“对 A * X = B” 矩阵的阶数不限

可通过修改 a 的值来改变A

a = np.array([[2, 1, 1], [3, 1, 2], [1, 2, 2]],dtype=float)

可通过修改 b 的值来修改B

b = np.array([[4],[6],[5]],dtype=float)

模块导入

import numpy as np

直接上代码

# 导入 numpy 模块
import numpy as np


# 行交换
def swap_row(matrix, i, j):
    m, n = matrix.shape
    if i >= m or j >= m:
        print('错误! : 行交换超出范围 ...')
    else:
        matrix[i],matrix[j] = matrix[j].copy(),matrix[i].copy()
    return matrix


# 变成阶梯矩阵
def matrix_change(matrix):
    m, n = matrix.shape
    main_factor = []
    main_col = main_row = 0
    while main_row < m and main_col < n:
        # 选择进行下一次主元查找的列
        main_row = len(main_factor)
        # 寻找列中非零的元素
        not_zeros = np.where(abs(matrix[main_row:,main_col]) > 0)[0]
        # 如果该列向下全部数据为零，则直接跳过列
        if len(not_zeros) == 0:
            main_col += 1
            continue
        else:
            # 将主元列号保存在列表中
            main_factor.append(main_col)
            # 将第一个非零行交换至最前
            if not_zeros[0] != [0]:
                matrix = swap_row(matrix,main_row,main_row+not_zeros[0])
            # 将该列主元下方所有元素变为零
            if main_row < m-1:
                for k in range(main_row+1,m):
                    a = float(matrix[k, main_col] / matrix[main_row, main_col])
                    matrix[k] = matrix[k] - matrix[main_row] * matrix[k, main_col] / matrix[main_row, main_col]
            main_col += 1
    return matrix,main_factor


# 回代求解
def back_solve(matrix, main_factor):
    # 判断是否有解
    if len(main_factor) == 0:
        print('主元错误,无主元！ ...')
        return None
    m, n = matrix.shape
    if main_factor[-1] == n - 1:
        print('无解！ ...')
        return None
    # 把所有的主元元素上方的元素变成0
    for i in range(len(main_factor) - 1, -1, -1):
        factor = matrix[i, main_factor[i]]
        matrix[i] = matrix[i] / float(factor)
        for j in range(i):
            times = matrix[j, main_factor[i]]
            matrix[j] = matrix[j] - float(times) * matrix[i]
    # 先看看结果对不对
    return matrix


# 结果打印
def print_result(matrix, main_factor):
    if matrix is None:
        print('阶梯矩阵为空！ ...')
        return
    m, n = matrix.shape
    result = [''] * (n - 1)
    main_factor = list(main_factor)
    for i in range(n - 1):
        # 如果不是主元列，则为自由变量
        if i not in main_factor:
            result[i] = 'x_' + str(i + 1) + '(free var)'
        # 否则是主元变量，从对应的行，将主元变量表示成非主元变量的线性组合
        else:
            # row_of_main表示该主元所在的行
            row_of_main = main_factor.index(i)
            result[i] = str(matrix[row_of_main, -1])
            for j in range(i + 1, n - 1):
                ratio = matrix[row_of_main, j]
                if ratio > 0:
                    result[i] = result[i] + '-' + str(ratio) + 'x_' + str(j + 1)
                if ratio < 0:
                    result[i] = result[i] + '+' + str(-ratio) + 'x_' + str(j + 1)
    print('方程的通解是:', )
    for i in range(n - 1):
        print('x_' + str(i + 1), '=', result[i])


# 得到简化的阶梯矩阵和主元列
def Handle(matrix_a, matrix_b):
    # 拼接成增广矩阵
    matrix_01 = np.hstack([matrix_a, matrix_b])
    print('增广矩阵为:')
    print(matrix_01)
    matrix_01, main_factor = matrix_change(matrix_01)
    print('阶梯矩阵为:')
    print(matrix_01)
    matrix_01 = back_solve(matrix_01, main_factor)
    print('方程的简化阶梯矩阵:')
    print(matrix_01)
    print('方程的主元列为：')
    print(main_factor)
    print_result(matrix_01, main_factor)
    return matrix_01, main_factor


if __name__ == '__main__':
    #a = np.array([[0, 1, 1], [0, 1, 0], [1, 0, 0]])
    a = np.array([[2, 1, 1], [3, 1, 2], [1, 2, 2]],dtype=float)
    b = np.array([[4],[6],[5]],dtype=float)
    Handle(a, b)
    print('*' * 20)

齐次线性方程组：

$$\begin{gathered} (1)只有零解：\iff 系数行列式D\ne 0 \\ 克莱姆法则：D\ne 0\Rightarrow 只有零解 \\ 极大线性无关向量组： \\ n个n维向量相关\iff 矩阵的行列式=0（并且同时交换一个向量组多个相同位置的分量不改变相关性） \end{gathered}$$
$$有非零解\iff R(A)<n,基础解系的个数为n-R(A)$$

非齐次线性方程组：

$$\begin{gathered} (1)有解：\iff R(A)=R(\bar{A}(增广矩阵)) \\ 唯一解\iff R(A)=R(\bar{A})=n \\ 无穷多解R(A)=R(\bar{A})<n \end{gathered}$$

用增广矩阵解方程组：

化简：
$$\left(\begin{array}{llcccc}1& 0& a& b& |& e\\ 0& 1& c& d& |& f\\ 0& 0& 0& 0& |& 0\end{array}\right)$$
$$\begin{gathered} 则： \\ {x}_1+a{x}_3+b{x}_4=e \\ {x}_2+c{x}_3+d{x}_4=f \end{gathered}$$

则特解为
$$\begin{gathered} \eta =\left(\begin{array}{l}e\\ f\\ 0\\ 0\end{array}\right) \\ 一般令自由变量为1 \\ {\xi }_1=\left(\begin{array}{l}a\\ c\\ 1\\ 0\end{array}\right){\xi }_2=\left(\begin{array}{l}b\\ d\\ 0\\ 1\end{array}\right) \end{gathered}$$
则原方程的解为：
$$x=\eta +k_1{\xi }_1+k_2{\xi }_2,(k_1,k_2为任意常数)$$

在矩阵的初等变换求逆中也用到了“增广的形式”

$$\left(\begin{array}{llc}A& |& E\end{array}\right)\stackrel{行变换}{⟶}\left(\begin{array}{llc}E& |& A^{-1}\end{array}\right)$$
$$\left(\begin{array}{l}A\\ -\\ E\end{array}\right)\stackrel{列变换}{⟶}\left(\begin{array}{l}E\\ -\\ A^{-1}\end{array}\right)$$