# 克萊姆法則

（重定向自克莱姆法则

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

## 基本方程

${\displaystyle Ax=c\,\qquad \qquad \qquad \qquad \qquad \qquad (1)}$

${\displaystyle x_{i}={\det(A_{i}) \over \det(A)}}$  (1)

${\displaystyle x_{i}={\Delta _{i} \over \Delta }\,}$

## 抽象方程

${\displaystyle R}$ 為一個環，${\displaystyle A}$ 就是一個包含${\displaystyle R}$ 的系數的${\displaystyle n\times n}$ 矩陣。所以：

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

## 證明概要

${\displaystyle A=\left(u_{1},u_{2},\cdots ,u_{n}\right)}$

${\displaystyle x^{*}=(x_{1},x_{2},\cdots ,x_{n})^{T}}$ ，即

${\displaystyle Ax^{*}=\sum _{k=1}^{n}x_{k}u_{k}=c}$

{\displaystyle {\begin{aligned}\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{aligned}}}

${\displaystyle x_{i}={\frac {\Delta _{i}}{\Delta }}}$

## 例子

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

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

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

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

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

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

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

### 微分幾何上的應用

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

${\displaystyle dx}$ ${\displaystyle dy}$ 代入${\displaystyle dF}$ ${\displaystyle dG}$ ，可得出：

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

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

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

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