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克莱默Cramer法则

克莱默法则

设有一个含有 $n$ 个末知数 $x_{1}, x_{2}, \dots, x_{n}, n$ 个线性方程的方程组

{\begin{matrix} a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} = b_{1}, \\ a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} = b_{2}, \\ \dots \\ a_{n 1} x_{1} + a_{n 2} x_{2} + \dots + a_{n n} x_{n} = b_{n}, \end{matrix} ⟷ A X = β

其中

A = (\begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{array}), x = (\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}), β = (\begin{matrix} b_{1} \\ b_{2} \\ ⋮ \\ b_{n} \end{matrix})

Cramer (克莱默) 法则

如果线性方程组 $A X = β$ 的系数行列式不等于零，即 $| A | \neq 0$ ，则方程组有唯一解：

x_{1} = \frac{D_{1}}{| A |}, x_{2} = \frac{D_{2}}{| A |}, \dots, x_{n} = \frac{D_{n}}{| A |},

其中 $D_{j} (j = 1, 2, \dots, n)$ 是把系数行列式的第 $j$ 列元素用 $β$ 的元素代替后得到的行列式.
证明因为 $| A | \neq 0$ ，所以 $A^{- 1}$ 存在. 令 $X = A^{- 1} β$ ，则有 $A X = A (A^{- 1} β) = β$ ，即 $X = A^{- 1} β$ 是线性方程组的解.
且由 $A^{- 1}$ 的唯一性可知，线性方程组的解是唯一的.
由求逆公式 $A^{- 1} = \frac{1}{| A |} A^{*}$ 可得 $X = A^{- 1} β = \frac{1}{| A |} A^{*} β$ ，

用克莱默法则求解线性方程组

{\begin{aligned} x_{1} - x_{2} - x_{3} & = - 1 \\ - 2 x_{1} + 2 x_{2} + x_{3} & = 1 \\ 2 x_{1} - x_{2} + 3 x_{3} & = 1 \end{aligned}

解：

| A | = | \begin{array}{ccc} 1 & - 1 & - 1 \\ - 2 & 2 & 1 \\ 2 & - 1 & 3 \end{array} | = | \begin{array}{ccc} 1 & - 1 & - 1 \\ 0 & 0 & - 1 \\ 0 & 1 & 5 \end{array} | = - | \begin{array}{ccc} 1 & - 1 & - 1 \\ 0 & 1 & 5 \\ 0 & 0 & - 1 \end{array} | = 1 \neq 0, D_{1} = | \begin{array}{ccc} - 1 & - 1 & - 1 \\ 1 & 2 & 1 \\ 1 & - 1 & 3 \end{array} | = | \begin{array}{ccc} - 1 & - 1 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array} | = - 2,

D_{2} = | \begin{array}{ccc} 1 & - 1 & - 1 \\ - 2 & 1 & 1 \\ 2 & 1 & 3 \end{array} | = | \begin{array}{ccc} 1 & - 1 & - 1 \\ 0 & - 1 & - 1 \\ 0 & 0 & 2 \end{array} | = - 2, D_{3} = | \begin{array}{ccc} 1 & - 1 & - 1 \\ - 2 & 2 & 1 \\ 2 & - 1 & 1 \end{array} | = | \begin{array}{ccc} 1 & - 1 & - 1 \\ 0 & 0 & - 1 \\ 0 & 1 & 3 \end{array} | = 1,

因此 $x_{1} = \frac{D_{1}}{| A |} = - 2, x_{2} = \frac{D_{2}}{| A |} = - 2, x_{3} = \frac{D_{3}}{| A |} = 1$ .

克莱姆法则的作用

1750 年, 瑞士的克莱姆发现了用行列式求解线性方程组的克莱姆 (Cramer) 法则。这个法则在表述上简洁自然, 思想深刻, 包含了对多重行列式的计算, 是对行列式与线性方程组之间关系的深刻理解。如果我们不能从几何上解释这个法则, 就难以领会向量、行列式和线性方程组之间的真正关系。

二阶克莱姆法则

对于二阶线性方程组: ${\begin{cases} a_{1} x + b_{1} y = c_{1} \\ a_{2} x + b_{2} y = c_{2} \end{cases}$ , 其克莱姆法则的解为

x = \frac{| \begin{array}{ll} c_{1} & b_{1} \\ c_{2} & b_{2} \end{array} |}{| \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} |}, y = \frac{| \begin{array}{ll} a_{1} & c_{1} \\ a_{2} & c_{2} \end{array} |}{| \begin{array}{ll} a_{1} & b_{1} \\ a_{1} & b_{2} \end{array} |}

三阶克莱姆法则

三阶线性方程组如下:

{\begin{cases} a_{1} x + b_{1} y + c_{1} z = d_{1} \\ a_{2} x + b_{2} y + c_{2} z = d_{2} \\ a_{3} x + b_{3} y + c_{3} z = d_{3} \end{cases}

其克莱姆法则的解为

x = \frac{| \begin{array}{lll} d_{1} & b_{1} & c_{1} \\ d_{2} & b_{2} & c_{2} \\ d_{3} & b_{3} & c_{3} \end{array} |}{| \begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array} |}, y = \frac{| \begin{array}{lll} a_{1} & d_{1} & c_{1} \\ a_{2} & d_{2} & c_{2} \\ a_{3} & d_{3} & c_{3} \end{array} |}{| \begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array} |}, z = \frac{| \begin{array}{lll} a_{1} & b_{1} & d_{1} \\ a_{2} & b_{2} & d_{2} \\ a_{3} & b_{3} & d_{3} \end{array} |}{| \begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array} |}

克莱姆法则的意义是可以用方程组的系数和常数项的行列式把方程组的解简洁地表达出来。但在实际工程应用中由于计算量较大, 常常采用高斯消元法来解大型的线性方程组。

例1 求方程$\left{ $\begin{array}{l} x_{1} + x_{2} + x_{3} + x_{4} = 1 2 x_{1} + 3 x_{2} + 4 x_{3} + 5 x_{4} = 1 4 x_{1} + 9 x_{2} + 16 x_{3} + 25 x_{4} = 1 8 x_{1} + 27 x_{2} + 64 x_{3} + 125 x_{4} = 1 \end{array}$ \right.$

解易知，其系数行列式是一个 4 阶 Vandermonde 行列式．

\begin{aligned} D & = | \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 2^{3} & 3^{3} & 4^{3} & 5^{3} \end{array} | = (3 - 2) (4 - 2) (5 - 2) (4 - 3) (5 - 3) (5 - 4) \\ = 12 \neq 0 . \end{aligned}

同理

\begin{aligned} D_{1} & = | \begin{array}{llll} 1 & 1 & 1 & 1 \\ 1 & 3 & 4 & 5 \\ 1^{2} & 3^{2} & 4^{2} & 5^{2} \\ 1^{3} & 3^{3} & 4^{3} & 5^{3} \end{array} | = (3 - 1) (4 - 1) (5 - 1) (4 - 3) (5 - 3) (5 - 4) \\ = 48, \\ D_{2} & = | \begin{array}{llll} 1 & 1 & 1 & 1 \\ 2 & 1 & 4 & 5 \\ 2^{2} & 1^{2} & 4^{2} & 5^{2} \\ 2^{3} & 1^{3} & 4^{3} & 5^{3} \end{array} | = (1 - 2) (4 - 2) (5 - 2) (4 - 1) (5 - 1) (5 - 4) \\ = - 72, \\ D_{3} & = | \begin{array}{llll} 1 & 1 & 1 & 1 \\ 2 & 3 & 1 & 5 \\ 2^{2} & 3^{2} & 1^{2} & 5^{2} \\ 2^{3} & 3^{3} & 1^{3} & 5^{3} \end{array} | = (3 - 2) (1 - 2) (5 - 2) (1 - 3) (5 - 3) (5 - 1) \\ = 48, \end{aligned}

\begin{aligned} \begin{aligned} D_{4} & = | \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 1 \\ 2^{2} & 3^{2} & 4^{2} & 1^{2} \\ 2^{3} & 3^{3} & 4^{3} & 1^{3} \end{array} | = (3 - 2) (4 - 2) (1 - 2) (4 - 3) (1 - 3) (1 - 4) \\ = - 12, \end{aligned} \\ 故 \\ x_{1} = \frac{D_{1}}{D} = 4, x_{2} = \frac{D_{2}}{D} = - 6, x_{3} = \frac{D_{3}}{D} = 4, x_{4} = \frac{D_{4}}{D} = - 1 . \end{aligned}

## 克莱默法则
设有一个含有 $n$ 个末知数 $x_1, x_2, \cdots, x_n, n$ 个线性方程的方程组

$$
\left\{\begin{array}{c}
a_{11} x_1+a_{12} x_2+\cdots+a_{1 n} x_n=b_1, \\
a_{21} x_1+a_{22} x_2+\cdots+a_{2 n} x_n=b_2, \\
\cdots \\
a_{n 1} x_1+a_{n 2} x_2+\cdots+a_{n n} x_n=b_n,
\end{array} \longleftrightarrow \boldsymbol{A X}=\boldsymbol{\beta}\right.
$$

其中

$$
\boldsymbol{A}=\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
a_{21} & a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right), \boldsymbol{x}=\left(\begin{array}{c}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{array}\right), \quad \boldsymbol{\beta}=\left(\begin{array}{c}
b_1 \\
b_2 \\
\vdots \\
b_n
\end{array}\right)
$$

### Cramer (克莱默) 法则
如果线性方程组 $A X=\beta$ 的系数行列式不等于零，即 $|A| \neq 0$ ，则方程组有唯一解：

$$
x_1=\frac{D_1}{|\boldsymbol{A}|}, x_2=\frac{D_2}{|\boldsymbol{A}|}, \cdots, x_n=\frac{D_n}{|\boldsymbol{A}|},
$$

其中 $D_j(j=1,2, \cdots, n)$ 是把系数行列式的第 $j$ 列元素用 $\beta$ 的元素代替后得到的行列式.
证明 因为 $|A| \neq 0$ ，所以 $A^{-1}$ 存在. 令 $X=A^{-1} \beta$ ，则有 $A X=A\left(A^{-1} \beta\right)=\beta$ ，即 $X=A^{-1} \beta$ 是线性方程组的解.
且由 $A^{-1}$ 的唯一性可知，线性方程组的解是唯一的.
由求逆公式 $A^{-1}=\frac{1}{|A|} A^{*}$ 可得 $X=A^{-1} \beta=\frac{1}{|A|} A^* \beta$ ，

### 用克莱默法则求解线性方程组

$$
\left\{\begin{aligned}
x_1-x_2-x_3 & =-1 \\
-2 x_1+2 x_2+x_3 & =1 \\
2 x_1-x_2+3 x_3 & =1
\end{aligned}\right.
$$

解：

$$
|\boldsymbol{A}|=\left|\begin{array}{ccc}
1 & -1 & -1 \\
-2 & 2 & 1 \\
2 & -1 & 3
\end{array}\right|=\left|\begin{array}{ccc}
1 & -1 & -1 \\
0 & 0 & -1 \\
0 & 1 & 5
\end{array}\right|=-\left|\begin{array}{ccc}
1 & -1 & -1 \\
0 & 1 & 5 \\
0 & 0 & -1
\end{array}\right|=1 \neq 0, \quad D_1=\left|\begin{array}{ccc}
-1 & -1 & -1 \\
1 & 2 & 1 \\
1 & -1 & 3
\end{array}\right|=\left|\begin{array}{ccc}
-1 & -1 & -1 \\
0 & 1 & 0 \\
0 & 0 & 2
\end{array}\right|=-2,
$$

$$
D_2=\left|\begin{array}{ccc}
1 & -1 & -1 \\
-2 & 1 & 1 \\
2 & 1 & 3
\end{array}\right|=\left|\begin{array}{ccc}
1 & -1 & -1 \\
0 & -1 & -1 \\
0 & 0 & 2
\end{array}\right|=-2, \quad D_3=\left|\begin{array}{ccc}
1 & -1 & -1 \\
-2 & 2 & 1 \\
2 & -1 & 1
\end{array}\right|=\left|\begin{array}{ccc}
1 & -1 & -1 \\
0 & 0 & -1 \\
0 & 1 & 3
\end{array}\right|=1,
$$

因此 $\quad x_1=\frac{D_1}{|\boldsymbol{A}|}=-2, \quad x_2=\frac{D_2}{|\boldsymbol{A}|}=-2, \quad x_3=\frac{D_3}{|\boldsymbol{A}|}=1$.

## 克莱姆法则的作用

1750 年, 瑞士的克莱姆发现了用行列式求解线性方程组的克莱姆 (Cramer) 法则。这个法则在表述上简洁自然, 思想深刻, 包含了对多重行列式的计算, 是对行列式与线性方程组之间关系的深刻理解。如果我们不能从几何上解释这个法则, 就难以领会向量、行列式和线性方程组之间的真正关系。
#### 二阶克莱姆法则 
对于二阶线性方程组: $\left\{\begin{array}{l}a_1 x+b_1 y=c_1 \\ a_2 x+b_2 y=c_2\end{array}\right.$, 其克莱姆法则的解为

$$
x=\frac{\left|\begin{array}{ll}
c_1 & b_1 \\
c_2 & b_2
\end{array}\right|}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|}, y=\frac{\left|\begin{array}{ll}
a_1 & c_1 \\
a_2 & c_2
\end{array}\right|}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_1 & b_2
\end{array}\right|}
$$

#### 三阶克莱姆法则

三阶线性方程组如下:

$$
\left\{\begin{array}{l}
a_1 x+b_1 y+c_1 z=d_1 \\
a_2 x+b_2 y+c_2 z=d_2 \\
a_3 x+b_3 y+c_3 z=d_3
\end{array}\right.
$$

其克莱姆法则的解为

$$
x=\frac{\left|\begin{array}{lll}
d_1 & b_1 & c_1 \\
d_2 & b_2 & c_2 \\
d_3 & b_3 & c_3
\end{array}\right|}{\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|}, y=\frac{\left|\begin{array}{lll}
a_1 & d_1 & c_1 \\
a_2 & d_2 & c_2 \\
a_3 & d_3 & c_3
\end{array}\right|}{\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|}, z=\frac{\left|\begin{array}{lll}
a_1 & b_1 & d_1 \\
a_2 & b_2 & d_2 \\
a_3 & b_3 & d_3
\end{array}\right|}{\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|}
$$

`例` 求方程$\left\{\begin{array}{l}
x_1+x_2+x_3+x_4=1 \\
2 x_1+3 x_2+4 x_3+5 x_4=1 \\
4 x_1+9 x_2+16 x_3+25 x_4=1 \\
8 x_1+27 x_2+64 x_3+125 x_4=1
\end{array}\right.$

解 易知，其系数行列式是一个 4 阶 Vandermonde 行列式．

$$
\begin{aligned}
D & =\left|\begin{array}{cccc}
1 & 1 & 1 & 1 \\
2 & 3 & 4 & 5 \\
2^2 & 3^2 & 4^2 & 5^2 \\
2^3 & 3^3 & 4^3 & 5^3
\end{array}\right|=(3-2)(4-2)(5-2)(4-3)(5-3)(5-4) \\
& =12 \neq 0 .
\end{aligned}
$$

同理

$$
\begin{aligned}
D_1 & =\left|\begin{array}{llll}
1 & 1 & 1 & 1 \\
1 & 3 & 4 & 5 \\
1^2 & 3^2 & 4^2 & 5^2 \\
1^3 & 3^3 & 4^3 & 5^3
\end{array}\right|=(3-1)(4-1)(5-1)(4-3)(5-3)(5-4) \\
& =48, \\
D_2 & =\left|\begin{array}{llll}
1 & 1 & 1 & 1 \\
2 & 1 & 4 & 5 \\
2^2 & 1^2 & 4^2 & 5^2 \\
2^3 & 1^3 & 4^3 & 5^3
\end{array}\right|=(1-2)(4-2)(5-2)(4-1)(5-1)(5-4) \\
& =-72, \\
& \\
D_3 & =\left|\begin{array}{llll}
1 & 1 & 1 & 1 \\
2 & 3 & 1 & 5 \\
2^2 & 3^2 & 1^2 & 5^2 \\
2^3 & 3^3 & 1^3 & 5^3
\end{array}\right|=(3-2)(1-2)(5-2)(1-3)(5-3)(5-1) \\
& =48,
\end{aligned}
$$

$$
\begin{aligned}
&\begin{aligned}
D_4 & =\left|\begin{array}{cccc}
1 & 1 & 1 & 1 \\
2 & 3 & 4 & 1 \\
2^2 & 3^2 & 4^2 & 1^2 \\
2^3 & 3^3 & 4^3 & 1^3
\end{array}\right|=(3-2)(4-2)(1-2)(4-3)(1-3)(1-4) \\
& =-12,
\end{aligned}\\
&\text { 故 }\\
&x_1=\frac{D_1}{D}=4, x_2=\frac{D_2}{D}=-6, x_3=\frac{D_3}{D}=4, x_4=\frac{D_4}{D}=-1 .
\end{aligned}
$$

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