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exam1_soln - Fall 2005 10.34 Exam I Friday Oct 7 2005 Read...

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Fall 2005 10.34 Exam I. Friday Oct. 7, 2005. Read through the entire exam before beginning work, and budget your time. Perform all calculations by hand, showing all steps. You may use a calcula- tion for simple multiplication and addition of numbers, however. 2 1 1 Problem 1 . Consider the matrix A = 1 3 1 1 1 2 1 ( a ) Compute the product Av for the vector v = . 2 3 Answer N Using the rule ( Av ) j = a jk v k , we have k 1 = Av = 2 1 1 1 2 ( ) 1 ( ) 1 ( )( ) + [ 2 1 ( ) 3 ( ) + ] 1 1 3 1 2 = 1 ( ) 1 ( ) 3 ( ) 2 + [ ( ) 1 ( ) 3 ( ) + ] = 10 1 1 2 3 1 ( ) 1 ( ) 1 ( ) 2 + [ ( ) 2 ( ) 3 ( ) + ] 7 ( b ) Compute the LU decomposition A = LU where L is a lower-triangular matrix and U is an upper-triangular matrix. Answer To generate A = LU we perform Gaussian elimination without partial pivoting. First, we zero the (2,1) component, calculating λ 21 a 21 1 = ------- = -- = 0.5 a 11 2 and then by performing the row operation 2 2 – λ 21 × 1 , we have the new matrix 2 1 1 2 1 1 , A ( 2 1 ) = 2 1 1 = [ 1 – ( 0.5 )( )] [ 3 – ( 0.5 )( )] [ 1 – ( 0.5 )( )] 0 2.5 1.5 1 1 2 1 1 2 Fall 2005 10.34 Exam I. Friday Oct. 7, 2005. October 6, 2005 1
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----------- ----------- ------ --------- ( 2 1 ) , a 31 1 Next, we zero the (3,1) component by calculating λ 31 = -- ( 2 1 ) = –-- = –0.5 and per- , 2 a 11 , forming the row operation on A ( 2 1 ) 3 3 – λ 31 × 1 , to yield 2 1 1 2 1 1 A ( 3 1 ) , = = 0 2.5 1.5 0 2.5 1.5 2 1 1 [ 1 ( –0.5 )( )] [ 1 – ( –0.5 )( )] [ 2 – ( –0.5 )( )] 0 1.5 1.5 ( 3 1 ) , a 32 1.5 3 2 3 Next, we zero the (3,2) component by calculating λ 32 = -- ( 3 1 ) = 2.5 = 5 2 = 5 -- = 0.6 , a 22 ,
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