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# Invers Matriks

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### Invers Matriks

1. 1. MODUL 5INVERS MATRIKPRAYUDI STT PLN
2. 2. PENGERTIAN INVERS MATRIK Matrik bujur sangkar A dikatakan mempunyai invers, jika terdapat matrikB sedemikian rupa sehingga :AB = BA = Idimana I matrik identitas B dikatakan invers matrik A ditulis A–1, maka, AA–1= A–1A = I A dikatakan invers matrik B ditulis B–1, maka, B–1B= BB–1= I Contoh ; AB = BA = I111230132653432321111230132100010001653432321
3. 3. TEKNIK MENGHITUNG INVERS Metode Adjoint matrik Metode operasi elementer baris Metode Perkalian Invers Matrik Elementer Metode partisi matrik Program Komputer – MATCADS, MATLAB WS OFICE EXCELL
4. 4. Metode Adjoint Matrik Andaikan A matrik bujur sangkar berordo (nxn), Cij=(-1)i+j Mij kofaktorelemen matrik aij, dan andaikan pula det(A)≠0 maka A mempunyai inversyaitu :adj(A)det(A)1A 1dimana,ijjiijnnnnnnnnMCCCCCCCCCCCCCCCCCAadj)1(...........................)(321333231323222121312111Kasus, n = 2 : maka22211211aaaaA11211222211222111 1aaaaaaaaA
5. 5. Kasus, n = 3333231232221131211aaaaaaaaaA3323133222123121113323133222123121111-MM-MM-MM-MM-Mdet(A)1CCCCCCCCCdet(A)1ACONTOH :554543432A det(A)= 11-21-26-51-55-9)-(812)-(10-16)-(1512)-(10-)16-10(20)-(15-16)-(1520)-(15-25)-(2043325432-54435342-54425453-54435543-5554(1)1A 1-
6. 6. KASUS : n = 4443424144333231342322212413121111-44434241343332312423222114131211MM-MM-M-MM-MMM-MM-M-MM-Mdet(A)1AaaaaaaaaaaaaaaaaACONTOH :Hitunglah invers matrik berikut ini :Ekspansi baris -1 :det(a)=M11-2M12+3M13-4M14=-10 – 2(5) + 3(9) – 4(2)= –16863475355324321AEkspansi baris-2 :det(A)=-2M21+3M22-5M23+5M24=-2(-6) –3(4) + 5(-6) –5(-1)= –1Ekspansi baris-3 :det(A)=3M31-5M32+7M33-4M34=3(-8) –5(3) + 7(6) –4(1)= –1Ekspansi baris-4 :det(A)=-3M41+6M42-8M43+6M44=-3(-7) +6(2) - 8(5) + 6(1)= –1
7. 7. INVERS : OPERASI ELEMENTER BARISOperasi Elementer baris yangdigunakan adalah :(1). Hj  kHj(2). Hj  Hi(3). Hj  Hj + kHj1000...1......00100001..................212222111211nnnniinnaaaaaaaaaaLangkah-langkah sebagai berikut(1). Bentuk matrik lengkap [A,I](2). Dengan serangkain operasi elelemterbaris reduksilah [A,I] menjadi matrikberbentuk [I,B](3). A–1 = Bnnnniinnbbbbbbbbbb..................1000...1......00100001212222111211nnnniinnbbbbbbbbbb..................AJadi,2122221112111-Operasi elementer barisGaouss-Jordan
8. 8. CONTOH :M.Asal2 3 4 1 0 03 4 5 0 1 04 5 5 0 0 1Iterasi-11 1.5 2 0.5 0 0 H1=(1/a11)H10 -0.5 -1 -1.5 1 0 H2=H2-(a21/a11)H10 -1 -3 -2 0 1 H3=H3-(a31/a11)H1Iterasi-21 1.5 2 0.5 0 00 1 2 3 -2 0 H2=(1/a22)H20 0 -1 1 -2 1 H3=H3-(a32/a22)H2Iterasi-31 1.5 2 0.5 0 00 1 2 3 -2 00 0 1 -1 2 -1 H3=(1/a33)H3
9. 9. Iterasi-41 1.5 0 2.5 -4 2 H1=H1-(a13/a33)H30 1 0 5 -6 2 H2=H2-(a23/a33)H30 0 1 -1 2 -1Iterasi-51 0 0 -5 5 -1 H1=H1-(a12/a22)H20 1 0 5 -6 20 0 1 -1 2 -1Lanjutan :1-21-26-51-55-AJadi, 1-
10. 10. Matrik Awal1 2 3 4 1 0 0 02 3 5 5 0 1 0 03 5 7 4 0 0 1 03 6 8 6 0 0 0 1Iterasi - 11 2 3 4 1 0 0 0 H1=(1/a11)H10 -1 -1 -3 -2 1 0 0 H2=H2-(a21/a11)H10 -1 -2 -8 -3 0 1 0 H3=H3-(a21/a11)H10 0 -1 -6 -3 0 0 1 H4=H4-(a41/a11)H1Iterasi - 21 2 3 4 1 0 0 00 1 1 3 2 -1 0 0 H2=(1/a22)H20 0 -1 -5 -1 -1 1 0 H3=H3-(a32/a22)H20 0 -1 -6 -3 0 0 1 H4=H4-(a42/a22)H2Iterasi-41 2 3 4 1 0 0 00 1 1 3 2 -1 0 00 0 1 5 1 1 -1 0 H3=(1/a33)H30 0 0 -1 -2 1 -1 1 H4=H4-(a43/a33)H3
11. 11. Iterasi-51 2 3 4 1 0 0 00 1 1 3 2 -1 0 00 0 1 5 1 1 -1 00 0 0 1 2 -1 1 -1 H4=(1/a44)H4Iterasi-61 2 3 0 -7 4 -4 4 H1=H1-a14*H40 1 1 0 -4 2 -3 3 H2=H2-a24*H40 0 1 0 -9 6 -6 5 H3=H3-a34*H40 0 0 1 2 -1 1 -1Iterasi-71 2 0 0 20 -14 14 -11 H1=H1-a13*H30 1 0 0 5 -4 3 -2 H2=H2-a23*H30 0 1 0 -9 6 -6 50 0 0 1 2 -1 1 -1Iterasi-81 0 0 0 10 -6 8 -7 H1=H1-a12*H20 1 0 0 5 -4 3 -20 0 1 0 -9 6 -6 50 0 0 1 2 -1 1 -11-11256-69-2-34-57-86-10AJadi, 1-
12. 12. PERKALIAN MATRIK ELEMENTER(1). Matrik elementer adalah matrikyang diperoleh dari operasielementer yang dikenakan padamatrik identitas.(2). Setiap matrik elementermempunyai invers, dan setiapmatrik bujur sangkar berordo(nxn) yang mempunyai inversekivalen baris terhadap matrikidentitas I.(3). Akibatnya, jika :EkEk–1Ek–2 …E2E1A = I, maka,A–1 = EkEk–1Ek–2 …E2E1Matrik elementer E diperoleh daritransformasi matrik identitas dimana padakolom ke-I diganti dengan normalitasvektor kolom :1......00...................0......00...................0......100......01....ikikikikiNNNNEiiinaiiiiiikaaaaN/.../1.../,,1,iiiik IAEEEN 121, ...:dimana
13. 13. CONTOHHitung invers matrik AJawab :Menghitung E1554543432A554A;543A;432A 321102-011.5-000.510/aa-01/aa-001/aE11311121111Menghitung E212-102-3034-102011.5-000.512-002-0031EE12-002-00311)(-1)/(-0.5-000.5-1/000.5-1.5/-1E1-0.5-1.5543102-011.5-000.5AEN122212
14. 14. Menghitung E3 dan Invers Matrik1-21-26-51-55-12-102-3034-1-002101-01EEE1-002101-011/(-1)00(2)/(-1)-10(-1)/(-1)-01E1-21-55412-102-3034-AEEN12333123Jadi Invers Matrik1-21-26-51-55-A 1-
15. 15. CONTOHHitung invers matrik AJawab :Menghitung E16863475355324321A1003-0103-0012-0001E3321AN111Menghitung E21003-011-1-001-20023-1003-0103-0012-00011000011-0001-00021EE1000011-0001-00021100/(-1)-001(-1)/(-1)-0001-1/0002/(-1)-1E01-1-265321003-0103-0012-0001AEN122212
16. 16. Menghitung E311-12-01-11012-10113-1003-011-1-001-20023-1000010000100001EEE11-0001-00011001011(-1)/(-1)-0001/(-1)0001/(-1)-1001/(-1)-01E1-1-1187531003-011-1-001-20023-AEEN12333123
17. 17. Menghitung E4 dan Invers Matrik1-11-256-69-2-34-57-86-1011-12-01-11012-10114-1-00051002-0107-001EEEEA1-00051002-0107-0011/(-1)0005/(-1)-100(-2)/(-1)-010(-7)/(-1)-001E1-52-7-645411-12-01-11012-10113-AEEEN12341-341234
18. 18. INVERS : PARTISI MATRIK (1)Partisi matrik A yang berordo (mxn)adalah sub matrik-sub matrik yangdiperoleh dari A dengan caramemberikan batasan-batasan garishorisontal diantara dua baris dan ataumemberikan batasan-batasan garisvertikal diantara dua kolom.CONTOH6863475355324321A6847A6353A5543A3221A:adalahAmatrikPartisi222112113155413343536322344334532ACONTOH
19. 19. INVERS : PARTISI MATRIK (2)Andaikan A matrik bujur sangkarberordo (nxn) yang mempunyaiinvers, yaitu : A–1 = B, dan partisinyamasing-masing adalah :Karena, AB=BA=I maka diperoleh :2221121122211211BBBBB;AAAAAI00IAAAABBBBI00IBBBBAAAA22211211222112112221121122211211Dari perkalian matrik diperoleh hasil :(1). A11 B11 + A12 B21 = I(2). A11 B12 + A12 B22 = 0(3). B21 A11 + B22 A21 = 0(4). B21 A12 + B22 A22 = IDengan asumsi, A11–1 ada, danB22 = L–1 adaMaka rumus untuk menghitung invermatriknya adalah :(1). B12 = –(A 11–1 A12)L–1(2). B21 = – L–1(A21 A11–1)(3). B11 = A11–1+(A11–1A12)L–1(A21 A11–1)(4). L = A22 – (A21A11–1A12)