Monday, February 2, 2015

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

11:52:00 PM Posted by Programming Seeker No Comments

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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();
}
}