# Jacobi Iteration Method Using C++ with Output

C++ program for solving system of linear equations using Jacobi Iteration Method.

## C++ Program for Jacobi Iteration

``````
#include<iostream>
#include<iomanip>
#include<math.h>

/* Arrange systems of linear
equations to be solved in
diagonally dominant form
and form equation for each
unknown and define here
*/
/* In this example we are solving
3x + 20y - z = -18
2x - 3y + 20z = 25
20x + y - 2z = 17
*/
/* Arranging given system of linear
equations in diagonally dominant
form:
20x + y - 2z = 17
3x + 20y -z = -18
2x - 3y + 20z = 25
*/
/* Equations:
x = (17-y+2z)/20
y = (-18-3x+z)/20
z = (25-2x+3y)/20
*/
/* Defining function */
#define f1(x,y,z)  (17-y+2*z)/20
#define f2(x,y,z)  (-18-3*x+z)/20
#define f3(x,y,z)  (25-2*x+3*y)/20

using namespace std;

/* Main function */
int main()
{
float x0=0, y0=0, z0=0, x1, y1, z1, e1, e2, e3, e;
int step=1;

/* Setting precision and writing floating point values in fixed-point notation. */
cout<< setprecision(6)<< fixed;

/* Input */
/* Reading tolerable error */
cout<<"Enter tolerable error: ";
cin>>e;

cout<< endl<<"Count\tx\t\ty\t\tz"<< endl;
do
{
/* Calculation */
x1 = f1(x0,y0,z0);
y1 = f2(x0,y0,z0);
z1 = f3(x0,y0,z0);
cout<< step<<"\t"<< x1<<"\t"<< y1<<"\t"<< z1<< endl;

/* Error */
e1 = fabs(x0-x1);
e2 = fabs(y0-y1);
e3 = fabs(z0-z1);

step++;

/* Set value for next iteration */
x0 = x1;
y0 = y1;
z0 = z1;
}while(e1>e && e2>e && e3>e);

cout<< endl<<"Solution: x = "<< x1<<", y = "<< y1<<" and z = "<< z1;
return 0;
}

```
```

## Output

```Enter tolerable error: 0.0000001

Count   x               y               z
1       0.850000        -0.900000       1.250000
2       1.020000        -0.965000       1.030000
3       1.001250        -1.001500       1.003250
4       1.000400        -1.000025       0.999650
5       0.999966        -1.000077       0.999956
6       1.000000        -0.999997       0.999992
7       0.999999        -1.000000       1.000000
8       1.000000        -1.000000       1.000000
9       1.000000        -1.000000       1.000000

Solution: x = 1.000000, y = -1.000000 and z = 1.000000
```