Gaussian Elimination to Solve Linear Equations


The article focuses on using an algorithm for solving a system of linear equations. We will deal with the matrix of coefficients. Gaussian Elimination does not work on singular matrices (they lead to division by zero).

Input: For N unknowns, input is an augmented 
       matrix of size N x (N+1). One extra 
       column is for Right Hand Side (RHS)
  mat[N][N+1] = {{3.0, 2.0,-4.0, 3.0},
                {2.0, 3.0, 3.0, 15.0},
                {5.0, -3, 1.0, 14.0}
Output: Solution to equations is:

Given matrix represents following equations
3.0X1 + 2.0X2 - 4.0X3 =  3.0
2.0X1 + 3.0X2 + 3.0X3 = 15.0
5.0X1 - 3.0X2 +    X3 = 14.0

There is a unique solution for given equations, 
solutions is, X1 = 3.0, X2 = 1.0, X3 = 2.0, 

Row echelon form: Matrix is said to be in r.e.f. if the following conditions hold:

  1. The first non-zero element in each row, called the leading coefficient, is 1.
  2. Each leading coefficient is in a column to the right of the previous row leading coefficient.
  3. Rows with all zeros are below rows with at least one non-zero element.

Reduced row echelon form: Matrix is said to be in r.r.e.f. if the following conditions hold –

  1. All the conditions for r.e.f.
  2. The leading coefficient in each row is the only non-zero entry in its column.

The algorithm is majorily about performing a sequence of operations on the rows of the matrix. What we would like to keep in mind while performing these operations is that we want to convert the matrix into an upper triangular matrix in row echelon form. The operations can be:

  1. Swapping two rows
  2. Multiplying a row by a non-zero scalar
  3. Adding to one row a multiple of another

The process:

  1. Forward elimination: reduction to row echelon form. Using it one can tell whether there are no solutions, or unique solution, or infinitely many solutions.
  2. Back substitution: further reduction to reduced row echelon form.


  1. Partial pivoting: Find the kth pivot by swapping rows, to move the entry with the largest absolute value to the pivot position. This imparts computational stability to the algorithm.
  2. For each row below the pivot, calculate the factor f which makes the kth entry zero, and for every element in the row subtract the fth multiple of corresponding element in kth row.
  3. Repeat above steps for each unknown. We will be left with a partial r.e.f. matrix.

Below is C++ implementation of above algorithm.

// C++ program to demostrate working of Guassian Elimination
// method
using namespace std;

#define N 3        // Number of unknowns

// function to reduce matrix to r.e.f.  Returns a value to 
// indicate whether matrix is singular or not
int forwardElim(double mat[N][N+1]);

// function to calculate the values of the unknowns
void backSub(double mat[N][N+1]);

// function to get matrix content
void gaussianElimination(double mat[N][N+1])
    /* reduction into r.e.f. */
    int singular_flag = forwardElim(mat);

    /* if matrix is singular */
    if (singular_flag != -1)
        printf("Singular Matrix.\n");

        /* if the RHS of equation corresponding to
           zero row  is 0, * system has infinitely
           many solutions, else inconsistent*/
        if (mat[singular_flag][N])
            printf("Inconsistent System.");
            printf("May have infinitely many "


    /* get solution to system and print it using
       backward substitution */

// function for elemntary operation of swapping two rows
void swap_row(double mat[N][N+1], int i, int j)
    //printf("Swapped rows %d and %d\n", i, j);

    for (int k=0; k<=N; k++)
        double temp = mat[i][k];
        mat[i][k] = mat[j][k];
        mat[j][k] = temp;

// function to print matrix content at any stage
void print(double mat[N][N+1])
    for (int i=0; i<N; i++, printf("\n"))
        for (int j=0; j<=N; j++)
            printf("%lf ", mat[i][j]);


// function to reduce matrix to r.e.f.
int forwardElim(double mat[N][N+1])
    for (int k=0; k<N; k++)
        // Initialize maximum value and index for pivot
        int i_max = k;
        int v_max = mat[i_max][k];

        /* find greater amplitude for pivot if any */
        for (int i = k+1; i < N; i++)
            if (abs(mat[i][k]) > v_max)
                v_max = mat[i][k], i_max = i;

        /* if a prinicipal diagonal element  is zero,
         * it denotes that matrix is singular, and
         * will lead to a division-by-zero later. */
        if (!mat[k][i_max])
            return k; // Matrix is singular

        /* Swap the greatest value row with current row */
        if (i_max != k)
            swap_row(mat, k, i_max);

        for (int i=k+1; i<N; i++)
            /* factor f to set current row kth elemnt to 0,
             * and subsequently remaining kth column to 0 */
            double f = mat[i][k]/mat[k][k];

            /* subtract fth multiple of corresponding kth
               row element*/
            for (int j=k+1; j<=N; j++)
                mat[i][j] -= mat[k][j]*f;

            /* filling lower triangular matrix with zeros*/
            mat[i][k] = 0;

        //print(mat);        //for matrix state
    //print(mat);            //for matrix state
    return -1;

// function to calculate the values of the unknowns
void backSub(double mat[N][N+1])
    double x[N];  // An array to store solution

    /* Start calculating from last equation up to the
       first */
    for (int i = N-1; i >= 0; i--)
        /* start with the RHS of the equation */
        x[i] = mat[i][N];

        /* Initialize j to i+1 since matrix is upper
        for (int j=i+1; j<N; j++)
            /* subtract all the lhs values
             * except the coefficient of the variable
             * whose value is being calculated */
            x[i] -= mat[i][j]*x[j];

        /* divide the RHS by the coefficient of the
           unknown being calculated */
        x[i] = x[i]/mat[i][i];

    printf("\nSolution for the system:\n");
    for (int i=0; i<N; i++)
        printf("%lf\n", x[i]);

// Driver program
int main()
    /* input matrix */
    double mat[N][N+1] = {{3.0, 2.0,-4.0, 3.0},
                          {2.0, 3.0, 3.0, 15.0},
                          {5.0, -3, 1.0, 14.0}


    return 0;


Solution for the system:


Time Complexity: Since for each pivot we traverse the part to its right for each row below it, O(n)*(O(n)*O(n)) = O(n3).

We can also apply Gaussian Elimination for calculating:

  1. Rank of a matrix
  2. Determinant of a matrix
  3. Inverse of an invertible square matrix

This article is contributed by Yash Varyani. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Please write to us at to report any issue with the above content.

Recommended Posts:

3.6 Average Difficulty : 3.6/5.0
Based on 3 vote(s)