一个矩阵的伴随矩阵不用现成的库（numpy）怎么求，首先看一下公式：
def remove_line(index,arr):yy=[]for i in range(len(arr)):hh=[]for j in range(len(arr)):if(i==index[0] or j==index[1]):passelse:hh.append(arr[i][j])if(hh!=[]):yy.append(hh)# yy.remove([])return yydd=[[0,4,3,5],[2,3,2,6],[3,4,5,7],[3,4,5,7]]for i in range(len(dd)):for j in range(len(dd)):tm_list=remove_line(tuple([i,j]),dd)print((i,j),tm_list)
的求法就是这样
怎么求代数余子式呢，就是求矩阵的行列式
def getInversion(numlist):count = 0for i in range(1,len(numlist)):subscript = numlist[i]for j in range(i):if subscript < numlist[j]:count += 1return countdef remove_line(index,arr):yy=[]for i in range(len(arr)):hh=[]for j in range(len(arr)):if(i==index[0] or j==index[1]):passelse:hh.append(arr[i][j])if(hh!=[]):yy.append(hh)# yy.remove([])return yydef permutation(dd,ilist,jlist,index):# global D# D = 0for i in range(index,len(jlist)):if index == len(jlist)-1:term = 1for ii in range(len(ilist)):i = ilist[ii]j = jlist[ii]term *= dd[i][j]if getInversion(jlist) % 2 == 0:Matrix.global_d_value += termelse:Matrix.global_d_value -= termreturntmp = jlist[index]jlist[index] = jlist[i]jlist[i] = tmppermutation(dd,ilist,jlist,index+1)tmp = jlist[index]jlist[index] = jlist[i]jlist[i] = tmpclass Matrix:"""docstring for Matrix"""global_d_value=0dd=[[1,2,3],[4,5,6],[7,8,9]]for i in range(len(dd)):for j in range(len(dd)):tm_list=remove_line(tuple([i,j]),dd)ilist=[i for i in range(len(tm_list))]jlist=[i for i in range(len(tm_list))]# i,j位置的余子式print((i,j),tm_list)Matrix.global_d_value=0permutation(tm_list,ilist,jlist,0)# i,j位置的代数余子式print((i,j),Matrix.global_d_value)
结果：
(0, 0) [[5, 6], [8, 9]](0, 0) -3(0, 1) [[4, 6], [7, 9]](0, 1) -6(0, 2) [[4, 5], [7, 8]](0, 2) -3(1, 0) [[2, 3], [8, 9]](1, 0) -6(1, 1) [[1, 3], [7, 9]](1, 1) -12(1, 2) [[1, 2], [7, 8]](1, 2) -6(2, 0) [[2, 3], [5, 6]](2, 0) -3(2, 1) [[1, 3], [4, 6]](2, 1) -6(2, 2) [[1, 2], [4, 5]](2, 2) -3[Finished in 0.1s]
有了这个就可以用伴随矩阵的方法求逆矩阵了
