math 175: numerical analysis ii lecturer: jomar fajardo rabajante 2 nd sem ay 2012-2013 imsp, uplb
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MATH 175: Numerical Analysis II
Lecturer: Jomar Fajardo Rabajante2nd Sem AY 2012-2013
IMSP, UPLB
Numerical Methods for Linear Systems
Review (Naïve) Gaussian EliminationGiven n equations in n variables.
• Operation count for elimination step:(multiplications/divisions)
• Operation count for back substitution:
3
3nO
2
2nO
Numerical Methods for Linear Systems
Overall (Naïve) Gaussian Elimination takes
Take note: we ignored here lower-order terms and we did not include row exchanges and additions/subtractions. WHAT MORE IF WE ADDED THESE STUFFS???!!! KAPOY NA!
323
323 nO
nO
nO
Numerical Methods for Linear Systems
Example: Consider 10 equations in 10 unknowns. The approximate number of operations is
If our computations have round-off errors, how would our solution be affected by error magnification? Tsk… Tsk…
3343
103
Numerical Methods for Linear Systems
Our goal now is to use methods that will efficiently solve our linear systems with minimized error
magnification.
1st Method: Gaussian Elimination with Partial Pivoting
• When we are processing column i in Gaussian elimination, the (i,i) position is called the pivot position, and the entry in it is called the pivot entry (or simply the pivot).
• Let [A|b] be an nx(n+1) augmented matrix.
1st Method: Gaussian Elimination with Partial Pivoting
STEPS:1. Begin loop (i = 1 to n–1):2. Find the largest entry (in absolute value) in
column i from row i to row n. If the largest value is zero, signal that a unique solution does not exist and stop.
3. If necessary, perform a row interchange to bring the value from step 2 into the pivot position (i,i).
1st Method: Gaussian Elimination with Partial Pivoting
4. For j = i+1 to n, perform
5. End loop.6. If the (n,n) entry is zero, signal that a unique
solution does not exist and stop. Otherwise, solve for the solution by back substitution.
jii,i
j,ij RRa
aR
1st Method: Gaussian Elimination with Partial Pivoting
Example:
8
12
1
1284
1244
221
8
1
12
1284
221
1244Original matrix (Matrix 0) Matrix 1
1st Method: Gaussian Elimination with Partial Pivoting
8
1
12
1284
221
1244
12
0
1244Matrix 1 Matrix 2
11
44 0
1st Method: Gaussian Elimination with Partial Pivoting
8
1
12
1284
221
1244
12
10
1244Matrix 1 Matrix 2
21
44 1
1st Method: Gaussian Elimination with Partial Pivoting
8
1
12
1284
221
1244
12
110
1244Matrix 1 Matrix 2
21
412 –1
1st Method: Gaussian Elimination with Partial Pivoting
8
1
12
1284
221
1244
2
12
110
1244Matrix 1 Matrix 2
11
412 –2
1st Method: Gaussian Elimination with Partial Pivoting
8
1
12
1284
221
1244
2
12
0
110
1244Matrix 1 Matrix 2
44
44 0
1st Method: Gaussian Elimination with Partial Pivoting
8
1
12
1284
221
1244
2
12
40
110
1244Matrix 1 Matrix 2
84
44 4
1st Method: Gaussian Elimination with Partial Pivoting
8
1
12
1284
221
1244
2
12
040
110
1244Matrix 1 Matrix 2
124
412 0
1st Method: Gaussian Elimination with Partial Pivoting
8
1
12
1284
221
1244
4
2
12
040
110
1244Matrix 1 Matrix 2
84
412 –4
1st Method: Gaussian Elimination with Partial Pivoting
2
4
12
110
040
1244
4
2
12
040
110
1244
Matrix 2 Matrix 3
1st Method: Gaussian Elimination with Partial Pivoting
2
4
12
110
040
1244
1
4
12
100
040
1244
Matrix 3 Final Matrix (Matrix 4)
1st Method: Gaussian Elimination with Partial Pivoting
1
4
12
100
040
1244
Final Matrix Back substitution:
1x
121244x
1212z4y4x
1y44y
1z1z
1st Method: Gaussian Elimination with Partial Pivoting
1
1
2
720
410
1290
8
2
12
000
1040
12191
a unique solution does not exist
1st Method: Gaussian Elimination with Partial Pivoting
• There are other pivoting strategies such as the complete (or maximal) pivoting. But complete pivoting is computationally expensive.