用Python实现三阶矩阵的求逆？ 求逆矩阵怎么用python源代码实现
python\u9006\u77e9\u9635\u600e\u4e48\u6c42\u5230\u5e95\u5e94\u8be5\u600e\u4e48\u6837\u53bb\u6c42\u9006\u77e9\u9635\u624d\u597d\u5462\uff1f
\u52a0\u4e0a\u5934\u6587\u4ef6
from numpy import *
\u77e9\u9635\u6709\u51e0\u4e2a\u7279\u6709\u7684\u5c5e\u6027\uff1a
(a) .T \uff0d\uff0d \u8fd4\u56de\u81ea\u8eab\u7684\u8f6c\u7f6e
(b) .H \uff0d\uff0d \u8fd4\u56de\u81ea\u8eab\u7684\u5171\u8f6d\u8f6c\u7f6e
(c) .I \uff0d\uff0d \u8fd4\u56de\u81ea\u8eab\u7684\u9006\u77e9\u9635
(d) .A \uff0d\uff0d \u8fd4\u56de\u81ea\u8eab\u6570\u636e\u76842\u7ef4\u6570\u7ec4\u7684\u4e00\u4e2a\u89c6\u56fe
你好，下面是一个对应的三阶矩阵求逆的代码
import warnings
warnings.filterwarnings("ignore")
matrix1 = [
[1,2,0,0],
[3,4,0,0],
[0,0,4,1],
[0,0,3,2],
]
matrix2 = [
[1,0,-1,2,1],
[3,2,-3,5,-3],
[2,2,1,4,-2],
[0,4,3,3,1],
[1,0,8,-11,4],
]
matrix3 = [
[1,0,-1,2,1,0,2],
[1,2,-1,3,1,-1,4],
[2,2,1,6,2,1,6],
[-1,4,1,4,0,0,0],
[4,0,-1,21,9,9,9],
[2,4,4,12,5,6,11],
[7,-1,-4,22,7,8,18],
]
def step0(m):
n = len(m)
l = []
for i in range(0,n):
l.append([])
for j in range(0,n):
if i == j:
l[i].append(1)
else:
l[i].append(0)
return l
def step1(m):
n = len(m)
"""交换操作记录数组 swap"""
swap = []
l = []
for i in range(0,n):
swap.append(i)
l.append([])
for j in range(0,n):
l[i].append(0)
"""对每一列进行操作"""
for i in range(0,n):
max_row = m[i][i]
row = i
for j in range(i,n):
if m[j][i] >= max_row:
max_row = m[j][i]
#global row
row = j
swap[i] = row
"""交换"""
if row != i:
for j in range(0,n):
m[i][j],m[row][j] = m[row][j],m[i][j]
"""消元"""
for j in range(i+1,n):
if m[j][i] != 0:
l[j][i] = m[j][i] / m[i][i]
for k in range(0,n):
m[j][k] = m[j][k] - (l[j][i] * m[i][k])
return (swap,m,l)
def step2(m):
n = len(m)
long = len(m)-1
l = []
for i in range(0,n):
l.append([])
for j in range(0,n):
l[i].append(0)
for i in range(0,n-1):
for j in range(0,long-i):
if m[long-i-j-1][long-i] != 0 and m[long-i][long-i] != 0:
l[long-i-j-1][long-i] = m[long-i-j-1][long-i] / m[long-i][long-i]
for k in range(0,n):
m[long-i-j-1][k] = m[long-i-j-1][k] - l[long-i-j-1][long-i] * m[long-i][k]
return (m,l)
def step3(m):
n = len(m)
l = []
for i in range(0,n):
l.append(m[i][i])
return l
def gauss(matrix):
n = len(matrix)
new = step0(matrix)
(swap,matrix1,l1) = step1(matrix)
(matrix2,l2) = step2(matrix1)
l3 = step3(matrix2)
for i in range(0,n):
if swap[i] != i:
new[i],new[swap[i]] = new[swap[i]],new[i]
for j in range(i+1,n):
for k in range(0,n):
if l1[j][i] != 0:
new[j][k] = new[j][k] - l1[j][i] * new[i][k]
for i in range(0,n-1):
for j in range(0,n-i-1):
if l2[n-1-i-j-1][n-1-i] != 0:
for k in range(0,n):
new[n-1-i-j-1][k] = new[n-1-i-j-1][k] - l2[n-1-i-j-1][n-i-1] * new[n-1-i][k]
for i in range(0,n):
for j in range(0,n):
new[i][j] = new[i][j] / l3[i]
return new
x1 = gauss(matrix1)
x2 = gauss(matrix2)
x3 = gauss(matrix3)
print (x1)
print (x2)
print (x3)
假设一个三阶矩阵
a = [ [1 0 0]，
[0 2 0]，
[0 0 3] ]
a的逆矩阵为np.linalg.inv(a)
如果求不出来证明没有逆矩阵
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