顺序 Gauss 消元法和列主元 Gauss 消元法求解python

以下是使用 Python 实现顺序 Gauss 消元法和列主元 Gauss 消元法的代码示例：

import numpy as np

# 顺序 Gauss 消元法
def gaussian_elimination(A, b):
    n = len(b)
    for i in range(n):
        # 找到主元
        for j in range(i+1, n):
            if abs(A[i][i]) < abs(A[j][i]):
                A[i], A[j] = A[j], A[i]
                b[i], b[j] = b[j], b[i]
        # 消元
        for j in range(i+1, n):
            ratio = A[j][i] / A[i][i]
            for k in range(i, n):
                A[j][k] -= ratio * A[i][k]
            b[j] -= ratio * b[i]
    # 回代
    x = np.zeros(n)
    for i in range(n-1, -1, -1):
        x[i] = (b[i] - np.dot(A[i][i+1:], x[i+1:])) / A[i][i]
    return x

# 列主元 Gauss 消元法
def gaussian_elimination_with_pivoting(A, b):
    n = len(b)
    for i in range(n):
        # 找到主元
        max_idx = i
        for j in range(i+1, n):
            if abs(A[j][i]) > abs(A[max_idx][i]):
                max_idx = j
        # 交换行
        A[i], A[max_idx] = A[max_idx], A[i]
        b[i], b[max_idx] = b[max_idx], b[i]
        # 消元
        for j in range(i+1, n):
            ratio = A[j][i] / A[i][i]
            for k in range(i, n):
                A[j][k] -= ratio * A[i][k]
            b[j] -= ratio * b[i]
    # 回代
    x = np.zeros(n)
    for i in range(n-1, -1, -1):
        x[i] = (b[i] - np.dot(A[i][i+1:], x[i+1:])) / A[i][i]
    return x

其中，`A` 是系数矩阵，`b` 是常数向量。函数 `gaussian_elimination` 实现顺序 Gauss 消元法，函数 `gaussian_elimination_with_pivoting` 实现列主元 Gauss 消元法。两个函数的返回值均为解向量 `x`。