算法学习笔记1.1.3 矩阵的逆

任务

给一个矩阵，求它的逆。

说明

将原矩阵A和一个单位矩阵R做成大矩阵(A,E)，用初等行变换将大矩阵中的A变为E，则得到(E,A^-1)的形式。

接口

void inverse(vector<double> A[], vector<double> C[], int N)

复杂度：O(n^3)
输入：

• A 原矩阵
• C 逆矩阵
• N 矩阵的逆

代码

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

inline vector<double> operator * (vector<double> a, double b) {
    int N = a.size();
    vector<double> res(N, 0);
    for (int i = 0; i < N; i ++)
        res[i] = a[i] * b;
    return res;
}
inline vector<double> operator - (vector<double> a, vector<double> b) {
    int N = a.size();
    vector<double> res(N, 0);
    for (int i = 0; i < N; i ++)
        res[i] = a[i] - b[i];
    return res;
}
inline void inverse(vector<double> A[], vector<double> C[], int N) {
    for (int i = 0; i < N; i ++)
        C[i] = vector<double>(N, 0);
    for (int i = 0; i < N; i ++)
        C[i][i] = 1;
    for (int i = 0; i < N; i ++) {
        for (int j = 0; j < N; j ++)
            if (fabs(A[j][i]) > 0) {
                swap(A[i], A[j]);
                swap(C[i], C[j]);
                break;
            }
        C[i] = C[i] * (1 / A[i][i]);
        A[i] = A[i] * (1 / A[i][i]);
        for (int j = 0; j < N; j ++)
            if (j != i && fabs(A[j][i]) > 0) {
                C[j] = C[j] - C[i] * A[j][i];
                A[j] = A[j] - A[i] * A[j][i];
            }
    }
}

// main for test
int main() {
    vector<double> A[3];
    vector<double> C[3];
    for (int i = 0; i < 3; i ++) {
        A[i] = vector<double>(3, 0);
        A[i][i] = 3;
        C[i] = vector<double>(3, 0);
        C[i][i] = 1;
    }
    inverse(A, C, 3);
    cout << "result:" << endl;
    for (int i = 0; i < 3; i ++) {
        for (int j = 0; j < 3; j ++) {
            cout << "\t" << C[i][j];
        }
        cout << endl;
    }
    return 0;
}
/*
output:
result:
        0.333333        0       0
        0       0.333333        0
        0       0       0.333333

*/