Tema 3: Resolución de sistemas mediante determinantes

UNIDAD3| B | = 2 ? 0 8 Existe B –1a ijÄÄÄ8 Adj (B) ÄÄÄ8 (Adj (B)) t 1ÄÄÄ8 (Adj (B)) t| B|(–2 –1 4) (–2 1 4) (–2 –2 2) (–2 –2 2)12 0 –4 8 –2 0 4 8 1 0 0 8 · 1 0 0 = B –12 0 –2 2 0 –2 4 4 –224 4 –2(3 –8 2)b) A · B = –2 6 –2 ; | A · B | = 4 ? 0 8 Existe (A · B) –12 –4 2a ijÄÄÄ8 Adj (AB) ÄÄÄ8 (Adj (AB)) t 1ÄÄÄ8 (Adj (AB)) t| AB|(4 0 –4)4 0 –4)4 8 4)4 8 4)1–8 2 4 8 8 2 –4 8 0 2 2 8 · 0 2 2 = (AB) –14 –2 2 (4 2 2 (–4 –4 24 (–4 –4 2(–2 –2 2)0 2 2)c) B –1 · A –1 1= · 1 0 0 · 0 0 1 = (A · B) –12 4 4 –2 (2 6 5(1 1 0)s26 Dada A = –1 1 2 , determina la matriz B que verifica B – I = A t A –1 .1 0 1(1 1 0)1 –1 1)A = –1 1 2 ; A t = 1 1 01 0 1 (0 2 1Calculamos A –1 :| A | = 4 ? 0 8 Existe A –1a ijÄÄÄ8 Adj (A) ÄÄÄ8 (Adj (A)) t 1ÄÄÄ8 (Adj (A)) t| A|1 –3 –1) (1 3 –1) (1 –1 2) (1 –1 2)1 1 –118 –1 1 1 8 3 1 –2 8 · 3 1 –2 = A –1(2 2 2 2 –2 2 –1 1 2 4 –1 1 2Calculamos A t · A –1 :1 –1 1) (1 –1 2) (–3 –1 6A t · A –1 11= · 1 1 0 · 3 1 –2 = · 4 0 04 (0 2 1 –1 1 2 4 5 3 –2|B| = A t · A –1 + I(–3 –1 6)1 0 0)1 –1 6)11B = · 4 0 0 + 0 1 0 = · 4 4 04 5 3 –2 (0 0 1 4 (5 3 2)Unidad 3. Resolución de sistemas mediante determinantes47