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线性代数之范德蒙德行列式及克莱姆法则

程序员文章站 2022-06-20 21:59:42
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二级目录三级目录1 范德蒙德行列式D=∣11...11x1x2...xn−1xn(x1)2(x2)2...(xn−1)2(xn)2...............(x1)n−2(x2)n−2...(xn−1)n−2(xn)n−2(x1)n−1(x2)n−1...(xn−1)n−1(xn)n−1∣=∏1⩽j

这里写目录标题

手动反爬虫：原博地址

 知识梳理不易，请尊重劳动成果，文章仅发布在CSDN网站上，在其他网站看到该博文均属于未经作者授权的恶意爬取信息

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1 范德蒙德行列式

D=∣11...11x1x2...xn−1xn(x1)2(x2)2...(xn−1)2(xn)2...............(x1)n−2(x2)n−2...(xn−1)n−2(xn)n−2(x1)n−1(x2)n−1...(xn−1)n−1(xn)n−1∣=∏1⩽j<i⩽n(xi−xj)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})

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−x1-x_{1}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=∣11...11x1x2...xn−1xn(x1)2(x2)2...(xn−1)2(xn)2...............(x1)n−2(x2)n−2...(xn−1)n−2(xn)n−2(x1)n−1(x2)n−1...(xn−1)n−1(xn)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}

=   ∣11...110x2−x1...xn−1−x1xn−x10(x2)2−x1x2...(xn−1)2−x1xn−1(xn)2−x1xn...............0(x2)n−2−x1(x2)n−3...(xn−1)n−2−x1(xn−1)n−3(xn)n−2−x1(xn)n−30(x2)n−1−x1(x2)n−2...(xn−1)n−1−x1(xn−1)n−2(xn)n−1−x1(xn)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}

=   ∣x2−x1...xn−1−x1xn−x1(x2)2−x1x2...(xn−1)2−x1xn−1(xn)2−x1xn............(x2)n−2−x1(x2)n−3...(xn−1)n−2−x1(xn−1)n−3(xn)n−2−x1(xn)n−3(x2)n−1−x1(x2)n−2...(xn−1)n−1−x1(xn−1)n−2(xn)n−1−x1(xn)n−2∣ = \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}

=   (x2−x1)(x2−x1)...(xn−x1)∣1...11x2...xn−1xn(x2)2...(xn−1)2(xn)2............(x2)n−2...(xn−1)n−2(xn)n−2(x2)n−1...(xn−1)n−1(xn)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)∏2⩽j<i⩽n(xi−xj)   =   ∏1⩽j<i⩽n(xi−xj) = \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})

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

∣11111x1x2x3x4x5(x1)2(x2)2(x3)2(x4)2(x5)2(x1)3(x2)3(x3)3(x4)3(x5)3(x1)4(x2)4(x2)4(x4)4(x5)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}
=   (x2−x1)(x3−x1)(x4−x1)(x5−x1)(x3−x2)(x4−x2)(x5−x2)(x4−x3)(x5−x3)(x5−x4)= \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})

小示例

∣111115−13242519416125−12786462518116256∣\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}
=(−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

{x1+x2+x3=7x1−x2+5x3=6−x1+x2+6x3=9      ⇒      D=∣1111−15−116∣\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}

如果系数行列式D!=0D != 0,，那么未知数的解就为xj=DjDx_{j} =\frac{D_{j}}{D}，其中DjD_{j}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

D1=∣7116−15916∣      D2=∣171165−196∣      D3=∣1171−16−119∣      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{ }

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）
x1=D1D=2.41      x2=D2D=3.23      x3=D3D=1.36x_{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

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。DiD_{i}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

本文地址：https://blog.csdn.net/lys_828/article/details/107120820

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