12.07.2015 Views

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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°¢£°¢£UNIDAD32. Discute y resuelve, en función <strong>de</strong>l parámetro a, el siguiente sistema <strong>de</strong>ecuaciones:° (a – 1)x + y = 0¢£ (a – 1)x + (a + 1)y = 0(a – 1)x + y = 0(a – 1)x + (a + 1)y = 0A =| A |1 1= (a – 1) = (a – 1) (a + 1 – 1) = a (a – 1) = 01 a + 1• Si a = 0, queda:||°¢£(a – 1 1a – 1 a + 1)a = 0a = 1–x + y = 0–x + y = 0y = x. Sistema compatible in<strong>de</strong>terminado.Soluciones: x = l, y = l• Si a = 1, queda:y = 02y = 0Sistema compatible in<strong>de</strong>terminado.Soluciones: x = l, y = 0• Si a ? 0 y a ? 1 8 ran (A) = 2El sistema solo tiene la solución trivial: x = 0, y = 0Página 881. Calcula la inversa <strong>de</strong> cada una <strong>de</strong> las siguientes matrices:1 –1 –1)2 –1 )A = –1 0 3 B =((–2 5 –31 –2Calculamos la inversa <strong>de</strong> la matriz A:| A | = –1 ? 0 8 Existe A –1a ijÄÄÄ8 Adj (A) ÄÄÄ8 (Adj (A)) t 1ÄÄÄ8 (Adj (A)) t| A|(–15 9 –5)8 –5 3–3 2 –1(–15 –9 –5) (–15 –8 –3) (15 8 3)8 –8 –5 –3 8 –9 –5 –2 8 9 5 2 = A –1–3 –2 –1 –5 –3 –1 5 3 1Unidad 3. <strong>Resolución</strong> <strong>de</strong> <strong>sistemas</strong> <strong>mediante</strong> <strong>de</strong>terminantes17

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