Answer to Question #249539 in Quantitative Methods for kofi

Question

Consider the following tridiagonal system of linear equations:

[3 2 0 0 [x₁ [12

2 3 2 0 x₂ = 17

0 2 3 2 x₃ 14

0 0 2 3] x₄ ] 7]

Solve this system using Thomas algorithm

Let 𝑥^(0)T=[0 0 0 0] and solve the system using Jacobi, and Gauss-Seidel for at least five iterations

Answer 1

Using Thomas algorithm:

$a_1=2/3,b_1=12/3=4$

$a_2=2/(3-2\cdot2/3)=6/5,b_2=(17-2\cdot4)/(3-2\cdot2/3)=27/5$

$a_3=2/(3-2\cdot6/5)=10/3,b_3=(14-2\cdot27/5)/(3-2\cdot6/5)=16/3$

$b_4=(7-2\cdot16/3)/(3-2\cdot10/3)=1$

$x_4=b_4=1$

$x_3=b_3-a_3x_4=16/3-10/3=2$

$x_2=b_2-a_2x_3=27/5-6\cdot2/5=3$

$x_1=b_1-a_1x_2=4-2\cdot3/3=2$

Using Jacobi method:

$x_1^{(1)}=12/3=4,x_2^{(1)}=17/3,x_3^{(1)}=14/3,x_4^{(1)}=7/3$

$x_1^{(2)}=1/3(12-2\cdot17/3)=2/9$

$x_2^{(2)}=1/3(17-2\cdot4-2\cdot14/3)=-1/9$

$x_3^{(2)}=1/3(14-2\cdot17/3-2\cdot7/3)=-2/3$

$x_4^{(2)}=1/3(7-2\cdot14/3)=-7/9$

$x_1^{(3)}=1/3(12+2/9)=110/27$

$x_2^{(3)}=1/3(17-2\cdot2/9+2\cdot2/3)=161/27$

$x_3^{(3)}=1/3(14+2/9+2\cdot7/9)=142/27$

$x_4^{(3)}=1/3(7+2\cdot2/3)=25/9$

$x_1^{(4)}=1/3(12-2\cdot161/27)=2/81$

$x_2^{(4)}=1/3(17-2\cdot110/27-2\cdot142/27)=-45/81=-5/9$

$x_3^{(4)}=1/3(14-2\cdot161/27-2\cdot25/9)=-94/81$

$x_4^{(4)}=1/3(7-2\cdot142/27)=-95/81$

$x_1^{(5)}=1/3(12+2\cdot5/9)=118/27=4.37$

$x_2^{(5)}=1/3(17-2\cdot2/81+2\cdot94/81)=1561/243=6.42$

$x_3^{(5)}=1/3(14+2\cdot5/9+2\cdot95/81)=1414/243=5.82$

$x_4^{(5)}=1/3(7+2\cdot94/81)=755/243=3.11$

Using Gauss-Seidel method:

$x_1^{(1)}=12/3=4$

$x_2^{(1)}=1/3(17-2\cdot4)=3$

$x_3^{(1)}=1/3(14-2\cdot3)=8/3$

$x_4^{(1)}=1/3(7-2\cdot8/3)=5/9$

$x_1^{(2)}=1/3(12-2\cdot3)=2$

$x_2^{(2)}=1/3(17-2\cdot2-2\cdot8/3)=23/9$

$x_3^{(2)}=1/3(14-2\cdot23/9-2\cdot5/9)=90/27=10/3$

$x_4^{(2)}=1/3(7-2\cdot10/3)=1/9$

$x_1^{(3)}=1/3(12-2\cdot23/9)=62/27$

$x_2^{(3)}=1/3(17-2\cdot62/27-2\cdot10/3)=155/81$

$x_3^{(3)}=1/3(14-2\cdot155/27-2\cdot1/9)=90/27=62/81$

$x_4^{(3)}=1/3(7-2\cdot62/81)=443/243$

$x_1^{(4)}=1/3(12-2\cdot155/81)=662/243$

$x_2^{(4)}=1/3(17-2\cdot662/243-2\cdot62/81)=2435/729$

$x_3^{(4)}=1/3(14-2\cdot2435/729-2\cdot443/243)=90/27=1.22$

$x_4^{(4)}=1/3(7-2\cdot1.22)=1.52$

$x_1^{(5)}=1/3(12-2\cdot2435/729)=1.77$

$x_2^{(5)}=1/3(17-2\cdot1.77-2\cdot1.22)=3.67$

$x_3^{(5)}=1/3(14-2\cdot3.67-2\cdot1.52)=90/27=1.21$

$x_4^{(5)}=1/3(7-2\cdot1.21)=1.53$