Matematika SMA/MA/SMK/MAK
Pengertian Adjoin Matriks Persegi Berordo 3
A. Pengertian Adjoin Matriks
Adjoin (atau adjugat) dari suatu matriks persegi A berordo 3×3 adalah transpose dari matriks kofaktor matriks tersebut.
Secara matematis, jika A adalah matriks persegi berordo 3, maka:
di mana C adalah matriks kofaktor dari A, dan CT adalah transpose dari matriks kofaktor tersebut.
Langkah-Langkah Menentukan Adjoin Matriks Ordo 3×3
Misalkan matriks A berordo 3×3:
- Tentukan Minor (Mij) setiap elemen: determinan dari submatriks 2×2 yang diperoleh dengan menghapus baris ke-i dan kolom ke-j.
- Tentukan Kofaktor (Cij) setiap elemen: Cij = (−1)i+j × Mij
- Susun Matriks Kofaktor C
- Transpose matriks kofaktor untuk mendapatkan adj(A) = CT
B. Pertanyaan Kunci
- • Apa perbedaan minor dan kofaktor suatu elemen matriks?
- • Mengapa adjoin didefinisikan sebagai transpose dari matriks kofaktor, bukan matriks kofaktor itu sendiri?
- • Bagaimana tanda (positif/negatif) kofaktor ditentukan oleh posisi elemen (i+j)?
- • Bagaimana hubungan adjoin dengan invers matriks?
C. Pola Tanda Kofaktor Ordo 3×3
Pola tanda (−1)i+j untuk matriks 3×3:
Artinya:
- • C₁₁ = +M₁₁, C₁₂ = −M₁₂, C₁₃ = +M₁₃
- • C₂₁ = −M₂₁, C₂₂ = +M₂₂, C₂₃ = −M₂₃
- • C₃₁ = +M₃₁, C₃₂ = −M₃₂, C₃₃ = +M₃₃
Hubungan Adjoin dengan Invers Matriks
Jika det(A) ≠ 0, maka:
Inilah mengapa adjoin sangat penting dalam menentukan invers matriks berordo 3.
D. Prosedur Detail Menghitung Adjoin
Diberikan matriks:
Langkah 1: Hitung semua Minor
M₁₁ = det4126 = (4)(6) − (1)(2) = 24 − 2 = 22
M₁₂ = det0156 = (0)(6) − (1)(5) = 0 − 5 = −5
M₁₃ = det0452 = (0)(2) − (4)(5) = 0 − 20 = −20
M₂₁ = det1326 = (1)(6) − (3)(2) = 6 − 6 = 0
M₂₂ = det2356 = (2)(6) − (3)(5) = 12 − 15 = −3
M₂₃ = det2152 = (2)(2) − (1)(5) = 4 − 5 = −1
M₃₁ = det1341 = (1)(1) − (3)(4) = 1 − 12 = −11
M₃₂ = det2301 = (2)(1) − (3)(0) = 2 − 0 = 2
M₃₃ = det2104 = (2)(4) − (1)(0) = 8 − 0 = 8
Langkah 2: Hitung Kofaktor (terapkan pola tanda)
C₁₁ = +22 = 22, C₁₂ = −(−5) = 5, C₁₃ = +(−20) = −20
C₂₁ = −0 = 0, C₂₂ = +(−3) = −3, C₂₃ = −(−1) = 1
C₃₁ = +(−11) = −11, C₃₂ = −2 = −2, C₃₃ = +8 = 8
Langkah 3: Susun Matriks Kofaktor
Langkah 4: Transpose → adj(A)
E. Contoh Soal
Tingkat Mudah
Soal 1.
Tentukan adj(A) jika:
Pembahasan:
Karena A diagonal, minor dihitung dari submatriks 2×2:
M₁₁ = (2)(3)−(0)(0) = 6, M₁₂ = (0)(3)−(0)(0) = 0, M₁₃ = (0)(0)−(2)(0) = 0
M₂₁ = (0)(3)−(0)(0) = 0, M₂₂ = (1)(3)−(0)(0) = 3, M₂₃ = (1)(0)−(0)(0) = 0
M₃₁ = (0)(0)−(2)(0) = 0, M₃₂ = (1)(0)−(0)(0) = 0, M₃₃ = (1)(2)−(0)(0) = 2
Kofaktor (dengan pola tanda):
C₁₁=6, C₁₂=0, C₁₃=0, C₂₁=0, C₂₂=3, C₂₃=0, C₃₁=0, C₃₂=0, C₃₃=2
adj(A) = CT =
Soal 2.
Tentukan adj(A) jika:
Pembahasan:
A adalah matriks identitas I₃.
Semua minor diagonal = 1, minor lainnya = 0.
Kofaktor = minor (karena tanda selalu +1 pada diagonal).
Matriks kofaktor = I₃, maka adj(I₃) = I₃T = I₃
Soal 3.
Tentukan adj(A) jika:
Pembahasan:
Matriks diagonal. Minor:
M₁₁ = (3)(1) = 3, M₂₂ = (2)(1) = 2, M₃₃ = (2)(3) = 6
Semua minor non-diagonal = 0.
Kofaktor diagonal: C₁₁=3, C₂₂=2, C₃₃=6
Soal 4.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (1)(1)−(0)(0) = 1, M₁₂ = (0)(1)−(0)(0) = 0, M₁₃ = (0)(0)−(1)(0) = 0
M₂₁ = (2)(1)−(0)(0) = 2, M₂₂ = (1)(1)−(0)(0) = 1, M₂₃ = (1)(0)−(2)(0) = 0
M₃₁ = (2)(0)−(1)(0) = 0, M₃₂ = (1)(0)−(0)(0) = 0, M₃₃ = (1)(1)−(2)(0) = 1
Kofaktor: C₁₁=1, C₁₂=0, C₁₃=0, C₂₁=−2, C₂₂=1, C₂₃=0, C₃₁=0, C₃₂=0, C₃₃=1
Soal 5.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (1)(1)−(0)(0) = 1, M₁₂ = (0)(1)−(0)(0) = 0, M₁₃ = (0)(0)−(1)(0) = 0
M₂₁ = (0)(1)−(1)(0) = 0, M₂₂ = (1)(1)−(1)(0) = 1, M₂₃ = (1)(0)−(0)(0) = 0
M₃₁ = (0)(0)−(1)(1) = −1, M₃₂ = (1)(0)−(0)(1) = 0, M₃₃ = (1)(1)−(0)(0) = 1
Kofaktor: C₁₁=1, C₁₂=0, C₁₃=0, C₂₁=0, C₂₂=1, C₂₃=0, C₃₁=−1, C₃₂=0, C₃₃=1
Tingkat Sedang
Soal 6.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (1)(1)−(4)(0) = 1, M₁₂ = (0)(1)−(4)(0) = 0, M₁₃ = (0)(0)−(1)(0) = 0
M₂₁ = (2)(1)−(3)(0) = 2, M₂₂ = (1)(1)−(3)(0) = 1, M₂₃ = (1)(0)−(2)(0) = 0
M₃₁ = (2)(4)−(3)(1) = 5, M₃₂ = (1)(4)−(3)(0) = 4, M₃₃ = (1)(1)−(2)(0) = 1
Kofaktor: C₁₁=1, C₁₂=0, C₁₃=0, C₂₁=−2, C₂₂=1, C₂₃=0, C₃₁=5, C₃₂=−4, C₃₃=1
Soal 7.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (3)(2)−(1)(1) = 5, M₁₂ = (1)(2)−(1)(0) = 2, M₁₃ = (1)(1)−(3)(0) = 1
M₂₁ = (1)(2)−(0)(1) = 2, M₂₂ = (2)(2)−(0)(0) = 4, M₂₃ = (2)(1)−(1)(0) = 2
M₃₁ = (1)(1)−(0)(3) = 1, M₃₂ = (2)(1)−(0)(1) = 2, M₃₃ = (2)(3)−(1)(1) = 5
Kofaktor: C₁₁=5, C₁₂=−2, C₁₃=1, C₂₁=−2, C₂₂=4, C₂₃=−2, C₃₁=1, C₃₂=−2, C₃₃=5
Soal 8.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (0)(3)−(1)(1) = −1, M₁₂ = (1)(3)−(1)(2) = 1, M₁₃ = (1)(1)−(0)(2) = 1
M₂₁ = (1)(3)−(2)(1) = 1, M₂₂ = (3)(3)−(2)(2) = 5, M₂₃ = (3)(1)−(1)(2) = 1
M₃₁ = (1)(1)−(2)(0) = 1, M₃₂ = (3)(1)−(2)(1) = 1, M₃₃ = (3)(0)−(1)(1) = −1
Kofaktor: C₁₁=−1, C₁₂=−1, C₁₃=1, C₂₁=−1, C₂₂=5, C₂₃=−1, C₃₁=1, C₃₂=−1, C₃₃=−1
Soal 9.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (1)(1)−(0)(2) = 1, M₁₂ = (2)(1)−(0)(0) = 2, M₁₃ = (2)(2)−(1)(0) = 4
M₂₁ = (3)(1)−(2)(2) = −1, M₂₂ = (1)(1)−(2)(0) = 1, M₂₃ = (1)(2)−(3)(0) = 2
M₃₁ = (3)(0)−(2)(1) = −2, M₃₂ = (1)(0)−(2)(2) = −4, M₃₃ = (1)(1)−(3)(2) = −5
Kofaktor: C₁₁=1, C₁₂=−2, C₁₃=4, C₂₁=1, C₂₂=1, C₂₃=−2, C₃₁=−2, C₃₂=4, C₃₃=−5
Soal 10.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (3)(4)−(2)(1) = 10, M₁₂ = (1)(4)−(2)(2) = 0, M₁₃ = (1)(1)−(3)(2) = −5
M₂₁ = (2)(4)−(1)(1) = 7, M₂₂ = (4)(4)−(1)(2) = 14, M₂₃ = (4)(1)−(2)(2) = 0
M₃₁ = (2)(2)−(1)(3) = 1, M₃₂ = (4)(2)−(1)(1) = 7, M₃₃ = (4)(3)−(2)(1) = 10
Kofaktor: C₁₁=10, C₁₂=0, C₁₃=−5, C₂₁=−7, C₂₂=14, C₂₃=0, C₃₁=1, C₃₂=−7, C₃₃=10
Tingkat Sulit
Soal 11.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (5)(7)−(−2)(3) = 35+6 = 41
M₁₂ = (4)(7)−(−2)(1) = 28+2 = 30
M₁₃ = (4)(3)−(5)(1) = 12−5 = 7
M₂₁ = (−1)(7)−(3)(3) = −7−9 = −16
M₂₂ = (2)(7)−(3)(1) = 14−3 = 11
M₂₃ = (2)(3)−(−1)(1) = 6+1 = 7
M₃₁ = (−1)(−2)−(3)(5) = 2−15 = −13
M₃₂ = (2)(−2)−(3)(4) = −4−12 = −16
M₃₃ = (2)(5)−(−1)(4) = 10+4 = 14
Kofaktor: C₁₁=41, C₁₂=−30, C₁₃=7, C₂₁=16, C₂₂=11, C₂₃=−7, C₃₁=−13, C₃₂=16, C₃₃=14
Soal 12.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (−2)(−3)−(1)(0) = 6, M₁₂ = (3)(−3)−(1)(5) = −9−5 = −14, M₁₃ = (3)(0)−(−2)(5) = 10
M₂₁ = (2)(−3)−(4)(0) = −6, M₂₂ = (−1)(−3)−(4)(5) = 3−20 = −17, M₂₃ = (−1)(0)−(2)(5) = −10
M₃₁ = (2)(1)−(4)(−2) = 2+8 = 10, M₃₂ = (−1)(1)−(4)(3) = −1−12 = −13, M₃₃ = (−1)(−2)−(2)(3) = 2−6 = −4
Kofaktor: C₁₁=6, C₁₂=14, C₁₃=10, C₂₁=6, C₂₂=−17, C₂₃=10, C₃₁=10, C₃₂=13, C₃₃=−4
Soal 13.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (4)(−1)−(−6)(7) = −4+42 = 38
M₁₂ = (−1)(−1)−(−6)(3) = 1+18 = 19
M₁₃ = (−1)(7)−(4)(3) = −7−12 = −19
M₂₁ = (−3)(−1)−(2)(7) = 3−14 = −11
M₂₂ = (5)(−1)−(2)(3) = −5−6 = −11
M₂₃ = (5)(7)−(−3)(3) = 35+9 = 44
M₃₁ = (−3)(−6)−(2)(4) = 18−8 = 10
M₃₂ = (5)(−6)−(2)(−1) = −30+2 = −28
M₃₃ = (5)(4)−(−3)(−1) = 20−3 = 17
Kofaktor: C₁₁=38, C₁₂=−19, C₁₃=−19, C₂₁=11, C₂₂=−11, C₂₃=−44, C₃₁=10, C₃₂=28, C₃₃=17
Soal 14.
Jika adj(A) diketahui, tentukan det(A) menggunakan sifat A × adj(A) = det(A) × I. Diberikan:
Pembahasan:
Hitung det(A) terlebih dahulu (ekspansi baris 1):
det(A) = 1[(5)(0)−(6)(8)] − 2[(4)(0)−(6)(7)] + 3[(4)(8)−(5)(7)]
= 1[0−48] − 2[0−42] + 3[32−35]
= −48 + 84 − 9 = 27
Sekarang hitung adj(A):
M₁₁=−48, M₁₂=−42, M₁₃=−3, M₂₁=−24, M₂₂=−21, M₂₃=−6, M₃₁=−3, M₃₂=−6, M₃₃=−3
Kofaktor: C₁₁=−48, C₁₂=42, C₁₃=−3, C₂₁=24, C₂₂=−21, C₂₃=6, C₃₁=−3, C₃₂=6, C₃₃=−3
Verifikasi: A × adj(A) = 27 × I₃ ✓
Soal 15.
Tentukan adj(A) jika:
Pembahasan:
M₁₁ = (−4)(2)−(6)(−8) = −8+48 = 40
M₁₂ = (3)(2)−(6)(7) = 6−42 = −36
M₁₃ = (3)(−8)−(−4)(7) = −24+28 = 4
M₂₁ = (5)(2)−(−1)(−8) = 10−8 = 2
M₂₂ = (−2)(2)−(−1)(7) = −4+7 = 3
M₂₃ = (−2)(−8)−(5)(7) = 16−35 = −19
M₃₁ = (5)(6)−(−1)(−4) = 30−4 = 26
M₃₂ = (−2)(6)−(−1)(3) = −12+3 = −9
M₃₃ = (−2)(−4)−(5)(3) = 8−15 = −7
Kofaktor: C₁₁=40, C₁₂=36, C₁₃=4, C₂₁=−2, C₂₂=3, C₂₃=19, C₃₁=26, C₃₂=9, C₃₃=−7
F. Rangkuman
- 1. Adjoin matriks A berordo 3 adalah transpose dari matriks kofaktornya: adj(A) = CT
- 2. Langkah: Hitung 9 minor → terapkan pola tanda → susun matriks kofaktor → transpose
- 3. Sifat penting: A × adj(A) = det(A) × I
- 4. Adjoin digunakan untuk menghitung invers: A−1 = (1/det A) × adj(A)
G. Latihan Soal
Kerjakan soal-soal berikut tanpa melihat pembahasan!
Tingkat Mudah
1.
Tentukan adj(A), A = 300020004
2.
Tentukan adj(A), A = 110010002
3.
Tentukan adj(A), A = 500050005
4.
Tentukan adj(A), A = 102010003
5.
Tentukan adj(A), A = 210020002
Tingkat Sedang
6.
Tentukan adj(A), A = 213121312
7.
Tentukan adj(A), A = 1−1230121−1
8.
Tentukan adj(A), A = 321142235
9.
Tentukan adj(A), A = 012103230
10.
Tentukan adj(A), A = 12−1213−131
Tingkat Sulit
11.
Tentukan adj(A), A = −3254−1678−2
12.
Tentukan adj(A), A = 6−53−24−781−9
13.
Jika det(A) = 5, tentukan adj(2A) untuk A = 101210011
14.
Tentukan adj(A) dan verifikasi A × adj(A) = det(A) × I untuk A = 2−3154−2−163
15.
Tentukan adj(AT) jika A = 1−2435−1−672