Lecture12 - 1 LECTURE 3.2 Overview: (covers section 3.5)...

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1 LECTURE 3.2 Dr. Vyas Overview: (covers section 3.5) Triangular Factorization : Direct and Indirect Factorization Examples/Demos TRIANGULAR FACTORIZATION: DIRECT METHOD 1. The purpose of triangular factorization is to speed up the computations for the calculations involving same coefficient matrix A but different vector B . Straightforward Gaussian elimination will have to carry out upper triangulation in its entirety every time B is changed. To eliminate such unnecessary repetition, triangular factorization technique is used. 2. A nonsingular matrix A has a triangular factorization if it can be expressed as a product of lower triangular matrix L (with unit diagonal elements) and an upper triangular matrix U . Mathematically, = A LU , 0 kk u . 3. In matrix form, for example, this may be written as 11 12 13 11 12 13 21 22 23 21 22 23 31 32 31 32 33 33 1 0 0 1 0 0 1 0 0 a a a u u u a a a m u u m m a a a u = (aka Doolittle Factorization ) 4. Suppose that the coefficient matrix A for the linear system = AX B has a triangular factorization = A LU , then = = AX B LUX B 5. Solution to = LUX B can be obtained by first defining Y = UX and then solving two systems: a) LY = B to determine Y ; b) UX = Y to determine X . If no row interchanges are needed, the method is called direct factorization . Example 1: Solve LY = B , UX = Y and verify that B = AX , given [ 4 10 5] ′ = - B and = A LU is provided as, 1 2 1 1 2 3 1 0 0 2 4 6 2 4 6 1 5 3 1 0 0 3 6 1 3 2 0 0 3 1 - - = Solution: First, use the forward substitution algorithm to determine Y via LY = B . Because finding the algorithm/formulae is part of your homework, it is not shown here. Instead a hand calculation procedure is clearly illustrated that should help you generate the algorithm and modify the back substitution program. 1 1 11 / y b l = 1 1 2 2 1 1 3 2 3 1 0 0 4 1 0 10 5 1 y y y -   =     . Start with the first equation to find, 1 4 y = - 1 2 1 2 (10 ) /1 10 ( 2) 12 y y = - = - - = ; 1 1 3 1 2 2 3 (5 )/1 5 2 4 3 y y y = - - = + - = Check 1: 4 0 0 4 - + + = - ; 1 2 4( ) 12 0 10 - + + = ; 1 1 2 3 4( ) 12( ) 3 5 - + + = Now solve the system of equations given by, UX = Y
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2 1 2 3 2 4 6 4 0 3 6 12 0 0 3 3 x x x - - = . Using the back substitution algorithm (method), 3 3/3 1 x = = , 2 (12 6) / 3 2 x = - = , 1 ( 4 ( 6) 4(2)) / 2 3 x = - - - - = - Check 2: 2( 3) 4(2) 6 4 - + - = - ; 3(2) 6(1) 12 + = ; 3(1) 3 = Check 3: 2( 3) 4(2) 6(1) 4 - + - = - ; 1( 3) 5(2) 3(1) 10 - + + = ; 1( 3) 3(2) 2(1) 5 - + + = Note that if we wanted to solve a different system such that only the B vector is changed, then we just have to solve
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Lecture12 - 1 LECTURE 3.2 Overview: (covers section 3.5)...

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