【线代&NumPy】第五章 – Vector课后练习 | 伴随矩阵求逆 | Cramer公式求联立方程 | 简述并提供代码
? 例1:
import numpy as np
def getMinorMatrix(A,i,j): # 删除矩阵A的i行和j列,并创建矩阵
n = len(A)
M = np.zeros((n-1, n-1))
for a in range(0,n-1):
k = a if (a < i) else a+1
for b in range(0, n-1):
l = b if (b < j) else b+1
M[a, b] = A[k, l]
return M
def determinant(M): # 计算
if len(M) == 2: # 2x2
return M[0,0]*M[1,1]-M[0,1]*M[1,0]
detVal = 0
for c in range(len(M)):
detVal += ((-1)**c)*M[0,c]*determinant(getMinorMatrix(M,0,c))
return detVal
A = np.array([[-4, 0, 2, -1, 0], [1, 3, -3, -1, 4], [2, 0, 1, 3, 0],
[-2, 1, -3, -1, 5], [1, -5, 1, 0, 5]])
print("A = ", A)
print("det(A) = ", determinant(A))
? 运行结果:
A = [[-4 0 2 -1 0]
[ 1 3 -3 -1 4]
[ 2 0 1 3 0]
[-2 1 -3 -1 5]
[ 1 -5 1 0 5]]
det(A) = -997.0
? 例2:
import numpy as np
def cofactor(A, i, j): # 辅助因子
(n,m) = A.shape
M = np.zeros((n-1, m-1))
for a in range(0, n-1):
k = a if (a < i) else a+1
for b in range(0, m-1):
l = b if (b < j) else b+1
M[a,b] = A[k, l]
return (-1)**(i+j)*np.linalg.det(M)
def inverseByAdjointMatrix(A): # 利用伴随矩阵计算A的逆矩阵
detA = np.linalg.det(A) # A的矩阵式计算
(n,m) = A.shape
adjA = np.zeros((n, m))
for i in range(0,n): # 生成伴随矩阵
for j in range(0, m):
adjA[j,i] = cofactor(A, i, j)
if detA != 0.0:
return (1./detA) * adjA
else:
return 0
A = np.array([[-4, 0, 2, -1, 0], [1, 3, -3, -1, 4], [2, 0, 1, 3, 0],
[-2, 1, -3, -1, 5], [1, -5, 1, 0, 5]])
print("A = ", A)
Ainv = inverseByAdjointMatrix(A)
print("A inverse = ", Ainv)
? 运行结果:
A = [[-4 0 2 -1 0]
[ 1 3 -3 -1 4]
[ 2 0 1 3 0]
[-2 1 -3 -1 5]
[ 1 -5 1 0 5]]
A inverse = [[-0.07321966 0.2106319 -0.03610832 -0.24573721 0.0772317 ]
[ 0.16950853 0.26579739 0.09729188 -0.14343029 -0.06920762]
[ 0.32397192 0.30090271 0.09127382 -0.35105316 0.11033099]
[-0.05917753 -0.24072217 0.32698094 0.28084253 -0.08826479]
[ 0.11935807 0.16349047 0.08625878 -0.02407222 0.09327984]]
import numpy as np
def solveByCramer(A, B): # # 利用Cramer公式求联立线性方程AX=B的解
X = np.zeros(len(B))
C = np.copy(A)
for i in range(0, len(B)):
for j in range(0, len(B)):
C[j,i] = B[j]
if i>0:
C[j,i-1] = A[j,i-1]
X[i] = np.linalg.det(C)/np.linalg.det(A)
return X
# AX = B의 해
A = np.array([[2,-1,5,1], [3,2,2,-6], [1,3,3,-1], [5,-2,-3,3]])
B = np.array([[-3], [-32], [-47], [49]])
X = solveByCramer(A, B)
print("A = ", A)
print("B = ", B)
print("X = ", X)
? 运行结果:
A = [[ 2 -1 5 1]
[ 3 2 2 -6]
[ 1 3 3 -1]
[ 5 -2 -3 3]]
B = [[ -3]
[-32]
[-47]
[ 49]]
X = [ 2. -12. -4. 1.]
参考文献
Introduction to Linear Algebra, International 4 th Edition by Gilbert Strang, Wellesley Cambridge Press.
百度百科[EB/OL]. []. https://baike.baidu.com/.
本篇完。