EXAM02-solutions

# EXAM02-solutions - Version 029 EXAM02 gilbert(54160 This...

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Version 029 – EXAM02 – gilbert – (54160) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points IF H is the set oF vectors 2 a a b 3 b in R 3 For all real a and b , then H is closed under vector addition. True or ±alse? 1. ±ALSE 2. TRUE correct Explanation: ±ix vectors v 1 = 2 a 1 a 1 b 1 3 b 1 , v 2 = 2 a 2 a 2 b 2 3 b 2 , in H . Then v 1 + v 2 = 2 a 1 a 1 b 1 3 b 1 + 2 a 2 a 2 b 2 3 b 2 = 2( a 1 + a 2 ) ( a 1 + a 2 ) ( b 1 + b 2 ) 3( b 1 + b 2 ) , in which case v 1 + v 2 is in H . Consequently, the statement is TRUE . 002 10.0 points IF eigenvectors oF an n × n matrix A are a basis For R n , then A is diagonalizable. True or ±alse? 1. TRUE correct 2. Explanation: An n × n matrix A is diagonalizable iF and only iF A has n linearly independent eigenvec- tors. On the other hand, n vectors in R n are a basis For R n iF and only iF the vectors are linearly independent. So iF eigenvectors oF A Form a basis For R n , they must be linearly independent, in which case A will be diagonalizable. Consequently, the statement is TRUE . 003 10.0 points Solve For X when A ( X + B ) = C , A = b 5 2 3 1 B , B = b 5 2 1 5 B , and C = b 1 2 1 5 B . 1. X = b 3 10 2 26 B 2. X = b 2 10 2 26 B 3. X = b 4 10 3 26 B correct 4. X = b 4 10 3 24 B 5. X = b 3 10 2 26 B 6. X = b 2 10 3 24 B Explanation:

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Version 029 – EXAM02 – gilbert – (54160) 2 By the algebra of matrices, X = A - 1 C B. But the inverse of any 2 × 2 matrix D = b d 11 d 12 d 21 d 22 B is given by D - 1 = b d 22 / Δ d 12 / Δ d 21 / Δ d 11 / Δ B with Δ = d 11 d 22 d 12 d 21 , so X = b 1 2 3 5 Bb 1 2 1 5 B b 5 2 1 5 B = b 1 12 2 31 B b 5 2 1 5 B . Thus X = b 4 10 3 26 B . 004 10.0 points An n × n matrix A is not invertible if 0 is an eigenvalue of A . True or False? 1. TRUE correct 2. FALSE Explanation: If λ = 0 is an eigenvalue of A , then | A λI | = | A | = 0 . But A is not invertible when | A | = 0. Consequently, the statement is TRUE . 005 10.0 points If H is the set of all vectors a b c d in R 4 where 2 a b = 3 c , a = 4 c 3 d , then there is a matrix A such that H = Nul( A ) .
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