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Unformatted text preview: Math 2J Lecture B Fall 2009 (Instructor: Professor R. C. Reilly) Answer Key for Practice Final Exam (1) Let A be the 2by3 matrix whose columns are the given vectors u , v , and w . That is, A = [ u v w ] = 3 2 5 1 5 2 . It was shown in class that the condition for the given vectors to be linearly independent is equivalent to the homoge neous system A x = having only one solution, namely x = . This, in turn, is equivalent to the general solution of the system having no free variables. However, since there are only two rows in the matrix A , the system A x = can have at most 2 lead variables. But the system has 3 variables in all, so there must be at least one free variable, so the vectors are not linearly independent. To find numbers a , b , and c as required by the problem, let us actually solve the system A x = by GaussJordan Reduction: [ A  ] = 3 2 5 1 5 2 1 5 2 3 2 5 1 5 2 0 17 1 1 5 2 0 1 1 17 1 0 29 17 0 1 1 17 From this last matrix, which is in reduced rowechelon form, one sees that the general solution of the system has one free variable, namely x 3 ; and the formula for this general solution is x = x 3  29 / 17 1 / 17 1 NOTE: We are free to use any nonzero value of x 3 to get the desired a , b and c . To simplify checking our answer, let us get rid of the fractions by choosing x 3 = 17, which gives us the special solution x =  29 1 17 . That is, we get a = 29, b = 1, c = 17. Clearly at least one of these numbers is not zero; indeed, this is true for all of them. Checking Our Answer : Note that 29 u v + 17 w = ( 29) 3 1 + ( 1) 2 5 + 17 5 2 = ( 29) ( 3) + ( 1) 2 + 17 ( 5) ( 29) 1 + ( 1) 5 + 17 2 = (2) Notice that one has lim n 12 n 3 7 n 2 + 9 n + 6 3 n + 2 = + and lim n (4 n 2 5 n + 2) = + . Thus, trying to compute the limit of the given sequence ( a 1 ,a 2 , ... ) using only the standard limit laws leads us to the indeterminante form  . However, if we do some preliminary algebra, we can simplify the problem. More precisely, write a n as a single fraction: a n = 12 n 3 7 n 2 + 9 n + 6 3 n + 2 (4 n 2 5 n + 2) = (12 n 3 7 n 2 + 9 n + 6) (3 n + 2)(4 n 2 5 n + 2) 3 n + 2 = (12 n 3 7 n 2 + 9 n + 6) (12 n 3 + 8 n 2 15 n 2 10 n + 6 n + 4) 3 n + 2 = 13 n + 2 3 n + 2 = 13 + 2 /n 3 + 2 /n Thus, one easiy computes that lim n a n = 13 / 3. In particular, the sequence does converge, and its value is 13 / 3. (3) The coefficient a 3 is given by a 3 = f 000 ( 1) / 3!, so we need only determine f 000 ( 1). To get f 000 ( x ) simply differentiate both sides of the differential equation f 00 ( x ) + 3 x 2 f ( x ) + 4 f ( x ) = 0 to get f 000 ( x ) + 6 xf ( x ) + 3 x 2 f 00 ( x ) + 4 f ( x ) = 0 ( * ) 1 But f ( 1) = 2 and f ( 1) = 3 (given facts); and we can substitute these values into the differential equation for f to get f 00 ( 1) + 3( 1) 2 ( 3) + 4 2 = 0. Thus, f 00 ( 1) = 9 8 = 1....
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This note was uploaded on 03/11/2012 for the course MATH 2J 44360 taught by Professor Donaldson,neil during the Spring '10 term at UC Irvine.
 Spring '10
 DONALDSON,NEIL

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