a.
Determinan
matriks A dengan Metode Sarrus
b. Adjoint matriks A dengan Metode Kofaktor
Alternatif Penyelesaian:
Determinannya
dengan metode Sarrus
= ((1)(–4)(–1)
+ (2)(2)(5) + (0)(3)(6)) –
((0)(–4)(5) + (1)(2)(6) + (2)(3)(–1))
= (4 +
20 + 0) – (0 + 12 + (–6))
= 24 – 6
= 18
b.
Metode Kofaktor
i)
Minor
ii)
Kofaktor
𝐾11 = (–1)2 ⋅ 43 = 1 ⋅ 43 = 43
𝐾12 = (–1)3 ⋅ 12 = –1 ⋅ 12 = –12
𝐾13 = (–1)4 ⋅ 15 = 1 ⋅ 15 = 15
𝐾21 = (–1)3 ⋅ 3 = –1 ⋅ 3 = –3
𝐾22 = (–1)4 ⋅ 1 = 1 ⋅ 1 = 1
𝐾23 = (–1)5 ⋅ 1 = –1 ⋅ 1 = –1
𝐾31 = (–1)4 ⋅ 8 = 1 ⋅ 8 = 8
𝐾32 = (–1)5 ⋅ 43 = -1 ⋅ 2 = –2
𝐾33 = (–1)6 ⋅ 3 = 1 ⋅ 3 = 3
Sumber
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