Python计算行列式

from itertools import combinations,permutations#检测#生成一个排序函数#计算逆序数#计算行列式Matrix = [[1,5,3],[7,4,3],[2,4,232]]#检测def test(Matrix):#检测是否是一个方阵for i in Matrix:if len(i)==len(Matrix):passfor ii in i:if type(ii)==type(101):passelse:try:b = 3/0except:print("请检查您输入的矩阵")print("不可靠的结果")breakelse:try:b = 3/0except:print("请检查您输入的矩阵")print("不可靠的结果")break#输出行列式阶数和n!sum = 1for i in range(len(Matrix)+1):sum *= ireturn len(Matrix),sumdef InversionNumber(l):sum = 0l0 = []for i in range(len(l)):if l[i] < l[i-1] and i-1 >= 0:l0.append(l[i-1])sum += 1for k in range(i-1):if l[i] < l[k]:sum += 1return (-1)**sumdef Permute(n):#返回一个列表,包含整个序列,最后一位表明(-1)^εl = []for k in range(1,n+1):l.append(k)L = list(permutations(l, len(l)))for i in L:L[L.index(i)]=list(i)for i in L:i.append(InversionNumber(i))return Ldef main(Matrix):l,S = test(Matrix)L = Permute(l)sum = 0for i in L:sum1 = 1for ii in range(len(Matrix)):sum1 *= Matrix[ii][i[ii]-1]sum1 *= i[-1]sum += sum1return sumif __name__ == "__main__":det = main(Matrix) 【Python计算行列式】