Calculus 2 - Midterm 2 Solutions

Calculus 2 - Midterm 2 Solutions - MAT1332 Spring/Summer...

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Unformatted text preview: MAT1332 Spring/Summer 2009 Midterm Exam 2 Solutions Problem 1 (4 points) Applying the elementary row operations to the augmented coefficient matrix we have - 4 5 1 4 10 5 2- 5 4 2 3 R 1 + R 2 / - 4 5 1 15 6 2- 5 4 2 3 - 5 4 * R 1 + R 2 / - 4 5 1 15 6 2- 9 4 3 4 3 1 3 * R 2, 4 3 * R 3 / - 4 5 1 5 2 2 3- 3 1 4 3 5 * R 2 + R 3 / - 4 5 1 5 2 2 3 11 5 22 5 So 11 z = 22 and z = 2. Then, since 5 y + 2 z = 2 3 we have 5 y + 4 = 2 3 and so y =- 2 3 . Finally, the first equation- 4 x + 5 y + z = 0 gives then 4 x = 2- 10 3 and x =- 1 3 . Problem 2 (5 points) det( A ) = (- 1)(- 3)(- 2)- (- 4)(- 3)(- 3) = 30. Thus, A is invertible and (2) is true. The trace is- 1- 3- 2 =- 6 and so (4) is also true. The eigenvalues of A can be found by det( A- I ) = (- 1- )(- 3- )(- 2- )- (- 4)(- 3- )(- 3) = (- 3- )( 2 + 3 - 10) = (- 3- )( - 2)( + 5). Thus, the eigenvalues are =- 3,2 and- 5. Thus, (2) and (4) are the only correct claims. Problem 3 (4 points) The correct answers are listed below: 1. ( ab ) c (b)- 42- 306 i 2. cd-| a | 1 (d) 37 + 6 i 3. a c (c) 1 2 + 2 3 i 4. c 3 d (b)- 7 25 + 1 25 i Problem 4 (6 points) We have det( A- I ) = (- 1- )(3- )- (- 4)(2) = 2- 2 + 5....
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This note was uploaded on 01/16/2011 for the course MAT 1332 taught by Professor Munteanu during the Fall '07 term at University of Ottawa.

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Calculus 2 - Midterm 2 Solutions - MAT1332 Spring/Summer...

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