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Unformatted text preview: Partial Solution Set, Leon Â§ 5.3 5.3.1 Find leastsquares solutions: (a) x 1 + x 2 = 3 2 x 1 3 x 2 = 2 x 1 + 0 x 2 = 1 Solution : We are trying to solve A x = b , where A = 1 1 2 3 and b = (3 , 1 , 2) T . Clearly b 6âˆˆ R ( A ). So we use the normal equation, A T A x = A T b , which becomes 5 5 5 10 x = 5 . The solution (unique, since A has rank 2) is Ë† x = (2 , 1) T . (c) x 1 + x 2 + x 3 = 4 x 1 + x 2 + x 3 = 0 x 2 + x 3 = 1 x 1 + x 3 = 2 Solution : The matrix equation is 1 1 1 1 1 1 1 1 1 0 1 x = 4 1 2 , which is inconsistent. The normal equations lead to the matrix equation, 3 0 1 0 3 1 1 1 4 x = 6 3 7 , so the solution is Ë† x = (1 . 6 , . 6 , 1 . 2) T . 5.3.2 For each solution Ë† x in exercise 5.3.1, 1. Determine p = A Ë† x . 2. Calculate r (Ë† x ). 3. Verify that r (Ë† x ) âˆˆ N ( A T )....
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This note was uploaded on 07/17/2011 for the course MATH 311 taught by Professor Anshelvich during the Spring '08 term at Texas A&M.
 Spring '08
 Anshelvich
 Linear Algebra, Algebra

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