克莱姆法则

一个线性方程组可以用矩阵与向量的方程来表示：

其中的${\displaystyle A}$ 是一个${\displaystyle n\times n}$ 的方块矩阵，而向量 ${\displaystyle x=(x_{1},x_{2},\cdots x_{n})^{T}}$  是一个长度为n的行向量(中国大陆为列向量）。${\displaystyle c=(c_{1},c_{2},\cdots c_{n})^{T}}$  也一样。

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

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

当中${\displaystyle x_{i}}$ 是列向量${\displaystyle x}$ 的第i行（行向量与列向量不一样，解释默认列向量）

当中${\displaystyle A_{i}}$ 是列向量${\displaystyle c}$ 取代了${\displaystyle A}$ 的第i列后得到的矩阵。为了方便，我们通常使用${\displaystyle \Delta }$ 来表示${\displaystyle \det(A)}$ ，用${\displaystyle \Delta _{i}}$ 来表示${\displaystyle \det(A_{i})}$ 。所以等式(1)可以写成为：

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

设${\displaystyle R}$ 为一个环，${\displaystyle A}$ 就是一个包含${\displaystyle R}$ 的系数的${\displaystyle n\times n}$ 矩阵。所以：

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

当中${\displaystyle \det(A)}$ 就是${\displaystyle A}$ 的行列式，以及${\displaystyle I}$ 就是单位矩阵。

对于${\displaystyle n}$ 元线性方程组 ${\displaystyle Ax=c}$ 

把系数矩阵 ${\displaystyle {\begin{smallmatrix}A\end{smallmatrix}}}$  表示成行向量的形式

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

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

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

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

考虑${\displaystyle \Delta _{i}}$ 的值，利用行列式的线性和交替性质，有

${\displaystyle {\begin{aligned}\Delta _{i}&=det\left(\cdots ,u_{i-1},c,u_{i+1},\cdots \right)\\&=det\left(\cdots ,u_{i-1},\sum _{k=1}^{n}x_{k}u_{k},u_{i+1},\cdots \right)\\&=\sum _{k=1}^{n}x_{k}\cdot det\left(\cdots ,u_{i-1},u_{k},u_{i+1},\cdots \right)\\&=x_{i}\cdot det\left(\cdots ,u_{i-1},u_{i},u_{i+1},\cdots \right)\\&=x_{i}\Delta \end{aligned}}}$ 

于是

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

运用克莱姆法则可以很有效地解决以下方程组。

已知：

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

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

使用矩阵来表示时就是：

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

当矩阵可逆时，x和y可以从克莱姆法则中得出：

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

以及

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

用3×3矩阵的情况亦差不多。

已知：

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

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

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

当中的矩阵表示为：

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

当矩阵可逆时，可以求出x、y和z：

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

微分几何上的应用

克莱姆法则在解决微分几何的问题时十分有用。

先考虑两条等式${\displaystyle F(x,y,u,v)=0\,}$ 和${\displaystyle G(x,y,u,v)=0\,}$ 。其中的u和v是需要考虑的变量，并且它们互不相关。我们可定义${\displaystyle x=X(u,v)\,}$ 和${\displaystyle y=Y(u,v)\,}$ 。

找出一条等式适合${\displaystyle \partial x/\partial u}$ 是克莱姆法则的简单应用。

首先，我们要计算${\displaystyle F}$ 、${\displaystyle G}$ 、${\displaystyle x}$ 和${\displaystyle y}$ 的导数：

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

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

${\displaystyle dx={\frac {\partial X}{\partial u}}du+{\frac {\partial X}{\partial v}}dv}$ 

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

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

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

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

因为${\displaystyle u}$ 和${\displaystyle v}$ 互不相关，所以${\displaystyle du}$ 和${\displaystyle dv}$ 的系数都要等于0。所以等式中的系数可以被写成：

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

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

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

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

现在用克莱姆法则就可得到：

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

用两个雅可比矩阵来表示的方程：

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

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

基本代数上的应用

克莱姆法则可以用来证明一些线性代数中的定理，当中的定理对环理论十分有用。

线性规划上的应用

克莱姆法则可以用来证明一个线性规划问题有一个基本整数的解。这样使得线性规划的问题更容易被解决。

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