Solving a simple matrix in row-reduced form in C++

Okay, I am pulling out all my hair on this one, though, as a noob, I am sure there are several problems. I want to take a matrix and, by sing elementary row operations, reduced it to row-reduced echelon form. We assume (1) it is solvable and (2) a unique solution. There is no checking for zeros or anything; it just does row operations. Here is the code:

#include <iostream>
#include <cstdlib>

using namespace std;

void printmatrix(float A[][4]);
void RowReduce (float A[][4]);


int main()
{


     float A[3][4] = {{5, -6, -7, 7}, {3, -2, 5, -17}, {2, 4, -3, 29}}; //answer should be {2, 4, -3}

    printmatrix(A);
    RowReduce(A);
 }

 void printmatrix(float A[][4]){ // Outputs the matrix

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]){
  int p=3;  //rows
  int q=4;  //columns

  int lead = 0; //the determines the column we are at which holds the diagonal, the basis for all elimination above and below

    cout << endl;
    while (lead<q-1)
    {
        for (int i=0; i<p; i++) //for each row . . .
        {
           if (i!=lead) // ignore the diagonal, and we will not have a tree rref as the diagonal will not be divided by itself. I can fix that.
                {
                    cout << A[lead][lead] << "  " << A[i][lead];
                    for (int j=0; j<q; j++)
                    {
                        A[i][j]=A[lead][lead]*A[i][j]; //here is the math . . . . probably where the problem is?
                        A[i][lead]=A[i][lead]*A[lead][j];
                        A[i][j]=A[i][j]-A[i][lead];
                    }
                cout << endl;
                }
    }
lead++;  // now go to the next pivot
cout << endl;

}

}

I tried doing it by hand, but what I get is, of course, the right answer, but this gets a diagonal matrix--which is great--but the wrong answer!