Metoda el. gaussa z wyświetleniem danych

#include<iostream>
#include<iomanip>
using namespace std;

int main(){

int n,i,j,k;
cout.precision(4);
cout.setf(ios::fixed);
cout<<"podaj liczbe niewiadomych: "<<endl;
cin>>n;
float a[n][n+1], x[n];
cout<<"wpisz niewiadome: "<<endl;
for(i=0;i<n;i++)
    for(j=0;j<=n;j++)
        cin>>a[i][j];

for(i=0;i<n;i++)
    for(k=i+1;k<n;k++)
        if(a[i][j]<a[k][i])
            for (j=0;j<=n;j++){
                double temp = a[i][j];
                a[i][j] = a[k][j];
                a[k][j]=temp;
            }
cout<<"Macierz po wykonaniu zmiany rzedow:"<<endl;
for(i=0;i<n;i++){
    for(j=0;j<=n;j++)
        cout<<a[i][j]<<setw(16);
        cout<<"\n";
}
for(i=0;i<n;i++)
    for(k=i+1;k<n;k++){
    double t = a[k][i]/a[i][i];
    for(j=0;j<=n;j++){
            a[k][j]=a[k][j]-t*a[i][j];
    }
}
cout<<"macierz po eliminacji gaussa: "<<endl;
for(i=0;i<n;i++){
    for(j=0;j<=n;j++)
        cout<<a[i][j]<<setw(16);
    cout<<"\n";
}
    for (i=n-1;i>=0;i--)                //back-substitution
    {                        //x is an array whose values correspond to the values of x,y,z..
        x[i]=a[i][n];                //make the variable to be calculated equal to the rhs of the last equation
        for (j=i+1;j<n;j++)
            if (j!=i)            //then subtract all the lhs values except the coefficient of the variable whose value                                   is being calculated
                x[i]=x[i]-a[i][j]*x[j];
        x[i]=x[i]/a[i][i];            //now finally divide the rhs by the coefficient of the variable to be calculated
    }
    cout<<"\nThe values of the variables are as follows:\n";
    for (i=0;i<n;i++)
        cout<<x[i]<<endl;            // Print the values of x, y,z,....
    return 0;

}