Answer each question: Section 7.3 The Jacobi and Gauss-Siedel Iterative Techniques 1) Find the first two iterations of the Jacobi method for the following linear systems, using x (0) = 0.: d. 4x 1 –x...

Section 7.3 The Jacobi and Gauss-Siedel Iterative Techniques
1)
Find the first two iterations of the Jacobi method for the following linear systems, using x(0)
= 0.:
d. 4x1 –x2- x4 = 0,
-x1 +4x2 –x3 _x5 =5,
-x2 – 4x3 – x6 = 0,
-x1 +4x4 –x5 = 6,
-x2 – x4 +4x5 –x6 = -2,
-x3 – x5 +4x6 =6
Solution:
Jacob method the sequence is
xk = Tx(k-1) + c




























































Given for initial approximation x(0) = 0, Then x(1) is given by

( )


( )


( )

( )


( )


( )


( )



( )


( )


( )

( )


( )


( )



( )


( )


( )


( )



( )


( )


( )
Similarly second iterate is calculated

( )

( )


( )

( )

( )

( )
The first two iteration values of the linear system are as follows:
k 0 1 2

( ) 0 0 0.6875

( ) 0 1.25 1.375

( ) 0 0 0.3125

( ) 0 1.5 1.625

( ) 0 0.5 0.1875

( ) 0 0
9)
The linear system 2x1 - x2 + x3 = -1,
2x1 +2x2 + 2x3 = 4,
-x1 –x2 + 2x3 = -5
Has the solution (1, 2, -1)t .
a. Show that p(Tj) = /2 > 1.
. Show that Jacobi method with x(0) = 0 fails to give a good approximation after 25
iterations.
c. Show that p(Tg) = 1/2
d. Use the Gauss-Seidel method with x(0) = 0 to approximate the solution to the linear
system to within 10-5 in the ɭ norm.
Solution:
a. 2x1 - x2 + x3 = -1,
2x1 +2x2 + 2x3 = 4,
-x1 –x2 + 2x3 = -5
A [



]
We split A = D+L+U
Where D is diagonal matrix
-L is lower triangular part of A
-U is upper triangular part of A
A = [



] [



] [



]
The equation Ax = b or (D-L-U)x = b is then transformed into
Dx = (L+U)x + b
x = D-1 (L+U)x + D-1b
Tj = D-1 (L+U)
L + U = [



]
D-1 = [



]
Tj = D-1 (L+U)
Tj =[



] [



]
Tj =[



]
det(Tj – λI) = [



] [



]
= [



]
= -λ(λ2 + 0.5) – 0.5(λ+0.5) – 0.5(-0.5+0.5λ)
= -λ(λ2 + 0.5) – 0.5 λ – 0.25 + 0.25 – 0.25 λ
= -λ(λ2 + 0.5 + 0.75)
= -λ(λ2 + 1.25)
[ρ(Tj)]2 = 1.25 = 5/4
ρ(Tj) = √5/2 > 1
. Jacob method the sequence is
xk = Tx(k-1) + c











Given for initial approximation x(0) = 0, Then x(1) is given by


















Similarly following iterations are done:
k
( )
( )
( )
0 0 0 0
1 -0.5 2 -2.5
2 1.75 5 -1.75
3 2.875 2 0.875
4 0.0625 -1.75 -0.0625
5 -1.3438 2 -3.3438
6 2.1719 6.6875 -2.1719
7 3.9297 2 1.9297
8 -0.4648 -3.8594 0.4648
9 2.6621 2 -4.6621
10 2.8311 9.3242 -2.8311
11 5.5776 2 3.5776
12 -1.2888 -7.1553 1.2888
13 -4.7220 2 -6.7220
14 3.8610 13.4441 -3.8610
15 8.1526 2 6.1526
16 -2.5763 -12.3051 2.5763
17 -7.9407 2 -9.9407
18 5.4703 19.8814 -5.4703
19 12.1759 2 10.1759
20 -4.5879 -20.3517 4.5879
21 -12.9698 2 -14.9698
22 7.9849 29.9397 -7.9849
23 18.4623 2 16.4623
24 -7.7311 -32.9246 7.7311
25 -20.8279 -22.8279
c. 2x1 - x2 + x3 = -1,
2x1 +2x2 + 2x3 = 4,
-x1 –x2 + 2x3 = -5
A [



]
We split A = D+L+U
Where D is diagonal matrix
-L is lower triangular part of A
-U is upper triangular part of A
A = [



] [



] [



]
The equation Ax = b or (D-L-U)x = b is then transformed into
(D-L)x = Ux + b
x = (D-L)-1 Ux + (D-L)-1b
Tg = (D-L)-1 U
D-L = [



] [



]
D-L = [



]
(D-L)-1=[



]
Tg =[



] [



]
Tg =[



]
det(Tg – λI) = [



] [



]
det(Tg – λI) =[



]
=-λ(0.5+ λ)2-0.5(0-0)-0.5(0-0)
=-λ(0.5+ λ)2
λ = 0.5
d. Gauss-Seidel method the sequence is


















Given for initial approximation x(0) = 0, we generate the Gauss-Seidel iterates in the
following Table:
k
( )

( )

( )

0 0 0 0
1 -0.5 2.5 -1.5
2 1.5 2 -0.75
3 0.875 1.875 -1.125
4 1 2.125 -0.9375
5 1.03125 1.90625 -1.03125
6 0.96875 2.0625 -0.98438
7...