c++利用类矩阵逆矩阵

要利用类实现矩阵逆矩阵的计算，需要以下步骤：

1. 定义一个矩阵类，包含矩阵的行数、列数和元素值等信息，以及实现矩阵加减、乘法、转置等基本运算。

2. 实现高斯-约旦消元法（Gauss-Jordan elimination）求解矩阵的逆矩阵。具体步骤如下：

- 将原矩阵和单位矩阵按行拼接成增广矩阵；
- 对增广矩阵进行初等行变换，将其变为左半部分为单位矩阵的形式；
- 右半部分即为所求的逆矩阵。

3. 实现矩阵行列式的计算，以判断矩阵是否可逆。

下面是一个简单的示例代码，仅供参考：

#include <iostream>
#include <vector>

using namespace std;

class Matrix {
public:
    Matrix(int r = 0, int c = 0) : rows(r), cols(c), data(r * c) {}
    int rows, cols;
    vector<double> data;

    double& operator()(int i, int j) {
        return data[i * cols + j];
    }

    friend Matrix operator+(const Matrix& a, const Matrix& b);
    friend Matrix operator*(const Matrix& a, const Matrix& b);
    friend Matrix transpose(const Matrix& a);
    friend double det(const Matrix& a);
    friend Matrix inverse(const Matrix& a);
};

Matrix operator+(const Matrix& a, const Matrix& b) {
    Matrix c(a.rows, a.cols);
    for (int i = 0; i < a.rows; i++) {
        for (int j = 0; j < a.cols; j++) {
            c(i, j) = a(i, j) + b(i, j);
        }
    }
    return c;
}

Matrix operator*(const Matrix& a, const Matrix& b) {
    Matrix c(a.rows, b.cols);
    for (int i = 0; i < a.rows; i++) {
        for (int j = 0; j < b.cols; j++) {
            for (int k = 0; k < a.cols; k++) {
                c(i, j) += a(i, k) * b(k, j);
            }
        }
    }
    return c;
}

Matrix transpose(const Matrix& a) {
    Matrix b(a.cols, a.rows);
    for (int i = 0; i < a.rows; i++) {
        for (int j = 0; j < a.cols; j++) {
            b(j, i) = a(i, j);
        }
    }
    return b;
}

double det(const Matrix& a) {
    if (a.rows != a.cols) {
        throw "not a square matrix";
    }
    int n = a.rows;
    Matrix b = a;
    double d = 1;
    for (int i = 0; i < n; i++) {
        int p = i;
        for (int j = i + 1; j < n; j++) {
            if (abs(b(j, i)) > abs(b(p, i))) {
                p = j;
            }
        }
        if (p != i) {
            swap(b.data.begin() + i * n, b.data.begin() + p * n);
            d = -d;
        }
        if (b(i, i) == 0) {
            return 0;
        }
        for (int j = i + 1; j < n; j++) {
            double f = b(j, i) / b(i, i);
            for (int k = i; k < n; k++) {
                b(j, k) -= f * b(i, k);
            }
        }
        d *= b(i, i);
    }
    return d;
}

Matrix inverse(const Matrix& a) {
    if (a.rows != a.cols) {
        throw "not a square matrix";
    }
    int n = a.rows;
    Matrix b(n, 2 * n);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            b(i, j) = a(i, j);
        }
        b(i, n + i) = 1;
    }
    for (int i = 0; i < n; i++) {
        int p = i;
        for (int j = i + 1; j < n; j++) {
            if (abs(b(j, i)) > abs(b(p, i))) {
                p = j;
            }
        }
        if (p != i) {
            swap(b.data.begin() + i * 2 * n, b.data.begin() + p * 2 * n);
        }
        if (b(i, i) == 0) {
            throw "matrix is not invertible";
        }
        double f = b(i, i);
        for (int j = i; j < 2 * n; j++) {
            b(i, j) /= f;
        }
        for (int j = 0; j < n; j++) {
            if (j != i) {
                f = b(j, i);
                for (int k = i; k < 2 * n; k++) {
                    b(j, k) -= f * b(i, k);
                }
            }
        }
    }
    Matrix c(n, n);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            c(i, j) = b(i, n + j);
        }
    }
    return c;
}

int main() {
    Matrix a(3, 3);
    a(0, 0) = 2; a(0, 1) = 1; a(0, 2) = 1;
    a(1, 0) = 4; a(1, 1) = -6; a(1, 2) = 0;
    a(2, 0) = -2; a(2, 1) = 7; a(2, 2) = 2;
    cout << "A =\n" << a.data[0] << " " << a.data[1] << " " << a.data[2] << "\n"
         << a.data[3] << " " << a.data[4] << " " << a.data[5] << "\n"
         << a.data[6] << " " << a.data[7] << " " << a.data[8] << endl;
    cout << "det(A) = " << det(a) << endl;
    Matrix b = inverse(a);
    cout << "A^-1 =\n" << b.data[0] << " " << b.data[1] << " " << b.data[2] << "\n"
         << b.data[3] << " " << b.data[4] << " " << b.data[5] << "\n"
         << b.data[6] << " " << b.data[7] << " " << b.data[8] << endl;
    Matrix c = a * b;
    cout << "A * A^-1 =\n" << c.data[0] << " " << c.data[1] << " " << c.data[2] << "\n"
         << c.data[3] << " " << c.data[4] << " " << c.data[5] << "\n"
         << c.data[6] << " " << c.data[7] << " " << c.data[8] << endl;
    return 0;
}

这段代码定义了一个矩阵类 Matrix，包含了矩阵的基本运算和逆矩阵计算。在 main 函数中，我们定义了一个 3x3 的矩阵 A，计算其行列式和逆矩阵，并验证 A 乘以 A 的逆矩阵是否得到单位矩阵。注意，这里的逆矩阵计算使用的是高斯-约旦消元法，而不是传统的伴随矩阵法，因为前者更简单、更高效。