def solve(A, b):
m, n = A.shape
# augmentedA = [A|b]
augmentedA = np.zeros((m, n + 1), float)
augmentedA[:, :n] = A
augmentedA[:, n] = b
R, rankAb, nre = LA.rref(augmentedA)
# compute the rank of A
rankA = 0
i = 0
j = 0
while (i < m and j < n):
pivot = R[i, j]
if (pivot == 0):
j += 1
else:
rankA += 1
i += 1
j += 1
print(rankA)
print(rankAb)
# rankA == rankAb == n : only solution
# rankA == rankAb < n : infinite solutions
# rankAb > rankA : no solution
if (rankA == rankAb):
if (rankA == n):
x = np.array(R[:n, n])
return x
else: # rankA<n
print('infinite solutions')
else: # rankAb>rankA
print('no solution')
def inv(M):
# check dimension of M
m, n = M.shape
if (m != n):
print('It is not a square matrix, no inverse define !')
else:
augmentedM = np.zeros((n, 2 * n), float)
augmentedM[:, :n] = M[:, :]
for i in range(n):
augmentedM[i, n + i] = 1
# reduced row echelon form
R, rank, nre = LA.rref(augmentedM)
# check if M is full rank or not
fullRank = 1
for i in range(n):
if (R[i, i] == 0):
fullRank = 0
break
if (fullRank == 0):
print('The matrix is singular. Try pseudo-inverse!')
else:
inverse = np.array(R[:, n:])
return inverse
def solve(A,b):
m,n = A.shape
# augmentedA = [A|b]
augmentedA = np.zeros((m,n+1),float)
augmentedA[:,:n] = A
augmentedA[:,n] = b
R, rankAb, nre = LA.rref(augmentedA)
# compute the rank of A
rankA = 0
i = 0
j = 0
while (i<m and j<n):
pivot = R[i,j]
if(pivot==0):
j += 1
else:
rankA += 1
i += 1
j += 1
print(rankA)
print(rankAb)
# rankA == rankAb == n : only solution
# rankA == rankAb < n : infinite solutions
# rankAb > rankA : no solution
if (rankA == rankAb):
if (rankA==n):
x=np.array(R[:n,n])
return x
else: # rankA<n
print('infinite solutions')
else: # rankAb>rankA
print('no solution')
def inv(M):
# check dimension of M
m,n = M.shape
if (m!=n):
print('It is not a square matrix, no inverse define !')
else :
augmentedM = np.zeros((n,2*n),float)
augmentedM[:,:n] = M[:,:]
for i in range(n):
augmentedM[i,n+i]=1
# reduced row echelon form
R, rank, nre = LA.rref(augmentedM)
# check if M is full rank or not
fullRank=1;
for i in range(n):
if(R[i,i]==0):
fullRank=0
break;
if (fullRank==0):
print('The matrix is singular. Try pseudo-inverse!')
else:
inverse = np.array(R[:,n:])
return inverse
# -*- coding: utf-8 -*- """ Created on Sat Oct 15 2015 Modified on Fri Oct 15 2015 @author: david """ import LA import numpy as np #A=np.array([[1,2,1],[4,5,2],[4,8,3]]) A=np.array([[1,2],[4,5],[7,8]]) #A=np.array([[0,2,3],[0,5,6],[0,8,9]]) #A=np.array([[1,2,3],[3,5,6]]) print(A) U=LA.upperTri(A) R=LA.rref(A) print(U) print(R) print(LA.det(A)) print(LA.inv(A)) b=np.array([3,6,9]) print(LA.solve(A,b))
# -*- coding: utf-8 -*- """ Created on Sat Oct 15 2015 Modified on Fri Oct 15 2015 @author: david """ import LA import numpy as np # A=np.array([[1,2,1],[4,5,2],[4,8,3]]) A = np.array([[1, 2], [4, 5], [7, 8]]) # A=np.array([[0,2,3],[0,5,6],[0,8,9]]) # A=np.array([[1,2,3],[3,5,6]]) print(A) U = LA.upperTri(A) R = LA.rref(A) print(U) print(R) print(LA.det(A)) print(LA.inv(A)) b = np.array([3, 6, 9]) print(LA.solve(A, b))
