KTU S4 EC,CE,ME,EEE
MA204,MAT204 ,
MA202 , MAT202
Lecture 1
Gauss Siedal Iteration
,HELLO!
I am Prijimol V B
Assistant Professor
Christ College of Engineering,
Irinjalakuda
,Gauss Seidal Iteration
1. Solve the system of equations by Gauss Seidel iteration starting with
(0,0,0)𝑇 , 10𝑥 + 𝑦 + 𝑧 = 6, 𝑥 + 10𝑦 + 𝑧 = 6, 𝑥 + 𝑦 + 10𝑧 = 6
Ans: The system of equations is diagonally dominant
Since 10 > 1 + |1| 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 > 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑦 + |𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑧|
10 > 1 + |1| 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑦 > 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 + |𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑧|
10 > 1 + |1| 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑧 > 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 + |𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑦|
, Therefore from first equation, 10𝑥 = 6 − 𝑦 − 𝑧
𝟏
𝒙= (𝟔 − 𝒚 − 𝒛)
𝟏𝟎
From second equation, 10𝑦 = 6 − 𝑥 − 𝑧
𝟏
𝒚= (𝟔 − 𝒙 − 𝒛)
𝟏𝟎
From third equation, 10𝑧 = 6 − 𝑥 − 𝑦
𝟏
𝒛= (𝟔 − 𝒙 − 𝒚)
𝟏𝟎
MA204,MAT204 ,
MA202 , MAT202
Lecture 1
Gauss Siedal Iteration
,HELLO!
I am Prijimol V B
Assistant Professor
Christ College of Engineering,
Irinjalakuda
,Gauss Seidal Iteration
1. Solve the system of equations by Gauss Seidel iteration starting with
(0,0,0)𝑇 , 10𝑥 + 𝑦 + 𝑧 = 6, 𝑥 + 10𝑦 + 𝑧 = 6, 𝑥 + 𝑦 + 10𝑧 = 6
Ans: The system of equations is diagonally dominant
Since 10 > 1 + |1| 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 > 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑦 + |𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑧|
10 > 1 + |1| 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑦 > 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 + |𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑧|
10 > 1 + |1| 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑧 > 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 + |𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑦|
, Therefore from first equation, 10𝑥 = 6 − 𝑦 − 𝑧
𝟏
𝒙= (𝟔 − 𝒚 − 𝒛)
𝟏𝟎
From second equation, 10𝑦 = 6 − 𝑥 − 𝑧
𝟏
𝒚= (𝟔 − 𝒙 − 𝒛)
𝟏𝟎
From third equation, 10𝑧 = 6 − 𝑥 − 𝑦
𝟏
𝒛= (𝟔 − 𝒙 − 𝒚)
𝟏𝟎
