MIT线代笔记

逆序数：

　　进行冒泡排序，统计期间交换次数

import copy
e =[[1,3,1],
    [2,1,0],
    [2,0,1]]

#求逆序数/冒泡法
def inverse_number(e):
    a=copy.deepcopy(e)
    c = [0,0,0]
    for n in range(len(a)):
        for i in range(len(a[n])):
            for j in range(3 - i):
                if j == 0:
                    continue
                elif a[n][j] < a[n][j - 1]:
                    k = a[n][j]
                    a[n][j] = a[n][j - 1]
                    a[n][j - 1] = k
                    c[n]+=1
    return c

 

行列式计算：

#求余子式
def yuzishi( e ):
    k=[0 for i in range(len(e))]
    for i in range(len(e)):
        b = copy.deepcopy(e)
        for m in range(len(b)):
            for n in range(len(b)):
                if n == i:
                    b[m].pop(n)
        b.pop(0)
        k[i]=b
    return k

##求行列式/递归
def get_A( a ):
    sum=0
    if len(a)==2:
        return a[0][0]*a[1][1]-a[1][0]*a[0][1]
    else:
        ge=yuzishi(a)
        for s in range(len(ge)):
            sum+=(-1)**s * get_A(ge[s])*a[0][s]
    return sum

 

矩阵求逆：

#伴随矩阵
def accompany(e):
    #定义返回数组
    k = [[ 0 for j in range(len(e))] for i in range(len(e))]
    #求余子式
    for m1 in range(len(e)):
        for m2 in range(len(e)):
            b = copy.deepcopy(e)
            for m3 in range(len(b)): 
                for m4 in range(len(b)):
                    if m4==m2:
                        b[m3].pop(m4)
            b.pop(m1)
            answer=get_A(b)
            k[m1][m2]=answer
    #代数余子式
    for m1 in range(len(e)):
        for m2 in range(len(e)):
            if (m1+m2)%2!=0:
                 k[m1][m2]= -k[m1][m2]
    #转置
    for m1 in range(len(e)):
        for m2 in range(len(e)):
            if m1>m2:
                m=k[m1][m2]
                k[m1][m2]=k[m2][m1]
                k[m2][m1]=m
    return k
#逆矩阵
def inverse(e):
    lAl=get_A(e)
    if lAl==0:
        return
    for m1 in range(len(e)):
        for m2 in range(len(e)):
            e[m1][m2]/=lAl
    return e