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

Marks : 8 M

Year : May 2014

Now,

$10x-5y-2z=3 \; \; \therefore 10x= 5y+2z+3 \; \; \; \therefore x= 0.5y+0.2z+0.3 \\ \; \\ \; \\ \; \\ 4x-10y+3z=-3 \; \; \therefore 10y = 4x+3z+3 \; \; \therefore y = 0.4x+0.3z+0.3 \\ \; \\ \; \\ \; \\ x+6y+10z=-3 \; \; \therefore 10z = -x-6y-3 \; \; \therefore z = -0.1x-0.6y-0.3 \\ \; \\ \; \\ \; \\ $

We initially assume the values as x(0)=5 , y(0)=5 & z(0)=5

$ \\ \; \\ $

First Iterations:

$ x^{(1)}= 0.5y^{(0)}+0.2z^{(0)}+0.3 \;=\; 0.5(5)+0.2(5)+0.3 \;=\; 3.8 \\ \; \\ \; \\ y^{(1)} = 0.4x^{(1)}+0.3z^{(0)}+0.3 \;=\; 0.4(3.8)+0.3(5)+0.3 \;=\; 3.32 \\ \; \\ \; \\ z^{(1)} = -0.1x^{(1)}-0.6y^{(1)}-0.3 \;=\; -0.1(3.8)-0.6(3.32)-0.3 \;=\; -2.672 \\ \; \\ \; \\ \; \\ \; \\ $

Second Iterations:

$ x^{(2)}= 0.5y^{(1)}+0.2z^{(1)}+0.3 \;=\; 0.5(3.32)+0.2(-2.672)+0.3 \;=\; 1.4256 \\ \; \\ \; \\ y^{(2)} = 0.4x^{(2)}+0.3z^{(1)}+0.3 \;=\; 0.4(1.4256)+0.3(-2.672)+0.3 \;=\; 0.0686 \\ \; \\ \; \\ z^{(2)} = -0.1x^{(2)}-0.6y^{(2)}-0.3 \;=\; -0.1(1.4256)-0.6(0.0686)-0.3 \;=\; -0.4837 \\ \; \\ \; \\ \; \\ \; \\ $

Third Iterations:

$ x^{(3)}= 0.5y^{(2)}+0.2z^{(2)}+0.3 \;=\; 0.5(0.686)+0.2(-0.4837)+0.3 \;=\; 0.2376 \\ \; \\ \; \\ y^{(3)} = 0.4x^{(3)}+0.3z^{(1)}+0.3 \;=\; 0.4(0.2376)+0.3(-2.4837)+0.3 \;=\; 0.2499 \\ \; \\ \; \\ z^{(3)} = -0.1x^{(3)}-0.6y^{(3)}-0.3 \;=\; -0.1(0.2376)-0.6(0.2499)-0.3 \;=\; -0.4737 \\ \; \\ \; \\ \; \\ \; \\ $

$\therefore$ Approximate values of x , y and z are 0.2376, 0.2499 and -0.4737 respectively.