# m340l quest exam 3 solutions - Version 100 EXAM03...

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Version 100 – EXAM03 – gilbert – (56780) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Use the fact that A = 1 4 9 7 1 2 4 1 5 6 10 7 1 0 1 5 0 2 5 6 0 0 0 0 to determine an orthogonal basis for Col( A ). 1. 4 2 6 , 1 1 5 2. 4 2 6 , 1 1 1 3. 1 1 5 , 1 4 1 4. 1 1 5 , 4 1 1 correct Explanation: The pivot columns of A provide a basis for Col( A ). But by row reduction, A 1 0 1 5 0 2 5 6 0 0 0 0 1 0 1 5 0 1 5 / 2 3 0 0 0 0 . Thus the pivot columns of A are a 1 = 1 1 5 , a 2 = 4 2 6 . We apply Gram-Schmidt to produce an or- thogonal basis: set u 1 = a 1 and u 2 = a 2 a 2 · u 1 u 1 2 u 1 = 4 2 6 ( 36) 27 1 1 5 = 4 2 6 + 4 / 3 4 / 3 20 / 3 = 8 / 3 2 / 3 2 / 3 . Consequently, the set of vectors 1 1 5 , 4 1 1 is an orthogonal basis for Col( A ). 002 10.0 points Simplify the expression ( u + 3 v ) · (3 u 4 v ) − ∥ 2 u + 2 v 2 for vectors u , v in R n . 1. −∥ u 2 3 u · v 16 v 2 correct 2. −∥ u 2 21 u · v + 16 v 2 3. −∥ u 2 3 u · v + 16 v 2 4. −∥ u 2 + 3 u · v 16 v 2 5. −∥ u 2 + 21 u · v 16 v 2 6. −∥ u 2 + 3 u · v + 16 v 2 Explanation: By linearity, ( u + 3 v ) · (3 u 4 v ) = u · (3 u 4 v ) + 3 v · (3 u 4 v ) = 3 u · u 4 u · v + 9 v · u 12 v · v , while 2 u + 2 v 2 = (2 u + 2 v ) · (2 u + 2 v ) = 2 u · (2 u + 2 v ) + 2 v · (2 u + 2 v ) = 4 u · u + 4 u · v + 4 v · u + 4 v · v .
Version 100 – EXAM03 – gilbert – (56780) 2 But v · u = u · v , u · u = u 2 , v · v = v 2 . So after expansion the expression becomes 3 u 2 + 5 u · v 12 v 2 (4 u 2 + 8 u · v + 4 v 2 ) = − ∥ u 2 3 u · v 16 v 2 . 003 10.0 points If v 1 and v 2 are linearly independent eigen- vectors of an n × n matrix A , then they cor- respond to distinct eigenvalues of A . True or False? 1. TRUE 2. FALSE correct Explanation: If two eigenvectors correspond to distinct eigenvalues then they are linearly indepen- dent. However, two linearly independent eigenvectors can also correspond to the same eigenvalue. In other words, the eigenspace for an eigenvalue λ can be multi-dimensional if the general solution of ( A λ I ) x = 0 has more than one free variable. Consequently, the statement is FALSE . 004 10.0 points