输入5×5矩阵a,完成下列要求:←(1) 输出矩阵a←(2) 将第2行和第5行元素对调后,输出新的矩阵a← (3)用对角线上的各元素分别去除各元素所在行,输出新的矩阵a,代码Python
时间: 2023-12-03 08:46:23 浏览: 298
以下是Python代码实现:
```python
# 输入5x5矩阵a
a = []
for i in range(5):
row = input(f"请输入第{i+1}行元素,用空格隔开:").split()
a.append(row)
# (1) 输出矩阵a
print("矩阵a为:")
for row in a:
print(row)
# (2) 将第2行和第5行元素对调后,输出新的矩阵a
a[1], a[4] = a[4], a[1]
print("交换后的矩阵a为:")
for row in a:
print(row)
# (3) 用对角线上的各元素分别去除各元素所在行,输出新的矩阵a
for i in range(5):
diagonal_element = a[i][i]
a[i] = [int(a[i][j])/int(diagonal_element) for j in range(5)]
print("对角线元素去除后的矩阵a为:")
for row in a:
print(row)
```
注意:此代码假设用户输入的都是合法的数字,没有进行输入验证。
相关问题
输入5*5矩阵a,完成下列要求: ①、输出矩阵a ②、将第2行和第5行元素对调后,输出新的矩阵a ③、输出矩阵a的转置矩阵
这里是一份Python代码,实现了上述要求:
```python
# 定义矩阵a
a = [[1,2,3,4,5],
[6,7,8,9,10],
[11,12,13,14,15],
[16,17,18,19,20],
[21,22,23,24,25]]
# ①、输出矩阵a
print("矩阵a:")
for row in a:
print(row)
# ②、将第2行和第5行元素对调后,输出新的矩阵a
a[1], a[4] = a[4], a[1]
print("交换后的矩阵a:")
for row in a:
print(row)
# ③、输出矩阵a的转置矩阵
transpose_a = [[a[j][i] for j in range(len(a))] for i in range(len(a[0]))]
print("矩阵a的转置矩阵:")
for row in transpose_a:
print(row)
```
输出结果为:
```
矩阵a:
[1, 2, 3, 4, 5]
[6, 7, 8, 9, 10]
[11, 12, 13, 14, 15]
[16, 17, 18, 19, 20]
[21, 22, 23, 24, 25]
交换后的矩阵a:
[1, 2, 3, 4, 5]
[21, 22, 23, 24, 25]
[11, 12, 13, 14, 15]
[16, 17, 18, 19, 20]
[6, 7, 8, 9, 10]
矩阵a的转置矩阵:
[1, 21, 11, 16, 6]
[2, 22, 12, 17, 7]
[3, 23, 13, 18, 8]
[4, 24, 14, 19, 9]
[5, 25, 15, 20, 10]
```
5 X 5 的矩阵a,完成下列要求:输出矩阵a将第2行和第5行元素对调后,再重新输出a
以下是代码实现:
```python
# 初始化矩阵a
a = [
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
[16, 17, 18, 19, 20],
[21, 22, 23, 24, 25]
]
# 输出原始矩阵a
print("原始矩阵a:")
for row in a:
print(row)
# 将第2行和第5行元素对调
a[1], a[4] = a[4], a[1]
# 输出交换后的矩阵a
print("交换后的矩阵a:")
for row in a:
print(row)
```
输出结果如下:
```
原始矩阵a:
[1, 2, 3, 4, 5]
[6, 7, 8, 9, 10]
[11, 12, 13, 14, 15]
[16, 17, 18, 19, 20]
[21, 22, 23, 24, 25]
交换后的矩阵a:
[1, 2, 3, 4, 5]
[21, 22, 23, 24, 25]
[11, 12, 13, 14, 15]
[16, 17, 18, 19, 20]
[6, 7, 8, 9, 10]
```
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