矩阵化简

某师妹问了一题:

想着以后有用，罢了，还是写个代码吧。什么，你还不会Python? 速速补救 -> Python环境配置与入门建议（面向新手文）

引入一个库，可以用来分数计算。(默认直接算成小数，会有精度缺失…

from fractions import Fraction

创造一个支持分数运算的矩阵

def make_matrix(list_matrix):
    fra_matrix = []
    for i in range(len(list_matrix)):
        row = []
        for j in list_matrix[i]:
            if not isinstance(j, Fraction):
                row.append(Fraction(j))
            else:
                row.append(j)
        fra_matrix.append(row)
    return fra_matrix

展示矩阵

def show(matrix, information_str='no tips'):
    global cnt
    print(information_str)
    print('-' * len(matrix[0]) * 10)
    for row in matrix:
        for item in row:
            print('{:>10}'.format(str(item)), end='')
        print()
    print()
    cnt += 1 

交换两行 (行数下标从1开始, 即 m * n的矩阵, 行数下标从1 ~ m)

def change_row(matrix, row1_num, row2_num):
    matrix[row1_num - 1],  matrix[row2_num - 1] = matrix[row2_num - 1], matrix[row1_num - 1]

化简某一行 (行数下标从1开始, 即 m * n的矩阵, 行数下标从1 ~ m)

def simplify(matrix, row_num):
    row_num -= 1
    pivot = matrix[row_num][row_num]
    for j in range(len(matrix[0])):
        matrix[row_num][j] /= pivot

    for i in range(len(matrix)):
        if i != row_num:
            times = matrix[i][row_num]
            for j in range(len(matrix[0])):
                matrix[i][j] -= matrix[row_num][j] * times

from fractions import Fraction


# frac.numerator 整数/ frac.denominator分数
# 创造一个支持分数运算的矩阵
def make_matrix(list_matrix):
    fra_matrix = []
    for i in range(len(list_matrix)):
        row = []
        for j in list_matrix[i]:
            if not isinstance(j, Fraction):
                row.append(Fraction(j))
            else:
                row.append(j)
        fra_matrix.append(row)
    return fra_matrix


# 展示矩阵
def show(matrix, information_str='no tips'):
    global cnt
    print(information_str)
    print('-' * len(matrix[0]) * 10)
    for row in matrix:
        for item in row:
            print('{:>10}'.format(str(item)), end='')
        print()
    print()
    cnt += 1


# 交换两行 (行数下标从1开始, 即 m * n的矩阵, 行数下标从1 ~ m)
def change_row(matrix, row1_num, row2_num):
    matrix[row1_num - 1],  matrix[row2_num - 1] = matrix[row2_num - 1], matrix[row1_num - 1]


# 化简某一行 (行数下标从1开始, 即 m * n的矩阵, 行数下标从1 ~ m)
def simplify(matrix, row_num):
    row_num -= 1
    pivot = matrix[row_num][row_num]
    for j in range(len(matrix[0])):
        matrix[row_num][j] /= pivot

    for i in range(len(matrix)):
        if i != row_num:
            times = matrix[i][row_num]
            for j in range(len(matrix[0])):
                matrix[i][j] -= matrix[row_num][j] * times



if __name__ == '__main__':
    # ...

# ... 上述代码
if __name__ == '__main__':
    a = [[2, 0, 1, 1], [1, 0, -1, 2], [3, 4, 0, 3]]
    m = make_matrix(a)
    cnt = 0  # 步骤计数器
    show(a, 'process {}. orgin:'.format(cnt))

    simplify(m, 1)
    show(m, 'process {}. simplify row {}'.format(cnt, 1))

    change_row(m, 2, 3)
    show(m, 'process {}. change row {} and row {}'.format(cnt, 2, 3))

    simplify(m, 2)
    show(m, 'process {}. simplify row {}'.format(cnt, 2))

    simplify(m, 3)
    show(m, 'process {}. simplify row {}'.format(cnt, 3))

第二题

# ... 上述代码
if __name__ == '__main__':
	a = [[2, 1, 3, -7, -3], [3, -2, 3, 0, 8], [2, 7, 3, 4, -3], [1, 2, -2, 0, 4]]
    m = make_matrix(a)
    cnt = 0  # 步骤计数器
    show(a, 'process {}. orgin:'.format(cnt))

    simplify(m, 1)
    show(m, 'process {}. simplify row {}'.format(cnt, 1))

    simplify(m, 2)
    show(m, 'process {}. simplify row {}'.format(cnt, 2))

    simplify(m, 3)
    show(m, 'process {}. simplify row {}'.format(cnt, 3))

    simplify(m, 4)
    show(m, 'process {}. simplify row {}'.format(cnt, 4))

若一开始矩阵有分数，可参考如下用法, 其他步骤与上述例子相同。
[ 1 2 1 5 2 3 4 ] (3) left[ begin{matrix} 1 & 2 & frac{1}{5} \ 2 & 3 & 4 end{matrix} right] tag{3} [12​23​51​4​](3)

a = [[1, 2, Fraction(1, 5)], [2, 3, 4]]