# C Program to Implement Gauss Jordan Elimination

This C program implements Gauss Jordan Elimination method. In linear algebra, Gaussian elimination is an algorithm for solving systems of linear equations.

Here is the source code of the C program to find solution of a system of linear equations. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

1. `#include<stdio.h>`
2. ` `
3. `void solution( int a[][20], int var );`
4. `int main()`
5. `{`
6. ` `
7. `    int a[ 20 ][ 20 ], var, i, j, k, l, n;`
8. `    printf( "\nEnter the number of variables:\n" );`
9. `    scanf( "%d", &var );`
10. ` `
11. `    for ( i = 0;i < var;i++ )`
12. `    {`
13. `        printf( "\nEnter the equation%d:\n", i + 1 );`
14. ` `
15. `        for ( j = 0;j < var;j++ )`
16. `        {`
17. `            printf( "Enter the coefficient of  x%d:\n", j + 1 );`
18. `            scanf( "%d", &a[ i ][ j ] );`
19. `        }`
20. ` `
21. `        printf( "\nEnter the constant:\n" );`
22. `        scanf( "%d", &a[ i ][ var] );`
23. `    }`
24. ` `
25. `    solution( a, var );`
26. `    return 0;`
27. `}`
28. ` `
29. ` `
30. ` `
31. `void solution( int a[ 20 ][ 20 ], int var )`
32. `{`
33. `    int k, i, l, j;`
34. ` `
35. `    for ( k = 0;k < var;k++ )`
36. `    {`
37. `        for ( i = 0;i <= var;i++ )`
38. `        {`
39. `            l = a[ i ][ k ];`
40. ` `
41. `            for ( j = 0;j <= var;j++ )`
42. `            {`
43. `                if ( i != k )`
44. `                a[i][j] = (a[k][k]*a[i][j])-(l*a[k][j]);`
45. `            }`
46. `        }`
47. `    }`
48. ` `
49. `    printf( "\nSolutions:" );`
50. ` `
51. `    for ( i = 0;i < var;i++ )`
52. `    {`
53. `        printf( "\nTHE VALUE OF x%d IS %f\n", i + 1, ( float ) a[ i ][ var ] / ( float ) a[ i ][ i ] );`
54. `    }`
55. ` `
56. `}`

```\$ gcc bubblesort.c -o bubblesort
\$ ./bubblesort

Enter the number of variables: 3

Enter the equation 1:
Enter the coefficient of  x1: 1
Enter the coefficient of  x2: 0
Enter the coefficient of  x3: 0

Enter the constant: 2

Enter the equation 2:
Enter the coefficient of  x1: 0
Enter the coefficient of  x2: 1
Enter the coefficient of  x3: 0

Enter the constant: 0

Enter the equation 3:
Enter the coefficient of  x1: 0
Enter the coefficient of  x2: 0
Enter the coefficient of  x3: 1

Enter the constant: -1

Solutions:
THE VALUE OF x1 IS 2.000000

THE VALUE OF x2 IS 0.000000

THE VALUE OF x3 IS -1.000000```

Sanfoundry Global Education & Learning Series – 1000 C Programs.