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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• Si a ? 1:ran (A) = ran (A') = 3, y el sistema es compatible <strong>de</strong>terminado.Para cada valor <strong>de</strong> a ? 1, tenemos un sistema con solución única.1 –1 –aa 2 4§ 0 a 1 § –a 3 – 3a +2x = =3(a – 1) 21 1 –a–3 a 4§ –1 0 1 § –a 2 + a – 1y = =3(a – 1) 2 3(a – 1) 21 –1 1–3 2 a§ –1 a 0 § –a 2 – 2a + 2z = =3(a – 1) 2 3(a – 1) 23(a – 1) 2 Unidad 3. <strong>Resolución</strong> <strong>de</strong> <strong>sistemas</strong> <strong>mediante</strong> <strong>de</strong>terminantes• Si a = 1:(1 –1 –1)A = –3 2 41 –18 | | = –1 ? 0 8 ran (A) = 2–1 1 1–3 2(1 –1 –1 1) |1 –1 1|A' = –3 2 4 1 ; –3 2 1 = –1 ? 0 8 ran (A') = 3–1 1 1 0 –1 1 0El sistema es incompatible.62
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