Bilher Adiguna Sihombing (10 0404 123)

METODE GAUSS-SEIDELSistem Persamaan :

3x + y – z = 5 (1)

4x + 7y – 3z = 20 (2)

2x – 2y + 5z = 10 (3)

Penyelesaian :

x=13

(5− y+z )

y=17

(20−4 x+3 z )

z=15

(10−2x+2 y )

Iterasi 1

y = 0 ; z = 0

x '=13

(5−0+0 )=¿1,667

∈x'=

|1,667−0|1,667

x 100 %=100 %

x’ = 1,667 ; z = 0

y '=17(20−4.1,66 7+3.0)=1,904

∈y'=

|1,904−0|1,904

x 100 %=100 %

x’ = 1,667 ; y’ = 1,904

z '=15(10−2.1,667+2.1,904 )=2,095

∈z'=

|2,095−0|2,095

x100 %=100 %

Iterasi 2

y’ = 1,904 ; z’ = 2,095

x ' '=13

(5−1,904+2,095 )=¿1,730

Bilher Adiguna Sihombing (10 0404 123)

∈x' '=

|1,730−1,667|1,730

x100 %=3,64 %

x’’ = 1,730 ; z’ = 2,095

y ' '=17(20−4.1,730+3.2 ,095) = 2,766

∈y' '=

|2,766−1,904|2,766

x100 %=31,16 %

x’’ = 1,730 ; y’’ = 2,766

z ' '=15

(10−2.1,730+2.2,766 )= 2,414

∈z' '=

|2,414−2,095|2,414

x100 %=13,21 %

Iterasi 3

y’’ = 2,766 ; z’’ = 2,414

x ' ' '=13

(5−2,766+2,414 )=¿1,549

∈x' ' '=

|1,549−1,730|1,549

x 100 %=11,68%

x’’’ = 1,549 ; z’’ = 2,414

y ' ' '=17(20−4.1,549+3.2,414) = 3,006

∈y' ' '=

|3,006−2,766|3,006

x 100 %=7,98%

x’’’ = 1,549 ; y’’’ = 3,006

z ' ' '=15

(10−2.1,549+2.3,006 )= 2,583

∈z' ' '=

|2,583−2,414|2,583

x 100 %=6,54 %

Iterasi 4

y’’’ = 3,006 ; z’’’ = 2,583

x iv=13

(5−3,006+2,583 )=¿1,526

Bilher Adiguna Sihombing (10 0404 123)

∈xiv=

|1,526−1,549|1,5 26

x100 %=1,51 %

xiv = 1,526 ; z’’’ = 2,583

y iv=17(20−4.1,5 26+3.2,583) = 3,092

∈yiv=

|3,092−3,006|3,0 92

x100 %=2,78 %

xiv = 1,526 ; yiv = 3,092

z iv=15

(10−2.1,5 26+2.3,0 92 )= 2,626

∈ziv=

|2 ,626−2 ,583|2 ,626

x 100 %=1,64 %

Iterasi 5

yiv = 3,092 ; ziv = 2,626

xv=13

(5−3,092+2,626 )=¿1,511

∈xv=

|1,511−1,526|1,511

x100 %=0,99 %

xv = 1,511 ; ziv = 2,626

yv=17(20−4.1,511+3.2,626) = 3,119

∈yv=

|3,119−3,092|3,119

x 100 %=0,86 %

xv = 1,511 ; yv = 3,119

zv=15

(10−2.1,5 11+2.3 ,119)= 2,643

∈zv=

|2,6 43−2 ,626|2,6 43

x 100 %=0,64 %

Iterasi 6

yv = 3,119 ; zv = 2,643

xv i=13

(5−3 ,119+2,6 43 )=¿1,508

Bilher Adiguna Sihombing (10 0404 123)

∈xv i=

|1,508−1,511|1,508

x 100 %=0 ,19 %

xvi = 1,508 ; zv = 2,643

yv i=17(20−4.1,5 08+3. 2,6 43) = 3,128

∈yv i=

|3,128−3 ,119|3 ,128

x 100 %=0 ,29 %

xvi = 1,508 ; yvi = 3,128

zv i=15

(10−2.1,5 08+2.3,1 28 )= 2,648

∈zv i=

|2,648−2,643|2,6 48

x100 %=0 ,19 %

Iterasi 7

yvi = 3,128 ; zv = 2,648

xvii=13

(5−3,128+2,648 )=¿1,507

∈xvii=

|1,507−1,508|1,507

x100 %=0,07 %

xvii = 1,507 ; zvi = 2,648

yvii=17(20−4.1,507+3.2,648) = 3,131

∈yvii=

|3,131−3,128|3,131

x100 %=0,09%

xvii = 1,507 ; yvii = 3,131

zvii=15

(10−2.1,507+2.3,131 )= 2,650

∈zvii=

|2,650−2,648|2,650

x100 %=0,08 %

Maka, solusi dari sistem persamaan diatas adalah :

x = 1,507

y = 3,131

z = 2,650