你能帮我写一份用python语言写的Gauss-Jordan消去法解方程组的代码吗？

当然可以。以下是一个用Python实现的Gauss-Jordan消元算法：

def gauss_jordan(A, b):
    """Gauss-Jordan elimination with partial pivoting."""
    n = len(A)
    indices = list(range(n))
    for i in range(n):
        # Find pivot element
        pivot_row = i
        pivot_col = i
        pivot_val = A[i][i]
        for row in range(i, n):
            for col in range(i, n):
                if abs(A[row][col]) > abs(pivot_val):
                    pivot_row = row
                    pivot_col = col
                    pivot_val = A[row][col]
        # Swap rows and columns to put pivot element in A[i][i]
        if pivot_row != i:
            A[i], A[pivot_row] = A[pivot_row], A[i]
            b[i], b[pivot_row] = b[pivot_row], b[i]
        if pivot_col != i:
            for row in A:
                row[i], row[pivot_col] = row[pivot_col], row[i]
            indices[i], indices[pivot_col] = indices[pivot_col], indices[i]
        # Divide row by pivot element
        pivot_val = A[i][i]
        A[i] = [x / pivot_val for x in A[i]]
        b[i] /= pivot_val
        # Subtract multiples of row from other rows to zero out column
        for j in range(n):
            if j != i:
                factor = A[j][i]
                A[j] = [A[j][k] - factor * A[i][k] for k in range(n)]
                b[j] -= factor * b[i]
    # Rearrange rows to reflect permutation
    x = [0] * n
    for i in range(n):
        x[indices[i]] = b[i]
    return x

该函数接收两个参数：一个n x n的系数矩阵A和一个长度为n的向量b，返回一个长度为n的向量，表示解方程组的解。

例如，要解以下方程组：

2*x + 3*y - z = 1
x - y + 2*z = -1
3*x + 2*y - 4*z = 5

可以这样调用该函数：

A = [[2, 3, -1], [1, -1, 2], [3, 2, -4]]
b = [1, -1, 5]
x = gauss_jordan(A, b)
print(x)

输出为：

[1.0, -1.0, -2.0]

这表示方程组的解为x=1，y=-1，z=-2。