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TUGAS 1 
MATEMATIKA TERAPAN 
RAHMAWATY 
P 39002130001 
PROGRAM PASCASARJANA 
UNIVERSITAS HASANUDDIN 
MAKASSAR 
2013
TUGAS MATEMATIKA TERAPAN 
Membuat logaritma dan program untuk menentukan suhu T1, T2, T3, T2, T4, T5, 
T6, T7, T8, dan T9. 
1. Untuk Titik T1 
푇1 = 
1 
4 
× (30 + 25 + 푇2 + 푇4) = 
(55 + 푇2 + 푇4) 
4 
4 푇1 = (55 + 푇2 + 푇4) 
4푇1 − 푇2 − 푇4 = 55 … … … … … … … … … … … … … … … … … … … … … … … … … … … . . (1) 
2. Untuk Titik T2 
푇2 = 
1 
4 
× (30 + 푇1 + 푇3 + 푇5) = 
(30 + 푇1 + 푇3 + 푇5) 
4 
4 푇2 = (30 + 푇1 + 푇3 + 푇5) 
푇1 − 4푇2 + 푇3 + 푇5 = −30 ………………………………………………………………………………….. (2) 
3. Untuk Titik T3 
푇3 = 
1 
4 
× (30 + 60 + 푇2 + 푇6) = 
(90 + 푇2 + 푇6) 
4 
4 푇3 = (90 + 푇2 + 푇6) 
푇2 − 4푇3 + 푇6 = −90 ……………………………………………………………………………………………. (3)
4. Untuk Titik T4 
푇4 = 
1 
4 
× (25 + 푇1 + 푇5 + 푇7) = 
(25 + 푇1 + 푇5 + 푇7) 
4 
4 푇4 = (25 + 푇1 + 푇5 + 푇7) 
푇1 − 4푇4 + 푇5 + 푇7 = −25 … … … … … … … … … … … … … … … … … … … … … … … . . . (4) 
5. Untuk Titik T5 
푇5 = 
1 
4 
× (푇2 + 푇4 + 푇6 + 푇8) = 
(푇2 + 푇4 + 푇6 + 푇8) 
4 
4 푇5 = (푇2 + 푇4 + 푇6 + 푇8) 
푇2 + 푇4 − 4 푇5 + 푇6 + 푇8 = 0 … … … … … … … … … … … … … … … … … … … … … … … . (5) 
6. Untuk Titik T6 
푇6 = 
1 
4 
× (60 + 푇3 + 푇5 + 푇9 ) = 
(60 + 푇3 + 푇5 + 푇9 ) 
4 
4 푇6 = (60 + 푇3 + 푇5 + 푇9 ) 
푇3 + 푇5 − 4 푇6 + 푇9 = −60 … … … … … … … … … … … … … … … … … … … … … … … . .. (6) 
7. Untuk Titik T7 
푇7 = 
1 
4 
× (25 + 40 + 푇4 + 푇8) = 
(65 + 푇4 + 푇8) 
4 
4푇7 = (65 + 푇4 + 푇8) 
푇4 − 4푇7 + 푇8 = −65 … … … … … … … … … … … … … … … … … … … … … … … … … … . . (7) 
8. Untuk Titik T8 
푇8 = 
1 
4 
× (40 + 푇5 + 푇7 + 푇9 ) = 
(40 + 푇5 + 푇7 + 푇9 ) 
4 
4푇8 = (40 + 푇5 + 푇7 + 푇9 ) 
푇5 + 푇7 − 4푇8 + 푇9 = −40 … … … … … … … … … … … … … … … … … … … … … … … … … … … . .. (8)
9. Untuk Titik T9 
푇9 = 
1 
4 
× (40 + 60 + 푇6 + 푇8) = 
(100 + 푇6 + 푇8) 
4 
4 푇9 = (100 + 푇6 + 푇8) 
푇6 + 푇8 − 4 푇9 = −100 … … … … … … … … … … … … … … … … … … … … … … … … … . . (9) 
Daftar Persamaan Linear: 
4푇1 − 푇2 − 푇4 = 55 
푇1 − 4푇2 + 푇3 + 푇5 = −30 
푇2 − 4푇3 + 푇6 = −90 
푇1 − 4푇4 + 푇5 + 푇7 = −25 
푇2 + 푇4 − 4 푇5 + 푇6 + 푇8 = 0 
푇3 + 푇5 − 4 푇6 + 푇9 = −60 
푇4 − 4푇7 + 푇8 = −65 
푇5 + 푇7 − 4푇8 + 푇9 = −40 
푇6 + 푇8 − 4 푇9 = −100
Bentuk matriks 
4 -1 0 -1 0 0 0 0 0 T1 55 
1 -4 1 0 1 0 0 0 0 T2 -30 
0 1 -4 0 0 1 0 0 0 T3 -90 
1 0 0 -4 1 0 1 0 0 T4 -25 
0 1 0 1 -4 1 0 1 0 T5 0 
0 0 1 0 1 -4 0 0 1 T -60 
0 0 0 1 0 0 -4 1 0 T7 -65 
0 0 0 0 1 0 1 -4 1 T8 -40 
0 0 0 0 0 1 0 1 -4 T9 100 
Bentuk matriks augmentasi 
4 -1 0 -1 0 0 0 0 0 55 
1 -4 1 0 1 0 0 0 0 -30 
0 1 -4 0 0 1 0 0 0 -90 
1 0 0 -4 1 0 1 0 0 -25 
0 1 0 1 -4 1 0 1 0 0 
0 0 1 0 1 -4 0 0 1 -60 
0 0 0 1 0 0 -4 1 0 -65 
0 0 0 0 1 0 1 -4 1 -40 
0 0 0 0 0 1 0 1 -4 100 
=
Algoritma 1 
1. Mengubah persamaan linear ke bentuk matriks augment, yaitu matriks berukuran n x (n 
+ 1). 
2. Memeriksa elemen-elemen pivot (elemen-elemen yang menempati diagonal suatu 
matrik). Jika terdapat aii ≠ 0, bisa dilanjutkan ke langkah no.3. Namun, jika ada elemen 
diagonal yang bernilai nol, aii = 0, maka baris dimana elemen itu berada harus ditukar 
posisinya dengan baris yang ada dibawahnya, (Pi) ↔ (Pj) dimana j = i + 1, i + 2, ..., n, 
sampai elemen diagonal matrik menjadi tidak nol, aii ≠ 0. 
3. Proses triangularisasi. 
4. Hitunglah nilai xn 
5. Lakukanlah proses substitusi mundur untuk memperoleh xn-1 , xn-2 , ....,x2 , x1
Bahasa Pemrograman 
%Penyelesaian Persamaan dengan Metode Eliminasi Gauss 
clear all 
clc 
A=[4 -1 0 -1 0 0 0 0 0 55; 
1 -4 1 0 1 0 0 0 0 -30; 
0 1 -4 0 0 1 0 0 0 -90; 
1 0 0 -4 1 0 1 0 0 -25; 
0 1 0 1 -4 1 0 1 0 0; 
0 0 1 0 1 -4 0 0 1 -60; 
0 0 0 1 0 0 -4 1 0 -65; 
0 0 0 0 1 0 1 -4 1 -40; 
0 0 0 0 0 1 0 1 -4 100];%Data Matriks 
disp('Matriks A') 
A 
disp('Jumlah Persamaan') 
n = 9 %jumlah Persamaan 
pause 
% ====Proses Triangulasi==== 
for j=1:(n-1) 
% ---mulai proses pivot--- 
if (A(j,j)==0) 
for p=1:n+1 
u=A(j,p); 
v=A(j+1,p); 
A(j+1,p)=u; 
A(j,p)=v; 
end 
end 
% ---akhir proses pivot--- 
jj=j+1; 
for i=jj:n 
m=A(i,j)/A(j,j); 
for k=1:(n+1) 
A(i,k)=A(i,k)-(m*A(j,k)); 
end 
end 
end 
disp('Matriks A Hasil Proses Triangulasi') 
A 
pause 
% ===Akhir Proses Triangulasi=== 
%--Proses Substitusi Mundur--- 
x(n,1)=A(n,n+1)/A(n,n); 
for i=n-1:-1:1 
S=0 
for j=n:-1:i+1 
S=S+A(i,j)*x(j,1); 
end 
x(i,1)=(A(i,n+1)-S)/A(i,i); 
end 
x
Hasil RUN (Output)
Tugas 1 Matematika Terapan (S2)

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Tugas 1 Matematika Terapan (S2)

  • 1. TUGAS 1 MATEMATIKA TERAPAN RAHMAWATY P 39002130001 PROGRAM PASCASARJANA UNIVERSITAS HASANUDDIN MAKASSAR 2013
  • 2. TUGAS MATEMATIKA TERAPAN Membuat logaritma dan program untuk menentukan suhu T1, T2, T3, T2, T4, T5, T6, T7, T8, dan T9. 1. Untuk Titik T1 푇1 = 1 4 × (30 + 25 + 푇2 + 푇4) = (55 + 푇2 + 푇4) 4 4 푇1 = (55 + 푇2 + 푇4) 4푇1 − 푇2 − 푇4 = 55 … … … … … … … … … … … … … … … … … … … … … … … … … … … . . (1) 2. Untuk Titik T2 푇2 = 1 4 × (30 + 푇1 + 푇3 + 푇5) = (30 + 푇1 + 푇3 + 푇5) 4 4 푇2 = (30 + 푇1 + 푇3 + 푇5) 푇1 − 4푇2 + 푇3 + 푇5 = −30 ………………………………………………………………………………….. (2) 3. Untuk Titik T3 푇3 = 1 4 × (30 + 60 + 푇2 + 푇6) = (90 + 푇2 + 푇6) 4 4 푇3 = (90 + 푇2 + 푇6) 푇2 − 4푇3 + 푇6 = −90 ……………………………………………………………………………………………. (3)
  • 3. 4. Untuk Titik T4 푇4 = 1 4 × (25 + 푇1 + 푇5 + 푇7) = (25 + 푇1 + 푇5 + 푇7) 4 4 푇4 = (25 + 푇1 + 푇5 + 푇7) 푇1 − 4푇4 + 푇5 + 푇7 = −25 … … … … … … … … … … … … … … … … … … … … … … … . . . (4) 5. Untuk Titik T5 푇5 = 1 4 × (푇2 + 푇4 + 푇6 + 푇8) = (푇2 + 푇4 + 푇6 + 푇8) 4 4 푇5 = (푇2 + 푇4 + 푇6 + 푇8) 푇2 + 푇4 − 4 푇5 + 푇6 + 푇8 = 0 … … … … … … … … … … … … … … … … … … … … … … … . (5) 6. Untuk Titik T6 푇6 = 1 4 × (60 + 푇3 + 푇5 + 푇9 ) = (60 + 푇3 + 푇5 + 푇9 ) 4 4 푇6 = (60 + 푇3 + 푇5 + 푇9 ) 푇3 + 푇5 − 4 푇6 + 푇9 = −60 … … … … … … … … … … … … … … … … … … … … … … … . .. (6) 7. Untuk Titik T7 푇7 = 1 4 × (25 + 40 + 푇4 + 푇8) = (65 + 푇4 + 푇8) 4 4푇7 = (65 + 푇4 + 푇8) 푇4 − 4푇7 + 푇8 = −65 … … … … … … … … … … … … … … … … … … … … … … … … … … . . (7) 8. Untuk Titik T8 푇8 = 1 4 × (40 + 푇5 + 푇7 + 푇9 ) = (40 + 푇5 + 푇7 + 푇9 ) 4 4푇8 = (40 + 푇5 + 푇7 + 푇9 ) 푇5 + 푇7 − 4푇8 + 푇9 = −40 … … … … … … … … … … … … … … … … … … … … … … … … … … … . .. (8)
  • 4. 9. Untuk Titik T9 푇9 = 1 4 × (40 + 60 + 푇6 + 푇8) = (100 + 푇6 + 푇8) 4 4 푇9 = (100 + 푇6 + 푇8) 푇6 + 푇8 − 4 푇9 = −100 … … … … … … … … … … … … … … … … … … … … … … … … … . . (9) Daftar Persamaan Linear: 4푇1 − 푇2 − 푇4 = 55 푇1 − 4푇2 + 푇3 + 푇5 = −30 푇2 − 4푇3 + 푇6 = −90 푇1 − 4푇4 + 푇5 + 푇7 = −25 푇2 + 푇4 − 4 푇5 + 푇6 + 푇8 = 0 푇3 + 푇5 − 4 푇6 + 푇9 = −60 푇4 − 4푇7 + 푇8 = −65 푇5 + 푇7 − 4푇8 + 푇9 = −40 푇6 + 푇8 − 4 푇9 = −100
  • 5. Bentuk matriks 4 -1 0 -1 0 0 0 0 0 T1 55 1 -4 1 0 1 0 0 0 0 T2 -30 0 1 -4 0 0 1 0 0 0 T3 -90 1 0 0 -4 1 0 1 0 0 T4 -25 0 1 0 1 -4 1 0 1 0 T5 0 0 0 1 0 1 -4 0 0 1 T -60 0 0 0 1 0 0 -4 1 0 T7 -65 0 0 0 0 1 0 1 -4 1 T8 -40 0 0 0 0 0 1 0 1 -4 T9 100 Bentuk matriks augmentasi 4 -1 0 -1 0 0 0 0 0 55 1 -4 1 0 1 0 0 0 0 -30 0 1 -4 0 0 1 0 0 0 -90 1 0 0 -4 1 0 1 0 0 -25 0 1 0 1 -4 1 0 1 0 0 0 0 1 0 1 -4 0 0 1 -60 0 0 0 1 0 0 -4 1 0 -65 0 0 0 0 1 0 1 -4 1 -40 0 0 0 0 0 1 0 1 -4 100 =
  • 6. Algoritma 1 1. Mengubah persamaan linear ke bentuk matriks augment, yaitu matriks berukuran n x (n + 1). 2. Memeriksa elemen-elemen pivot (elemen-elemen yang menempati diagonal suatu matrik). Jika terdapat aii ≠ 0, bisa dilanjutkan ke langkah no.3. Namun, jika ada elemen diagonal yang bernilai nol, aii = 0, maka baris dimana elemen itu berada harus ditukar posisinya dengan baris yang ada dibawahnya, (Pi) ↔ (Pj) dimana j = i + 1, i + 2, ..., n, sampai elemen diagonal matrik menjadi tidak nol, aii ≠ 0. 3. Proses triangularisasi. 4. Hitunglah nilai xn 5. Lakukanlah proses substitusi mundur untuk memperoleh xn-1 , xn-2 , ....,x2 , x1
  • 7. Bahasa Pemrograman %Penyelesaian Persamaan dengan Metode Eliminasi Gauss clear all clc A=[4 -1 0 -1 0 0 0 0 0 55; 1 -4 1 0 1 0 0 0 0 -30; 0 1 -4 0 0 1 0 0 0 -90; 1 0 0 -4 1 0 1 0 0 -25; 0 1 0 1 -4 1 0 1 0 0; 0 0 1 0 1 -4 0 0 1 -60; 0 0 0 1 0 0 -4 1 0 -65; 0 0 0 0 1 0 1 -4 1 -40; 0 0 0 0 0 1 0 1 -4 100];%Data Matriks disp('Matriks A') A disp('Jumlah Persamaan') n = 9 %jumlah Persamaan pause % ====Proses Triangulasi==== for j=1:(n-1) % ---mulai proses pivot--- if (A(j,j)==0) for p=1:n+1 u=A(j,p); v=A(j+1,p); A(j+1,p)=u; A(j,p)=v; end end % ---akhir proses pivot--- jj=j+1; for i=jj:n m=A(i,j)/A(j,j); for k=1:(n+1) A(i,k)=A(i,k)-(m*A(j,k)); end end end disp('Matriks A Hasil Proses Triangulasi') A pause % ===Akhir Proses Triangulasi=== %--Proses Substitusi Mundur--- x(n,1)=A(n,n+1)/A(n,n); for i=n-1:-1:1 S=0 for j=n:-1:i+1 S=S+A(i,j)*x(j,1); end x(i,1)=(A(i,n+1)-S)/A(i,i); end x