Matrix in BASIC.pdf
29
1' IAE E l|, ,1 J i i'l tl.r i irl.j i,,1' ':.; \ Peniumlahan Dua lUlalriks rf:.i:': "1' 'I' I r11,.',i:' Syarat penjumlahan dua matriks adalah harus mempunyai ordo yang salna. : :: : ':; l isalkan diketahui matrlks A dan matriks B mempunyai ordo 2x3 A(2,3) = [ffi:ll ffi3i fll3i] B(2,3) : [BE:ll il#i BHii] Dimisalkan A(2:3) + B@,3) = C(2,3)' er2 ?\ : Ia(t,t)+B(1,1) A(12)+B(12) A(1,3)+S(I,S)] \'t-"" - [ a€,r)+B(2,1) A(2a)+B(2p) A(2,3)+eeJ) I rrrr2\ _ [c(r,t) c(12) c(t.g)l \'r-iJ, - L c1z,ty qzzl ctagl J ,i"'' Diperoleh persamaaru C(l,t)=A(l,l)+B(1,1) C(r,2)=611.2)+B(12)
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Transcript of Matrix in BASIC.pdf
1'
IAE
E l|, ,1 J
i i'l tl.r i irl.ji,,1' ':.; \
Peniumlahan Dua lUlalriksrf:.i:': "1'
'I' I r11,.',i:'
Syarat penjumlahan dua matriks adalah harus mempunyai ordo yangsalna. : :: : ':;
l isalkan diketahui matrlks A dan matriks B mempunyai ordo 2x3
A(2,3) = [ffi:ll ffi3i fll3i]B(2,3) : [BE:ll il#i BHii]
Dimisalkan A(2:3) + B@,3) = C(2,3)'
er2 ?\ : Ia(t,t)+B(1,1) A(12)+B(12) A(1,3)+S(I,S)]\'t-"" - [ a€,r)+B(2,1) A(2a)+B(2p) A(2,3)+eeJ) I
rrrr2\ _ [c(r,t) c(12) c(t.g)l\'r-iJ, - L c1z,ty qzzl ctagl J
,i"''Diperoleh persamaaru
C(l,t)=A(l,l)+B(1,1)C(r,2)=611.2)+B(12)
C(1,3) = A(1,3) + B(1,3) li'' . 'i:
C(2'1)=A(2'l)+B(2'1)C(2,2r=N2,2r+8(221
Kesimpulan umum yang dapat diperoleh untuk dimasul*an ke dalam
Progran adelah:'
Ctl,.l)'=A(Ifir)+B(l'J)' "tCODITOHDiketahui:
A(2.3) : [; 3 ',; B(2,3) = [? 3 t)
Malo C(2,3) : A(2,3) + B(2,3)
.rr,,) =[3 3 i] .[? l;]c(2,3) = [AI? 3ll '^ii]=I", iZ'l]
PROGRAIUI FLOWCHART
PEIIJlJIILf,llR tIDUA IIAIRISS
F0[ l=1 l0*[
tOR J:l I0 3
((I,il : . .
lil,il + Bil;J)
tOR l:1 I0 ?
F0[ J:l l0 3
tOR I:t I0 2
USTTIIG PROGRAIUI :
Io rxBrilT cmt(rsia0 nEM pnoorulu pruannagrr DUA l[arn'Irs60 DrM A(A,O,B(8,6),C(8,5)40 N,EM BACA UAIBffi8 AE0fOBI:1[0300FOB,I-1106?0 READ A(I'II)80 NEXT J90 }IEXT ITOO DATA 8,6,?,8,63
u0 nEu BAc,a uAtalrs B
U0 I0BI=1lOB160fOBtI=1T05140 IEAD B(In )160 I{EXil J160 NEXT I1?0 DATA 0,?,4,1,9,8
lS0Bi
USTING PROGRAIUI
10 IJ8,IM CHR$(I6)e0 nEM PBo0BAI[ PEILIITMLAHAN DUA IIATBIKS
s0 DrM A(e,5),8(e,5),C(e,6)40 N,EM BACA TIATBIKS A DAI{ MAM;ITS B
60r0BI:110860FOBtI:I106?0 nEAD A(I,cr),8(I,J)80 NEXI e,
MNEIOI100 DATA 0,0,6,?,?,4,8,1,6,9,4,8
I10 8,EM PROS.ES DAN CETAI( C : A + B
UlOEOBI=IMl50FOBcI=1IP6I40 C(I,cr) : A(I,.I) + B(I,J)160 TPBIM USIIIG "++ ";C(Id);160 MXr J170 LPRIIfl : LPRIMI8O NE}f, I1190 ESD
HASIL PROGRAIUI:
11 13 11
912 6
Program penjumlahan dua matriks di atas menggunakan RFAD-DATA(kalau ada read pasti ada data). Cara ini kurang efektif, karena dimensinyatertentu. Kalau kita memakai ordo yang lain, maka dimensinyp harusdiubah dan datanya juga diubah.Untuk lebih efektif, maka datanya kita masukkan rnelalui keyboard,demikian juga ordonya.
PROGRAIUI FLOTVC}IART
PEilJUiltAr{4ilDl,A IlRIRII(S
FOR I:1 I0 I
FOB l:t I0 I
C(I;JI:
eil,J) { gil,J)
FOR I:t I0 I
FOR J=l I0 l(
11
I
ti
lil
I
tlAilITG PROGRAII
r cls10 LPHiIm CEnaGs)80 NEU PAOGBAII IEI{dUUI.AIIAII DUA II{IBIEE50 IBIIIT 'lf,AglrtrtrAil 0BD0 t[{tBIf,B : A,'{0 ilPm "rrIrULAE BAIXB :"!60 IilPm ".rUllIAH f,OIOU :";f,60 DIII A(B.K)"B(B,K),C(B,E)?O PNNT80 I.PHM 'I[T[BIf,8 A J'90 IXBIIITI0O PBIilT 'llAgIIf,trAI[ EIAUEI{ UAIniIf,B A :"II0FORI=lI0Bf20f0nJ=1T0f,f50 PniIfT '?Antr8 ";I., trOml[ "xr;1{0 NPUI A(In )fEO LPBIIIT ITBIilO "++ ";A(I,.I);100 xitxl e,
U0 IXBITT : LPBIMIEO MAT II90 L!S|II{T : LPniIItT200 IJBIM 'tf,AfBlf,S B :"310 ITBITTA8O PBINT 'If,A8l'f,f,All EI,EUEI{ MA[BI[8 B :"480!0BI=1108A0fOBJ=110f,260 PBIilT "BARtrs ";I;" f,OIOII "xr;260 IITPLT B(I,J)870 LPBIIff IrgIII0 "++ ,'!([.I);280 NEX! .'800 IX8WT : IJBIM8OO NEXT I5I0 LPRII{T : LPBIMS30 LPBNT "PEI{.rUIIIAIIAN [fAmIfB A DNI tlAtR|If,B B :"580nCl[?n08EBC=A*B5{0fORI:l. l0B5E0!0RJ=11Of,560 C(I,er) - A(I,qr) + B(I,.r)6?0 LPBIXT IIBIilG "++ ";C(I,J);880 tEXr J890 LPBIIU : LPRIM400 I{txT I4r0 EID
}IAsTL PROGRAIUI
MASUKKAN ORDO MATRIKS : AJUMLAH BARIS :? 2JUMLAH KOLOM :? 3
MATRIKS A :
MASUKKAN ELEMEN MATRIKS A :
BARISlKOLOMl?5BARISlKOLOM2?6BARISlKOLOM3?7BARIS2KOLOMl?8BARIS2KOLOM2?3BARIS2KOLOM3?4
MATRIKS B :
MASUKKAN ELEMEN MATRIKS B :
BARISlKOLOM2?7BARISlKOLOM3?4BARIS2KOLOMl?1BARIS2KOLOM2?9BARIS2KOLOM3?2
PENJUMLAHAN MATRIKS A DAN MATRIKS B :
11 13 11
912 6
,1
Diperoleh persamaan:
B(1,1)=KxA(1,1)B(1'2):KxA(1'2) :
B(1'3)=KxA(1'3)B(2'l)=KxA(2'1)B(2,21 = K x A(2'2)B(2'3)=KxA(2'3)
Kesimpulan umum yang daPat diperoleh untuk di'masukkan ke dalam
Program adalah:
B(I'J)=KxA(l'J)
CONTOHDiketahui:
t.1 3 tlK = 3 A(2,3) = li s 6l
B(2'3) = K x
=3x
A(2,3)
lz 3 llLa s 6l
Perkalian Skalar DenganMatriksSetiap Elemen Matriks dikalikan dengan bilangan skalar.
Diketahui:Bilangan skalar = K
A(2,3) = [fllli N,i'jlfll3i]
Maka K x A(2,3) = K x hffi fll3i fll3l ]
- [r x A(l,l) K x A(1,2) r x elrsylLX x e12,t) K x N2,2) K x A(2,3)l
Dimisalkan B(2,3) = K x A(2,3)
B(2'3) = [38,1i Bll3i 383i]
10
I e e sl= Ltz 15 rgl
PROGRAIUI FLOIilCHART
PEBI{f,LIAII Sl(ALIRDE}IGAII IlAIR Il{S
USTING PROGRAITI:
r0 LPBIIIT CEBt(lB)AO NEU PEffiAIIAII UATBIXS DEI{GAIT BIIASGAI{ 8IAIABs0 DIU A(a,0,8(2,8)
il ffi#'uurorw Bf,ALAB = ";KOO IJBITT70 LPn|III 'lf,AfBIxB A I'80!08I=I10290fOBtI=IT05I00 nEAD A(I,J)lro LHirM rr8truG "++ f*G,o,U0 !\[EXT eI t180 IJBIM : LPRIM t
1{0 l[ExT Il8o DATA 3,5,1,4,6,6
lOO I,PBII[!1?O IXB|IM '?EN,trAI,IAII MAIT|ITS A DE}IOAN BII,AIIGAN Sf,N,AB f, :''U0fOBI=1T0219OI0BJ=I105800B(I,J)=f,*A(I'J)210 LPniIM IlEIlt0 "++ "!(L.I);2A0 l{E[r J880 LPRIM : LPBIM840 NIXT I860 El{D
}IASIL PROGRAIUI
BILANGAN SKALAR = 3MATR]KS A :
2 3.1456PERKALIAN MATRIKS A DENGAN BILANGAN SKALAR K :
69312 15 18
Agar bersifat universal, maka READ-DATA diubah dengan input, di manadata-datanya dimasukkan melalui keyboard.
FOR J:l I0 3
8(1,il = t(tt(l,J)
t2 r3
I
l
I
PROGRAM FLOWC}IART
IISTIIIG PROGRAM
10 clas0 LPniIM CIIRS(16)
60 N8M EROGBAM PEN AIJAN MfltsIls A DENGA}I BII,ANGAN STAI,AR T40 INPITT "!,IIA8IIIIKNI BIIANCTAII SKALAB = ";K
t4
60 LPBIM'tsIIANCAI{ SMIAB = ":K00 PniII\lT
70 PBIM ''MA8I'KAN OX,DO I/[,[IIRISB A :''80 INPITT "JIIMLAH 8ABI8 = "lBA90 INPIIT ".rUIIIAII K0U)M = ";80100 Dn[ A(BAr0),8(BA,r0)110 I.PtsIM ''MAIBItrS A I'UO I,PBIM1g0FOBI=11OBA1{O FOB cI : I 1O K0$0 PBIilT "BABI8 "tr;" KOLOM "d;160 INn r A(I,J)I70 LPBIM U9ING "++ "A(I'.I);U0 MXf q,
I90 LPB;IM : LPRIMAOO IIEXI IeIO IJBIM : LPMMS8O I,PBIM ''PEN,trAIJAN MATBITS A DENGAN BIT. 8I(AIAR K : ''850FOBI=1T0BA440 FOB tI : I 1O K0
A60B(I,J) = K*A(I,J)e60 LPBIM USING "++ "iB(I,.I)lE?O $EXT Je80 LPAIIfI : LPBIMEgO NEXT I600 El{D
I
HASIL PROGRAM
MASUKKAN BILANGAN SKALAR = ? 3BILANGAN SKALAR = 3MASUKKAN ORDO MATRIKS A :
JUMLAHBARIS:?2JUMLAHKOLOM=?3MATRIKS A :
BARIS 1 KOLOM 1? 2.BARISlKOLOM2?3BARISlKOLOM3?1BARIS2KOLOMl?4BARIS2KOLOM2?5BARIS2KOLOM3?6
I}IPUIBIL SI(ALAR:
l(
FOB I:l I0 B
Bfl,J) = Nt(A(l,J)
15
I
ili
i
L. lrt***
lf==-:-.
:li,.it
PERKALIAN MATRIKS A DENGAN BIL. SK.ALAR'K :
12 15 18
DAD
Lii
1irir
1111
illilliiili
iri',
I
rl i
Transpose Matriks
Bila setiap elemen baris dijadikan kolom, maka rndtriks yang terjadi
Diketahui:
A(2.3) = [fllif Xr3ifll3i]Dimisallon transpose matriks
.A(2,3), adalah B(3,2).
Maka:
I a(t,t) A(2,1)'l r
a1r2) = IA(1.2) N2,21I16,1r,3) A(231J
Sudah diketahui balnva:
f B(t,t) B(1,2) IB(3.2) = | 842,11 g(Z.Z)
II s(g,r) B(3p) J
Diperoleh persamaan: :
B(l'l) = A(l,l)A1t2) = A(2,1)
ii 16lr
B(2.1) = A(l'2)8(2,21 = N2,2)B(3'l) : A(l'3)B(3'2) : A(2'3)
Kesimpulan umum yang dapat diperoleh unuk dimasukkan ke dalam
B(IrJ) : 6P'1;
cot{ToHDiketahui:
A(2,3) = [? I Z]I'g 71
B(3,2)=N=ll 8lL2 2J
PROGRAITI FLOWCHI\RT
INA||SPOSE lIAIRIl(S
USTIIYG PROGRAI{:
10 LBPIIn CHn$(16)
EO NEM PEOG&AM IBANSPOSE
80 DrM A(A,5),8(6,8)40 LPRINT "MATBII(fI A :"50FOBI=1T0860FOBtI:IlOB?O BEAD A(I,J)80 LPBIM USING "++ "S(I'.I);90 B(J,I) : A(I,.I)100 NEXI eI
lIO I,PRIM : I,PBIMu!0 lfExr I
PR I III,
B (I,J)
B (1,0: A (J,l)
FOR I:1 I0 3
FOB J:l I0 2
Dlllc (2.3)I (3;2)
0B I:1 l0 2
fl
ri
rao DATA 6,1,8,7,83
140 LPniIM : LPBIM160 LPBIIIT 'TIANBPOSE l[{lBIf,B A :"160 I.PBIM1?0fOnI=1K)5I80EOBJ=1K)B190 LPBIIIT ITBING "++ ";B(L,I)I800 t{Exr Je10 LPB;IIIT : {,PBIM280 llExI IAgO EIID
}IASIL PROGRATT
MATRIKS A :
312782
TRANSPOSE MATRIKS A :
371822
Agar dapat bersifat universal, maka sebaiknya data dimasukkan melalui
keyboard (INPUD.
PROGRAM FLOWCIIART
IRAIISPOSE IIAIRIl(S
I
I:l I0 B
J:l I0 l(
B(l,J)=A(J,l)
F0f, I:1 I0 I
I
I
I
USTIIIG PROGRAIUI .
r gI8' "
10 J4B.INI CUnSlro;a0 inEil tf,ocMu rBAt{8PosE8I ,PBIXE "I[A8I'[XAIT ON,DO L[AM,If,8 A I'EA INIM.''JI'I[I,AII BASIS I{AIE|ITS A = ''!as:[wuf "etrItrf,Iall rolou uarBIrB A = "SE4 M,IM60 D.E[ A(8,tr),8(K,B)40 IJHUI '!IATBIf,S A :"4I FB|ITfT 'UASUKAI{ EI.EMEN If,ATX,IXS A :''00fOBI=1fl)B60 10X",{ .= }r1O tr?O PtstrlE'?ABI8 ";I;"f,OI0M "tilt?r u{Pt I A(I,J)s0 LPf,trtIf I,SII{O "++ "S(t,.1)'
l
90 B(cIJ) = A(Id)rco -x$q-rr1!.0 IXEIIIE: I?BIMuo NEIIT'I '.
I4O Iflinfui : I,PBIMrgir Ilnnm 'minsPogB MATaIf,B A :,'
160 r,PROii '"
l
170 !08 I,= I 1'0 f,U0fOBtI=1[0B$0 unililT I]SII{0 "## "!(I,J);800 NEXI q, l
e10 LPBIIfl : LPBIMe80 ilExr r
AOO EI{D
HASJL PR9GRAM ,,
MASUKKAN ORDO MATRIKS A :
JUMLAH BARIS MATRIKS A'= ? 2JUMLAH KOLOM MATRIKS A = ? 3
MATRIKS A :
MASUKKAN ELEMEN MATRIKS A :
BARISlKOLOMl?5BARISlKOLOM2?4BARISlKOLOM3?8BARIS2KOLOMl?2
22
BARIS2KOLOM2?1BARIS2KOLOM3?2
TRANSFOSE MATRIKS A :
524182 r :I\. '
;':' .il :,-t
'L :
. 'di i:
,.,' a:'
lit .t:t..,,\, '+ :'''i:" "'!'-t'l
?.
,l t-i
., . ,i.+ r* ri$.x;.r:
,r ;ii ''i it:,,i ..::
i., ,l i
I I ;j ;iii i ,,r
i:,, .,,, ,';.{ril
;?ts
i'
BADC(2, I ) = A(2, I )x B( I, I )+A(2,2) x B(2, I )+A(2,3) x B(3, I )
C(2,21 = A(2, I ) x B (1,21 + N2,2) x B(2,2) +A(2,3) x B(3,2)
Kesimpulan umum yang dapat diambil dimasukkan ke dalam programadalah:
C(lJ) : A(I, I ) x B( I J)+A(1,2) x B(2,J)+A(1,3)x B(3J)Perkalian Dua Matriks
JilG kita mau mengalikan dua mabiks, maka harus tahu persyaratan
perkalian, Yaihr:'- gunyuf kobm pada mahiks Pertama harus sama dengan banyak
baris pada matriks kedua.
- setiap uaris pada matriks pertama harus dikalikan pada setiap kolom
pada matriks kedua.
A(m,n)*B(n,k):C(m'k)
c(2,2, = [ ff I I I I 3[ Il]ifl I iil x 3[3:liil[ i :ii I itiii,-
eiz, t i x s( l, I )+A(2,2) x B(2, 1 )+A(?'11', !q'1 I
eiz, t i x a( t,2l+ N2,2lxBl2,2l + N2'3) x B(3'2) J
c(2,2) = [;tl:ll :tl3i]
Diperoleh Persamaan:
C(l.l ) = A(1,1 )x B(1, I )+A(1,2)xB(2,1 )+A(l'3)x B(3'l )ll
C(l 2) = A(1,1 )x B(1 21+e(l 2)x B(22)+A(l'3)x B(32)
24
COIYTOHDiketahui:
A(2,3) =
c(2,21 =
c(2,2) =
A(2,3) x B(3,2)
lz 21B(3,2) = | I ql
15, 3J
+ 2x6 lxz + 3x4+ 5x6 4x2 + Ox4
ll
Ir 3 z1Le o sl
tl a 3l .
I txz + 3xlLqxl + oxl
l7 21
Lt 3l+ 2x3l+ srsl
I t+ 3+t2Lza+ o+go
lzz zolLsa zsl
2+12+ 61a+ o+ts I
PROGRAM FLOWC}IART
rER,{ALIAH DU,I IIflIRII{5
25
PB I III
C (I,J)
F0fl I:1 I0 ?
F0B I:1 I0 ?
F0B J=l I0 3
F0R J:l I0 3
c (1,.I) :
F0[ I=t I0 2
F0H I=1 I0 ?
F0R J:l I0 ?
F0[ J:l I0 ?
USTITYG PROGRAM
10 LPBIM Cr{B$(16)30 BEM PNOCRAM PEB;rAIJAN DUA MATR|IKS
80 DIM A(e,5),8(8,e),C(A,A)40 PBII{T "MATRIKS A :"60 PBIM60fOBI=1T08?0FOncI=1T0680 nEAD A(I,,r)90 PBIM U8IN0 "++ "$(I,.I);lOO I{EXT JU0 PAII{T : PBIMIEO l{Hif, I1S0 DATA 5,6,9,e,6,7
140 PBIM : PBIM160 ?B;IIU "IIIATBIKS B :"160 PBINT
I70 I0&I=1T06I80f0&J=lIOP190 EEAD B(I,cr)
e00 PniIM UgINe "++ "!(L.I);ElO NEXT J280 PruM : PMlfI8A0 IIEXT I
2627
f,
a40 DATA 8,1,8,4,6,6
880 PBII{T : PBIMa80FOBI:1I088?0FOBrtr=1102eSO CO,J) = A(I,1)*B(lier) + A(I,S)'rB(a'.,) + A(I'5)*B(5'J)
890 NEXI J600 NEXT I6LO PBIIIT "PEBKALIAN lf{frurc A DENGAN l[AIAiIf,S B :"
6A0 PS|IM
660FOBI=IT08C40fOBtI=1T03S60 PAIM IIEING "+++ ";C(I'J);s00 l{Ext J670 PniIM : PRIM580 NEf,T I890 EtrD
HASIL PROGRAIUI
MATR]KS A :
359267
3465PERKALTAN MATRIKS A DENGAN MATRIKS B :
75 68
64 61
Cara yang lebih singkatBentuk
C(l,J) = A(I, 1 ) x B( r J) +A(1,2) x B(2J)+A(I.3) x B(3J)
Kalau sampai 10 looping, maka bentuk di atas tidak efisien lagi, sepertidiperlihatkan di bawah ini:
C(IJ) : A(1, l)x B( lJ) + A(1, 2)x B( z,Jl + A(I,3)x B( 3J) + A(1,
4)x B( 4J) + A(r,5)x B( 5J) + A(1,6)x B( 6J) + A(I,7)x B( 7J) +A(t, 8)x B( 8,J) + A(1, 9)x B( 9J) + A(1,1O)x B(loJ)
Karena benhrkA(l,l)xB(lJ) , A(12)xB(2Jl danA(1,3)xB(3J) dari C(lJ)= A0,1)xB(1J) + A(1,2)xB(2J) + A0,3)xB(3J mempunyai bentukyang sama dan dit6mbah lagi bilangan yang ada merupakan bilanganasli, maka bilangan asli tersebut (1,2,3) dapat dibentuk menjadi loopingFOR...NE)ff yang baru.(Misalnya: For K = 1 to 3 ..... Next K).
Dengan demikian bentuk di atas dapat disederhanakan hingga menjadi:
C(I,J) = C(IJ) + A(I,K) x B(KJ)
PROGRAII FLOWC}IART
MATRIKS B :
21
i
i
I 23mengalami ffga looPlng
28
PERXALIAII DUA
F0fi I=1 I0 2
0[ J:l I0 3 0B l=t I0 3
J=l lll 2
29
q -ti @-
I
lil,
40 I,PBIM ''MATBIIIB A I'60 I.PBII{T
60rOBI:110370fOBtI=1T08so nEAD A(I,J)90 LPBINT IISING "++ "$(L.I);IOO I\IEXT J110 LPBIIfT : ISEIMUO I{ErI IIgO DATA 6,6,9,2,6,?
I4O I,PBIM : I,PRIMIEO I,PBII\IT ''trIIf,TXIKB B :''IOO IJBINT 1
l?0FOBI=1T08180fOBJ=1I02I90 nEAD B(I,J)800 LPBMT ITSING "++ "|B(I,,I);810 I{EIltr eI
840 LPAiMT : LPBIIII8t0 Nuxr r840 DATA e.L5,4,8,6
880 LPniIillT : LPBIMSOO IJBTNT "PEN,KIIIAN MATRITS A DEIIGAN IIAIB,IKS B I'8?O I,PBIM880r0BI:1$2890[0BrI=1T08A00FORf,=IT065I0 C(IID : C(I,J) + A(I,tr)*B(&J)a80 ltBxt r060 LPBII{T U8fi0 "+++ ";C([,tr);640 r{EXT J6EO IXruM : LPBIMAOO NEXT Ic?0 El\lD
HASIL PROGRAIUI
MATR]KS A :
359LISflNG PROGRAM
ro LPBIM CEn$(ls)AO REU TNOOBAU PENTNIAI{ DUA UAMiITS
so DrM A(a$)3(C,A),C(A*)
30
267
MATRIKS B :
21
PR I
'IIc il',J)
0B l=t I0 2
0B J:l l0 2
X=t I0 3
3l
3465
PERKALIAN MATRIKS A DENGAN MATRIKS B :
75 68
64 61
Agar program dapat bersifat universal, maka sebaiknya data dimasukkan
m-elalui tteyUoara (dengan statement INP(D
PROGRAIT'I FLOWCHART
tOX I:t l0 Bt
t0[ J:l l0 tt2
tOR t(:1 I0 l(l
DIII
f, (El,l{t)I (82;l(2)c (Et;l(2)
0[ I:t I0 81
USTIIIIG PROGRAITI
I CIsr.0 LPBIM CHB0(16)
20 NEM PROGRAII PEN,KAIJAN DUA UATBIKS
60 P$M "tr[Aglrf,f,A[ 0RD0 I/[AIBIKS A :" I
40 NruT "rIIrtr[LAII BAHS MAIBIII8 A : ";B1
60 INPITT "tIIItrILAII IOIOU MATBiIfg 4 = ";IO
60 I,PBIITT
?O IBII{T 't/lAgl'trtr4l[ ON,DO UAMiIf,B B J'
S0 ilP[r'T "JIIIIIAII BABIS tlAIBIffi B = "'F8
90 INIUT ""I'MI,AII KOTOU MAIEIKS '
_ I';K8
lOO IJBIMuo Ir Kl <> B8 IIIEN 50
180 DEI A(BI,K1),8(B3S2),C(BLm)r5O IXBIIIT 'Xf,A[BIf,B A I'140 I,PB|IM
$O PniIM 't[AgIrKtrAlI EIIMEN MAIBIKS A :"160 FOB I = l' 10 B1
t?0 fOB rI = I T0 f,Ifgg mgp1 '96319 "{;"K0I0M "ilir90 INIUT A(I,J)200 IJniIM II9INO "++ "*(I''I);eLO NEXI qI
e80 LPRIM : LPRIM
eso I[Ext I -
A/m I,PBM: LPBIMP60 LPBIIfl '!f,AlIBIK8 B :"A60 I,PBII{T
2?O PBIM "III,A$UKKAN EI.EMEN I/IAIBIKS B :''
AS0fORI=1T0B8a90r0BJ:110K8600 PBIIIT 'i411,5 "J;'t0IOM "IIi5r0 INPUT B(I,sr)
6eo If-Bnm IT8ING "++ "!(L.I);560 NH(T eI
640 LPBII{T : LPBIMSEO ilEXT I600 LPBIIIT : LPBIMg?O I.IB|IXIT '?EN,trAI,IATI MAM|ITS A DEI{GAN IIAIN|ItrB B :''
6S0 I,PRIM590tr'OBI=11081400 POB tI = 1 T0 X2
410F08f,:tI0Kl480 C(k) = C([.r) + A(I,K) * B(r'iI)450 NErt r440 LPRIM USINO "+++ ";C(I'J);46Q NEXT J
34
400 LPniIM : LPBIM4?O NEXT I480 EIID
}IAsIL PROGRAM
MASUKKAN ORDO MATRIKS A :
JUMLAH BARIS MATRIKS A=? 2JUMLAH KOLOM MATRIKS R = ? IMASUKKAN ORDO MATRIKS B :
JUMLAH BARIS MATRIKS B = ? 3JUMLAH KOLOM MATRIKS B = ? 2
MATRIKS A :
MASUKKAN ELEMEN MATRIKS A :
BARISlKOLOMl?3BARISlKOLOM2?5BARISlKOLOM3?9BARIS2KOLOMl?2BARIS2KOLOM2?.6BARIS2KOLOM3?7
MATRIKS B :
MASUKKAN ELEMEN MATRIKS B :
BARISlKOLOMl?2BARISlKOLOM2?1BARIS2KOLOMl?3BARIS2KOLOM2?4'BARIS3KOLOMl?6BARIS3KOLOM2?5
PERKALIAN MATRIKS A DENGAN MATRIKS B
75 68
il61
;t
lililiriL
.1,,illr
I
1
i
i
i
35
DAE
Determinan Matriks
Syarat determinan adalah matiksnya harus bujur sangkar.
Mencari determinan ordo 2x2:
a - [e(t,l) e(t,z)lFt2x2 - Le(z,t) 4zz)l
L{ = A(1,1) x A(2,2) - A(1'2) x A(2'l)
coNToH
DETEB}I II{AI{HAL T
USTTIYG PROGRAIUI
r0 Dru A(8,2)A0fOBI=1102S0l0BcI=1I08{0 8,EAD A(I,ar)EO NEXI J60 I{EXT I?O DATA 9,0,4380 x = A(1,1) * A (83) - ql3) * A(8,1)90 IXBITT 'DEf,lnUIlIAW llAlnrc A = "I(lo0 tto
itI
IilI
lili
]]1
tiliti
i
i
I
l
^: [?
lAl =9.8-5.4=72-20:52
s'lal
0B I=1 l0 2
L*36 3?
11 '1"*
}IASIL PROGRAIUI
DeTnUITRNMATRIKSA=52
Mencari determinan ordo 3x3 ada dua cara:
1. Cara Sarrus2. Cara Kofaktor
Cara Sam.re:
A(1,3)'lA(23) |
A(3,S; J
lA(1,1) A(1,2)A3"3 : lalz,ty N2,21
Ln(g,t) A(3,2)
lAl : IA(1,1) x A(2.2) x A(3,3) +A(1,2) x A(2,3) x A(3,1)+ A(1,3) x A(2,1) x A(3,2) I - [ A(1,3) x A(2,2) x A(3,1)+ A(1,1) x A(2,3) x A(3,2) + A(1,2) x A(2,1) x A(3,3) l
Kofaktor:K(ii) = (-l)r+i t\
A(3,1) A(3,2) A(3,3) = *1 [A(2,2)xA(3,3)-A(2,3)xA(3,2)]: A(2,2)xA(3,3) - A(2,3)xA(3,2)
A(1,1) A(1,2) A(1,3A(2,r) A(2,2tarz,3l K(1'1) = (-r)r+r I ffi:l l[3,3i
lAl : A(1,1)xK(1,1) + A(1,2)xK(1,2) + A(1'3)xK(l'3)lAl : A(1,1)x I A(2,2)xA(3,3) - A(2,3)xA(3'2) ] +
A(1,2)x [ -A(2,1)xA(3,3) + A(2,3)xA(3,1) ] +A(1,3)x I A(2,1)xA(3,2t - A(2,2)xA(3,1) ]
lal : + A(1,1)xA(2,2)xA(3,3) - A(1,1)xA(2'3)xA(3'2)- A(1,2)xA(2,1)xA(3,3) + A(1,2)xA(2,3)xA(3,1)+ A(1,3)xA(2,1)xA(3,2) - A(1,3)xA(2,2)xA(3,1)
lAl = + A(1,1)xA(2,2)xA(3,3) + A(1,2)xA(2'3)xA(3'1)+ A(1,3) xA(2,1 ) xA(3,2) - A(t,S) xA(2,2) xA(3, 1 )- A( 1,1 )xA(2,3) xA(3,2) - A( 1,2) xA(2, I ) xA(3,3)
lAl = [ A(1,1)xA(2,2)xA(3,3) + A(1,2)xA(2,3)xA(3,1)+ A(1,3)xA(2,1)xA(3,2) I - t A(1,3)xA(2,2)xA(3,1)+ A(1,1)xA(2,3)xA(3,2) + A(1,2)xA(2,1)xA(3,3) l
Cara pemecahan rumusan determinan ordo 3x3Misalkan lnl = X
x : I A(1,1)xA(2,2)xA(3,3) + A(1,2)xA(2,3)xA(3,1)+ A(1,3)xA(2,1)xA(3,2) I - t A(1,3)xA(2,2)xA(3,1)+ A(1,1)xA(2,3)xA(3,2) + A(1,2)xA(2,1)xA(3,3) l
JikaX-C-D,maka:
C = A(1,1)xA(2,2)xA(3,3)A(1,2)xA(2,3)xA(3,1)A(1,3) xA(2,1)xA(3,2) +
C_ c + A(1J)xA(2J+1)xA(3J+1)MN
A(1,3)xA(2,2)xA(3,1)A(1,1)xA(2,3)xA(3,2)A(1,2)xA(2,1) xA(3,3) +
D + A(1 J+2)xA(2J+1) xA(3J)NM
p=
lAl :A(1,!) A(1'2) A(1,3)A(2,1) N2,2' A(2,3)A(3,1) A(32) A(3,3)
A(1,1) A(1,2)A(2,1) N2,2'A(3,1) A(3,2)
+++
A(1,1) A(1,2) A(1,:A(2,1) A(2,2)erz,3l K(l,2) = (-1)r*2 I ffi:li l[3:3i I
A(B,r) A(3,2) A(3,3) =
_l llr,i,l6ii,? r0l?;i5^ti;1,
A(1,1) A(1,2) A(1,:A(z,r) a(2,2tetz]lr K(1'3) = (-t)r+s I ffi:ll l[3:3] I
A(3,1) A(3,2) A(3,3) = *1 [A(2,1)xA(3,2)-A(2,2)xA(3,1)]= A(2,1)xA(3,2, - A(2,2)xA(3,1)
38
Diambil:
M=J*1N=J*2
fft,I'
C : C + A(lJ)xA(2"tYt)xA(3,N)
D - D + A(1,N)xA(2Jtt)xA(3,J)
Ketentuan:
- JilG'M : 4, maka M = 1
-JikaN:4,makaN=1-JikaN:5,makaN:2Perhitungan determinan dengan cara Sarrus:
lAl :
lAl : (1xlx4 + 2x5x3 + 3x4x2l * (3xlx3 + 1x5x2 +2x4x4l
:(4+30+241 -(9+10+32)-a
Perhitungan determinan dengan cara kofat<tor:
PROGRAIUI FLOWC}IART
t24132
3'l
;l23ts24
r1 2
^: lo IL3 2
I*e
212
I43
I43
I43
1524
4534
4132
I
40
?i Ktt=(-l;r*r24
i i Ktz=(-1;r*z24
= +l (1x4 - 5x2l=4-10-; -61
= -l (4x4 - 5x3)--l(16-ls)
a- -l
= +l (4x2 - I x3)= I -, 3_E_J
', Kra = (-l1r*e4
lAl = r(l,l)xK(l,1) + A(1,2)xK(l,2) + A(1,3)xK(l,3)
lel = t x (-6) + 2 x (-l) + 3 x (5)--6+-2+15=J
DEIEnililill
FOX I=1 I0 3
0R J:l l0 3
0B il=l I0 3
ll = Jtlll = JtZ
4t
ft
I
1L1lL
,IL
USTIIYG PROGRAIII
r0 LPB,IM CHB$(16)EO BEM PBOOBAM DEIEBTIII{AII OB,DO gX8
80 DrM A(S,S)40 L'PBIM "IIAIBIKS A :"60F0&I:11O660FORJ=ITO6?o BEAD A(I,J,80 LPBIM IISI}IG "++ "S(LD;90 I{dXT qI
100 LPRIM : LPBIM110 NEXT Iu0 DAIA r,3,9J,1,6,5,A,4
ll3 il*='r=*'r*'160N:cr+8160IFM=,tTIIENM=1t?0 IF N = 4 TIIEN N = 1 EISE IF N : 6 TIIEN N = 2
180 C : C + A(rn). * A(a,M) * A(6J{)190 D = D + A(l,t{) * A(8,M)'r A(8,J)EOO NEXT J810X:c-DPeO IJBIM "DEIE8ffINAI{ A(6,6) = ";XA5O EI{D
}IAsIL PROGRAIUI
MATRIKS A :
123
321
DETERMINAN A(3,3) = 7
Agar bersifat universal, maka sebaiknya data dimasukkankeyboard (dengan statement INPUT).
USTING PROGRAM
t0 crs80 INIIrT "ordo matrlko bgur Bangar = ",N
42
s0 Dru A(nD"P(N-2)"EP([-e)"Dr(e),KB(il-e)40 FOn I = I S0 N : FOB q, : 1 T0 N : PRIM'tatrlg "f;"
kolom ";tI;: INPII A(I,J) : NEl$ J : PBIM : NEICI I50 r0B I : 1 I0 N-P : P(F):f,1F,t) : f,P(F)=F : (B(E)=l : NEXI F60 Df,(I)=a : DK(!)=g : DK:1 : N0=170FORG=lION-880 If [P(G):NO TIIEN N0=N0+I : G0f0 70
90 r{ExT G
100 Itr DK<6 TIIEN DK(DK):NO : DK:DK*I : N0=N0*1 : G010 70
110 DK:A(N-1,DK(1)) x A(N,Dr(8)) - A(N-I,DK(2)) * A(N,DK(l))feO fOB X : I 1! N€ : DK:DK*P(X) : I{EX(T X180 D=D+DK140 FOB F : N-8 1! t STEP -1 : IP:[P(F)160 f,P:BP+l : IF f,P>}I TIIEN 810
160mF:lTIIEN?001?0 FOB G = F-I T0 1 $tsP -I180 II f,P:8P(c) IIIEN 160
190 t{ExT G
e00 rR(f):Et(F)+1 : ZI=F : ZP'=W : G08ID 640 : GOI0 880
e10 NEXI F : GOTO 500
eeO IF f = N-2 TIIEN 60 EISE N0=I450 FOB H : F+I 10 N-8
2/mFOBtI=IT0FP60 Ir rP(cr):N0 TIIEN N0:N0+1 : COtg ?40
e60 $ExT J??0 XR(H):I : Zl:ll : Z8=N0 : GQSID 6?0 : N0:N0+1280 trEXr H
e90 G0T0 60
500 tP&Xm "dot€rnlna,n = "iD610 EI\]D
SzO W(ZD:M: P(ZI)=A(Z1,Z2) * ( r)^(XC(ZI)+r) : B,EfiIRl{
HASIL PROGRAM
ordo matriks buiur sangkar : 3barislkoloml?1barislkolom2?2barislkolom3?3
baris2koloml?4baris2kolom2?1baris2kolom3?5
l
melalui
43
baris3koloml?3baris3kolom2T2baris3kotom3?4
a '.determinan = 7
DAD
E
Penyelesaian invers matriks secara analitis ada tiga cara, yaihJi ' r',.
1. Eliminasi biasa dari
A.A-r=In , ., 1..
2. Operasi baris etementer yang disebut juga elimirysf ,Giuss
.t9rd1.
r loBE r l[^t,..|- [ut^-'1
3. Mabiks Mjoint
A-r = fr. aairer i1'1i':' .i : , ', r
Dalam hal ini kita akan mempergunekan cara kedua, yaitu dengart
operasi baris elementer (OBE) t,
:,,,. _, '
n* = 12 3 ] renhrlonlah inveis matriks A "4 4Xt
-
Dengan cara OBE ( A I lz ) ( lz I A-r)
urcl I r----J I
L2
bzt(-21
--€b12(-l)
o -l lbm(;)I ol ,o 2J
-1'l3l2J
OBE r1 0lo ILs o
11 0lo 1
Le o
A-l :
0-ti
I
1i_li-1
0I0
-1.l
3l
Io
lq 3 Ilzzltl i I
tl?l
: li ol2lo rli I I olr I -t z)tl
?lolrl
o'lb3(2)O l----,rJ
o 1us1-|yO l---+zJ
b2(2',)
-J
103110
0o1
t_t_
I rr t2 13'l: lzt 22 zslLrr s2 ssJ
I-3-1u4I
2
I-t
-itl 6-r = [],-ri Izl
CONTOH
Agx3 =lz 0 r'lL? I ?l
Tentukan lnvers matriks A
Cara penyelesaian inverse matriks dengan program:Misalkan matriks yang diketahui:
Xers
Di mana11 -+ Indeks dari element X(1,1)12 + Indeks dari element X(1,2)13 + lndeks dari element X(1,3)2l + lndeks dari element X(2,1)22 + Indeks dari elemert X(2,21
23 -+ lndeks dari element X(2,3)31 + lndeks dari element X(3,1)32 + Indeks dari element X(3,2)33 + lndeks dari element X(3,3)
M=l
I
liiltLll
rlllril,
I
iii
I
I
Dengan cara OBE
OBE(Alla)---+(IslA-r)
!2
-liI
0Io
lo01oo
!2
oo
12 o Ils I oL1 o 1
[r o ils r o[1 o l
0I0
oIo
oI0
!2
-rLo
!2
-rlI-2
o'lblG)
? l--0l b2r(-3)O l----------.---rJ
0lb31(-l)
?.l----
t4 15 1624 25 2634 35 36
11 t2 132t 22 2331 32 33
I?
-ria
!-iI
l'r olo lL1 o
rl olo IL6 o
l'r olo lLo o
!-i-1
11
11
t2
11
13
1l
t4
11
15
11
16
1l
l1-(21x I l) 22-(ztxtzl 23-(21x 13) 24-Qlxl4 25-(2lxl5'; 26-(21x16)
tl -(31x 11 32-(3r x l2) 33-(31x 13) 34-(31x14 35-(31x ls) 36-(31xlQ
47
Xi 1 'r'""'.'
Perhitungan secara program:14 15 t624 23 2634 35 36
11 t2 132t 22 2t3l ,2 33
I o o]o I olo o lJ
ollool
12 0A3"3=13 I
Ll o
l'l 12ol-----lsrJ [r
illrLlllr
:
Nl=2
M:3
t4 15 1624 25 2634 35 36
1l t2 t32l 22 2t31 32 33
X(l'l) : 2 X(l'2) : g X(1'3) = 1
X(2'l) : 3 X(2,2Y = 1 X2'3) = 0X(3'l) : 1 X(3'2) = g X(3'3) = t
Untuk M = 1
D=X(IttJt)={t,t)=f,K = 1 -+ X(1,1) = X(1,1) I D = 2li2 = |K = 2 -+ X(1,2) = X(1,2) / D = OD = O
K = 3 + X(1,3) = X(1,3) / D - lP, = O.5K.= 4 + X(1,4) = X(1,4) I D = ll2 : O.5K = 5 -+ X(1,5) = X(1,5) / D = OP = O
K = 6 --r X(1,6) = X(1,6) / D = Ol2 : O
l'r o o.tls I oL1 o 1
"=rlM=tJ-
o.5 0 lD'lo I olo o rJB=l) apakah
) B:M?M: lJ 1t: ty
v+
Kesimpulan:
(rJ)-(urr)x(MJ)X(IJ) : X(lJ) - X0,M) x X(Ir!,,J)
JikaB:t (B-baris)K:J (K-kolom)
Ivlaka:
X(B,K) = X(B,K) - X(BJUI) x X(M.K)
48
XB,K) = XB,K) - X(B,IUI) x X(Iut K)
X(B,K) = X(B,K) - C x X(ttl"K)
K : I + X(2,1) ] *(3,r1 : : X T,r.r,
K = 2 + X(2,2) :O?,r, _ : X },r.r,
keluar jalur
t--+ C = X(BJI)
= X(f,,t) = 3
-o
apakahB:M?(2 *tl
=t
I t -(12x21) tl-(12x22) l3-(12x23) t4-(t2x24) l5-(12x25) l6-(12x26)
2t
22
22
22?22
24
22
25
22
26
22
lt-(32x2t 32-(32x22) 33-(32x23) 3r.-(32x24l 35-(32x25) 36-(32x26)
ll-(t3x3l l2-(l3x32l l3-(l3x33l 14-(l3x34l l5-(13x35) r6-(r3x36)
2t-(23x31) 22-(23x32) 23-(23x33) 24-(23x34) 25-(23x35) 26-@3x36)
3t
33
32
33
33
33
u33
35
33
36
33
l,lrliLi
K = 3 -+ x(2,3) : O3,r' _ : X rg,r,
K = 4 _>x(2,41: X(24) _: X rg,n,
K : 5 + X(2,5) : X(2,5) - C x X(l's)= I -3xO =
K = 6 -+ X(2,6) = X(2,6) - c x x(1,6): 0 -3x0 :Q
1ll
= -1.5
= -1.5
I
t---+ C : X(B,M)
X(3'1) = 1
=$
-o= 0.5
: -1.5
-0
:l
Untuk t{ : 2
D=X(M,lYl)=X(2,2):1
K = I -+ X(2,1) = X(2,1) / D :K:2-+X(2,21 =X(2,2, /D =K : 3 -+ X(2,3) : X(2'3) I D =K=4+X(2,4)=X(2'4)ID:K=5+X(2,5):X(2'5)lD:K=6?X(2,6)=X(2'6)/D=
o.5 0-1.5 1
-0.s 0
0/l=0lll=l
-l.5ll = -1.5-1.5/1 =' -1.5
lll:l0/1 =0
t--+ C : X(BJvt)
: X(1,2) = 0
-0
= O.5
= 0.5
=Q
=Q
I
i
I
ililr
illiii11
illr.il'
lt
tlll,ll
iI
lt
rl
I
tt!l,t-
o.5 0 0'l-1.5 101o o 1J
rl 0 0.5lo I -r.5Lr o 1
" = 3l
M = 1T
o'lol1J
l'1 0 0.slo r -r.5Lo o o.s
B=1\ apakahl g=M?
Nr=z) o+zlapakahB:M?(3 +1)
X(B.K) : X(B,K) - C x X(Itl'K)
K : I + X(3,1) =:
*,?,r, _ t X f,r'r,
K = 2 + x(3,2) I ,(3,r1 _r X X1'2)
K = 3 + X(3,3) ::
*,?,r, _ t X fg,r,
K = 4 + X(3,4) : x(3J) : : X Ig,n,K = 5 + X(3,5)
:: *,3'U, : : X f(r'U,
K = 6 + X(3,6) : *(?,:) _ : X
x(1,6)
x(B,K) = X(B,K) - C x X(M'K)
K = I + X(1,1) ] x(l,rl : t X x2,r)
K = 2 + x(1,2) ] "(l'zl _ 3 X !r,r,K = 3 + x(1,3)
=: *,1:3,
_ 3 X x(23)
K = 4 + x(1,4) ==
*,1:3, _ 3 X
x(2'4)
K = 5 + x(1,5) I *(l'ul : 3 X T,r'u,
K = 6 --+ X(r,6) : *,1,., _ 3 X
x(2,6)
=t
o'lolrJ
r1 0lo lLo o
oIo
rl olo lLo o
0.5-1.5
o.5
o.5-1.5-o.5
o.5-1.5
o.5
0.5 0-1.5 I-o.5 o
o'lol1J
51
I'Llrl
apakahB:M?Q=2)
apakahB=M?'(3 +2)
":rl^=rl" = rln=zlX(B,K) = X(B,K) - C x X(tul,K)
K = I + x(3,1) =:
*,3,r, _ 3 X 6(2,1)
R = 2 + X(3,2):: ",3,r, _ 3 X !r,r,
K : 3 + X(3,3) = X(3,3) - C x X(2,3), : O.5 -0x-1.5
K = 4 --+ X(3'4) :=
*,1'3]u_ 3 I :,i.'f,
K : 5 + x(3,5) =:
*,3,u, _ 3 X fr,u,
K : 5 + x(3,6) ] *(?,el : 3 X x(2,6)
Ontuk M : 3
D = X(M,IUI) = X(3,3) = .0.5
K=l-X(3,1):X(3.1)lD:K:2 + X(3,2) : X(3,2) / D =K:3+X(3,3):X(3,3)lD:K=4*X(3,4):X(3,4)lD:K = 5 -- X(3,5) : X(3,5) I D :K:6-X(3,6)=X(3,6)lD=
52
v-+ keluar jalur
t.----r C = X(B,M)
: X(3,2) = 0
:Q
-0
: 0.5
= -1.5
:Q
=l
: I 'l aPakahI e:M?t_:3) (l +3)
B = 2'l apakahl_e=M?
M = 3 ) Q +s)
t+ C = X(BJvl)
= X(1,3) = 0.5
t-+ C = X(B,M)
= X(2,3) = -1.5
o'lol2J
o 0.51 -1.5o1
l1loLo
B
M
0.5 0-1.5 I-1 0
Ii
I
X(B,K) = X(B,K) - C x X(M'K)
K = I '+ X(1,1) _:
x(1,1) _ 3.J f$'r, = t
K = 2 + X(1,2) I *(l,rl _:Jff'r, _ oK : 3 + x(1,3)
:: *,1:3,
_ 3J },ir, _ oK = 4 + X(r,4): ol:1, _ tJI,:'f, = rK = 5 + X(r,5)
==
*,1,u, : 3.J ff,u, _ oK : 6 + x(1,6) I x(l,el : ila,t'., = _t
o.s o ol-1.s I ol-o.s o 1J
rl o o.slo r -r.5L6 o o.5
1 0 -rl-1.s I Ol-102)
Ir o olo 1 -r.5Lo o I
O/O.5 = 0O/O.5 = 0
O.5/O.5 = I-0.5/0.5 = -l
0/0.5 = 0110.5 = 2
X(B,K) = X(B,K) - C x X(M'K)
K = I + X(2,1) ] x(3,r1 : i_;.#r;rl = e
53
li I Try'r
lri,L*,
K=2--->x(2,2)=O?,r' _i_i.#irl : IK : 3 --' X(2'3)
:= *('_'i]u_
i_L#';'l : e
K : 4 + x(2,4) : *(1f]u_ i_i.#r;-lr = _3
K : 5 *; X(2,5) :: ",?,u, _ i_i,.#Tl _ I
K = 6 + x(2,6) =:
r,fi,u' _ i_L#1.1 _ 3
USTING PROGRAM
10 LPBIM CnnOtrOlEO N,EM PBOGBAM MENC,ARI INVENS MATAIKS A60 cls40 PBINT "MASIIIffiAN OB,DO II[A[X,II(S A :''50 INPIIT "JIIMLAH BAfiIS MflnIKS A = ";I60 INPIE "erIrMLAH K0IOM MAIAIKS A = "iI70 If I <> eI fiIEN I080cl:Ix?90 DIM X(I,tI)rcOFOnB=1 T0 I110 X(B'B+I) = ]IzO I{EXT B
160 PBIM ''MA,SIJtrKAN EI,EIIEN MAIBIKE A :''I40FOBB:1I0I160FOBK=1T0I160 PR;IM "BABIS ";B;"K0IOM "j(;170 INPITT X(B,K)180 NEIIT K190 PBIMEOO NEXT B
54
AI0 LPBINT "I\IATHIKS A ";TAB(I6*I);"MATA,Itr8 SAIITAII (IDENffiAS)"
1 0 -1'l-3 I 3l-1 0 2l
I r o -1'lA-I=l-3 1 3lL-r o zJ
Ir o olo I oL6 o 1
2EO TPBII{T
3S0fORB=l.T0I840fOBK=1TOJe60 rJBrM x(B,r),A60 MXT Ke70 LPBIM : LPBIME8O NEXT B
e90 LPB|IM : LPBIMS00rcBM=1.T0 I6L0 DX = X(M,M)6A0FOBK=lT0cIs60 x(M,r) : x(M,K) / DX
640 NEXI r350F08,B=1T0I660IFB:MTIIEN4l0670 Dx : x(B,M)580FOBK:1T0rIs90 X(B,K) = X(B,f,) - DX * X(M,r)4OO NEXT K410 NEXT B
4AO NEXT M450 LPBII{T "MAIA,ItrS SATUAN ( IDEIfTITA,9 )";IA8(16*I);
"IMIEBS MATBIKS IT'440 I,PRINT
460fOnB=1T0I480FORf,=1T0.I470 LPBII{T X(B,r),480 NEXT K.490 LPBIM : LPRIM600 NEI|T B
610 END
HASIL PROGRAIUI
Contoh 1, matriks 2 x 2;
MASUKKAN ORDO MATRIKS A :
JUMLAH BARIS MATRIKS A = ? 2JUMLAH KOLOM MATRIKS A = ? 2MASUKKAN ELEMEN. MATRIKS A :
BARISlKOLOMl?4BARISlKOLOM2?3
55
BARIS2KOLOM1?2 ]iBARIS2KOLOM2?2
MATRIKS A ; MATRIKS SATUAN (IDENTITAS)
4.OO 3.OO 1.OO O.OO
2.00 2.00 0.00 1.00
MATRIKS SATUAN ( IDENTITAS ) ; ]NVERS MATRIKS A
1.00 0.00 1.00 -1.500.00 1.@ -1.00 2.00
Contoh 2, matriks 3 x 3:
MASUKKAN ORDO MATRIKS A :
JUMLAH BARIS MATRIKS A = ? 3JUMLAH KOLOM MATRIKS A = ? 3MASUKKAN ELEMEN MATRIKS A :
BARISlKOLOMl?2BARISlKOLOM2?OBARISlKOLOM3?1BARIS2KOLOMl?3BARIS2KOLOM2?1BARIS2KOLOM3?O
BARIS3KOLOMl?1BARISSKOLOM2?OBARIS3KOLOM3?1
MATRIKS A ; MATRIKS SATUAN (IDENTITAS)
2.00 0.00 1.00 1.00 0.00 0.00
3.00 1.00 0.00 0.00 1.00 0.00
1.00 0.00 1.00 0.00 0.00 1.00
MATRIKS SATUAN ( IDENTITAS ) ; INVERS MATRIKS A
1.00 0.00 0.00 1.00 0.00 -1.000.00 1.00 0,@ -3.00 1.00 3.00 t, .
0.00 0.00 1.00 -1.00 0.00 2;00 , ' 1
1,,
rilii'i
56
