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Diketahui matriks A= (1 3 2 2 6 2 5 9 4). Tentukan berikut
Pertanyaan
Diketahui matriks A= [[1, 3, 2], [2, 6, 2], [5, 9, 4]]. Tentukan berikut ini. a. det A b. adj(A) c. A^-1
Solusi
Verified
det A = -12, adj(A) = [[6, 6, -6], [2, -6, 2], [-12, 6, 0]], A^-1 = [[-1/2, -1/2, 1/2], [-1/6, 1/2, -1/6], [1, -1/2, 0]]
Pembahasan
Untuk matriks A= [[1, 3, 2], [2, 6, 2], [5, 9, 4]]: a. Determinan (det A): | 1 3 2 | det A = | 2 6 2 | | 5 9 4 | det A = 1(6*4 - 2*9) - 3(2*4 - 2*5) + 2(2*9 - 6*5) = 1(24 - 18) - 3(8 - 10) + 2(18 - 30) = 1(6) - 3(-2) + 2(-12) = 6 + 6 - 24 = -12 b. Adjoin (adj A): Determinan minor: M11 = (6*4 - 2*9) = 24-18 = 6 M12 = -(2*4 - 2*5) = -(8-10) = 2 M13 = (2*9 - 6*5) = 18-30 = -12 M21 = -(3*4 - 2*9) = -(12-18) = 6 M22 = (1*4 - 2*5) = 4-10 = -6 M23 = -(1*9 - 3*5) = -(9-15) = 6 M31 = (3*2 - 2*6) = 6-12 = -6 M32 = -(1*2 - 2*2) = -(2-4) = 2 M33 = (1*6 - 3*2) = 6-6 = 0 Matriks Kofaktor (C): [[ 6, 2, -12] [ 6, -6, 6] [-6, 2, 0]] Adjoin A adalah transpose dari matriks kofaktor: adj(A) = C^T = [[ 6, 6, -6], [ 2, -6, 2], [-12, 6, 0]] c. Invers (A^-1): A^-1 = (1 / det A) * adj(A) = (1 / -12) * [[ 6, 6, -6], [ 2, -6, 2], [-12, 6, 0]] = [[-6/12, -6/12, 6/12], [-2/12, 6/12, -2/12], [ 12/12, -6/12, 0/12]] = [[-1/2, -1/2, 1/2], [-1/6, 1/2, -1/6], [ 1, -1/2, 0]]
Buka akses pembahasan jawaban
Topik: Matriks, Invers Matriks, Adjoin, Determinan
Section: Sistem Persamaan Linear, Operasi Matriks
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