Unit 2.1 Gauss Elemination Method

//Mera takrika

#include <stdio.h>
void p(){
    printf("\n");
}
int main()
{
    int n,i,j,k;
    float c,a[5][5],x[10];

    printf("Enter the value of n\n");
    scanf("%d",&n);
    for(i=1 ; i<=n ; i++){
        for(j=1 ; j<=n+1 ; j++){
            scanf("%f",&a[i][j]);
        }
    }
    for(i=1 ; i<=n ; i++){
        for(j=1 ; j<=n ; j++){
            if(i<j){
                c = a[j][i]/a[i][i];
                for(k=1 ; k<=n+1 ; k++){
                    a[j][k] = a[j][k] - (c*a[i][k]);
                }
            }
        }
    }
    p();
    for(i=1 ; i<=n ; i++){
        for(j=1 ; j<=n+1 ; j++){
            printf("%f ",a[i][j]);
        }
        p();
    }
    // now doing back substitution

    x[3] = a[3][4]/a[3][3];
    x[2] = (a[2][4]- (x[3] * a[2][3]))/a[2][2];
    x[1] = (a[1][4] - ( (x[2]*a[1][2]) + (x[3]*a[1][3]) )) /a[1][1];

    for(i=1;i<=n;i++)
    {
        printf("\nx[%d] = %f",i,x[i]);
    }
    return 0;
}

//Mam ka tarika

// AFTER CHANGES

// Algorithm
start
declare the variables and read the order of the matrix n
take the cofficients of the linear equation as

Do for i=1 to n
    Do for j=1 to n+1
        read a[i][j]
    end for j
end for i

Do for i=1 to n-1
    Do for j=i to n
        Do for k=i+1 to n+1
            a[j][k] = a[j][k]- ((a[j][k]/a[i][i]) * a[i][k])
        End for k
    End for j
End for i

Compute x[n] = a[n][n+1]/a[n][n]

Do for i=n-1 to 1
    sum =0
    Do for j = i+1 to n
        sum = sum + a[i][j] * x[j]
    End for j
    x[i] = 1/a[i][i] * (a[i][n+1]-sum)
End for i
Display the result x[k]
Stop


// Program

#include <stdio.h>

#include <stdio.h>
int main()
{
    int n,i,j,k;
    float sum=0,a[5][5],x[10];

    printf("Enter the value of n\n");
    scanf("%d",&n);
    for(i=1 ; i<=n ; i++){
        for(j=1 ; j<=n+1 ; j++){
            scanf("%f",&a[i][j]);
        }
    }

    for(i=1 ; i<n ; i++){
        for(j=i+1 ; j<=n ; j++){
            for(k=i+1 ; k<=n+1 ; k++){
                a[j][k] = a[j][k] - ((a[j][i]/a[i][i]) * a[i][k]);
            }
        }
    }
    x[n]=a[n][n+1]/a[n][n];
    for(i=n-1;i>=1;i--)
    {
        sum=0;
        for(j=i+1;j<=n;j++)
        {
            sum=sum+(a[i][j]*x[j]);
        }
        x[i]=(1/a[i][i])*((a[i][n+1]-sum));
    }

    for(i=1;i<=n;i++)
    {
        printf("\nx[%d] = %f",i,x[i]);
    }

    return 0;
}






// BEFORE CHANGES

// Algorithm
start
declare the variables and read the order of the matrix n
take the cofficients of the linear equation as

Do for k=1 to n
    Do for j=1 to n+1
        read a[k][j]
    end for j
end for k

Do for k=1 to n-1
    Do for i=k to n
        Do for j=k+1 to n+1
            a[i][j] = a[i][j]- ((a[i][j]/a[k][k]) * a[k][j])
        End for j
    End for i
End for k

Compute x[n] = a[n][n+1]/a[n][n]

Do for k=n-1 to 1
    sum =0
    Do for j = k+1 to n
        sum = sum + a[k][j] * x[j]
    End for j
    x[k] = 1/a[k][k] * (a[k][n+1]-sum)
End for k
Display the result x[k]
Stop


// Program

#include <stdio.h>

#include <stdio.h>
int main()
{
    int n,i,j,k;
    float sum=0,a[5][5],x[10];

    printf("Enter the value of n\n");
    scanf("%d",&n);
    for(k=1 ; k<=n ; k++){
        for(j=1 ; j<=n+1 ; j++){
            scanf("%f",&a[k][j]);
        }
    }

    for(k=1 ; k<n ; k++){
        for(i=k+1 ; i<=n ; i++){
            for(j=k+1 ; j<=n+1 ; j++){
                a[i][j] = a[i][j] - ((a[i][k]/a[k][k]) * a[k][j]);
            }
        }
    }
    x[n]=a[n][n+1]/a[n][n];
    for(k=n-1;k>=1;k--)
    {
        sum=0;
        for(j=k+1;j<=n;j++)
        {
            sum=sum+(a[k][j]*x[j]);
        }
        x[k]=(1/a[k][k])*((a[k][n+1]-sum));
    }

    for(k=1;k<=n;k++)
    {
        printf("\nx[%d] = %f",k,x[k]);
    }

    return 0;
}

/*
Enter the value of n
3
2 1 4 12
8 -3 2 23
4 11 -1 33
x[1] = 3.357143
x[2] = 1.857143
x[3] = 0.857143
*/