a_{12} & a_{13}\\ b_{2} & a_{22} & a_{23} \\ b_{3} & a_{32} & a_{33}\end{vmatrix}, \\ D_2=\begin{vmatrix} a_{11} & b_{1} & a_{13}\\ a_{21} & b_{2} & a_{23} \\ a_{31} & b_{3} & a_{33}\end{vmatrix}, D_3 =\begin{vmatrix} a_{11} & a_{12} & b_{1}\\ a_{21} & a_{22} & b_{2} \\ a_{31} & a_{32} & b_{3}\end{vmatrix} \quad (D_1,D_2,D_3的意义与上同)D=∣∣∣∣∣∣a11a21a31a12a22a32a13a23a33∣∣∣∣∣∣=0,D1=∣∣∣∣∣∣b1b2b3a12a22a32a13a23a33∣∣∣∣∣∣,D2=∣∣∣∣∣∣a11a21a31b1b2b3a13a23a33∣∣∣∣∣∣,D3=∣∣∣∣∣∣a11a21a31a12a22a32b1b2b3∣∣∣∣∣∣(D1,D2,D3的意义与上同)

若D≠0D\neq0D=0，则有唯一解：
x1=D1D,x2=D2D,x3=D3Dx_1=\frac{D_1}{D},x_2=\frac{D_2}{D},x_3=\frac{D_3}{D} x1=DD1,x2=DD2,x3=DD3