Determinant (1)

Primary

$$\begin{cases} a_{11}x_1 + a_{12}x_2 = b_1 \\ a_{21}x_1 + a_{22}x_2 = b_2 \end{cases}$$

消去后当$a_{11}a_{22} - a_{12}a_{21} \neq 0$时得到：

$$x1 = \frac {b_1 a_{22} - a_{12} b_2}{a_{11} a_{22} - a_{12} a_{21}} , x2 = \frac {a_{11} b_2 - b_1 a_{21}}{a_{11}a_{22} - a_{12}a_{21}}$$

按方程组中的顺序排好行列：

$$\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}$$

分母即是行列式的主对角线”积” - 副对角线”积”
如果定义：

$$D = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}, D_1 = \begin{vmatrix} b_{1} & a_{12} \\ b_{2} & a_{22} \end{vmatrix}, D_2 \begin{vmatrix} a_{11} & b_{1} \\ a_{21} & b_{2} \end{vmatrix},$$

则有：

$$x_1 = \frac{D_1}{D}, x_2 = \frac{D_2}{D}$$

example

$$\begin{cases} 3x_1 - 2x_2 = 1 \\ 6x_1 + 5x_2 = -4 \end{cases}$$ $$D = \begin{vmatrix} 3 & -2 \\ 6 & 5 \end{vmatrix} = 27 \neq 0 D_1 = \begin{vmatrix} 1 & -2 \\ -4 & 5 \end{vmatrix} D_2 = \begin{vmatrix} 3 & 1 \\ 6 & -4 \end{vmatrix}$$ $$x_1 = \frac{D_1}{D} = \frac{13}{27}, x_2 = \frac{D_2}{D} = \frac{-30}{27}$$

n Factorial determinant

In 3 factorial determinant:

$$\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}a_{22}a_{33} + a_{13}a_{21}a_{32} + a_{12}a_{23}a_{31} - a_{13}a_{22}a_{31} - a_{12}a_{21}a_{33} - a_{11}a_{23}a_{32}$$

带负号系数的组合都是逆序数为奇数的组合，用p1p2p3来代表某组合矩阵位置有

$$\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix} = \sum_{}{(-1)^ta_{1p_1}a_{2p_2}a_{3p_3}}$$

t表示p1p2p3组合时的逆序数，取和即使为矩阵的表示

More generalizate

$$\sum_{}{(-1)^ta_{1p_1}a_{2p_2}...a_{np_n}}$$

简记作$det(a{ij})$, 数$a{ij}$为D的(i, j)元

Prove

$$\begin{vmatrix} & & & \lambda_1 \\ & & \lambda_2 & \\ & ... & & \\ \lambda_n & & & \end{vmatrix} = (-1)^{\frac{n(n-1)}{2}}\lambda_1\lambda_2...\lambda_n$$

符号代表着副对角线的元素，对$\lambdai$可以用$a{i, n-i+1}$来表示。

$$\begin{vmatrix} & & & \lambda_1 \\ & & \lambda_2 & \\ & ... & & \\ \lambda_n & & & \end{vmatrix} = \begin{vmatrix} & & & a_{1n} \\ & & a_{2,n-1} & \\ & ... & & \\ a_{n1} & & & \end{vmatrix} = (-1)^{\frac{n(n-1)}{2}}a_{1n}a_{2,n-1}...a{n1}$$

对n n-1 n-2 … 1 组合的元素求逆序数

$$t = 0 + 1 + 2 ... n-1 = \frac{n(n-1)}{2}$$