Tema 3: Resolución de sistemas mediante determinantes

°§¢§£UNIDAD3Página 871. Discute y resuelve:° x + y + az = 0° x + y = k§§a) ¢ ax – y = –1b) ¢ kx – y = 13§§£ x + 4y + 6z = 0£ 5x + 3y = 16a) x + y + az = 0ax – y = –1x + 4y + 6z = 01 1 a)1 1 a 0)A = a –1 0 A' = a –1 0 –1(1 4 6 (1 4 6 0| A | = 4a 2 5 ± √25 + 96 5 ± √121– 5a – 6 = 0 8 a = = =88• Si a = 2, queda:1 1 2 0)1 1A' = 2 –1 0 –1 | | = –3 ? 0 8 ran (A) = 2(1 4 6 02 –114243A|1 1 0|2 –1 –11 4 0= 3 ? 0 8 ran (A' ) = 3 ? ran (A)El sistema es incompatible.• Si a = –3/4, queda:(1 1 –3/4 0)A' = –3/4 –1 0 –11 1 –1|| = ? 0 8 ran (A) = 21 4 6 0–3/4 –1 41442443A|1 1 0|–3/4 –1 –1 = 3 ? 0 8 ran (A' ) = 3 ? ran (A)1 4 0El sistema es incompatible.• Si a ? 2 y a ? –3/4 8 ran (A) = ran (A' ) = n.° de incógnitas = 3, el sistemaes compatible determinado. Lo resolvemos:| |0 1 a1 0 a–1 –1 0 |a –1 0 |0 4 6 6 – 4a1 0 6 a – 6x = = ; y = = ;4a 2 – 5a – 6 4a 2 – 5a – 6 4a 2 – 5a – 6 4a 2 – 5a – 6|1 1 0a –1 –1 |1 4 0 3z = =4a 2 – 5a – 6 4a 2 – 5a – 66 – 4aa – 6Solución: x = , y = , z =4a 2 – 5a – 6 4a 2 – 5a – 65 ± 11834a 2 – 5a – 6a = 2–3a = — 4Unidad 3. Resolución de sistemas mediante determinantes15