# 克萊姆法則

$\mathbf{A} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

## 基本方程

$Ax = c\, \qquad \qquad \qquad \qquad \qquad \qquad (1)$

 $x_i = { \det(A_i) \over \det(A)}$ $(1)\,$

$x_i = { \Delta_i \over \Delta }\,$

## 抽象方程

R為一個環，A就是一個包含R的系數的n×n矩陣。所以：

$\mathrm{Adj}(A)A = \mathrm{det}(A)I\,$

## 證明概要

$A = \left( u_1, u_2, \cdots, u_n\right)$

$x^* = (x_1, x_2, \cdots, x_n)^T$，即

$A x^* = \sum_{k=1}^n x_k u_k = c$

\begin{align} \Delta_i &= det\left(\cdots, u_{i-1}, c, u_{i+1}, \cdots\right) \\ & = det\left(\cdots, u_{i-1}, \sum_{k=1}^n x_k u_k, u_{i+1}, \cdots\right) \\ & = \sum_{k=1}^n x_k \cdot det\left(\cdots, u_{i-1}, u_k, u_{i+1}, \cdots\right) \\ & = x_i \cdot det\left(\cdots, u_{i-1}, u_i, u_{i+1}, \cdots\right) \\ & = x_i \Delta \end{align}

$x_i = \frac{\Delta_i}{\Delta}$

## 例子

$ax + by = {\color{red}e}\,$
$cx + dy = {\color{red}f}\,$

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} {\color{red}e} \\ {\color{red}f} \end{bmatrix}$

$x = \frac { \begin{vmatrix} \color{red}{e} & b \\ \color{red}{f} & d \end{vmatrix} } { \begin{vmatrix} a & b \\ c & d \end{vmatrix} } = { {\color{red}e}d - b{\color{red}f} \over ad - bc}$

$y = \frac { \begin{vmatrix} a & \color{red}{e} \\ c & \color{red}{f} \end{vmatrix} } { \begin{vmatrix} a & b \\ c & d \end{vmatrix} } = { a{\color{red}f} - {\color{red}e}c \over ad - bc}$

$ax + by + cz = {\color{red}j}\,$
$dx + ey + fz = {\color{red}k}\,$
$gx + hy + iz = {\color{red}l}\,$

$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} {\color{red}j} \\ {\color{red}k} \\ {\color{red}l} \end{bmatrix}$

$x = \frac { \begin{vmatrix} {\color{red}j} & b & c \\ {\color{red}k} & e & f \\ {\color{red}l} & h & i \end{vmatrix} } { \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} }$、   $y = \frac { \begin{vmatrix} a & {\color{red}j} & c \\ d & {\color{red}k} & f \\ g & {\color{red}l} & i \end{vmatrix} } { \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} }$   以及   $z = \frac { \begin{vmatrix} a & b & {\color{red}j} \\ d & e & {\color{red}k} \\ g & h & {\color{red}l} \end{vmatrix} } { \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} }$

### 微分幾何上的應用

$dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy +\frac{\partial F}{\partial u} du +\frac{\partial F}{\partial v} dv = 0$
$dG = \frac{\partial G}{\partial x} dx + \frac{\partial G}{\partial y} dy +\frac{\partial G}{\partial u} du +\frac{\partial G}{\partial v} dv = 0$
$dx = \frac{\partial X}{\partial u} du + \frac{\partial X}{\partial v} dv$
$dy = \frac{\partial Y}{\partial u} du + \frac{\partial Y}{\partial v} dv$

$dF = \left(\frac{\partial F}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial u} +\frac{\partial F}{\partial u} \right) du + \left(\frac{\partial F}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial v} +\frac{\partial F}{\partial v} \right) dv = 0$
$dG = \left(\frac{\partial G}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial u} +\frac{\partial G}{\partial u} \right) du + \left(\frac{\partial G}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial v} +\frac{\partial G}{\partial v} \right) dv = 0$

$\frac{\partial F}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial u} = -\frac{\partial F}{\partial u}$
$\frac{\partial G}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial u} = -\frac{\partial G}{\partial u}$
$\frac{\partial F}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial v} = -\frac{\partial F}{\partial v}$
$\frac{\partial G}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial v} = -\frac{\partial G}{\partial v}$

$\cfrac{\partial x}{\partial u} = \cfrac{\begin{vmatrix} -\cfrac{\partial F}{\partial u} & \cfrac{\partial F}{\partial y} \\ -\cfrac{\partial G}{\partial u} & \cfrac{\partial G}{\partial y}\end{vmatrix}}{\begin{vmatrix}\cfrac{\partial F}{\partial x} & \cfrac{\partial F}{\partial y} \\ \cfrac{\partial G}{\partial x} & \cfrac{\partial G}{\partial y}\end{vmatrix}}$

$\cfrac{\partial x}{\partial u} = - \cfrac{\left(\cfrac{\partial\left(F, G\right)}{\partial\left(y, u\right)}\right)}{\left(\cfrac{\partial\left(F, G\right)}{\partial\left(x, y\right)}\right)}$