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 ∣ a 11 a 12 a 21 a 22 ∣ \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix} ∣∣∣∣a11a21a12a22∣∣∣∣称为二阶行列式，它的值为
a 11 a 22 − a 12 a 21 a_{11}a_{22}-a_{12}a_{21} a11a22−a12a21，记为 D D D，即
D = ∣ a 11 a 12 a 21 a 22 ∣ = a 11 a 22 − a 12 a 21 D=\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：
{ a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 （ 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
{ x 1 = b 1 a 22 − b 2 a 12 a 11 a 22 − a 12 a 21 x 2 = b 2 a 11 − b 1 a 21 a 11 a 22 − a 12 a 21 （ a 11 a 22 − a 12 a 21 ≠ 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
D 1 = ∣ b 1 a 12 b 2 a 22 ∣ = b 1 a 22 − b 2 a 12 D 2 = ∣ a 11 b 1 a 21 b 2 ∣ = a 11 b 2 − b 1 a 21 D_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 = ∣ a 11 a 12 a 21 a 22 ∣ ≠ 0 D=\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}\neq0 D=∣∣∣∣a11a21a12a22∣∣∣∣=0时，方程组的解惟一且可以表示为：
x 1 = D 1 D , x 2 = D 2 D x_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 ∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ \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
a 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 a_{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：
{ 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 \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 = ∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ ≠ 0 , D 1 = ∣ b 1 a 12 a 13 b 2 a 22 a 23 b 3 a 32 a 33 ∣ , D 2 = ∣ a 11 b 1 a 13 a 21 b 2 a 23 a 31 b 3 a 33 ∣ , D 3 = ∣ a 11 a 12 b 1 a 21 a 22 b 2 a 31 a 32 b 3 ∣ ( D 1 , D 2 , D 3 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 ≠ 0 D\neq0 D=0，则有唯一解：
x 1 = D 1 D , x 2 = D 2 D , x 3 = D 3 D x_1=\frac{D_1}{D},x_2=\frac{D_2}{D},x_3=\frac{D_3}{D} x1=DD1,x2=DD2,x3=DD3