【线性代数（6）】范德蒙德行列式及克莱姆法则

朱雀 2023-02-22 13:50 342阅读 0赞

### 这里写目录标题 ###

*  1 范德蒙德行列式
 *   *  简单证明
     *  5阶行列范德蒙德展开
     *  小示例
 *  2 克莱姆法则

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# 1 范德蒙德行列式 #

D = ∣ 1 1 . . . 1 1 x 1 x 2 . . . x n − 1 x n ( x 1 ) 2 ( x 2 ) 2 . . . ( x n − 1 ) 2 ( x n ) 2 . . . . . . . . . . . . . . . ( x 1 ) n − 2 ( x 2 ) n − 2 . . . ( x n − 1 ) n − 2 ( x n ) n − 2 ( x 1 ) n − 1 ( x 2 ) n − 1 . . . ( x n − 1 ) n − 1 ( x n ) n − 1 ∣ = ∏ 1 ⩽ j < i ⩽ n ( x i − x j ) D = \\begin\{vmatrix\}1&1 &...& 1& 1\\\\x\_\{1\}&x\_\{2\} &... & x\_\{n-1\}&x\_\{n\} \\\\(x\_\{1\})^\{2\}&(x\_\{2\})^\{2\} &... & (x\_\{n-1\})^\{2\}&(x\_\{n\})^\{2\}\\\\... & ...&... &...&... \\\\ (x\_\{1\})^\{n-2\}&(x\_\{2\})^\{n-2\} &... & (x\_\{n-1\})^\{n-2\}&(x\_\{n\})^\{n-2\}\\\\(x\_\{1\})^\{n-1\}&(x\_\{2\})^\{n-1\} &... & (x\_\{n-1\})^\{n-1\}&(x\_\{n\})^\{n-1\}\\end\{vmatrix\} = \\prod\_\{1\\leqslant j <i \\leqslant n \}(x\_\{i\} - x\_\{j\}) D=∣∣∣∣∣∣∣∣∣∣∣∣1x1(x1)2...(x1)n−2(x1)n−11x2(x2)2...(x2)n−2(x2)n−1..................1xn−1(xn−1)2...(xn−1)n−2(xn−1)n−11xn(xn)2...(xn)n−2(xn)n−1∣∣∣∣∣∣∣∣∣∣∣∣=1⩽j<i⩽n∏(xi−xj)

## 简单证明 ##

①当n=2时，显然成立

②当n>2时，采用由下往上相邻两行的上一行乘以 − x 1 -x\_\{1\} −x1加到下一行，将第一列全部化为0

③这时候就变成了加边法后的结果了，直接可以去掉“加的边”，此时行列式的阶数就由原来的n阶变成n-1阶了

④然后发现每一列都存在着公因式，因此可以进行提取因式

⑤最后就发现提取的公因式正是j=1时候的表达式相乘，而后面的是在j>1时候的表达式，因此就得证

D = ∣ 1 1 . . . 1 1 x 1 x 2 . . . x n − 1 x n ( x 1 ) 2 ( x 2 ) 2 . . . ( x n − 1 ) 2 ( x n ) 2 . . . . . . . . . . . . . . . ( x 1 ) n − 2 ( x 2 ) n − 2 . . . ( x n − 1 ) n − 2 ( x n ) n − 2 ( x 1 ) n − 1 ( x 2 ) n − 1 . . . ( x n − 1 ) n − 1 ( x n ) n − 1 ∣ D = \\begin\{vmatrix\}1&1 &...& 1& 1\\\\x\_\{1\}&x\_\{2\} &... & x\_\{n-1\}&x\_\{n\} \\\\(x\_\{1\})^\{2\}&(x\_\{2\})^\{2\} &... & (x\_\{n-1\})^\{2\}&(x\_\{n\})^\{2\}\\\\... & ...&... &...&... \\\\ (x\_\{1\})^\{n-2\}&(x\_\{2\})^\{n-2\} &... & (x\_\{n-1\})^\{n-2\}&(x\_\{n\})^\{n-2\}\\\\(x\_\{1\})^\{n-1\}&(x\_\{2\})^\{n-1\} &... & (x\_\{n-1\})^\{n-1\}&(x\_\{n\})^\{n-1\}\\end\{vmatrix\} D=∣∣∣∣∣∣∣∣∣∣∣∣1x1(x1)2...(x1)n−2(x1)n−11x2(x2)2...(x2)n−2(x2)n−1..................1xn−1(xn−1)2...(xn−1)n−2(xn−1)n−11xn(xn)2...(xn)n−2(xn)n−1∣∣∣∣∣∣∣∣∣∣∣∣

=     ∣ 1 1 . . . 1 1 0 x 2 − x 1 . . . x n − 1 − x 1 x n − x 1 0 ( x 2 ) 2 − x 1 x 2 . . . ( x n − 1 ) 2 − x 1 x n − 1 ( x n ) 2 − x 1 x n . . . . . . . . . . . . . . . 0 ( x 2 ) n − 2 − x 1 ( x 2 ) n − 3 . . . ( x n − 1 ) n − 2 − x 1 ( x n − 1 ) n − 3 ( x n ) n − 2 − x 1 ( x n ) n − 3 0 ( x 2 ) n − 1 − x 1 ( x 2 ) n − 2 . . . ( x n − 1 ) n − 1 − x 1 ( x n − 1 ) n − 2 ( x n ) n − 1 − x 1 ( x n ) n − 2 ∣ = \\text\{ \} \\text\{ \} \\text\{ \}\\begin\{vmatrix\}1&1 &...& 1& 1\\\\0&x\_\{2\} -x\_\{1\}&... & x\_\{n-1\}-x\_\{1\}&x\_\{n\}-x\_\{1\} \\\\0&(x\_\{2\})^\{2\} -x\_\{1\}x\_\{2\}&... & (x\_\{n-1\})^\{2\} -x\_\{1\}x\_\{n-1\}&(x\_\{n\})^\{2\}-x\_\{1\}x\_\{n\}\\\\... & ...&... &...&... \\\\ 0&(x\_\{2\})^\{n-2\}-x\_\{1\}(x\_\{2\})^\{n-3\} &... & (x\_\{n-1\})^\{n-2\}-x\_\{1\}(x\_\{n-1\})^\{n-3\}&(x\_\{n\})^\{n-2\}-x\_\{1\}(x\_\{n\})^\{n-3\}\\\\0&(x\_\{2\})^\{n-1\} -x\_\{1\}(x\_\{2\})^\{n-2\}&... & (x\_\{n-1\})^\{n-1\}-x\_\{1\}(x\_\{n-1\})^\{n-2\}&(x\_\{n\})^\{n-1\}-x\_\{1\}(x\_\{n\})^\{n-2\}\\end\{vmatrix\} =   ∣∣∣∣∣∣∣∣∣∣∣∣100...001x2−x1(x2)2−x1x2...(x2)n−2−x1(x2)n−3(x2)n−1−x1(x2)n−2..................1xn−1−x1(xn−1)2−x1xn−1...(xn−1)n−2−x1(xn−1)n−3(xn−1)n−1−x1(xn−1)n−21xn−x1(xn)2−x1xn...(xn)n−2−x1(xn)n−3(xn)n−1−x1(xn)n−2∣∣∣∣∣∣∣∣∣∣∣∣

=     ∣ x 2 − x 1 . . . x n − 1 − x 1 x n − x 1 ( x 2 ) 2 − x 1 x 2 . . . ( x n − 1 ) 2 − x 1 x n − 1 ( x n ) 2 − x 1 x n . . . . . . . . . . . . ( x 2 ) n − 2 − x 1 ( x 2 ) n − 3 . . . ( x n − 1 ) n − 2 − x 1 ( x n − 1 ) n − 3 ( x n ) n − 2 − x 1 ( x n ) n − 3 ( x 2 ) n − 1 − x 1 ( x 2 ) n − 2 . . . ( x n − 1 ) n − 1 − x 1 ( x n − 1 ) n − 2 ( x n ) n − 1 − x 1 ( x n ) n − 2 ∣ n − 1 = \\text\{ \} \\text\{ \} \\text\{ \}\\begin\{vmatrix\}\\\\x\_\{2\} -x\_\{1\}&... & x\_\{n-1\}-x\_\{1\}&x\_\{n\}-x\_\{1\} \\\\(x\_\{2\})^\{2\} -x\_\{1\}x\_\{2\}&... & (x\_\{n-1\})^\{2\} -x\_\{1\}x\_\{n-1\}&(x\_\{n\})^\{2\}-x\_\{1\}x\_\{n\}\\\\ ...&... &...&... \\\\ (x\_\{2\})^\{n-2\}-x\_\{1\}(x\_\{2\})^\{n-3\} &... & (x\_\{n-1\})^\{n-2\}-x\_\{1\}(x\_\{n-1\})^\{n-3\}&(x\_\{n\})^\{n-2\}-x\_\{1\}(x\_\{n\})^\{n-3\}\\\\(x\_\{2\})^\{n-1\} -x\_\{1\}(x\_\{2\})^\{n-2\}&... & (x\_\{n-1\})^\{n-1\}-x\_\{1\}(x\_\{n-1\})^\{n-2\}&(x\_\{n\})^\{n-1\}-x\_\{1\}(x\_\{n\})^\{n-2\}\\end\{vmatrix\}\_\{n-1\} =   ∣∣∣∣∣∣∣∣∣∣x2−x1(x2)2−x1x2...(x2)n−2−x1(x2)n−3(x2)n−1−x1(x2)n−2...............xn−1−x1(xn−1)2−x1xn−1...(xn−1)n−2−x1(xn−1)n−3(xn−1)n−1−x1(xn−1)n−2xn−x1(xn)2−x1xn...(xn)n−2−x1(xn)n−3(xn)n−1−x1(xn)n−2∣∣∣∣∣∣∣∣∣∣n−1

=     ( x 2 − x 1 ) ( x 2 − x 1 ) . . . ( x n − x 1 ) ∣ 1 . . . 1 1 x 2 . . . x n − 1 x n ( x 2 ) 2 . . . ( x n − 1 ) 2 ( x n ) 2 . . . . . . . . . . . . ( x 2 ) n − 2 . . . ( x n − 1 ) n − 2 ( x n ) n − 2 ( x 2 ) n − 1 . . . ( x n − 1 ) n − 1 ( x n ) n − 1 ∣ n − 1 = \\text\{ \} \\text\{ \} \\text\{ \}(x\_\{2\} -x\_\{1\})(x\_\{2\} -x\_\{1\})...(x\_\{n\} -x\_\{1\})\\begin\{vmatrix\}1 &...& 1& 1\\\\x\_\{2\} &... & x\_\{n-1\}&x\_\{n\} \\\\(x\_\{2\})^\{2\} &... & (x\_\{n-1\})^\{2\}&(x\_\{n\})^\{2\}\\\\...&... &...&... \\\\ (x\_\{2\})^\{n-2\} &... & (x\_\{n-1\})^\{n-2\}&(x\_\{n\})^\{n-2\}\\\\(x\_\{2\})^\{n-1\} &... & (x\_\{n-1\})^\{n-1\}&(x\_\{n\})^\{n-1\}\\end\{vmatrix\}\_\{n-1\} =   (x2−x1)(x2−x1)...(xn−x1)∣∣∣∣∣∣∣∣∣∣∣∣1x2(x2)2...(x2)n−2(x2)n−1..................1xn−1(xn−1)2...(xn−1)n−2(xn−1)n−11xn(xn)2...(xn)n−2(xn)n−1∣∣∣∣∣∣∣∣∣∣∣∣n−1

=     ( x 2 − x 1 ) ( x 2 − x 1 ) . . . ( x n − x 1 ) ∏ 2 ⩽ j < i ⩽ n ( x i − x j )     =     ∏ 1 ⩽ j < i ⩽ n ( x i − x j ) = \\text\{ \} \\text\{ \} \\text\{ \}(x\_\{2\} -x\_\{1\})(x\_\{2\} -x\_\{1\})...(x\_\{n\} -x\_\{1\}) \\prod\_\{2\\leqslant j <i \\leqslant n \}(x\_\{i\} - x\_\{j\}) \\text\{ \} \\text\{ \} \\text\{ \}= \\text\{ \} \\text\{ \} \\text\{ \} \\prod\_\{1\\leqslant j <i \\leqslant n \}(x\_\{i\} - x\_\{j\}) =   (x2−x1)(x2−x1)...(xn−x1)2⩽j<i⩽n∏(xi−xj)   =   1⩽j<i⩽n∏(xi−xj)

## 5阶行列范德蒙德展开 ##

①当j=1时，i = 2,3,4,5

②当j=2时，i = 3,4,5

③当j=3时，i = 4,5

④当j=4时，i = 5

⑤最后展开一共十项

∣ 1 1 1 1 1 x 1 x 2 x 3 x 4 x 5 ( x 1 ) 2 ( x 2 ) 2 ( x 3 ) 2 ( x 4 ) 2 ( x 5 ) 2 ( x 1 ) 3 ( x 2 ) 3 ( x 3 ) 3 ( x 4 ) 3 ( x 5 ) 3 ( x 1 ) 4 ( x 2 ) 4 ( x 2 ) 4 ( x 4 ) 4 ( x 5 ) 4 ∣ \\begin\{vmatrix\}1&1 &1& 1& 1\\\\x\_\{1\}&x\_\{2\} &x\_\{3\}& x\_\{4\}&x\_\{5\} \\\\(x\_\{1\})^\{2\}&(x\_\{2\})^\{2\} &(x\_\{3\})^\{2\}& (x\_\{4\})^\{2\}&(x\_\{5\})^\{2\}\\\\ (x\_\{1\})^\{3\}&(x\_\{2\})^\{3\} &(x\_\{3\})^\{3\} & (x\_\{4\})^\{3\}&(x\_\{5\})^\{3\}\\\\(x\_\{1\})^\{4\}&(x\_\{2\})^\{4\} &(x\_\{2\})^\{4\} & (x\_\{4\})^\{4\}&(x\_\{5\})^\{4\}\\end\{vmatrix\} ∣∣∣∣∣∣∣∣∣∣1x1(x1)2(x1)3(x1)41x2(x2)2(x2)3(x2)41x3(x3)2(x3)3(x2)41x4(x4)2(x4)3(x4)41x5(x5)2(x5)3(x5)4∣∣∣∣∣∣∣∣∣∣  
 =     ( x 2 − x 1 ) ( x 3 − x 1 ) ( x 4 − x 1 ) ( x 5 − x 1 ) ( x 3 − x 2 ) ( x 4 − x 2 ) ( x 5 − x 2 ) ( x 4 − x 3 ) ( x 5 − x 3 ) ( x 5 − x 4 ) = \\text\{ \} \\text\{ \} \\text\{ \} (x\_\{2\}-x\_\{1\})(x\_\{3\}-x\_\{1\})(x\_\{4\}-x\_\{1\})(x\_\{5\}-x\_\{1\})(x\_\{3\}-x\_\{2\})(x\_\{4\}-x\_\{2\})(x\_\{5\}-x\_\{2\})(x\_\{4\}-x\_\{3\})(x\_\{5\}-x\_\{3\})(x\_\{5\}-x\_\{4\}) =   (x2−x1)(x3−x1)(x4−x1)(x5−x1)(x3−x2)(x4−x2)(x5−x2)(x4−x3)(x5−x3)(x5−x4)

## 小示例 ##

∣ 1 1 1 1 1 5 − 1 3 2 4 25 1 9 4 16 125 − 1 27 8 64 625 1 81 16 256 ∣ \\begin\{vmatrix\}1&1 &1& 1& 1\\\\5&-1 &3& 2&4 \\\\25&1 &9& 4&16\\\\125&-1&27&8&64\\\\ 625&1&81 &16&256\\end\{vmatrix\} ∣∣∣∣∣∣∣∣∣∣15251256251−11−111392781124816141664256∣∣∣∣∣∣∣∣∣∣  
 = ( − 1 − 5 ) ( 3 − 5 ) ( 2 − 5 ) ( 4 − 5 ) ( 3 + 1 ) ( 2 + 1 ) ( 4 + 1 ) ( 2 − 3 ) ( 4 − 3 ) ( 4 − 2 ) = − 4320 =(-1-5)(3-5)(2-5)(4-5)(3+1)(2+1)(4+1)(2-3)(4-3)(4-2)=-4320 =(−1−5)(3−5)(2−5)(4−5)(3\+1)(2\+1)(4\+1)(2−3)(4−3)(4−2)=−4320  
python代码实现

import numpy as np
    from numpy.linalg import  *
    mydet=np.array([[1,1,1,1,1],[5,-1,3,2,4],[25,1,9,4,16],[125,-1,27,8,64],[625,1,81,16,256]])
    print(det(mydet))

→ 输出的结果为：(验证无误)

-4320.000000000022

# 2 克莱姆法则 #

注意**克莱姆法则使用的前提条件：方程的个数等于未知数的个数**

系数行列式：就是把方程组里面未知数的系数拿出来组成的行列式，比如下面方程组的系数行列式如下

\{ x 1 + x 2 + x 3 = 7 x 1 − x 2 + 5 x 3 = 6 − x 1 + x 2 + 6 x 3 = 9        ⇒        D = ∣ 1 1 1 1 − 1 5 − 1 1 6 ∣ \\begin\{cases\} x\_\{1\}+x\_\{2\} + x\_\{3\} =7 \\\\ x\_\{1\}-x\_\{2\}+5x\_\{3\}=6\\\\ -x\_\{1\} +x\_\{2\} +6x\_\{3\} =9 \\end\{cases\} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\Rightarrow \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} D = \\begin\{vmatrix\}1&1 &1\\\\1&-1 &5\\\\-1&1 &6\\end\{vmatrix\} ⎩⎪⎨⎪⎧x1\+x2\+x3=7x1−x2\+5x3=6−x1\+x2\+6x3=9      ⇒      D=∣∣∣∣∣∣11−11−11156∣∣∣∣∣∣

如果系数行列式 D ! = 0 D != 0 D!=0,，那么未知数的解就为 x j = D j D x\_\{j\} =\\frac\{D\_\{j\}\}\{D\} xj=DDj，其中 D j D\_\{j\} Dj是将未知数所在的列元素换成方程组等号右侧的值，然后其他列的元素就是原系数行列式元素不变，注意观察每列的变化

D 1 = ∣ 7 1 1 6 − 1 5 9 1 6 ∣        D 2 = ∣ 1 7 1 1 6 5 − 1 9 6 ∣        D 3 = ∣ 1 1 7 1 − 1 6 − 1 1 9 ∣        D\_\{1\} = \\begin\{vmatrix\}7&1 &1\\\\6&-1 &5\\\\9&1 &6\\end\{vmatrix\} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \}D\_\{2\} = \\begin\{vmatrix\}1&7 &1\\\\1&6 &5\\\\-1&9&6\\end\{vmatrix\} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \}D\_\{3\} = \\begin\{vmatrix\}1&1 &7\\\\1&-1 &6\\\\-1&1&9\\end\{vmatrix\} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} D1=∣∣∣∣∣∣7691−11156∣∣∣∣∣∣      D2=∣∣∣∣∣∣11−1769156∣∣∣∣∣∣      D3=∣∣∣∣∣∣11−11−11769∣∣∣∣∣∣

故可以算出未知数的解：（很明显发现D在分母上，所以要求系数行列式不为0）  
 x 1 = D 1 D = 2.41        x 2 = D 2 D = 3.23        x 3 = D 3 D = 1.36 x\_\{1\} = \\frac\{D\_\{1\}\}\{D\}=2.41 \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \}x\_\{2\} = \\frac\{D\_\{2\}\}\{D\}= 3.23\\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \} \\text\{ \}x\_\{3\} = \\frac\{D\_\{3\}\}\{D\}=1.36 x1=DD1=2.41      x2=DD2=3.23      x3=DD3=1.36

综上，克莱姆法则的使用中：①方程的个数与未知数个数相等；②系数行列式的值不为0；③计算量是非常大的，可借助编程实现

import numpy as np
    
    d = np.reshape([1,1,1,1,-1,5,-1,1,6],(3,3))
    d1 = np.reshape([7,1,1,6,-1,5,9,1,6],(3,3))
    d2 = np.reshape([1,7,1,1,6,5,-1,9,6],(3,3))
    d3 = np.reshape([1,1,7,1,-1,6,-1,1,9],(3,3))
    
    x1 = np.linalg.det(d1)/np.linalg.det(d)
    x2 = np.linalg.det(d2)/np.linalg.det(d)
    x3 = np.linalg.det(d3)/np.linalg.det(d)
    print(x1,x2,x3)
    #2.409090909090908 3.2272727272727266 1.3636363636363635

**定理1： 如果多元方程组的右侧全为0（也称齐次方程组），至少有零解**，因为直接将方程组最后的结果替换到相应的列。 D i D\_\{i\} Di行列式的值均为0，也就是分子为0

**定理2：对于齐次方程组，且系数行列式不为0，那么方程组只有零解**

**定理3：齐次方程组（方程个数等于未知数个数）有非零解**     ⟺    D = 0 \\iff D = 0 ⟺D=0

[Link 1]: https://blog.csdn.net/lys_828/article/details/107120820