Lecture 19 - Determinant of 3x3 matrix abc Augmentation a b...

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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
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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 ) ( \ = = = = -
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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
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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
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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' - 3.6667 .... 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
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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 2 )/3=1.0
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