Ch E 310 - Fall 10 - Lecture 6

Ch E 310 - Fall 10 - Lecture 6 - Lecture 6 September 9,...

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Lecture 6 – September 9, 2010 Agenda: Review of linear algebra (video series): http://academicearth.org/courses/linear-algebra Gauss(ian) Elimination calculation of the determinant backward substitution to find x Gauss-Jordan Iterative Methods: Jacobi Gauss-Seidel
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Methods for Solving Linear Systems Summary of exact methods for solving A x = b Cramer’s Rule (ratio of determinants) A -1 b (matrix inverse: A\b or inv(A)*b ) Gauss Elimination (today’s topic) For most simple systems, any of the above will work well Things get trickier when the system is large (where there is a large computational “cost”)
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Linear Systems: Gauss Elimination Eliminating unknowns and back-substituting is one of the first methods we learn for solving small linear systems (usually in a course on linear algebra, which ISU doesn’t require… ugh) This same method can be used for larger systems as well What results from this procedure is an upper triangular matrix Converting a full matrix A to an upper triangular matrix U does not change its determinant (and interesting feature that we can exploit), however the time required to find the determinant is O ( n ) instead of O ( n · n !) – this is a HUGE savings!
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We can multiply rows by scalars and subtract from one another to generate zeros in the lower left This process can be repeated until we have an upper triangular matrix U that can be back substituted to find the unknowns “Naïve” refers to the fact that the forward elimination is done without manipulating the order of the equations in any way The determinant of U and A are the same, but det( U ) is easier to calculate: 96 . 5 0 0 0 69 . 0 38 . 6 0 0 5 . 1 3 5 . 6 0 1 4 9 4 Naïve (Simple) Gauss Elimination 6 2 4 1 5 9 2 8 3 9 7 6 1 4 9 4 A = naïve Gauss U =   n i ii u U 1 det det( A ) = det( U ) = 4 × 6.5 × -6.38 × 5.96 = 989 >> prod(diag(U))
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59 73 79 58 6 2 4 1 5 9 2 8 3 9 7 6 1 4 9 4 4 3 2 1 x x x x Naïve (Simple) Gauss Elimination Consider the system we solved last lecture by Cramer’s rule: An alternate way to solve this system is to perform Gauss elimination on a new matrix containing A and b Once we have one row (an equation) with one unknown, we can back substitute to find all other unknown variables
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This note was uploaded on 09/21/2011 for the course CH E 310 taught by Professor Staff during the Spring '08 term at Iowa State.

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Ch E 310 - Fall 10 - Lecture 6 - Lecture 6 September 9,...

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