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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°§¢§£UNIDAD3Resolvemos ahora el sistema para a = 3:x – y + z = 33x + 2y – z = 92x + 3y – 2z = 6Sabemos que el sistema es compatible in<strong>de</strong>terminado.Eliminamos la 3. a ecuación, pasamos z al segundomiembro y lo resolvemos aplicando la regla <strong>de</strong> Cramer:x – y = 3 – z3x + 2y = 9 + z§ 9 + l § 2§ 3 § 23 – l –11 3 – l15 – l §3 9 + l § 4lx = = , y = = , z = l1 –1 55 52. Determina para qué valores <strong>de</strong> a existe la matriz inversa <strong>de</strong> M. Calcula dichamatriz inversa para a =2.(2 1 –a)M = 2a 1 –12 a 1La matriz tendrá inversa si su <strong>de</strong>terminante es distinto <strong>de</strong> 0.|2 1 –a| |1 1 –a|| M| = 2a 1 –1 = 2 a 1 –1 = 2(1 – a 3 – 1 + a + a – a) = 2(–a 3 + a)2 a 1 1 a 1| M | = 0 8 –2(a 3 – a) = 0 8 –2a(a 2 – 1) = 0M tiene inversa si a ? 0, a ? 1 y a ? –1.Para a = 2:2 1 –2)M = 4 1 –1 ; |M| = –12(2 2 1M 11= 3; M 12= –6; M 13= 6M 21= –5; M 22= 6; M 23= –2M 31= 1; M 32= –6; M 33= –2(3 –6 6)(3 –5 1)(M ij) = 8 (M ij) t = 8 M –1 1–5 6 –2–6 6 –6 = –1 –6 –26 –2 –212(M ij) t(–1/4 5/12 –1/12)M –1 = 1/2 –1/2 1/2–1/2 1/6 1/6°¢£a = 0a = 1a = –1Unidad 3. <strong>Resolución</strong> <strong>de</strong> <strong>sistemas</strong> <strong>mediante</strong> <strong>de</strong>terminantes59

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