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

Question

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() {

    // 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;

    }

}

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!

Solution

The main error in you code is that you are calculating the divisor or multiplier within the for loop. You should calculate them before iterating over the cells.

Hint: debugging is easier if the code is well formated and the variables have meaningful names.

See the implementation of RowReduce():

#include <iostream>
#include <cstdlib>
#include <iomanip>

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 << setw(7) << setprecision(4) << A[i][j] << " ";
            }
            cout << endl;
    }

    cout << endl;
}

void RowReduce(float A[][4])
{
    const int nrows = 3; // number of rows
    const int ncols = 4; // number of columns

    int lead = 0; 

    while (lead < nrows) {
        float d, m;

        for (int r = 0; r < nrows; r++) { // for each row ...
            /* calculate divisor and multiplier */
            d = A[lead][lead];
            m = A[r][lead] / A[lead][lead];

            for (int c = 0; c < ncols; c++) { // for each column ...
                if (r == lead)
                    A[r][c] /= d;               // make pivot = 1
                else
                    A[r][c] -= A[lead][c] * m;  // make other = 0
            }
        }

        lead++;
        printmatrix(A);
    }
}

The output:

  5      -6      -7       7 
  3      -2       5     -17 
  2       4      -3      29 


  1    -1.2    -1.4     1.4 
  0     1.6     9.2   -21.2 
  0     6.4    -0.2    26.2 

  1       0     5.5   -14.5 
  0       1    5.75  -13.25 
  0       0     -37     111 

  1       0       0       2 
  0       1       0       4 
  0       0       1      -3

Comments

[user5179531]

Very nice, thanks. Not sure, however, what is wrong with mine. I also tried to do mine as I would have done by hand. Yours is a different means to get the same ends, but I am not sure still where I went wrong. Thanks for your effort though!

[Jan Veselý]

Is there a way to make it work for : 1 2 3 4 / 0 0 1 1 / 0 0 1 1 /