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Calculus 2 - Midterm 2 Solutions

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

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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 0 4 10 5 2 - 5 4 2 3 R 1 + R 2 / - 4 5 1 0 0 15 6 2 - 5 4 2 3 - 5 4 * R 1 + R 2 / - 4 5 1 0 0 15 6 2 0 - 9 4 3 4 3 1 3 * R 2, 4 3 * R 3 / - 4 5 1 0 0 5 2 2 3 0 - 3 1 4 3 5 * R 2 + R 3 / - 4 5 1 0 0 5 2 2 3 0 0 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 - λ I ) = ( - 1 - λ )( - 3 - λ )( - 2 - λ ) - ( - 4)( - 3 - λ )( - 3) = ( - 3 - λ )( λ 2 + 3 λ - 10) = ( - 3 - λ )( λ - 2)( λ + 5). Thus, the eigenvalues are λ = - 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
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(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 - λ I ) = ( - 1 - λ )(3 - λ ) - ( - 4)(2) = λ 2 - 2 λ + 5. By the quadratic formula, this has roots λ = 2 ± p 4 - 20 2 = 2 ± 4 p - 1 2 = 1 ± 2 i . Thus, the eigenvalues of A are 1 + 2 i and 1 - 2 i .
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