Metode Adjoint Matriks
Andaikan A matrik bujur sangkar berordo (nxn), Cij=(-1)i+j Mij kofaktor elemenmatrik aij, dan andaikan pula det(A)≠0 maka A mempunyai invers yaitu :
- Kasus, n = 3
det(A)= 1
- Kasus, n = 4
Hitunglah invers matrik berikut ini
Ekspansi baris -1 :
det(A)=M11-2M12+3M13-4M14
=-10 – 2(5) + 3(9) – 4(2)
= –1
Ekspansi baris-2 :
det(A)=-2M21+3M22-5M23+5M24
=-2(-6) –3(4) + 5(-6) –5(-1)
= –1
Ekspansi baris-3 :
det(A)=3M31-5M32+7M33-4M34
=3(-8) –5(3) + 7(6) –4(1)
= –1
Ekspansi baris-4 :
det(A)=-3M41+6M42-8M43+6M44
=-3(-7) +6(2) - 8(5) + 6(1)
= –1
Operasi Baris Elementer (OBE)
Operasi Elementer baris yang digunakan adalah :
(1). Hj ß kHj
(2). Hj ß Hi
(3). Hj ß Hj + kHj
Langkah-langkah sebagai berikut
(1). Bentuk matrik lengkap [A,I]
(2). Dengan serangkain operasi elelemter baris reduksilah [A,I] menjadi matrik berbentuk
[I,B]
(3). A–1
= B
Operasi elementer baris
Gaouss-Jordan
=
CONTOH :
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M.Asal
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2
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3
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4
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1
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0
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0
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3
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4
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5
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0
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1
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0
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4
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5
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5
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0
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0
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1
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Iterasi-1
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1
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1.5
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2
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0.5
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0
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0
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H1=(1/a11)H1
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0
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-0.5
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-1
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-1.5
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1
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0
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H2=H2-(a21/a11)H1
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0
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-1
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-3
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-2
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0
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1
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H3=H3-(a31/a11)H1
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Iterasi-2
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1
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1.5
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2
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0.5
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0
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0
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0
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1
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2
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3
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-2
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0
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H2=(1/a22)H2
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0
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0
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-1
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1
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-2
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1
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H3=H3-(a32/a22)H2
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Iterasi-3
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1
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1.5
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2
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0.5
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0
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0
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0
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1
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2
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3
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-2
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0
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0
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0
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1
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-1
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2
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-1
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H3=(1/a33)H3
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Iterasi-4
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1
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1.5
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0
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2.5
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-4
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2
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H1=H1-(a13/a33)H3
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0
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1
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0
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5
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-6
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2
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H2=H2-(a23/a33)H3
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0
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0
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1
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-1
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2
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-1
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Iterasi-5
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1
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0
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0
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-5
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5
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-1
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H1=H1-(a12/a22)H2
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0
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1
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0
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5
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-6
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2
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0
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0
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1
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-1
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2
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-1
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