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Review Operasi Matriks
- 1. Review Operasi Matriks Menghitung invers matriks? Determinan? Matriks Singular?
- 3. Determinan Hanya untuk square matrices Jika determinan = 0 matriks singular, tidak punya invers bcad dc ba dc ba −≡≡ det 123213312132231321 321 321 321 det cbacbacbacbacbacba ccc bbb aaa −+−+−≡
- 5. Sistem Persamaan Linear Simultaneous Linear Equations
- 6. Metode Penyelesaian • Metode grafik • Eliminasi Gauss • Metode Gauss – Jourdan • Metode Gauss – Seidel • LU decomposition
- 7. Metode Grafik 2 -2 = − 2 4 11 21 2 1 x x Det{A} ≠ 0 ⇒ A is nonsingular so invertible Unique solution
- 8. Sistem persamaan yang tak terselesaikan = 5 4 42 21 2 1 x x No solution Det [A] = 0, but system is inconsistent Then this system of equations is not solvable
- 9. Sistem dengan solusi tak terbatas Det{A} = 0 ⇒ A is singular infinite number of solutions = 8 4 42 21 2 1 x x Consistent so solvable
- 10. Ill-conditioned system of equations A linear system of equations is said to be “ill- conditioned” if the coefficient matrix tends to be singular
- 11. Ill-conditioned system of equations • A small deviation in the entries of A matrix, causes a large deviation in the solution. = 47.1 3 99.048.0 21 2 1 x x = 47.1 3 99.049.0 21 2 1 x x = 1 1 2 1 x x ⇒ = 0 3 2 1 x x ⇒
- 12. Gaussian Elimination Merupakan salah satu teknik paling populer dalam menyelesaikan sistem persamaan linear dalam bentuk: Terdiri dari dua step 1. Forward Elimination of Unknowns. 2. Back Substitution [ ][ ] [ ]CXA =
- 13. Forward Elimination Tujuan Forward Elimination adalah untuk membentuk matriks koefisien menjadi Upper Triangular Matrix 7.000 56.18.40 1525 −−→ 112144 1864 1525
- 14. Forward Elimination Persamaan linear n persamaan dengan n variabel yang tak diketahui 11313212111 ... bxaxaxaxa nn =++++ 22323222121 ... bxaxaxaxa nn =++++ nnnnnnn bxaxaxaxa =++++ ...332211 . . . . . .
- 20. Warning.. Dua kemungkinan kesalahan -Pembagian dengan nol mungkin terjadi pada langkah forward elimination. Misalkan: 655 901.33099.26 7710 321 123 21 =+− =−+ =− xxx xxx xx - Kemungkinan error karena round-off (pengulangan?)
- 21. Contoh Dari sistem persamaan linear − − − 515 6099.23 0710 3 2 1 x x x 6 901.3 7 = Akhir dari Forward Elimination − − 1500500 6001.00 0710 3 2 1 x x x 15004 001.6 7 = − − − 6 901.3 7 515 6099.23 0710 − − 15004 001.6 7 1500500 6001.00 0710
- 22. Kesalahan yang mungkin terjadi Back Substitution 99993.0 15005 15004 3 ==x 5.1 001.0 6001.6 3 2 −= − − = x x 3500.0 10 077 32 1 −= −+ = xx x = − − 15004 001.6 7 1500500 6001.00 0710 3 2 1 x x x
- 23. Contoh kesalahan Bandung-kan solusi exact dengan hasil perhitungan [ ] − − = = 99993.0 5.1 35.0 3 2 1 x x x X calculated [ ] −= = 1 1 0 3 2 1 x x x X exact
- 24. Improvements Menambah jumlah angka penting Mengurangi round-off error Tidak menghindarkan pembagian dengan nol Gaussian Elimination with Partial Pivoting Menghindarkan pembagian dengan nol Mengurangi round-off error
- 25. Pivoting pka Eliminasi Gauss dengan partial pivoting mengubah tata urutan baris untuk bisa mengaplikasikan Eliminasi Gauss secara Normal How? Di awal sebelum langkah ke-k pada forward elimination, temukan angka maksimum dari: nkkkkk aaa .......,,........., ,1+ Jika nilai maksimumnya Pada baris ke p, ,npk ≤≤ Maka tukar baris p dan k.
- 26. Partial Pivoting What does it Mean? Gaussian Elimination with Partial Pivoting ensures that each step of Forward Elimination is performed with the pivoting element |akk| having the largest absolute value. Jadi, Kita selalu mengecek apakah angka diagonalnya memberikan nilai absolut terbesar
- 27. Partial Pivoting: Example Consider the system of equations 655 901.36099.23 7710 321 321 21 =+− =++− =− xxx xxx xx In matrix form − − − 515 6099.23 0710 3 2 1 x x x 6 901.3 7 = Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping
- 28. Partial Pivoting: Example Forward Elimination: Step 1 Examining the values of the first column |10|, |-3|, and |5| or 10, 3, and 5 The largest absolute value is 10, which means, to follow the rules of Partial Pivoting, we don’t need to switch the rows = − − − 6 901.3 7 515 6099.23 0710 3 2 1 x x x = − − 5.2 001.6 7 55.20 6001.00 0710 3 2 1 x x x ⇒ Performing Forward Elimination
- 29. Partial Pivoting: Example Forward Elimination: Step 2 Examining the values of the first column |-0.001| and |2.5| or 0.0001 and 2.5 The largest absolute value is 2.5, so row 2 is switched with row 3 = − − 5.2 001.6 7 55.20 6001.00 0710 3 2 1 x x x ⇒ = − − 001.6 5.2 7 6001.00 55.20 0710 3 2 1 x x x Performing the row swap
- 30. Partial Pivoting: Example Forward Elimination: Step 2 Performing the Forward Elimination results in: = − 002.6 5.2 7 002.600 55.20 0710 3 2 1 x x x
- 31. Partial Pivoting: Example Back Substitution Solving the equations through back substitution 1 002.6 002.6 3 ==x 1 5.2 55.2 2 2 = − = x x 0 10 077 32 1 = −+ = xx x = − 002.6 5.2 7 002.600 55.20 0710 3 2 1 x x x
- 32. Partial Pivoting: Example [ ] −= = 1 1 0 3 2 1 x x x X exact [ ] −= = 1 1 0 3 2 1 x x x X calculated Compare the calculated and exact solution The fact that they are equal is coincidence, but it does illustrate the advantage of Partial Pivoting