Monday, February 2, 2015

Augmented Matrix Solution By reducing it to Row Echelon Form in C++

11:42:00 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 C++)....

#include<iostream.h>
#include<conio.h>
class row{
int i,j,k,n,a;
float A[20][20],c,x[10],sum;
public:
row(){

sum=0.0;
cout<<"\nEnter the order of matrix: ";
cin>>n;
a=n;
cout<<"\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){cout<<"\nEnter the Value of B: (1st row):- \n ";}
cout<<"A["<<i<<"]["<<j<<"] : ";
cin>>A[i][j];
}
}
}

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(){
cout<<"\nThe solution is: \n";
for(i=1; i<=n; i++)
{
cout<<"\nx"<<i<<" = "<<x[i]<<"\t";
}
}

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

};

void main()
{
clrscr();
row obj;
obj.operation();
obj.display();
obj.result();
getch();

}