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Unformatted text preview: Math 425 Exam 1 Jerry L. Kazdan March 3, 2011 12:00 1:20 Directions This exam has three parts, Part A, short answer, has 1 problem (12 points). Part B has 5 shorter problems (7 points each, so 35 points). Part C has 3 traditional problems (15 points each so 45 points). Total is 92 points. Closed book, no calculators or computers but you may use one 3 00 5 00 card with notes on both sides. Part A: Short Answer (1 problems, 12 points). 1. Let S and T be linear spaces and A : S T be a linear map. Say V and W are particular solutions of the equations A V = Y 1 and A W = Y 2 , respectively, while Z 6 = 0 is a solution of the homogeneous equation A Z = 0. Answer the following in terms of V , W , and Z . a) Find some solution of A X = 3 Y 1 . Solution : X = 3 V b) Find some solution of A X =- 5 Y 2 . Solution : X =- 5 W c) Find some solution of A X = 3 Y 1- 5 Y 2 . Solution : X = 3 V- 5 W d) Find another solution (other than Z and 0) of the homogeneous equation A X = 0. Solution : X = 2 Z e) Find two solutions of A X = Y 1 . Solution : X = V and X = V + Z f) Find another solution of A X = 3 Y 1- 5 Y 2 . Solution : X = 3 V- 5 Y 2 + Z Part B: Short Problems (5 problems, 7 points each so 35 points) B1. U = (1 , 1 , , 1) and V = (- 1 , 2 , 1 ,- 1) are orthogonal vectors in R 4 . Write the vector X = (1 , 1 , 1 , 2) in the form X = a U + b V + W , where a, b are scalars and W is a vector perpendicular to U and V . Solution : Since k U k = 3 and k V k = 7 , then U := U / 3 and V := V / 7 are orthonormal vectors in the same directions as U and V , respectively. Well write X in the form X = U + V + W , (1) where W is a vector perpendicular to both U and V and hence U and V ....
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