在c++中求解一个行简化形式的简单矩阵

好吧，我在这上面扯了我所有的头发，尽管，作为一个新手，我确信有几个问题。我想取一个矩阵，通过初等行变换，把它化简成行简化阶梯形。我们假设(1)它是可解的，(2)它是唯一解。不需要检查零或其他任何东西;它只是行变换。下面是代码:

#include <iostream>
#include <cstdlib>
using namespace std;
void printmatrix(float A[][4]);
void RowReduce (float A[][4]);
int main() {
    // answer should be { 2, 4, -3 }
    float A[3][4] = {
        { 5, -6, -7,   7 },
        { 3, -2,  5, -17 },
        { 2,  4, -3,  29 }
    };
    printmatrix(A);
    RowReduce(A);
}
// Outputs the matrix
void printmatrix(float A[][4]) { 
    int p = 3;
    int q = 4;
    for (int i = 0; i < p; i++) {
        for (int j = 0; j < q; j++) {
            cout << A[i][j] << " ";
        }
        cout << endl;
    } 
}
void RowReduce (float A[][4]){
    //rows
    int p = 3;  
    //columns
    int q = 4;  
    // the determines the column we are at which holds the diagonal,
    // the basis for all elimination above and below
    int lead = 0; 
    cout << endl;
    while ( lead < q - 1 ) {
        // for each row . . .
        for (int i = 0; i < p; i++)  {
            // ignore the diagonal, and we will not have a tree rref
            // as the diagonal will not be divided by itself. I can fix that.
            if ( i != lead )  {
                cout << A[lead][lead] << "  " << A[i][lead];
                for (int j = 0; j < q; j++) {
                    //here is the math . . . . probably where the problem is?
                    A[i][j]    = A[lead][lead] * A[i][j]; 
                    A[i][lead] = A[i][lead]    * A[lead][j];
                    A[i][j]    = A[i][j]       - A[i][lead];
                }
                cout << endl;
            }
        }
        // now go to the next pivot
        lead++;  
        cout << endl;
    }
}

我试着手工做，但我得到的，当然是正确的答案，但这得到的是一个对角矩阵——这很好——但答案是错误的!