def submatrix(A,i,j):#矩阵A第i行第j列元素的余矩阵p=len(A)#矩阵的行数q=len(A[0])#矩阵的列数C=[[A[x][y] for y in range(q) if y!=j] for x in range(p) if x!=i]#列表推导式return C
def det(A):#按第一行展开递归求矩阵的行列式p=len(A)#矩阵的行数q=len(A[0])#矩阵的列数if(p==1 and q==1):return A[0][0]else:value=0for j in range(q):value+=((-1)**(j+2))*A[0][j]*det(submatrix(A,0,j))return value
def INVERSE(A):p=len(A)#矩阵的行数q=len(A[0])#矩阵的列数C=copy.deepcopy(A)d=det(A)print(d)for i in range(p):for j in range(q):C[i][j]=((-1)**(i+j+2))*det(submatrix(A,j,i))C[i][j]=C[i][j]/dprint(C)#不使用numpy实现矩阵操作
def ADD(A,B):p=len(A)#矩阵的行数q=len(A[0])#矩阵的列数C=copy.deepcopy(A)for i in range(p):for j in range(q):C[i][j]=A[i][j]+B[i][j]print(C)
def SUBTRACT(A,B):p=len(A)#矩阵的行数q=len(A[0])#矩阵的列数C=copy.deepcopy(A)for i in range(p):for j in range(q):C[i][j]=A[i][j]-B[i][j]print(C)
def SCALA_MULTIPLY(A,num):p=len(A)#矩阵的行数q=len(A[0])#矩阵的列数C=copy.deepcopy(A)for i in range(p):for j in range(q):C[i][j]=A[i][j]*numprint(C)
def MULTIPLY(A,B):p=len(A)#矩阵A的行数q=len(A[0])#矩阵A的列数=矩阵B的行数r=len(B[0])#矩阵B的列数C=[[0 for j in range(r)] for i in range(p)]for i in range(p):for j in range(r):for k in range(q):C[i][j]+=A[i][k]*B[k][j]print(C)
def IDENTITY(dim):#生成对角矩阵C=[[0 for j in range(dim)] for i in range(dim)]for i in range(dim):C[i][i]=1print(C)
def TRANSPOSE(A):p=len(A)#矩阵的行数q=len(A[0])#矩阵的列数C=copy.deepcopy(A)for i in range(p):for j in range(q):C[i][j]=A[j][i]print(C)本文链接：https://my.lmcjl.com/post/13763.html
展开阅读全文
4 评论