t1_solutions - York University Faculty of Arts, Faculty of...

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Unformatted text preview: York University Faculty of Arts, Faculty of Science Math 1025 Class Test 1 SOLUTIONS Instructions: 1. Time allowed: 50 minutes 2. There are 5 questions on 5 pages. 3. Answer all questions. 4. Your work must justify the answer you give. 5. No calculators or other aids permitted. Question 1 2 3 4 5 Total Points 13 7 7 7 6 40 Marks MATH 1025 1. (a) (8 points) Find all solutions to Test 1 Page 1 January 24, 2005 2x1 + 6x2 − 3x3 + 12x4 3x1 + 9x2 − 7x3 + 13x4 Answer: Reduce the augmented coefficient 13 2 6 39 1 0 0 3 0 0 matrix. −2 −3 −7 5 12 13 5 2 −2 x1 + 3x2 − 2x3 + 5x4 = = = 4 9 11 . From which 4 1 −1 1 3 −2 5 4 0 0 1 2 1 00 0 00 13096 0 0 1 2 1 00000 −2 1 −1 x1 x2 x3 x4 = = = = 6 − 9t − 3s s 1 − 2t t. 4 9 11 (b) (2 points) Is there a solution with x2 = 1 and x4 = 0? Find one or explain why not. Answer: Yes. Set s = 1 and t = 0 to get x1 = 3, x2 = 1, x3 = 1, x4 = 0. (c) (3 points) Is there a solution with x3 = 0 and x4 = 1? Find one or explain why not. Answer: No. To obtain x4 = 1 we need t = 1. To obtain x3 = 0 we need t = 1 . 2 MATH 1025 Test 1 Page 2 January 24, 2005 2. (a) (4 points) Find the row echelon form of Answer: Reduction gives 1 2 5 1 2 5 3 1 5 2 5 12 3 1 k 3 1 k 3 1 5 2 5 12 If k = 5 the row echelon form obtained is 13 0 1 00 If k ￿= 5 the row echelon form obtained is 13 0 1 00 (b) (3 points) For which values of k does 1 2 5 1 3 2 3 0 −5 1 −5 0 −10 2 k − 15 1 3 2 3 0 1 −1 1 5 0 −10 2 k − 15 13 2 3 0 1 −1 1 . 5 0 0 0 k−5 2 −1 5 0 3 1. 0 3 1. 1 2 −1 5 0 have (a) a unique solution (b) an infinite number of number of solutions (c) no solutions? Answer: (a) never (b) k = 5 (c) k ￿= 5 the system 32 x 3 1 5 y = 1 5 12 z k MATH 1025 3. 2 0 Let A = 0 0 1 0 Test 1 Page 3 January 24, 2005 (a) (4 points) Write A−1 as a product of elementary matrices. Answer: −1 −1 001 1 00 A−1 = 0 1 0 −2 1 0 100 0 01 1 001 100 0 2 = 0 1 0 2 1 0 0 1 100 001 00 0 1 −2 0 1 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0. 0 2 0 0 0 0. 1 0 1 0 −1 0 0 1 (b) (3 points) Determine A−1 . Answer: Multiply the matrices above to obtain 1 001 2 A−1 = 0 1 0 1 100 0 001 1 1 0. = 1 00 2 0 0 1 0 1 0 MATH 1025 4. (7 points) Find A−1 , where Test 1 0 2 . −3 Page 4 January 24, 2005 Answer: Reduce to obtain 1 3 A= −2 −2 −7 5 from which 1 −2 0 1 0 0 3 −7 2 0 1 0 −2 5 −3 0 0 1 1 −2 0 1 00 0 −1 2 −3 1 0 0 1 −3 2 0 1 1 −2 0 1 0 0 0 1 −2 3 −1 0 0 1 −3 2 0 1 1 −2 0 1 00 0 1 −2 3 −1 0 0 0 −1 −1 1 1 1 −2 0 1 0 0 0 1 −2 3 −1 0 00 1 1 −1 −1 1 −2 0 1 0 0 0 1 0 5 −3 −2 0 0 1 1 −1 −1 1 0 0 11 −6 −4 0 1 0 5 −3 −2 0 0 1 1 −1 −1 A− 1 11 5 = 1 −6 −3 −1 −4 −2 −1 MATH 1025 5. The matrix Test 1 −1 −1 A= −4 0 3 13 1 0 −1 Page 5 January 24, 2005 satisfies A3 − A2 − A − 2I = 0. (a) (3 points) Write A−1 as a polynomial expression in A. Hint: A3 − A2 − A = 2I . Answer: 13 (A − A2 − A) 2 12 (A − A − I )A 2 from which = = I I 1 1 1 1 A−1 = (A2 − A − I ) = A2 − A − I . 2 2 2 2 (b) (3 points) Use the expression you gave in (a) to find A−1 . Note: No credit will be given for solutions which do not use (a). Answer: Now −1 0 1 −1 0 1 −3 13 0 −1 3 0 = −2 9 A2 = −1 3 −4 13 −1 −4 13 −1 −5 26 1 0 −A = 1 −3 4 −13 −1 0 −I = 0 −1 0 0 −3 13 2 −1 5 A −A−I = −1 13 3 13 −2 2 −1 −1 5 A = 2 2 − 1 13 2 2 −2 1 −3 −1 0 1 0 0 −1 −3 −1 −3 −3 2 −1 . 2 −3 2 The end ...
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