13.5. ITERATIVE METHODS FOR LINEAR SYSTEMS 345

Multiplying by the inverse of the matrix on the left2 this yieldsxr+11

xr+12

xr+13

xr+14

 = −

0 1

3 0 0

0 − 112

14 0

0 130 − 1

1015

0 − 160

120 − 1

10



xr1xr2xr3xr4

+

1351213304760

As before, I will be totally unoriginal in the choice of x1. Let it equal the zero vector.

Therefore, x2 =(

13

512

1330

4760

)T. Now

x3 = −

0 1

3 0 0

0 − 112

14 0

0 130 − 1

1015

0 − 160

120 − 1

10



x2︷ ︸︸ ︷1351213304760

+

1351213304760

 =

. 194

. 343

. 306

. 846

 .

Continuing this way,

x4 = −

0 1

3 0 0

0 − 112

14 0

0 130 − 1

1015

0 − 160

120 − 1

10



. 194

. 343

. 306

. 846

+

1351213304760

 =

. 219

. 368 75

. 283 3

. 858 35

and so

x5 = −

0 1

3 0 0

0 − 112

14 0

0 130 − 1

1015

0 − 160

120 − 1

10



. 219

. 368 75

. 283 3

. 858 35

+

1351213304760

 =

. 210 42

. 376 57

. 277 7

. 861 15

 .

This is fairly close to the answer. You could continue doing these iterates and it appearsthey converge to the solution. Now consider the following example.

Example 13.5.5 Use the Gauss Seidel method to solve the system1 4 0 0

1 4 1 0

0 2 5 1

0 0 2 4



x1

x2

x3

x4

 =

1

2

3

4

The exact solution is given by doing row operations on the augmented matrix. When this

is done the solution is seen to be(

6.0 −1. 25 1.0 0.5).The Gauss Seidel iterations

are of the form1 0 0 0

1 4 0 0

0 2 5 0

0 0 2 4



xr+11

xr+12

xr+13

xr+14

 = −

0 4 0 0

0 0 1 0

0 0 0 1

0 0 0 0



xr1xr2xr3xr4

+

1

2

3

4

2As in the case of the Jacobi iteration, the computer would not do this. It would use the iteration

procedure in terms of the entries of the matrix directly. Otherwise all benefit to using this method is lost.