Lecture 19 - Determinant of 3x3 matrix abc def a b c ab defde ghigh ve defde ghi Augmentation a b c ab ghigh abcab defde ghigh-ve MATLABs Methods

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Augmentation Determinant of 3 x 3 matrix +ve -ve a b c a b d e f d e g h i g h a b c a b d e f d e g h i g h - a b c d e f g h i a b c a b d e f d e g h i g h
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
MATLAB’s Methods Forward slash ( / ) Back-slash ( \ ) Multiplication by the inverse of the quantity under the slash b * A inv x b A x b A x b Ax 1 ) ( \ = = = = -
Background image of page 2
Gauss Elimination Manipulate equations to eliminate one of the unknowns Develop algorithm to do this repeatedly The goal is to set up upper triangular matrix Back substitution to find solution (root) = nn n 3 33 n 2 23 22 n 1 13 12 11 a a a a a a a a a a U
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Basic Gauss Elimination Direct Method (no iteration required) Forward elimination Column-by-column elimination of the below-diagonal elements Reduce to upper triangular matrix Back-substitution
Background image of page 4
Gauss Elimination Method Solve x in: - 2 1 2 1 4 1 1 1 3 3 2 1 x x x 3 2 1 10 12 2 - - - - - - - - - = - 10 2 1 2 12 1 4 1 2 1 1 3 Soln. i. Augmented matrix : [A | b] = ii. Reduction (to reduce A to an upper triangular matrix) ' 3 1'*(2/3) - 3 ' 2 ) 3 / 1 ( * 1 2 ' 1 1 row row row row row row row row - 3 1 1 2 ....1' 0 1.3333 11.3333 . ...2' 0 0.3333 2.6667 8.6668 . ...3' - ....1 ....2 ....3 " 3 ) 3.6667 0.3333 2'*( 3'- " 2 ' 2 " 1 ' 1 R R R R R R R " 3 .... " 2 .... " 1 .... 6365 . 7 5455 . 2 0 0 3333 . 11 3333 . 1 6667 . 3 0 2 1 1 3 - * Now A has become an upper-triangular matrix pivot element
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Gauss Elimination Method iii. Back substitution * Solve for x 3 : x 3 = 7.6365 / 2.5454 = 3.0 (from 3") * Solve for x 2 : x 2 = (11.3333-1.3333*3.0)/3.6667=2.0 * Solve for x 1 : x 1 = (2+ x 3 - x
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/05/2011 for the course EGM 3344 taught by Professor Raphaelhaftka during the Fall '09 term at University of Florida.

Page1 / 21

Lecture 19 - Determinant of 3x3 matrix abc def a b c ab defde ghigh ve defde ghi Augmentation a b c ab ghigh abcab defde ghigh-ve MATLABs Methods

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online