Gaussian Elimination Notes

# Gaussian Elimination Notes - Chapter 04.06 Gaussian...

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Chapter 04.06 Gaussian Elimination After reading this chapter, you should be able to: 1. solve a set of simultaneous linear equations using Naïve Gauss elimination, 2. learn the pitfalls of the Naïve Gauss elimination method, 3. understand the effect of round-off error when solving a set of linear equations with the Naïve Gauss elimination method, 4. learn how to modify the Naïve Gauss elimination method to the Gaussian elimination with partial pivoting method to avoid pitfalls of the former method, 5. find the determinant of a square matrix using Gaussian elimination, and 6. understand the relationship between the determinant of a coefficient matrix and the solution of simultaneous linear equations. How is a set of equations solved numerically? One of the most popular techniques for solving simultaneous linear equations is the Gaussian elimination method. The approach is designed to solve a general set of n equations and n unknowns 1 1 3 13 2 12 1 11 ... b x a x a x a x a n n = + + + + 2 2 3 23 2 22 1 21 ... b x a x a x a x a n n = + + + + . . . . . . n n nn n n n b x a x a x a x a = + + + + ... 3 3 2 2 1 1 Gaussian elimination consists of two steps 1. Forward Elimination of Unknowns: In this step, the unknown is eliminated in each equation starting with the first equation. This way, the equations are reduced to one equation and one unknown in each equation. 2. Back Substitution: In this step, starting from the last equation, each of the unknowns is found. Forward Elimination of Unknowns: In the first step of forward elimination, the first unknown, 1 x is eliminated from all rows below the first row. The first equation is selected as the pivot equation to eliminate 1 x . So, 04.06.1

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04.06.2 Chapter 04.06 to eliminate 1 x in the second equation, one divides the first equation by 11 a (hence called the pivot element) and then multiplies it by 21 a . This is the same as multiplying the first equation by 11 21 / a a to give 1 11 21 1 11 21 2 12 11 21 1 21 ... b a a x a a a x a a a x a n n = + + + Now, this equation can be subtracted from the second equation to give 1 11 21 2 1 11 21 2 2 12 11 21 22 ... b a a b x a a a a x a a a a n n n - = - + + - or 2 2 2 22 ... b x a x a n n = + + where n n n a a a a a a a a a a 1 11 21 2 2 12 11 21 22 22 - = - = This procedure of eliminating 1 x , is now repeated for the third equation to the th n equation to reduce the set of equations as 1 1 3 13 2 12 1 11 ... b x a x a x a x a n n = + + + + 2 2 3 23 2 22 ... b x a x a x a n n = + + + 3 3 3 33 2 32 ... b x a x a x a n n = + + + . . . . . . . . . n n nn n n b x a x a x a = + + + ... 3
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## This note was uploaded on 05/01/2011 for the course CHBE 2120 taught by Professor Gallivan during the Spring '07 term at Georgia Institute of Technology.

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Gaussian Elimination Notes - Chapter 04.06 Gaussian...

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