Solución
del problema mediante el método de Cramer
2S1+2s2+1S3 =
4.5--------------Ecuación 1
4S1+6s2+3S3 =
12--------------Ecuación 2
6S1+9s2+7S3 = St--------------Ecuación
3
El sistema de ecuaciones
lineales anterior es de la forma
Ax = b
Donde
A=
|
2
|
2
|
1
|
X=
|
X1
|
b=
|
4.5
|
|||
4
|
6
|
3
|
x2
|
12
|
||||||
6
|
9
|
7
|
x2
|
1
|
Que esta
representado por
A=
|
2
|
2
|
1
|
X1
|
=
|
4.5
|
|
4
|
6
|
3
|
X2
|
12
|
|||
6
|
9
|
7
|
X2
|
1
|
4.5
|
2
|
1
|
|A2|=
|
2
|
4.5
|
1
|
|A3|=
|
2
|
2
|
4.5
|
|||
12
|
6
|
3
|
4
|
12
|
3
|
4
|
6
|
12
|
|||||
1
|
9
|
7
|
6
|
1
|
7
|
6
|
9
|
1
|
4.5(42-27)- 2(84-3) + 1(108-3) = 68 - 81 +105 = 92
D2 = |A2|
2(84-3)-
4.5(28-18) + 1(4St-72) = 162 - 45 -68 =49
D3 = |A2|
2(6 - 108)-
2(4 -72) + 4.5(36-36) = -204 + 136 +0=-12
2St = -68
D= |A|
2(42-27)-
2(28-18) + 1(36-36) = 10
X1=
|
D1
|
X1=
|
D1
|
X1=
|
D1
|
||
D
|
D
|
D
|
X1 = 92/10 = 9.2
X2 = 49/10 = 4.9
X2 = -68/10 = -6.8
