(选做题) 求两个3*3矩阵的和、乘积并输出。
时间: 2024-02-23 16:03:22 浏览: 32
好的,以下是两个3*3矩阵的和、乘积的Python代码:
```python
# 定义两个3*3的矩阵
matrix1 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
matrix2 = [[9, 8, 7], [6, 5, 4], [3, 2, 1]]
# 矩阵求和
sum_matrix = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
for i in range(3):
for j in range(3):
sum_matrix[i][j] = matrix1[i][j] + matrix2[i][j]
print("矩阵求和结果:", sum_matrix)
# 矩阵求乘积
product_matrix = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
for i in range(3):
for j in range(3):
for k in range(3):
product_matrix[i][j] += matrix1[i][k] * matrix2[k][j]
print("矩阵求乘积结果:", product_matrix)
```
输出结果如下:
```
矩阵求和结果: [[10, 10, 10], [10, 10, 10], [10, 10, 10]]
矩阵求乘积结果: [[30, 24, 18], [84, 69, 54], [138, 114, 90]]
```
希望对你有所帮助!
相关问题
C语言编程求两个3*3矩阵的和、乘积并输出。
```c
#include <stdio.h>
int main() {
int a[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
int b[3][3] = {{9, 8, 7}, {6, 5, 4}, {3, 2, 1}};
int sum[3][3], product[3][3];
int i, j, k;
// 计算矩阵和
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
sum[i][j] = a[i][j] + b[i][j];
}
}
// 计算矩阵乘积
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
product[i][j] = 0;
for (k = 0; k < 3; k++) {
product[i][j] += a[i][k] * b[k][j];
}
}
}
// 输出矩阵和
printf("矩阵和:\n");
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("%d ", sum[i][j]);
}
printf("\n");
}
// 输出矩阵乘积
printf("矩阵乘积:\n");
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("%d ", product[i][j]);
}
printf("\n");
}
return 0;
}
```
使用c++编程求两个3*3矩阵的和
以下是使用C语言编写的求两个3*3矩阵的和的程序:
```c
#include <stdio.h>
int main()
{
int matrix1[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
int matrix2[3][3] = {{9, 8, 7}, {6, 5, 4}, {3, 2, 1}};
int result[3][3];
int i, j;
// 计算矩阵和
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
result[i][j] = matrix1[i][j] + matrix2[i][j];
}
}
// 输出结果
printf("矩阵1:\n");
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("%d ", matrix1[i][j]);
}
printf("\n");
}
printf("矩阵2:\n");
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("%d ", matrix2[i][j]);
}
printf("\n");
}
printf("矩阵和:\n");
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("%d ", result[i][j]);
}
printf("\n");
}
return 0;
}
```
输出结果:
```
矩阵1:
1 2 3
4 5 6
7 8 9
矩阵2:
9 8 7
6 5 4
3 2 1
矩阵和:
10 10 10
10 10 10
10 10 10
```