克萊姆法則

克萊姆法則（Cramer's rule，或「克拉瑪公式」）是一個線性代數中的定理，用行列式來計算出線性等式組中的所有解。這個定理因加百列·克萊姆（1704年 - 1752年）的卓越使用而命名。在計算上，並非最有效率之法，所以在很多條等式的情況中沒有廣泛應用。不過，這定理在理論性方面十分有用。

基本方程

一個線性方程組可以用矩陣与向量的方程來表示：

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

其中的$A$$A$是一个$n\times n$$n\times n$的方塊矩陣，而向量 $x=(x_{1},x_{2},\cdots x_{n})^{T}$$x=( x_1, x_2, \cdots x_n )^T$ 是一个长度为n的列向量。$c=(c_{1},c_{2},\cdots c_{n})^{T}$$c=( c_1, c_2, \cdots c_n )^T$ 也一样。

克莱姆法则说明：如果$A$$A$是一个可逆矩陣（ $\det {A}\neq 0$$\det{A} \neq 0$ ），那么方程(1)有解 $x=(x_{1},x_{2},\cdots x_{n})^{T}$$x=( x_1, x_2, \cdots x_n )^T$，其中

$x_{i}={\det(A_{i}) \over \det(A)}$$x_i = { \det(A_i) \over \det(A)}$

$(1)\,$$(1)\,$

當中$A_{i}$$A_{i}$是被列向量$c$$c$取代了$A$$A$的第i列的列向量后得到的矩阵。為了方便，我們通常使用$\Delta$$\Delta$來表示$\det(A)$$\det(A)$，用$\Delta _{i}$$\Delta_i$來表示$\det(A_{i})$$\det(A_i)$。所以等式(1)可以寫成為：

$x_{i}={\Delta _{i} \over \Delta }\,$$x_i = { \Delta_i \over \Delta }\,$。

抽象方程

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

當中det(A)就是A的行列式，以及I就是單位矩陣。

證明概要

对于 ${\begin{smallmatrix}n\end{smallmatrix}}$${\begin{smallmatrix}n\end{smallmatrix}}$ 元线性方程组 $Ax=c$$Ax=c$

把系数矩阵 ${\begin{smallmatrix}A\end{smallmatrix}}$${\begin{smallmatrix}A\end{smallmatrix}}$ 表示成列向量的形式

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

由于系数矩阵可逆，故方程组一定有解$x^{*}=A^{-1}c$$x^{*}=A^{{-1}}c$.

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

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

考虑 ${\begin{smallmatrix}\Delta _{i}\end{smallmatrix}}$${\begin{smallmatrix}\Delta _{i}\end{smallmatrix}}$ 的值，利用行列式的線性和交替性質，有

${\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}}$${\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}}$

于是

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

例子

运用克萊姆法則可以很有效地解決以下方程组。

已知：

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

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

使用矩陣來表示時就是：

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

当矩阵可逆时，x和y可以從克萊姆法則中得出：

$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}$$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}$

以及

$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}$$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}$

用3×3矩陣的情況亦差不多。

已知：

$ax+by+cz={\color {red}j}\,$$ax + by + cz = {\color{red}j}\,$

$dx+ey+fz={\color {red}k}\,$$dx + ey + fz = {\color{red}k}\,$

$gx+hy+iz={\color {red}l}\,$$gx + hy + iz = {\color{red}l}\,$

當中的矩陣表示為：

${\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}}$$\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}$

当矩阵可逆时，可以求出x、y和z：

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

克萊姆法則在解決微分幾何的问题时十分有用。

先考慮兩條等式$F(x,y,u,v)=0\,$$F(x, y, u, v) = 0\,$和$G(x,y,u,v)=0\,$$G(x, y, u, v) = 0\,$。其中的u和v是需要考虑的变量，并且它们互不相关。我們可定義$x=X(u,v)\,$$x = X(u, v)\,$和$y=Y(u,v)\,$$y = Y(u, v)\,$。

找出一條等式適合$\partial x/\partial u$$\partial x/\partial u$是克萊姆法則的簡單應用。

首先，我們要計算F、G、x和y的導數：

$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$$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$$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$$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$$dy = \frac{\partial Y}{\partial u} du + \frac{\partial Y}{\partial v} dv$

將dx和dy代入dF和dG，可得出：

$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$$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$$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$

因為u和v互不相关，所以du和dv的系數都要等於0。所以等式中的系數可以被寫成：

${\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 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 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 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}}$$\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}$

現在用克萊姆法則就可得到：

${\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{\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}}$

用兩個雅可比矩陣來表示的方程：

${\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)}}$$\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)}$

用類似的方法就可以找到${\frac {\partial x}{\partial v}}$$\frac{\partial x}{\partial v}$、${\frac {\partial y}{\partial u}}$$\frac{\partial y}{\partial u}$以及${\frac {\partial y}{\partial v}}$$\frac{\partial y}{\partial v}$。

基本代數上的應用

克萊姆法則可以用來證明一些線性代數中的定理，當中的定理對環理論十分有用。

線性規劃上的應用

克萊姆法則可以用來證明一個線性規劃問題有一個基本整數的解。這樣使得線性規劃的問題更容易被解決。

外部链接

• Online calculator to solve a system of ecuations using the Cramer's Rule(不能用了)
• Systems Solver