帮我写一个求解Ax=b的C++函数代码，该函数是矩阵类中的成员函数

好的，下面是一个求解Ax=b的C++函数代码，该函数是矩阵类中的成员函数：

#include <iostream>
#include <vector>

using namespace std;

class Matrix {
private:
    int rows, cols;
    vector<vector<double>> data;
public:
    Matrix(int r, int c) : rows(r), cols(c), data(vector<vector<double>>(r, vector<double>(c))) {}

    void set(int i, int j, double val) {
        data[i][j] = val;
    }

    void print() {
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                cout << data[i][j] << " ";
            }
            cout << endl;
        }
    }

    vector<double> solve(vector<double>& b) {
        if (rows != cols || rows != b.size()) {
            throw "Invalid input!";
        }

        vector<vector<double>> A(rows, vector<double>(cols + 1));

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                A[i][j] = data[i][j];
            }
            A[i][cols] = b[i];
        }

        for (int i = 0; i < rows; i++) {
            int pivot = i;

            for (int j = i + 1; j < rows; j++) {
                if (abs(A[j][i]) > abs(A[pivot][i])) {
                    pivot = j;
                }
            }

            if (pivot != i) {
                swap(A[i], A[pivot]);
            }

            for (int j = i + 1; j <= cols; j++) {
                A[i][j] /= A[i][i];
            }

            for (int j = i + 1; j < rows; j++) {
                for (int k = i + 1; k <= cols; k++) {
                    A[j][k] -= A[j][i] * A[i][k];
                }
            }
        }

        vector<double> x(cols);

        for (int i = cols - 1; i >= 0; i--) {
            x[i] = A[i][cols];

            for (int j = i + 1; j < cols; j++) {
                x[i] -= A[i][j] * x[j];
            }
        }

        return x;
    }
};

int main() {
    Matrix A(3, 3);
    A.set(0, 0, 2);
    A.set(0, 1, -1);
    A.set(0, 2, 0);
    A.set(1, 0, -1);
    A.set(1, 1, 2);
    A.set(1, 2, -1);
    A.set(2, 0, 0);
    A.set(2, 1, -1);
    A.set(2, 2, 2);

    vector<double> b{1, 2, 3};

    try {
        vector<double> x = A.solve(b);

        for (int i = 0; i < x.size(); i++) {
            cout << "x" << i + 1 << " = " << x[i] << endl;
        }
    } catch (const char* msg) {
        cerr << msg << endl;
    }

    return 0;
}

这个函数使用了高斯-约旦消元法来求解Ax=b，其中A是一个矩阵，b是一个列向量，x是一个列向量，表示方程组Ax=b的解。在代码中，我们首先将A和b组合成一个增广矩阵，然后使用高斯-约旦消元法将其化为行阶梯形式，最后通过回带法求解x。