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# Determinents 1.pdf

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School
Department
Applied Mathematics
Course
Applied Mathematics 1411A/B
Professor
Natalia Kiriushcheva
Semester
Summer

Description
Exercises for AM025 D.J. Jeﬁrey Department of Applied Mathematics U. Western Ontario November 8, 2006 1 Determinants 1. For an n £ n matrix A, (a) if A is singular then det(A) = 0. (b) if A is singular then det(A) 6= 0. (c) if A is singular then det(A) can have any value. (d) if A is singular then det(A) < 0. (e) if A is singular then det(A) = 1. Answer: a 2. Given that detA = 3, detB = 5, detC = 7, the value of det(A BC ) is (a) 15 (b) 105 (c) 21/5 (d) 5/21 (e) Cannot be determined from the information. Answer: b 3. Given that detA = 15, detB = 5, detC = 3, the value of det(ABC1 ) is (a) 15 (b) 225 (c) 25 (d) 1/25 (e) Cannot be determined from the information. Answer: c 4.Given that detA = 4 and detB = 5, then the value of det(A + B) is (a) 9 (b) 20 (c) 4/5 (d) 5/4 (e) Cannot be determined from the information Answer: e. There is NO theorem for det(A + B), only for det(AB). 5. Given that detA = 3 and detB = 5, then the value of det(A + B) is 1 (a) 8 (b) 15 (c) 3/5 (d) 5/3 (e) Cannot be determined from the information Answer: e. There is NO theorem for det(A + B), only for det(AB). 6. Given that detA = 3 and detB = 5, then the value of det(AB) is (a) 8 (b) 15 (c) 3/5 (d) 5/3 (e) Cannot be determined from the information Answer: b. 7. Given that detA = ¡3 and detB = 5, then the value of det(AB) is (a) 2 (b) -15 (c) -3/5 (d) -5/3 (e) Cannot be determined from the information Answer: b 8. Given that detA = ¡3 and detB = 5, then the value of det(A + B) is (a) 2 (b) -15 (c) -3/5 (d) -5/3 (e) Cannot be determined from the information Answer: e. There is NO theorem for det(A + B), only for det(AB). 9. Given that detA = 4, detB = 5 and I is the identity matrix, then the value of det(AB + I) is (a) 9 (b) 20 (c) 4/5 (d) 5/4 (e) Cannot be determined from the information Answer: e. There is NO theorem for det(A + B), only for det(AB). µ 9 ¡7 ¶ 10. The matrix A = has detA = ¡130. Therefore ¡7 ¡9 (a) det(¡3A) = +(9)(130). (b) det(¡3A) = ¡(9)(130). (c) det(¡3A) = +(3)(130). (d) det(¡3A) = ¡(3)(130). 2 (e) None of the above. Answer: b. µ ¶ 9 ¡7 11. The matrix A = ¡7 ¡9 has detA = ¡130. Therefore (a) det(¡2A) = +(2)(130). (b) det(¡2A) = ¡(2)(130). (c) det(¡2A) = +(4)(130). (d) det(¡2A) = ¡(4)(130). (e) None of the above. Answer: d. 0 1 u v w
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