GAUSS JORDAN ELIMINATION

Solve The Given Linear Equations Systems By Using Gauss Jordan Elimination Method

Source code :-

#include <stdio.h>
int main()
{
    int i, j, k, n;
    float a[20][20], x[20], r;
    printf("Enter the number of equations: ");
    scanf("%d", &n);
    // Input the matrix.
    printf("\nEnter co-efficient matrix [A] row wise:\n");    
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < n; j++)
        {
            printf("\na[%d][%d]= ", i + 1, j + 1);
            scanf("%f", &a[i][j]);
        }
        printf("\n");
    }
    printf("\n\nEnter column matrix [B] row wise:\n");
    for (i = 0, j = n; i < n; i++)
    {
        printf("\nb[%d]= ", i + 1);
        scanf("%f", &a[i][j]);
    }
    printf("\n");
    // transform coefficient matrix into normalize form
    for (k = 0; k < n; k++)
    {
        for (i = 0; i < n; i++)
        {
            if (i == k)
                continue;
            r = a[i][k] / a[k][k];
            for (j = 0; j <= n; j++)
            {
                a[i][j] = a[i][j] - r * a[k][j];
            }
        }
        // Steps display start. [optional]
        printf("\nStep %d:-\n\n", k + 1);
        for (i = 0; i < n; i++)
        {
            printf("|");
            for (j = 0; j < n; j++)
            {
                printf("%5.2f", a[i][j]);
            }
            printf(" |\t| %5.2f |\n", a[i][j]);
        }
        printf("\n");
        // Steps display stop.
    }
    for (i = 0; i < n; i++)
    {
        x[i] = a[i][n] / a[i][i];
        a[i][i] = a[i][i] / a[i][i];
    }
    printf("\nStep %d:-\n\n", k + 1);
    for (i = 0; i < n; i++)
    {
        printf("|");
        for (j = 0; j < n; j++)
        {
            printf("%5.2f", a[i][j]);
        }
        printf(" |\t| %5.2f |\n", x[i]);
    }
    printf("\n");
    printf("\nThe solution is:\n");
    for (i = 0; i < n; i++)
    {
        printf("x[%d]=%5.2f\n", i + 1, x[i]);
    }
    return 1;
}

Input-Output :-

abhi@hp-15q-laptop:~$ gcc gauss_jordan.c
abhi@hp-15q-laptop:~$ ./a.out

Enter the number of equations: 3

Enter co-efficient matrix [A] row wise:      

a[1][1]= 2

a[1][2]= -1

a[1][3]= 3


a[2][1]= 1

a[2][2]= 1

a[2][3]= 1


a[3][1]= 1

a[3][2]= -1

a[3][3]= 1



Enter column matrix [B] row wise:

b[1]= 9

b[2]= 6

b[3]= 2


Step 1:-

| 2.00-1.00 3.00 |      |  9.00 |
| 0.00 1.50-0.50 |      |  1.50 |
| 0.00-0.50-0.50 |      | -2.50 |


Step 2:-

| 2.00 0.00 2.67 |      | 10.00 |
| 0.00 1.50-0.50 |      |  1.50 |
| 0.00 0.00-0.67 |      | -2.00 |


Step 3:-

| 2.00 0.00 0.00 |      |  2.00 |
| 0.00 1.50 0.00 |      |  3.00 |
| 0.00 0.00-0.67 |      | -2.00 |


Step 4:-

| 1.00 0.00 0.00 |      |  1.00 |
| 0.00 1.00 0.00 |      |  2.00 |
| 0.00 0.00 1.00 |      |  3.00 |


The solution is:
x[1]= 1.00
x[2]= 2.00
x[3]= 3.00