Metodos numéricos  Ecs. lineales Ejemplo 4 |
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Ejercicio 4
Resuelva el siguiente sistema de ecuaciones lineales usando el método de
Gauss - Jordan
3.141x1 + 6.954x2 + 3x3 = 43.098
12x1 + 0.3x2 - 1.85x3 = 19.65
- 6.66x1 - 0.94x2 - x3 = - 20.98
Representación matricial
|
3.141 |
6.954 |
3 |
43.098 |
|
|
12 |
0.3 |
-1.85 |
19.65 |
|
|
-6.66 |
-0.94 |
-1 |
-20.98 |
|
Iteración 1.
|
1 |
2.213944 |
0.955109 |
13.721107 |
|
|
0 |
-26.267328 |
-13.31130 |
-145.003284 |
|
|
0 |
13.804867 |
5.361025 |
70.432572 |
|
Iteración 2
|
1 |
0 |
-0.166833 |
1.499494 |
|
|
0 |
1 |
0.506762 |
5.520290 |
|
|
0 |
0 |
-1.634757 |
-5.774297 |
|
Iteración 3.
|
1 |
0 |
0 |
2.088782 |
|
|
0 |
1 |
0 |
3.730302 |
|
|
0 |
0 |
1 |
3.532205 |
|
Solución
X1= 2.088782
X2= 3.730302
X3= 3.532205
Sustitución
3.141(2.088782) + 6.954(3.730302) + 3(3.532205) = 43.09799937
12(2.088782) + 0.3(3.730302) – 1.85(3.532205) = 19.64989535
-6.66(2.088782) – 0.94(3.730302) - 3.532205= - 20.949977