Tema 3: Resolución de sistemas mediante determinantes

UNIDAD3(2 0 5) (x)7)1 1 –2 · y = –2 Calculamos A –1 ( | A | = 16 ? 0 8 existe A –1 ):–1 1 1 z (–114243 { {A · X = Ba ijÄÄÄ8 Adj (A) ÄÄÄ8 (Adj (A)) t 1ÄÄÄ8 (Adj (A)) t| A|(3 –1 2) (3 1 2)3 5 –5)3 5 –5)–5 7 218 5 7 –2 8 1 7 9 8 1 7 9 = A –1–5 –9 2 –5 9 2 (2 –2 216 (2 –2 2Por tanto:A · X = B 8 X = A –1· 3 5 –5)7) (16) (1)1B = 1 7 9 · –21= –16 = –116 (2 –2 2 (–11616 1(x) (1)Luego y = –1 ; es decir: x = 1, y = –1, z = 1z 135 Resuelve la ecuación siguiente:1 1 2)2 0 0)1 1 0)X 3 4 6 – 1 1 2 = 0 1 0(4 2 9 (2 0 1 (0 1 21 1 2)2 0 0)1 1 0)Sea A = 3 4 6 , B = 1 1 2 y C = 0 1 0 . Entonces:(4 2 9 (2 0 1 (0 1 2X · A – B = C 8 X · A = C + B 8 X · A · A –1 = (C + B) · A –1 8 X = (C + B) ·A –11 1 0)2 0 0)3 1 0)C + B = 0 1 0 + 1 1 2 = 1 2 2(0 1 2 (2 0 1 (2 1 31 1 2)A = 3 4 6 8 | A | = 1 ? 0 8 Existe A –1(4 2 9Calculamos A –1 :a ijÄÄÄ8 Adj (A) ÄÄÄ8 (Adj (A)) t 1ÄÄÄ8 (Adj (A)) t| A|(24 3 –10) (24 –3 –10)24 –5 –2)24 –5 –2)5 1 –2 8 –5 1 2 8 –3 1 0 8 –3 1 0 = A –1–2 0 1 –2 0 1 (–10 2 1 (–10 2 1Unidad 3. Resolución de sistemas mediante determinantes53