矩阵相加

难度：⭐⭐

1. 题目描述

题目描述

随机两个矩阵进行相加

2. 关键点

• 数学矩阵概念

3. 代码实现

点击显示代码

import random


def new_jz(row_len=2, col_len=3):
    """随机矩阵创建函数"""
    jz = []
    for i in range(row_len):
        row = [random.randint(-10, 10) for i in range(col_len)]
        jz.append(row)
    return jz


def jz_add(jz1, jz2):
    """矩阵相加"""
    row_len = len(jz1)
    col_len = len(jz1[0])
    jz3 = []
    for i in range(row_len):
        jz3.append([0]*col_len)
        for j in range(col_len):
            jz3[i][j] = jz1[i][j] + jz2[i][j]
    return jz3


jz1 = new_jz()
jz2 = new_jz()
jz3 = jz_add(jz1, jz2)

print(f'矩阵1为：{jz1}')
print(f'矩阵2为：{jz2}')
print(f'两者之和为：{jz3}')

4. 运行示例

矩阵1为：[[9, 8, -4], [-2, 4, -1]]
矩阵2为：[[4, -5, 1], [-8, -3, 6]]
两者之和为：[[13, 3, -3], [-10, 1, 5]]

5. 进阶思考

1. 矩阵相加函数中加入俩矩阵是否是同型矩阵