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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 = 5x + z = 6y + z = 72x + y + z =11Tenemos que:1 1 0 5)1 0 1 6A' =0 1 1 7(2 1 1 1114243A||FILAS1 1 0 5|||(1.ª)1 1 0 51 0 1 6 (2.ª) – (1.|A'|= = a ) 0 –1 1 1=0 1 1 7 (3.ª)0 1 1 72 1 1 11 (4.ª) – 2 · (1. a ) 0 –1 1 1|–1 1 1| |1 1 0|= 1 1 7 = 0 y 1 0 1 = –2 ? 0–1 1 1 0 1 1Luego ran (A) = ran (A') = n.° <strong>de</strong> incógnitas = 3.El sistema es compatible <strong>de</strong>terminado.Para resolverlo, po<strong>de</strong>mos prescindir <strong>de</strong> la 4. a ecuación:§ 7 1 1 §§ 0 7 1 §§ 0 1 7 §5 1 05 5 01 1 56 0 11 6 11 0 6x =–4–6–8= = 2; y = = = 3; z = = = 4–2 –2–2 –2–2 –2Solución: x = 2, y = 3, z = 4Página 95Discusión <strong>de</strong> <strong>sistemas</strong> <strong>mediante</strong> <strong>de</strong>terminantess12 Discute los siguientes <strong>sistemas</strong> según los valores <strong>de</strong>l parámetro m:° 7y + 5z = –7§a) ¢ 3x + 4y + mz = –1b)§£ 7x + 5z = 7° 2x + y – z = 1§c) ¢ x – 2y + z = 3d)§£ 5x – 5y + 2z = m°§¢§£°§¢§£mx + y – z = 1x – 2y + z = 13x + 4y – 2z = –3x + y + z = 62x + 2y + mz = 6mx = 0a)7y + 5z = –73x + 4y + mz = –17x + 5z = 70 7 5 –7)A' = 3 4 m –1(7 0 5 714243A28Unidad 3. <strong>Resolución</strong> <strong>de</strong> <strong>sistemas</strong> <strong>mediante</strong> <strong>de</strong>terminantes

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