Diketahui B = [(2 0) (0 1)], C = [(2 1) (-3 1)]. Jika diketahui matriks A berordo 2 x 2 sehingga AB + AC = [(4 2) (-3 1)] maka determinan (2A^(-1). B^(-1)) adalah... (A) 4. (B) 1. (C) 1/6. (D) 2. (E) 1/4.

Jawaban yang benar adalah 2 ⅕, tidak ada pilihan jawaban yang benar Pembahasan Ingat! 1. Sifat distributif pada matriks : AB + AC = A(B + C) 2. Perkalian dua matriks : misal A = [(a b) (c d)] dan B = [(e f) (g h)], maka A + B = [(a + e b + f) (c + g d + h)] 3. Perkalian dua matriks : misal A = [(a b) (c d)] dan B = [(e f) (g h)], maka A × B = [(ae+bg af+bh) (ce+dg cf+dh)] 4. Jika matriks A = B, maka setiap elemen yang seletak pada matriks A dan B bernilai sama 5. Misal A = [(a b) (c d)], maka det A = ad – bc 6. Misal A = [(a b) (c d)], maka A^(–1) = 1/det A [(d –b) (–c a)] 7. Misal A = [(a b) (c d)] dan I = [(1 0)(0 1)], maka A.I = A dengan I adalah matriks identitas 8. Misal A = [(a b) (c d)] dan k adalah konstan, makan k.A = [(ak bk) (ck dk)] Diketahui B = [(2 0) (0 1)], C = [(2 1) (-3 1)] matriks A berordo 2 x 2 AB + AC = [(4 2) (-3 1)] Misalkan, A = [(a b) (c d)], maka AB + AC = [(4 2) (–3 1)] A(B + C) = [(4 2) (–3 1)] A([(2 0) (0 1)] + [(2 1) (–3 1)]) = [(4 2) (–3 1)] A([(2+2 0+1) (0+(–3) 1+1)]) = [(4 2) (–3 1)] [(a b) (c d)][(4 1) (–3 2)] = [(4 2) (–3 1)] [(4a+(–3b) 1a+2b) (4c+(–3)d 1c+2d)] = [(4 2) (–3 1)] [(4a–3b a+2b) (4c–3d c+2d)] = [(4 2) (–3 1)] 4a – 3b = 4....(1) a + 2b = 2—> a = 2 – 2b....(2) 4c – 3d = –3....(3) c + 2d = 1—> c = 1 – 2d.....(4) Substitusi (2) ke (1) 4a – 3b = 4 4(2 – 2b) – 3b = 4 8 – 8b – 3b = 4 –11b = 4 – 8 –11b = –4 b = –4/(–11) b = 4/11 a = 2 – 2b a = 2 – 2(4/11) a = 22/11 – 8/11 a = 14/11 Substitusi (4) ke (3) 4c – 3d = –3 4(1 – 2d) – 3d = –3 4 – 8d – 3d = –3 –11d = –3 – 4 –11d = –7 d = –7/(–11) d = 7/11 c = 1 – 2d c = 1 – 2(7/11) c = 11/11 – 14/11 c = –3/11 Jadi, A = [(14/11 4/11) (–3/11 7/11)] A^(–1) = 1/det A [(7/11 –4/11) (3/11 14/11)] = 1/(14/11 × 7/11 – 4/11 × (–3/11)) [(7/11 –4/11) (3/11 14/11)] = 1/(98/121 + 12/121) [(7/11 –4/11) (3/11 14/11)] = 1/(110/121) [(7/11 –4/11) (3/11 14/11)] = 121/110 [(7/11 –4/11) (3/11 14/11)] = 11/10 [(7/11 –4/11) (3/11 14/11)] = [(7/11 × 11/10 –4/11 × 11/10) (3/11 × 11/10 14/11 × 11/10)] = [(7/10 –2/5) (3/10 7/5)] B^(–1) = 1/det B [(1 0) (0 2)] = 1/(2 × 1 – 0 × 0) [(1 0)(0 2)] = 1/2 [(1 0)(0 2)] =[(1×½ 0×½)(0×½ 2× ½)] =[(½ 0)(0 1)] 2.A^(-1). B^(-1) = 2 . [(7/10 –2/5) (3/10 7/5)] . [(½ 0)(0 1)] = [(7/10 × 2 –2/5 × 2) (3/10 × 2 7/5 × 2)]. [(½ 0)(0 1)] = [(7/5 –4/5) (3/5 14/5)]. [(½ 0)(0 1)] = [(7/5(½)+(–4/5)(0) 7/5(0)+(–4/5)(1)) (3/5(½)+(14/5)(0) (3/5)(0)+(14/5)(1))] = [(7/10 –4/5)(3/10 14/5)] det(2A^(-1). B^(-1)) = 7/10 × 14/5 – (–4/5) × 3/10 = 98/50 + 12/50 = 110/50 = 11/5 = 2 ⅕ Jadi, det(2A^(-1). B^(-1)) adalah 2 ⅕ Oleh karena itu, tidak ada pilihan jawaban yang benar