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# finals - SOLUTIONS TO THE 18.02 FINAL EXAM BJORN POONEN...

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Unformatted text preview: SOLUTIONS TO THE 18.02 FINAL EXAM BJORN POONEN December 14, 2010, 9:00am-12:00 (3 hours) 1) For each of (a)-(e) below: If the statement is true, write TRUE. If the statement is false, write FALSE. (Please do not use the abbreviations T and F.) No explanations are required in this problem. (a) (5 pts.) If f ( x, y ) is a continuously differentiable function on R 2 , and ∂ 2 f ∂x 2 = 0 at every point of R 2 , then there exist constants a and b such that f ( x, y ) = ax + b for all x and y . Solution: FALSE. The values of a and b could depend on y . The function f ( x, y ) = y is a counterexample. (b) (5 pts.) The annulus R in R 2 defined by 9 ≤ x 2 + y 2 ≤ 16 is simply connected. Solution: FALSE. The circle of radius 3 . 5 centered at the origin gives a closed curve in R that cannot be shrunk to a point within R . (c) (5 pts.) If f ( x, y ) is a function whose second derivatives exist and are continuous everywhere on R 2 , and f (0 , 0) = f x (0 , 0) = f y (0 , 0) = f xx (0 , 0) = f yy (0 , 0) = 0 and f xy (0 , 0) 6 = 0, then f has a saddle point at (0 , 0). Solution: TRUE. This follows from the second derivative test. In the notation of that test, we are given A = 0, B 6 = 0, and C = 0, so AC- B 2 < 0, so f has a saddle point at (0 , 0). (d) (5 pts.) If A is a 3 × 3 matrix, and b is a column vector in R 3 , and the square system A x = b has more than one solution x , then det A = 0. Solution: TRUE. If det A 6 = 0, then there would have been one solution, namely x = A- 1 b . (e) (5 pts.) If F is a continuously differentiable 3D vector field on R 3 such that curl F = everywhere, and A and B are two points in R 3 , then the value of R C F · d r is the same for every piecewise smooth path C from A to B . Solution: TRUE. The region R 3 is simply connected, so all six conditions for conserva- tiveness are equivalent, and these are two of them. 1 2) (a) (5 pts.) For which pairs of real numbers ( a, b ) does the matrix A = a b 0 0 1 1 0 0 have an inverse? Solution: We have det A = 0 + b + 0--- 0 = b, so A is invertible if and only if b 6 = 0. (There is no condition on a .) (b) (10 pts.) For such pairs ( a, b ), compute A- 1 . (Its entries may depend on a and b . Suggestion: Once you have the answer, check it!) Solution: Compute nine 2 × 2 determinants to get the matrix of minors: - 1- b b a Apply the checkerboard signs to get the matrix of cofactors: 1 b b- a Take the transpose to get the adjoint matrix: 0 0 b 1 0- a b Divide by det A to get the inverse matrix: 1 1 /b- a/b 1 ....
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## This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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finals - SOLUTIONS TO THE 18.02 FINAL EXAM BJORN POONEN...

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