Tema 3: Resolución de sistemas mediante determinantes

UNIDAD3Comprobación:1 2 1) (3 –6 –1)1 0 0)A · A –1 = 0 1 0 0 1 0 = 0 1 0(2 0 3 –2 4 1 (0 0 1• |B | = 2 + 6 – 6 = 21 0 0B 11= |3 3 1= –2; B 12= – = 6; B 13= = –21 1|| 2 1|| 2 1|1 2 2B 21= – |0 0 1= –1; B 22= = 2; B 23= – = 01 1|| 2 1|| 2 1|1 2 2B 31= |0 0 1= 3; B 32= – = –6; B 33= = 21 3|| 0 3|| 0 1|(–2 6 –2)(–2 –1 3)Adj (B) = –1 2 0 8 [Adj (B)] t = 6 2 –63 –6 2–2 0 2(–2 –1 3) (–1 –1/2 3/2)B –1 1= 6 2 –6 = 3 1 –32–2 0 2 –1 0 1Comprobación:2 1 0) (–1 –1/2 3/2)1 0 0)B · B –1 = 0 1 3 3 1 –3 = 0 1 0(2 1 1 –1 0 1 (0 0 1a) AX = B 8 A –1 AX = A –1 B 8 X = A –1 B(3 –6 –1)2 1 0) (4 –4 –19)X = 0 1 0 0 1 3 = 0 1 3–2 4 1 (2 1 1 –2 3 13b) XB = A 8 XBB –1 = AB –1 8 X = AB –11 2 1) (–1 –1/2 3/2) (4 3/2 –7/2)X = 0 1 0 3 1 –3 = 3 1 –3(2 0 3 –1 0 1 –5 –1 6PARA RESOLVER18 Estudia y resuelve estos sistemas homogéneos:° 9x + 3y + 2z = 0° x + y – z = 0§§ 3x – y + z = 0a) ¢ 12x – 3y – 2z = 0b) ¢§8x + y + 4z = 0£ x – 2y + z = 0§£ x + 2y – 2z = 0Unidad 3. Resolución de sistemas mediante determinantes35