This document provides examples of solving a system of linear equations using the Jacobi and Gauss-Seidel iterative methods. For the Jacobi method, the given 3x3 system is iteratively solved, reaching the final solution of X1=2.000, X2=3.000, X3=4.000 after 7 iterations. For the Gauss-Seidel method on the same system, the iterative process converges to the same solution after only 3 iterations.

1. -527685-544195KATHERINE LORENA SILVA ALONSOCODIGO: 2073612METODOS NUMERICOS <br />EXAMPLE OF ITERATIVES METHODS<br />JACOBI METHOD<br />Solving by the method of Jacobi, the following system of equations<br />6x1 + 2x2 + x3 = 22<br />- x1 + 8x2 + 2x3 = 30<br />x1 – x2 + 6x3 = 23<br />SOLUTION <br />x1k+1=1a11b1-a12x2-a13x3=16(22-2x2k-x3k<br />x2(k+1)=1830+x1k-2x3k<br />x3(k+1)=16(23-x1k+x2k)<br />x(0)=(b1b2b3a11b22a33)T=(226,308,236)T<br />For k=0 it is<br />x11=1622-2x20-x30<br />x21=1830+x10-2x30<br />x31=1623-x10+x20<br />Substituting the values:<br />x11=1622-2308-236=1.778<br />x21=1830+226-2236=3.250<br />x31=1623-226+308=3.847<br />Consequently<br />x1=1.778 3.250 3.847T<br />Now For k=1<br />x12=1622-2(3.250)-1.942=1.942<br />x22=1830+1.778-2(3.847)=3.011<br />x32=1623-1.778+3.250=4.079<br />It is <br />x2=1.942 3.011 4.079T<br />Following these steps will have then:<br />For k=2<br />x3=1.983 2.973 4.012T<br />For k=3<br />x4=2.007 2.995 3.998T<br />For k=4<br />x5=2.002 3.001 3.998T<br />For k=5<br />x6=2.000 3.001 4.000T<br />For k=6<br />x7=2.000 3.000 4.000T<br />With what the final solution is obtained:<br />X1=2.000 X2=3.000 X4=4.000<br />GAUSS SEIDEL METHOD<br />Solve the following system of linear equations using the Gauss-Seidel.<br />6x1 + 2x2 + x3 = 22<br />- x1 + 8x2 + 2x3 = 30<br />x1 – x2 + 6x3 = 23<br />SOLUTION<br />Solve for x1, x2 and x3 of the first, second and third equation respectively<br />x1=1622-2x2-x3<br />x2=1830+x1-2x3<br />x3=1623-x1+x2<br />Now, we express these terms punts recursive<br />x1k+1=16(22-2x2k-x3k<br />x2(k+1)=1830+x1k-2x3k<br />x3(k+1)=16(23-x1k+x2k)<br />We have the vector of initial approximations<br />x(0)=(22/6,30/8,23/6)T<br />If k=0, <br />x10=226, x20=308, x30=236<br />x11=1622-2(308)-236=1.778<br />x21=1830+1.778-2236=3.014<br />x31=1623-1.778+3.014=4.039<br />Now x11=1.778, x21=3.014, x31=4.039<br />x1-x2=(1.778-3.667, 3.014-3.75, 4.039-3.833=(1.889, 0.736, 0.206)<br />In the following iterations, the results are:<br />k=1 With x12=1.989, x22=2.989, x32=4.000<br />k=2 with <br />x13=2.004, x23=3.001, x33=4.000<br />K=3 with <br />x14=2.000, x24=3.000, x34=4.000<br />