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Tema 3: Resolución de sistemas mediante determinantes

Tema 3: Resolución de sistemas mediante determinantes

Tema 3: Resolución de sistemas mediante determinantes

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°§¢§£d) x + y + z = 6 1 1 1 6)2x + 2y + mz = 6 A' = 2 2 m 6mx = 0(m 0 0 014243A|1 1 1|m = 0| A | = 2 2 m = m (m – 2) = 0m 0 0m = 21 1 1 6)1 1• Si m = 0 8 A' = 2 2 0 6 | | ? 0 8 ran (A) = ran (A') = 2(0 0 0 02 0El sistema es compatible in<strong>de</strong>terminado.1 1 1 6)• Si m = 2 8 A' = 2 2 2 6(2 0 0 02|22 0||1 1 6|? 0; 2 2 6 ? 0 8 ran (A) = 2 ? ran (A') = 32 0 0El sistema es incompatible.°§¢§£• Si m ? 0 y m ? 2 8 ran (A) = ran (A') = 3. El sistema es compatible<strong>de</strong>terminado.s13 Discute los siguientes <strong>sistemas</strong> homogéneos en función <strong>de</strong>l parámetro a:° 2x – ay + 4z = 0§a) ¢ x + y + 7z = 0b)§£ x – y + 12z = 0°§¢§£x – z = 0ay + 3z = 04x + y – az = 0a) 2x – ay + 4z = 0x + y + 7z = 0x – y + 12z = 0Los <strong>sistemas</strong> homogéneos son siempre compatibles porque ran (A) = ran (A').Pue<strong>de</strong>n tener solución única o infinitas soluciones. Estudiamos el rango <strong>de</strong> A:|2 –a 4|| A | = 1 1 7 = 24 – 4 – 7a – 4 + 14 + 12a = 5a + 30 = 0 8 a = –61 –1 121 1• Si a = –6 8 ran (A) = ran (A') = 2, porque ? 0.1 –1El sistema es compatible in<strong>de</strong>terminado.• Si a ? –6 8 ran (A) = ran (A') = 3. El sistema es compatible <strong>de</strong>terminado.||30Unidad 3. <strong>Resolución</strong> <strong>de</strong> <strong>sistemas</strong> <strong>mediante</strong> <strong>de</strong>terminantes

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