Metodos numéricos  Ecs. lineales Ejemplo 2 |
 pdf |
Ejercicio 2
Resuelva el siguiente sistema de ecuaciones lineales usando el método de
Gauss - Jordan
3x1 - 5x3 + 6x4 = 24
x1 + 9x2 - 7x3 - 10x4 = - 66
5x1 + 3x2 + 9x4 = 91
x1 - x2 + x3 - x4 = 7
Representación matricial
|
3 |
0 |
-5 |
6 |
24 |
|
|
1 |
9 |
-7 |
-10 |
-66 |
|
|
5 |
3 |
0 |
9 |
91 |
|
|
1 |
-1 |
1 |
-1 |
7 |
|
Iteración 1.
|
1 |
0 |
-1.666 |
2 |
8 |
|
|
0 |
9 |
-5.334 |
-12 |
-74 |
|
|
0 |
3 |
8.333 |
-1 |
51 |
|
|
0 |
-1 |
2.6666 |
-3 |
-1 |
|
Iteración 2
|
1 |
0 |
-1.666 |
2 |
8 |
|
|
0 |
1 |
-.592592 |
-1.33333 |
-8.222222 |
|
|
0 |
0 |
10.11099 |
2.99999 |
75.66666 |
|
|
0 |
0 |
2.074074 |
-4.33333 |
-9.222222 |
|
Iteración 3.
|
1 |
0 |
0 |
2.494509 |
20.47266 |
|
|
0 |
1 |
0 |
-1.157507 |
-3.787501 |
|
|
0 |
0 |
1 |
0.296706 |
7.483599 |
|
|
0 |
0 |
0 |
-4.948723 |
-24.74376 |
|
Iteración 4.
|
1 |
0 |
0 |
0 |
8.000042 |
|
|
0 |
1 |
0 |
0 |
2.000067 |
|
|
0 |
0 |
1 |
0 |
6.00006 |
|
|
0 |
0 |
0 |
1 |
5.000029 |
|
Solución.
X1=8, X2=2, X3=6, X4=5
Sustitución
3(8) - 5(6) + 6(5) = 24
8 + 9(2) - 7(6) - 10(5) = - 66
5(8) + 3(2) + 9(5) = 91
8 – 2 + 6 – 5 = 7