082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

316 Chapter 8 Matrix algebra
We find
x (1) = − 1 8 = −0.1250
y (1) = − 16 5 = −3.2000
z (1) = − 7 4 = −1.7500
Then,
x (2) = 1 8 (−3.2000) + 1 8 (−1.7500) − 1 8 = −0.7438
y (2) = 1 5 (−0.1250) + 1 5 (−1.7500) − 16 5 = −3.5750
z (2) = 1 4 (−0.1250) + 1 4 (−3.2000) − 7 4 = −2.5813
Finally,
x (3) = 1 8 (−3.5750) + 1 8 (−2.5813) − 1 8 = −0.8945
y (3) = 1 5 (−0.7438) + 1 5 (−2.5813) − 16 5 = −3.8650
z (3) = 1 4 (−0.7438) + 1 4 (−3.5750) − 7 4 = −2.8297
To apply the Gauss--Seidel iteration to Equation (8.11), the most recent approximation
isusedateach stage leading to
x (n+1) = 1 8 y(n) + 1 8 z(n) − 1 8
y (n+1) = 1 5 x(n+1) + 1 5 z(n) − 16 5
z (n+1) = 1 4 x(n+1) + 1 4 y(n+1) − 7 4
Startingfromx (0) =y (0) =z (0) = 0,wefind
x (1) = − 1 8 = −0.1250
y (1) = 1 5 (−0.1250) + 1 5 (0) − 16 5 = −3.2250
z (1) = 1 4 (−0.1250) + 1 4 (−3.2250) − 7 4 = −2.5875