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 nOnOnO

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…

3343103

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 RRaa

R

1st Method: Gaussian Elimination with Partial Pivoting

Example:

8121

12841244221

8112

12842211244

Original matrix (Matrix 0) Matrix 1

1st Method: Gaussian Elimination with Partial Pivoting

8112

12842211244

120

1244Matrix 1 Matrix 2

11

44 0

1st Method: Gaussian Elimination with Partial Pivoting

8112

12842211244

12101244

Matrix 1 Matrix 2

21

44 1

1st Method: Gaussian Elimination with Partial Pivoting

8112

12842211244

121101244

Matrix 1 Matrix 2

21

412 –1

1st Method: Gaussian Elimination with Partial Pivoting

8112

12842211244

2121101244

Matrix 1 Matrix 2

11

412 –2

1st Method: Gaussian Elimination with Partial Pivoting

8112

12842211244

212

01101244

Matrix 1 Matrix 2

44

44 0

1st Method: Gaussian Elimination with Partial Pivoting

8112

12842211244

212

401101244

Matrix 1 Matrix 2

84

44 4

1st Method: Gaussian Elimination with Partial Pivoting

8112

12842211244

212

0401101244

Matrix 1 Matrix 2

124

412 0

1st Method: Gaussian Elimination with Partial Pivoting

8112

12842211244

4212

0401101244

Matrix 1 Matrix 2

84

412 –4

1st Method: Gaussian Elimination with Partial Pivoting

2412

1100401244

4212

0401101244

Matrix 2 Matrix 3

1st Method: Gaussian Elimination with Partial Pivoting

2412

1100401244

1412

1000401244

Matrix 3 Final Matrix (Matrix 4)

1st Method: Gaussian Elimination with Partial Pivoting

1412

1000401244

Final Matrix Back substitution:

1x 121244x

1212z4y4x1y44y

1z1z

1st Method: Gaussian Elimination with Partial Pivoting

112

7204101290

8212

000104012191

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.