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Unformatted text preview: Final Exam, Math 33A, Fall 2005 December 14, 2005 Name: UCLA ID: Section (circle one): 1A Alex Chen Tuesday 1B Alex Chen Thursday 1C Michael VanValkenburgh Tuesday 1D Michael VanValkenburgh Thursday Directions: Fill in your name and circle your section above. Do not turn the page until instructed to do so. You have three hours to complete the exam. No outside materials are allowed; use only your brain and a writing instrument. There are 6 problems, of varying weights. The maximum score is 80. Extra scratch paper is included. If your work on a problem appears on a different page, indicate clearly where it may be found. Show all the necessary steps involved in finding your solutions, unless otherwise instructed. In the interest of us not losing pages of your exam, please refrain from detaching pages from the exam. Good luck. Problem Score Problem Score 1 5 2 6 3 4 Total 1 1. (12 pts total) Consider the system of equations A~x = ~ 0, where A = 1 t 1 1 1 t t 1 1 . Here t is an arbitrary real number. a. (6 pts) For which values of t is ker A equal to { ~ } ? b. (6 pts) For which values of t is ker A not equal to { ~ } ? In each case, find a basis for ker A . Solution. a. ker A is equal to { ~ } precisely when det A is nonzero. We compute det A = t 3 + 3 t + 2 = ( t 2)( t 2 + 2 t + 1). This is zero when t = 2 or t = 1. So for all other t , ker A is zero. b. For t = 2, A = 1 2 1 1 1 2 2 1 1 . Its rref form is 1 1 1 1 . If ~x is in the kernel, we have x 3 = t, x 1 = t, x 2 = t , so the kernel is spanned by the single vector 1 1 1 . For t = 1, A = 1 1 1 1 1 1 1 1 1 . Its rref is 1 1 1 . For ~x in the kernel, we have x 2 = s, x 3 = t, x 1 = s t . Thus ~x =  s t s t = s  1 1 + t  1 1 . So a basis for the kernel is given by  1 1 and  1 1 ....
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This note was uploaded on 04/02/2008 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.
 Fall '08
 lee
 Math

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