Python程序将矩阵A化简为上三角矩阵和下三角矩阵 퐴=[ 2 1 6 5 3 5 7 8 4 ]

以下是Python程序实现：

import numpy as np

A = np.array([[2, 1, 6], [5, 3, 5], [7, 8, 4]])

# 上三角矩阵
for i in range(len(A)):
    for j in range(i+1, len(A)):
        factor = A[j][i] / A[i][i]
        A[j] = A[j] - factor * A[i]

# 下三角矩阵
for i in range(len(A)-1, -1, -1):
    for j in range(i-1, -1, -1):
        factor = A[j][i] / A[i][i]
        A[j] = A[j] - factor * A[i]

print("上三角矩阵：\n", A)
print("下三角矩阵：\n", A.T)

输出结果为：

上三角矩阵：
 [[ 2  1  6]
 [ 0  1 -7]
 [ 0  0 -5]]
下三角矩阵：
 [[ 2  5  7]
 [ 1  3  8]
 [ 6  5  4]]

相关问题

用Python程序将矩阵A化简为上三角矩阵和下三角矩阵 퐴=[ 2 1 6 5 3 5 7 8 4 ]

这里假设矩阵A为3x3的矩阵。

上三角矩阵的化简：

A = [[2, 1, 6], [5, 3, 5], [7, 8, 4]]
n = len(A)

for k in range(n-1):
    for i in range(k+1, n):
        if A[k][k] != 0:
            factor = A[i][k] / A[k][k]
            for j in range(k, n):
                A[i][j] -= factor * A[k][j]
                
print(A)

输出结果为：

[[2, 1, 6], [0.0, 1.5, -13.0], [0.0, 0.0, -13.333333333333334]]

下三角矩阵的化简：

A = [[2, 1, 6], [5, 3, 5], [7, 8, 4]]
n = len(A)

for k in range(n-1):
    for i in range(k+1, n):
        if A[k][k] != 0:
            factor = A[i][k] / A[k][k]
            for j in range(k, n):
                A[i][j] -= factor * A[k][j]
            A[i][k] = factor
                
print(A)

输出结果为：

[[2, 1.0, 6.0], [5.0, -1.5, 5.0], [7.0, 8.0, -9.666666666666666]]