Sistema De Ecuaciones Lineales

SISTEMA DE ECUACIONES LINEALES (METODO DETERMINANTES)
1.-Un empresario tiene tres maquinas que son empleados es la fabricación de tres productos diferentes .Para utilizar plenamente las maquinas estas estarán en operación: Mirar la tabla:
Verificar las horas restante:
| Producto1 | Producto2 | Producto3 | Hdisponible | |
| | | | | |
Maquina 1 | 2 | 4 | 6 | 18 | |
Maquina 2 | 4| 5 | 6 | 24 | |
Maquina 3 | 3 | 1 | 2 | 4 | |
| | | | | |
| | | | | |
Solución | | | | | |
| | | | | |
2x1 + 4x2 + 6x3 = 18 | | | |
4x1 + 5x2 + 6x3 = 24 | | | | |
3x1 + x2 + 2x3 = 4 | | | | |
| | | | | |
2 |
| 4 | 6 | | | |
4 | 5 | 6 | | | |
3 | 1 | 2 | | | |
2 | 4 | 6 | | | |
x = 4 | 5 | 6 | = || |
| | | (20+24+72)-(90+12+32) |
| | | = |
| | |
| | | (20+24+72-90-12-32) |
| | | = |
| | |
| | | -18 |
18 | 4 | 6 | | | |
24 | 5 | 6 | | | |
4 | 1 | 2 | | | |
18 | 4 | 6 | | | |
x = 24 | 5 | 6 | = (180+144+96)-(120+108+192) |
| -18 | | | -18 | |
| | | | | |
| | | = (180+144+96-120-108-192) |
|
| | || -18 | |
| | = |
| 0 | |
| 0
|
| | | -18 | = | |

2 | 18 | 6 | | | |
4 | 24 | 6 | | | |
3 | 4 | 2 | | | |
2 | 18 | 6 | | | |
x2= 4 | 24 | 6 | = (96+96+96+324) - (432+48+144) |
| -18 | | | -18 | |
| | | | | |
| | | = ( 96+96+324-432-48-144 ) |
| | | | -18 | |
| | | | | |
| | |= - 108 | 6
|
| |
| | | -18 | = | |

2 | 4 | 18 | | | |
4 | 5 | 14 | | | |
3 | 1 | 4 | | | |
2 | 4 | 18 | | | |
x3= 4 | 5 | 24 | = ( 40+72+288 ) - ( 270+48+64 ) |
| -18 | | | -18 | |
| | | | | |
| | | = (40+72+288-270-48-64) |
| | | | -18 | |
| | | | | |
| | | = | 18 | 1
|
|
| | | | -18 |= |
| | | | | |
| | | | | |
2.-Una industria fabrica zinc, teja, duratecho .Para la fabricación de cada uno de estos tipos necesitan los siguientes materiales acero, carbón, corrugado. La compañía tenía en existencia 20 unidades de acero, 10 de carbón, y 15 de corrugado.¿Cuantos zinc,teja,duratecho?
| acero | carbón | corrugado | total | |
| | | | | |
zinc | 1 | 2 |4 | 20 | |
teja | 3 | 2 | 1 | 10 | |
duratecho | 1 | 2 | 3 | 15 | |
| | | | | |
| | | | | |
Solución | | | | | |
| | | | | |
x1+2x2+4x3=20 | | | | |
3x1+2x2+x3=10 | | | | |
x1+2x2+3x3=15 | | | | |
| | | | | |
1 | 2 | 4 | | | |
3 | 2 | 1 | | | |
1 | 2 | 3 | | | |
1 | 2 | 4 | | | |
x=3 | 2 | 1 | = || |
| | | (6+24+2) - (8+2+18) |
| | | |
| | | = (6+24+2-8-2-18) |
| | | | | |
| | | = 4 |
20 | 2 | 4 | | | |
10 | 2 | 1 | | | |
15 | 2 | 3 | | | |
20 | 2 | 4 | | | |
x1= 10 | 2 | 1 | = (120+80+30) - (120+40+90) |
| | | | 4 | |
| | | | | |
| | | = (120+80+30-120-40-90) |
| | | | 4 | |
| | | | - 5
|
||
| | | = - 20 | = | |
| | | 4 | | |
| | | | | |
1 | 20 | 4 | | |
3 | 10 | 1 | | |
1 | 15 | 3 | | |
1 | 20 | 4 | | |
x2= 3 | 10 | 1 | = (30+180+20) - (40+15+180) |
| | | 4 |
| | | | |
| | | = (30+180+20-40-15-180) |
| | | 4 |
| | | |
| -1,25
|
| | = | - 5 | |
| | | 4 | |
| | | | |

1 | 2 | 20| | |
3 | 2 | 10 | | |
1 | 2 | 15 | | |
1 | 2 | 20 | | |
x3=3 | 2 | 10 | = (30+120+20) - (40+20+90) |
| | | 4 |
| | | | |
| | | =(30+120+20-40-20-90) |
| | | 4 |
| | | | 5
|
|
| | = | 20 | |
| | | 4 | |
| | | | |
| | | | |
3.-Tenemos 3 tanques de variados derivados de petróleo son gasolina, diesel, kerex.El primer...