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

Marks : 8 M

Year : May 2013

10$x_1+x_2+x_3=12 \; \; or \; \; 10x_1=-x_2-x_3+12 \\ \; \\ \; \\ \therefore x_1=-0.1x_2-0.1x_3+1.2 \\ \; \\ \; \\ 2x_1+10x_2+x_3=13 \;,\; \; or \; \; 10x_2 = -2x_1 - x_3 + 13 \;,\; \\ \; \\ \; \\ x_2 = -0.2x_1 - 0.1x_3 + 1.3 \;,\; \\ \; \\ \; \\ 2x_1+2x_2+10x_3=14, \; \; \; or \; \; 10x_3 = -2x_1 - 2x_2 + 14, \\ \; \\ \; \\ x_3 = -0.2x_1 - 0.2x_2 + 1.4, \\ \; \\ \; \\ $

We initially assume the values of
$ x_1^{(0)} =x_2^{(0)}=x_3^{(0)}=5 $

[ $ x_1^{(0)} $ stands for initial approximation of $ x_1$ & so on], which we call as ‘poor approximations’.

Now using these initial values, we will go on calculating values of $ x_1$ , $ x_2$ & $ x_3$ at iterations.

For that, we shall make table as follows:-

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

First Iterations:

$ x_1^{(1)} -0.1x_2^{(0)} -0.1x_3^{(0)} +1.2 \;=\; x_1^{(1)} \\ \; \\ \; \\ x_2^{(1)} -0.2x_1^{(0)} - 0.1x_3^{(0)} + 1.3 \;=\; x_2^{(1)} \\ \; \\ \; \\ x_3^{(1)} -0.2x_1^{(0)} - 0.2x_2^{(0)} + 1.4, \;=\; x_3^{(1)} \\ \; \\ \; \\ $

Explanation of the table:

While calculating $ x_1^{(1)}$ , we have to use assumed values, i.e. $ x_2^{(0)}$ and $ x_3^{(0)}$ in the equation.

Now, while calculating $ x_2^{(1)}$ , we have newly calculated value, of $ x_1^{(1)}$ . So, we will use that (instead of $ x_1^{(0)}$ ). But, $ x_3^{(1)}$ value is ‘not available’ while calculating $ x_2^{(1)}$ . Hence, we shall use $ x_3^{(0)}$ value and then calculate $ x_2^{(1)}$ . To calculate $ x_3^{(1)}$ , we have $ x_1^{(1)}$ and $ x_2^{(1)}$ ‘available’ so we’ll use those values (instead of $ x_1^{(0)}$ & $ x_2^{(0)}$ ) and Calculate . $ x_3^{(1)}$

First iteration:

$ \\ \; \\ $

$ x_1^{(1)} \;=\; -0.1(5) -0.1(5) +1.2 \;=\; 0.2 \\ \; \\ \; \\ x_2^{(1)} \;=\; -0.2(0.2) - 0.1(5) + 1.3 \;=\; 0.76 \\ \; \\ \; \\ x_3^{(1)} \;=\; -0.2(0.2) - 0.2(0.76) + 1.4, \;=\; 1.208 \\ \; \\ \; \\ \; \\ \; \\ $

Second Iteration:

To calculate $x_1^{(1)}$ , we shall use $x_2^{(1)}$ & $x_3^{(1)}$ . To find $x_2^{(2)}$ , we’ll use $x_2^{(1)}$ & $x_3^{(1)}$ . To find $x_3^{(2)}$ , we’ll use $x_1^{(2)}$ & $x_2^{(2)}$ .

$ \\ \; \\ \therefore x_1^{(2)} \;=\; -0.1(0.76) -0.1(1.208) +1.2 \;=\; 1.0032 \\ \; \\ \; \\ x_2^{(2)} \;=\; -0.2(1.0032) - 0.1(1.208) + 1.3 \;=\; 0.97856 \\ \; \\ \; \\ x_3^{(2)} \;=\; -0.2(1.0032) - 0.2(0.97856) + 1.4, \;=\; 1.00365 \\ \; \\ \; \\ \; \\ \; \\ $

Third Iteration:

$ \\ \; \\ \therefore x_1^{(3)} \;=\; -0.1(0.97856) -0.1(1.00365) +1.2 \;=\; 1.00178 \\ \; \\ \; \\ x_2^{(3)} \;=\; -0.2(1.00178) - 0.1(1.00365) + 1.3 \;=\; 0.99928 \\ \; \\ \; \\ x_3^{(3)} \;=\; -0.2(1.00178) - 0.2(0.99928) + 1.4, \;=\; 0.99979 \\ \; \\ \; \\ \; \\ \; \\ $

$ \therefore$ Approximate values of $ x_1$ , $ x_2$ and $ x_3$ are 1.00178, 0.99928 and 0.99979 respectively.