# EXAM03-solutions - Version 074 EXAM03 gilbert(54620 This...

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Version 074 – EXAM03 – gilbert – (54620) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points By diagonalizing the matrix A = b 3 4 2 3 B , compute f ( A ) For the polynomial f ( x ) = 3 x 3 + 3 x 2 x 2 . 1. f ( A ) = b 7 8 4 5 B correct 2. f ( A ) = b 7 8 4 5 B 3. f ( A ) = b 7 8 4 5 B 4. f ( A ) = b 7 8 4 5 B Explanation: IF A can be diagonalized by A = PDP - 1 = P b d 1 0 0 d 2 B P - 1 , then f ( A ) = Pf ( D ) P - 1 = P b f ( d 1 ) 0 0 f ( d 2 ) B P - 1 . Now A can be diagonalized iF we can fnd an eigenbasis oF R 2 oF eigenvectors v 1 , v 2 oF A corresponding to eigenvalues λ 1 , λ 2 , For then: A = P b λ 1 0 0 λ 2 B P - 1 , P = [ v 1 v 2 ] . But det[ A λI ] = v v v v 3 λ 4 2 3 λ v v v v = 8 (3 λ )(3 + λ ) = λ 2 1 = 0 , i.e. , λ 1 = 1 and λ 2 = 1. Corresponding eigenvectors are v 1 = b 2 1 B , v 2 = b 1 1 B , so P = b 2 1 1 1 B , P - 1 = b 1 1 1 2 B . Thus f ( A ) = b 2 1 1 1 Bb f (1) 0 0 f ( 1) 1 1 1 2 B . Now f (1) = 3 x 3 + 3 x 2 x 2 v v v x =1 = 3 , while f ( 1) = 3 x 3 + 3 x 2 x 2 v v v x = - 1 = 1 . Consequently, f ( A ) = b 2 1 1 1 3 0 0 1 1 1 1 2 B = b 7 8 4 5 B . 002 10.0 points ±ind the distance From y = 1 11 2 to the plane in R 3 spanned by u 1 = 2 2 1 , u 2 = 1 2 2 . 1. dist = 2 2. dist = 5 3. dist = 3 correct
Version 074 – EXAM03 – gilbert – (54620) 2 4. dist = 4 5. dist = 1 Explanation: The plane in R 3 spanned by u 1 , u 2 is the subspace W = Span { u 1 , u 2 } , and each y in R 3 has a unique orthogonal decomposition y = proj W y + ( y proj W y ) where y proj W y is in the orthogonal com- plement W . But then dist( y , W ) = b y proj W y b . Now u 1 , u 2 are non-zero othogonal vectors, so form a basis for W such that proj W y = p y · u 1 b u 1 b 2 P u 1 + p y · u 2 b u 2 b 2 P u 2 . But when y = 1 11 2 , u 1 = 2 2 1 , u 2 = 1 2 2 , we see that p y · u 1 b u 1 b 2 P u 1 = 2 2 2 1 , while p y · u 2 b u 2 b 2 P u 2 = 3 1 2 2 . Consequently, proj W y = 2 2 2 1 3 1 2 2 = 1 10 4 , and so y proj W = 1 11 2 1 10 4 = 2 1 2 .