1920t1math1030cw3sol.pdf - 2019 MATH1030 Classwork 3 Last updated • online exercise The following webpage will be set up on Thursday Sep 26 Link

1920t1math1030cw3sol.pdf - 2019 MATH1030 Classwork 3 Last...

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2019 MATH1030 Classwork 3 Last updated: September 27, 2019 online exercise The following webpage will be set up on Thursday Sep 26 . Link: Login: your student ID Password: your student ID (please change it right after you login) 0.2 pt for each question set . Midterm: Oct 23 Wed 7:30-9:30. Important, Venue changed : LSB LT2, 3, 5, 6 Hand in 2a, 4b, 5b, 6, 7, 11, 12b, 13, 16. For section A try also 14, 15, 17. Then do as many questions as possible. You can get full credit as long as you show you effort. You can discuss the questions with your classmates or TA. But don’t copy answers directly. Answer will be posted online. If you cannot finish the class work, take the question paper home and try to finish the rest of them . CW2 Solution has been posted on math1030 class webpage: For password protected area (lecture notes or exercises): login : math1030 password : like1030 Questions Notation : Given a n × n matrix A , we use the notation A 0 = I n , A 1 = A , A 2 = AA , A 3 = AAA and so on. Obviously A k +1 = A k A = AA k for each integer k 0. 1. Let A = 1 3 5 7 2 4 6 8 1 1 1 1 , u = 1 10 100 1000 (a) Write down [ A ] 23 , [ A ] 14 , [ u ] 2 . (b) Compute A u . Answer. (a) [ A ] 23 = 6, [ A ] 14 = 7, [ u ] 2 = 10. (b) A u = 7531 8642 1111 2. Solve each of the following equations for x , or explain why a solution does not exist. 1
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(a) 2 1 2 3 0 4 2 - 3 1 1 2 0 1 x = - 1 1 0 0 5 - 2 (b) x 1 3 4 2 1 - 1 + 4 3 6 0 1 1 = 7 12 6 6 4 2 Answer. (a) The left hand side of the equation is equal to - 1 1 0 0 5 4 - 3 x . Hence x must satisfy the equation 4 - 3 x = - 2, implying that x = 2. (b) The left hand side of the equation is equal to x + 4 3 x + 3 4 x + 6 2 x x + 1 - x + 1 . Equating the (1 , 1) entries of both sides gives the equation x + 4 = 7, so that x = 3. Equating the (2 , 3) entries of both sides gives the equation - x + 1 = 2, which is not satisfied by x = 3. Hence, the original equation has no solution. 3. Compute the following. (a) 1 2 - 1 1 1 1 1 1 - 1 - 1 0 0 2 3 (b) 1 1 3 5 - 1 1 0 - 2 1 1 1 - 1 1 1 0 1 1 2 0 0 (c) 2 1 2 3 4 0 1 0 1 - 1 - 1 1 0 - 3 0 1 1 2 1 1 1 1 0 - 1 - 1 - 1 1 1 2 1 - 1 0 0 - 1 Answer. (a) 1 1 3 5 - 1 - 1 3 4 1 1 1 2 2
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(b) 4 8 - 1 0 2 4 (c) 1 1 - 2 - 3 - 5 - 4 4. (a) Let A = 1 0 1 0 1 1 0 0 2 , B = 1 1 2 1 0 0 0 1 1 . Find AB and BA . (b) Let a, b, c, d be real numbers and let A = 1 0 a b 0 1 c d , B = - a - b - c - d 1 0 0 1 . Find AB and BA . Answer. (a) AB = 1 2 3 1 1 1 0 2 2 BA = 1 1 6 1 0 1 0 1 3 (b) AB = 0 0 0 0 BA = - a - b - a 2 - bc - ab - bd - c - d - ac - cd - bc - d 2 1 0 a b 0 1 c d 5. (a) Let A = 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 . Find A 2 . (b) Let B = 1 0 1 5 1 3 1 - 1 2 , C = - 5 1 1 7 - 1 - 2 6 - 1 - 1 .
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