# Lecture11 - 1 LECTURE 3.1C Dr Vyas Overview(Covers section...

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Unformatted text preview: 1 LECTURE 3.1C Dr. Vyas Overview: (Covers section 3.4) • Gaussian Elimination and Pivoting • Limitations of/Improvements to Gaussian Elimination Method • Examples/Demos I. Gaussian Elimination and Pivoting 1. Using the back substitution algorithm, we can find the solution to the upper triangular system of linear equations. Now, the next issue to consider is to be able to find an algorithm that generates an upper triangular system of linear equations given a set of linear equations. 2. Gaussian elimination and pivoting method helps us solve the set of linear equations by using an algorithm to transform the set of given linear equations into an upper triangular system of linear equations. Thereafter, back substitution algorithm can be employed to find the solution. 3. Theorem 3.7: The following operations applied to a linear system yield an equivalent system: 1) Interchanges : The order of two equations can be interchanged 2) Scaling : Multiplying an equation with a nonzero constant 3) Replacement : An equation can be replaced by the sum of itself and a nonzero multiple of any other equation. 4. Theorem 3.9: If [ ] i j N N a × = A is a non-singular matrix, then there exists a system = U X Y , equivalent to = AX B , where U is an upper triangular matrix with kk u ≠ . After & U Y are constructed, back substitution can be used to solve = U X Y for X . 5. At this juncture, it is better to start with an example that serves to reinforce symbolic (as opposed to numeric) theoretic algebra later. Example 1 (self-study revision) : Find the cubic 2 3 4 3 2 1 y x x x x x x x = + + + that passes thru the points (0,0), (1,1), (2, 2), & (3, 2) . Solution: If the cubic passes thru the points given then the coordinates of these points must satisfy the equation of the cubic. Therefore, the following equations result. 4 3 2 1 2 3 9 27 x x x x = + + + 4 3 2 1 2 2 4 8 x x x x = + + + 4 3 2 1 1 x x x x = + + + 4 3 2 1 x x x x = + + + Rearranging, the system of equations can be written as (noting that 4 x = ): 1 2 3 1 2 3 1 2 3 27 9 3 2 (1) 8 4 2 2 (2) 1 (3) x x x x x x x x x + + = + + = + + = KKK KKK KKK 2 To reduce the above system of equations to upper triangular system of linear equations, we first start with selecting a “pivot” point. Typically, the pivot point is chosen to be upper left most coefficient (in this case 27). This means we will leave the term 1 27 x unchanged and “pivot/transform/turn” the system of equation around it. Next, eliminate the terms below the pivot point in the equation (2) and (3). To do this, we multiply equation (1) by a factor of (8/27) and subtract it from the equation (2). Likewise, multiply equation (1) by a factor of (1/27) and subtract it from equation (3) The process is below 8 8 16 1 2 3 3 9 27 1 1 2 1 2 3 3 9 27 (8 8) (4 ) (2 ) 2 (2) (1 1) (1 ) (1 ) 1 (3) x x x x x x- +- +- =-- +- +- = - KKK KKK The new set of transformed equations is: 1 2 3 10 38 4 1 2 3 3 9 27 8 25 2 1 2 3 3 9 27 27 9 3 2 (1) (2) (3) x x x x x...
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## This note was uploaded on 04/14/2008 for the course MATH 353 taught by Professor Vyas during the Spring '08 term at University of Delaware.

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Lecture11 - 1 LECTURE 3.1C Dr Vyas Overview(Covers section...

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