Python实现用初等行变换将矩阵化为最简行形式_python 初等行变换

import fractions

def row_echelon(A):
    """
    将矩阵A化为行阶梯矩阵
    """
    lead = 0
    rowCount = len(A)
    columnCount = len(A[0])
    for r in range(rowCount):
        if lead >= columnCount:
            return
        i = r
        while A[i][lead] == 0:
            i += 1
            if i == rowCount:
                i = r
                lead += 1
                if columnCount == lead:
                    return
        A[i], A[r] = A[r], A[i]
        lv = A[r][lead]
        A[r] = [fractions.Fraction(mrx, lv) for mrx in A[r]]
        for i in range(rowCount):
            if i != r:
                lv = A[i][lead]
                A[i] = [iv - lv*rv for rv,iv in zip(A[r],A[i])]
        lead += 1

# 定义一个3x3的矩阵
A = [
    [4,1,2],
    [13,-2,3],
    [0,3,2]
]


# 将矩阵化为行最简形式
row_echelon(A)

# 以分数形式打印结果
for row in A:
    print([str(fractions.Fraction(x).limit_denominator()) for x in row])
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