0
32kviews
Solve the following equations by Gauss-Siedel Method: 27x+6y-z=85, 6x+15y+2z=72, x+y+54z=110

Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 8 M

Year : Dec 2014

1 Answer
1
6.4kviews

The given equations can be written as follows:

$ x= \dfrac{-6}{27} y + \dfrac{1}{27} z+ \dfrac{85}{27} \\ \; \\ y= -\dfrac{6}{15} x- \dfrac{2}{15} z+ \dfrac{72}{15} \\ \; \\ z=-\dfrac{1}{54} x- \dfrac{1}{54} y+ \dfrac{110}{54} \\ \; \\ \; \\ \; \\ $

We will make use of following Iteration formulae :

$ x^{(k+1)}= \dfrac{-6}{27} y^{(k)} + \dfrac{1}{27} z^{(k)} + \dfrac{85}{27} \\ \; \\ y^{(k+1)}= -\dfrac{6}{15} x^{(k+1)}- \dfrac{2}{15} z^{(k)} + \dfrac{72}{15} \\ \; \\ z^{(k+1)}=-\dfrac{1}{54} x^{(k+1)}- \dfrac{1}{54} y^{(k+1)}+ \dfrac{110}{54} \\ \; \\ \; \\ \; \\ $

We will start from k=0. Hence, initially x, y and z values are equal to zero. Now, we will go on incrementing k value by 1.

$ \therefore x^{1} \;=\; \dfrac{85}{27} \;=\; 3.148 \\ \; \\ \; \\ y^{1} \;=\; -\dfrac{6}{15} \times (3.148) \;-\; \dfrac{2}{15}(0) + \dfrac{72}{15} \;=\; 3.541 \\ \; \\ \; \\ z^{1} \;=\; -\dfrac{1}{54} (3.148)- \dfrac{1}{54} (3.541)+ \dfrac{110}{54} \;=\; 1.913 \\ \; \\ \; \\ \; \\ \; \\ $

Likewise, we will calculate till values of x, y, z become unchanged for further iterations.

k x y z
1 3.148 3.541 1.913
2 2.432 3.572 1.926
3 2.426 3.573 1.926
4 2.426 3.573 1.926

$ \\ \; \\ \; \\ $

Hence, x=2.426, y=3.573 and z=1.926

Please log in to add an answer.