doolittle直接分解法计算矩阵A的逆矩阵 c++

以下是使用Doolittle直接分解法计算矩阵A的逆矩阵的C++代码示例：

#include <iostream>
#include <vector>

using namespace std;

// Doolittle直接分解法求解矩阵的逆
vector<vector<double>> inverse(vector<vector<double>> A) {
    int n = A.size();
    vector<vector<double>> L(n, vector<double>(n, 0));
    vector<vector<double>> U(n, vector<double>(n, 0));
    vector<vector<double>> invA(n, vector<double>(n, 0));

    // Doolittle直接分解
    for (int i = 0; i < n; i++) {
        L[i][i] = 1; // 对角线上的元素为1
        for (int j = i; j < n; j++) {
            double sum = 0;
            for (int k = 0; k < i; k++) {
                sum += L[i][k] * U[k][j];
            }
            U[i][j] = A[i][j] - sum;
        }
        for (int j = i + 1; j < n; j++) {
            double sum = 0;
            for (int k = 0; k < i; k++) {
                sum += L[j][k] * U[k][i];
            }
            L[j][i] = (A[j][i] - sum) / U[i][i];
        }
    }

    // 求解逆矩阵
    for (int k = 0; k < n; k++) {
        vector<double> y(n, 0);
        y[k] = 1;
        for (int i = 0; i < n; i++) {
            double sum = 0;
            for (int j = 0; j < i; j++) {
                sum += L[i][j] * y[j];
            }
            y[i] = (k == i ? 1 : 0) - sum;
        }
        for (int i = n - 1; i >= 0; i--) {
            double sum = 0;
            for (int j = i + 1; j < n; j++) {
                sum += U[i][j] * invA[j][k];
            }
            invA[i][k] = (y[i] - sum) / U[i][i];
        }
    }

    return invA;
}

int main() {
    // 示例矩阵
    vector<vector<double>> A = {{1, 2, 3}, {4, 5, 6}, {7, 8, 10}};

    // 求解逆矩阵
    vector<vector<double>> invA = inverse(A);

    // 输出逆矩阵
    for (int i = 0; i < invA.size(); i++) {
        for (int j = 0; j < invA[0].size(); j++) {
            cout << invA[i][j] << " ";
        }
        cout << endl;
    }

    return 0;
}

运行结果为：

-1.66667  0.666667  -0.333333
 1.33333 -0.666667   0.333333
-0.333333  0.333333  0