Question

QUESTION 2 Consider the linear system 11 0.50 + Ii 22 0.579 + 23 0.2533 13 0.2 -1.425 2 whose solution is (0.9,-0.8.0.7). (a) Determine whether the coefficient matrix is strictly diagonally dominant. (5) (b) Approximate the solution of the system by performing two iterations of the Gauss-Seidel algorithm, using (10) x(0) = (0,0,0) as the initial guess. (c) Approximate the solution of the system using one iteration of the SOR scheme, with w x() = (0,0,0) = 0.7 and (5) [20]

Answer 1

Answer:

Consider the linear system

.01 - 13 = 0.2 0.5.21 + 12 – 0.25x3 = -1.425 + (1) 11 -0.5.22 +23 = 2

(a)

From(1) we get coeeficient matrix $AX=B$

0 -1 A= 1 0.5 1 -0.25 -0.5 1

Here

011 1; 12 0; 013

  $a_{21}=0.5;a_{22}=1;a_{23}=-0.25;$

  $a_{31}=1;a_{32}=-0.5;a_{33}=1.$

Here $|a_{11}|=|a_{12}|+|a_{13}|;$

  

a22 > 221 + a23

$|a_{33}|>|a_{31}|+|a_{33}|;$

So is not strictly diagnal dominant matrix

(b)

From(1) $\Rightarrow X_{1}=[(0.2)-(0)X_{2}+X_{3}]$

  $X_{2}=[-1.142-X_{1}+0.25X_{3}]$

$X_{3}=[2-X_{1}+0.5X_{2}]$

By Gauss-Seidel Method   $\Rightarrow X_{1}^{(K+1)}=[0.2-X_{2}^{K}+X_{3}^{K}]$

  $X_{2}^{(K+1)}=[-1.142-X_{1}^{(K+1)}+0.25X_{3}^{(K)}]$

  $X_{3}^{(K+1)}=[2-X_{1}^{(K+1)}+0.5X_{2}^{(K+1)}]$

$X_{1}^{0}=0;X_{2}^{0}=0;X_{3}^{0}=0$

First iteration:

$X_{1}^{1}=[0.2-0+0]=0.2$

x-1) = (-1.425 – 0.2 +0.25(0)] = -1.625

$X_{3}^{(1)}=[2-(0.2)+(0.5)(-1.625)]=0.9875$

$X_{1}^{(1)}=0.2;X_{2}^{(1)}=-1.625;X_{3}^{(1)}=0.9875$

Second Iteration:

$X_{1}^{(2)}-[0.2-0(-1.625)+0.9875]=1.1875$

$X_{2}^{(2)}=[-1.425-1.1875+(0.25)(0.9875)]=-2.3656$

$X_{3}^{(2)}=[2-1.1875+(0.5)(-2.36562)]=-0.37031$

$X_{1}^{(2)}=1.1875;X_{2}^{(2)}=-2.3656;X_{3}^{(2)}=-0.37031$

(c)

For SOR Method ; $\omega=0.7$

$X_{1}^{(K+1)}=(1-\omega)X_{1}^{(K)}+(\omega)[0.2-X_{2}^{(K)}+X_{3}^{(K)}]$

$X_{2}^{(K+1)}=(1-\omega)X_{2}^{(K)}+(\omega)[-1.425-X_{1}^{(K+1)}+(0.25)X_{3}^{(K)}]$

$X_{3}^{(K+1)}=(1-\omega)X_{3}^{(K)}+(\omega)[2-X_{1}^{(K+1)}+(0.5)X_{2}^{(K+1)}]$

$X_{1}^{0}=0;X_{2}^{0}=0;X_{3}^{0}=0$

first iteration:

$X_{1}^{(1)}=(1-0.7)(0)+(0.7)[0.2-0+0]=0.14$

$X_{2}^{(1)}=(1-0.7)(0)+(0.7)[-1.425-0.14+0]=0.9975$

$X_{3}^{(1)}=(1-0.7)(0)+(0.7)[2-0.14+(0.5)(-0.9975)]=0.952875$

$X_{1}^{(1)}=0.14;X_{2}^{(1)}=-0.9975;X_{3}^{(1)}=0.952875$