Gauss-Jordan Elimination method

Program in C

==================================================================================================
Program to find solutions of a system of linear equations by
Gauss-Jordan Elimination Method 
(The coefficient matrix must be Diagonally Dominant)
LANGUAGE :: C  (Compiler: GNU GCC Compiler)
==================================================================================================

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
    int n;
    printf("Enter the number of variables: ");
    scanf("%d",&n);
    float a[n][n+1];       //The augmented matrix
    int i,j,k;
    float solution,multiplier;
    printf("\nEnter the augmented matrix:\n");
    for (i = 0;i < n;i++){
        for (j = 0;j < n+1;j++){
            scanf("%f",&a[i][j]);    //reading the matrix
        }
    }
    printf("\n");
    //------------ to get the diagonal matrix -------------------|
    for (k = 0;k < n;k++){
        for (i=0; i < n; i++){
            if(i!= k){
                multiplier = a[i][k]/a[k][k];
                for (j=0; j < n+1; j++){
                    a[i][j] = a[i][j] - a[k][j]*multiplier;
                }
            }
        }
    }
    //-------------- Printing the diagonal matrix -----------|
    printf("\nThe diagonal matrix:\n");
    for (i = 0;i < n;i++){
        for (j = 0;j < n+1;j++){
            printf("\t%f",a[i][j]);
        }
        printf("\n");
    }
    // -------------------Printing the Solutions---------------------|
    printf("\nThe Solutions are:\n");
    for (i = 0;i < n;i++){
        solution = a[i][n]/a[i][i];
        printf("\t x%d = %f",i+1,solution);
    }
    printf("\n");
    return 0;
}

========================= OUTPUT ============================
Enter the number of variables: 3

Enter the augmented matrix:
1 1 1 9
7 -5 4 9
2 3 8 10


The diagonal matrix:
        1.000000        0.000000        0.000000        6.130435
        0.000000        -12.000000      0.000000        -60.521740
        0.000000        0.000000        5.750000        -12.500000

The Solutions are:
         x1 = 6.130435   x2 = 5.043478   x3 = -2.173913

Program in Fortran 95

===================================================================================
Program to find solutions of a system of linear equations by
Gauss-Jordan Elimination Method 
(The coefficient matrix must be Diagonally Dominant)
LANGUAGE :: FORTRAN 95  
Compiler :: GNU Fortran
===================================================================================

PROGRAM gaussJordanElimination
  IMPLICIT NONE

  INTEGER :: n,i,j,k
  REAL,ALLOCATABLE :: a(:,:)            !The augmented matrix 'a'
  REAL multiplier, solution
  WRITE(*,*)'Enter the number of variables'
  READ(*,*)n
  ALLOCATE (a(n,n+1))
  WRITE(*,*)'Enter the augmented matrix:'
  DO i=1,n
      READ(*,*)(a(i,j),j=1,n+1)         !reading the matrix
  END DO
  !!------------ to get the diagonal matrix -------------------|
  DO k = 1,n
      DO i=1,n
        IF(i.NE.k) THEN
          multiplier = a(i,k) / a(k,k)
          DO j=1,n+1
              a(i,j) = a(i,j) - a(k,j)*multiplier
          END DO
        ENDIF
      END DO
  END DO
  !!-------------- Printing the diagonal matrix -----------|
  WRITE(*,'(/)')
  WRITE(*,*)'The diagonal matrix:'
  DO i = 1,n
      WRITE(*,'(*(2X,F10.6))')(a(i,j),j=1,n+1)
  END DO
  !! -------------------Printing the Solutions---------------------|
  WRITE(*,'(/)')
  WRITE(*,*)'The Solutions are:'
  DO j = 1,n
      solution = a(j,n+1)/a(j,j)
      WRITE(*,'(2X,A1,I1,A1,F10.6,2X)')'x',j,'=',solution
  END DO
END PROGRAM

===================---------- OUTPUT ----------===================
 Enter the number of variables
 3
 Enter the augmented matrix:
 5 -2 3 18
 1 7 -3 -22
 2 -1 6 22


 The diagonal matrix:
    5.000000   -0.000000    0.000000    5.000000
   -0.000000    7.400000    0.000000  -14.800001
   -0.000000    0.000000    4.702703   14.108109


 The Solutions are:
  x1=  1.000000
  x2= -2.000000
  x3=  3.000000