Augmented Matrix Solution By reducing it to Row Echelon Form in Java

Augmented Matrix Solution By reducing it to Row Echelon Form And then Calculate the result by backward substitution (Code in Java) 



package echolon;
import java.util.Scanner;
/**
 *
 * @author usman
 */
public class Echolon {
     Scanner usman=new Scanner(System.in);
    int i,j,k,n,a;
    float [][] A=new float[20][20];
    float  sum;
    float c;
    float [] x=new float[10];
    
    public  Echolon(){
    sum=0;
    System.out.println("\nEnter the order of matrix: ");
    n=usman.nextInt();
    a=n;
    System.out.println("\nEnter the elements of augmented matrix row-wise:\n\n");
    for(i=1; i<=n; i++)
    {
for(j=1; j<=(n+1); j++)
{ 
            if(a<j){System.out.println("\nEnter the Value of B: (1st row):- \n ");}
   System.out.println("A["+i+"]["+j+"] : ");
            
   A[i][j]=usman.nextFloat();
}
    }
    }

    void operation() {
    for(j=1; j<=n; j++)
    {
for(i=1; i<=n; i++)
{
   if(i>j)
   {
c=A[i][j]/A[j][j];
for(k=1; k<=n+1; k++)
{
   A[i][k]=A[i][k]-c*A[j][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]=(A[i][n+1]-sum)/A[i][i];
    }
    }

    void result(){
    System.out.println("\nThe solution is: \n");
    for(i=1; i<=n; i++)
    {
System.out.println("\nx"+i+" =  "+x[i]+"\t");
    }
    }

void display()
{
System.out.println("\n The Matrix is :-\n ");
for(int i=1;i<=n;i++){
System.out.println("\n");
for(int j=1;j<=n+1;j++){
System.out.println(A[i][j]+"     ");
}}
}


public static void main(String args[]){
Echolon obj=new Echolon();
obj.operation();
obj.display();
obj.result();
}
}