M 346 - Fall 2011 Midterm #1 - Solutions

M 346 - Fall 2011 Midterm #1 - Solutions - M346 Midterm...

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M346 Midterm Exam 1 Solutions I. Let L : R 2 R 2 be de ned by L ( x ) = (15 x 1 + 24 x 2 , 8 x 1 13 x 2 ) T , where x = ( x 1 , x 2 ) T . Let E be the standard basis of R 2 and B = { (2 , 1) T , ( 3 , 2) T } be an alternate basis. 1. (10 points) Find the change-of-basis matrices P EB and P BE . Answers: P EB = [ 2 3 1 2 ] and P BE = P 1 EB = [ 2 3 1 2 ] 2. (5 points) If v = (1 , 1) T , nd [ v ] B . Answers: [ v ] B = P BE [ v ] E = [ 2 3 1 2 ][ 1 1 ] = [ 1 1 ] 3. (10 points) Find the matrices [ L ] E and [ L ] B . Answers: Since L ( e 1 ) = (15 , 8) T = 15 e 1 8 e 2 and L ( e 2 ) = (24 , 13) T = 24 e 1 13 e 2 , we have [ L ] E = [ 15 24 8 13 ] Since L ( b 1 ) = (6 , 3) T = 3 b 1 and L ( b 2 ) = (3 , 2) T = b 2 , we have [ L ] B = [ 3 0 0 1 ] 4. (5 points) Let x ( n ) R 2 be the sequence of vectors de ned by x ( n + 1) = L ( x ( n )) and x (0) = (2 , 1) T . De ne y ( n ) = [ x ( n )] B . Find y (3) . Answers:
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This note was uploaded on 11/07/2011 for the course M 346 taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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M 346 - Fall 2011 Midterm #1 - Solutions - M346 Midterm...

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