MATH 251 MATH 251: Multivariate Calculus MATH 251 FALL 2005 EXAM-I FALL 2005 EXAM-I EXAMINATION COVER PAGE Professor Moseley PRINT NAME ( ) Last Name, First Name MI (What you wish to be called) ID # EXAM DATE Friday, September 16, 2005 I swear and/or affirm that all of the work presented on this exam is my own and that I have neither given nor received any help during the exam. SIGNATURE INSTRUCTIONS DATE 1. Besides this cover page, there are 10 pages of questions and problems on this exam. MAKE SURE YOU HAVE ALL THE PAGES. If a page is missing, you will receive a grade of zero for that page. Read through the entire exam. If you cannot read anything, raise your hand and I will come to you. 2. Place your I.D. on your desk during the exam. Your I.D., this exam, and a straight edge are all that you may have on your desk during the exam. NO CALCULATORS! NO SCRATCH PAPER! Use the back of the exam sheets if necessary. You may remove the staple if you wish. Print your name on all sheets. 3. Explain your solutions fully and carefully. Your entire solution will be graded, not just your final answer. SHOW YOUR WORK! Every thought you have should be expressed in your best mathematics on this paper. Partial credit will be given as deemed appropriate. Proofread your solutions and check your computations as time allows. GOOD LUCK!! REQUEST FOR REGRADE Please regrade the following problems for the reasons I have indicated: (e.g., I do not understand what I did wrong on page.) Scores page points score 1 14 2 17 3 12 4 5 5 8 6 8 7 10 8 8 9 10 10 8 11 12 13 14 15 16 17 18 (Regrades should be requested within a week of the date the exam is returned. Attach additional sheets as necessary to explain your reasons.) I swear and/or affirm that upon the return of this exam I have written nothing on this exam except on this REGRADE FORM. (Writing or changing anything is considered to be cheating.) Date Signature 19 20 21 22 Total 100 Total 100
MATH 251 EXAM I Fall 2005 Prof. Moseley Page 1 PRINT NAME ( ) SS No. 1 i 1 i 1 0 Let α= 2, A =, and B =. Compute the following: 1 0 i 1+ i Circle the correct answer from those listed below. 1. (1 pt.) A =. A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. ABC. 2. (1 pt.) A T =. A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. ABC. 3. (1 pt.) A* =. A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. ABC. 4. (2 pt.) αa =. A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. ABC. 5. (2 pts.) A+B =. A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. ABC. 6. (3 pts.) AB =. A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. ABC. Possible answers this page. 1 i 1 i 1 0 2 i 1 i 1 i 1+ i 1 i 1 i A., B., C., D., E., 1 0 i 1+ i 1+ i 1+ i 1 0 1 0 1 i 1 2 i 2 i 1 i 1 i 2 2i 2 2 2i 2 i AB., AC., AD., AE., BC., 1+ i 0 1 0 1 1 1 0 2 0 2 i 2 i 0 1 i 1 2i 1 i 1 i 1 i 1 i 1 i BD., BE., CD., CE., DE., 1 0 1 1 i 1 0 2 0 1 2 ABC. None of the above. Possible points this page = 10. POINTS EARNED THIS PAGE
MATH 251 EXAM I Fall 2005 Prof. Moseley Page 2 Matrix algebra. Circle the correct answer from the choices below to fill in the blank.. Using the abbreviated (tensor) notation for a matrix discussed in class, let A = [a ij ], B=[b ij ], C=[c ij ], D=[d ij ], and E=[e ij ] be nxn square matrices. 7. ( 2 pt.) If α is a scalar and C = αa, then c ij =. A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD. CE. DE. 8. ( 2 pt.) If D = A + B, then d ij =. A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD. CE. DE. 9. ( 3 pts.) If E = AB, then e ij =.A. B. C. D. E. AB. AC. AD. AE. BC. Possible Answers for questions 7, 8, and 9. BD. BE. CD. CE. DE. A. αa ij, B. βa ij, C. b ij a ij, D. b ij + a ij, E.a ij /b ij, AB.= a b, AC. a b, AD. a ij b ij, AE. a ij, BC. a ij +c ij BD. b ij, BE. b ij d ij, CD. b ij + e ij, CE. b ij a ij, DE. None of the above n i1 ij ij n k 1 ik kj (10 pts.) True or False. Matrix Algebra. Circle True or False, but not both. If I cannot read your answer, it is WRONG. 10. True or False, Matrix addition is associative. 11.True or False, Matrix addition is not commutative. 12. True or False, α,βr and AR m n, α(βa) = (αβ)a.. 13.True or False, Multiplication of square matrices is associative. 14. True or False, Multiplication of square matrices is commutative. 15.True or False, If A and B are invertible square matrices, then (AB) -1 exists and (AB) -1 = A -1 B -1. 16.True or False, If A is an invertible square matrix, then (A -1 ) -1 exists and (A -1 ) -1 = A. 17.True or False, If A and B are square matrices, then (AB) T exists and (AB) T = A T B T. 18.True or False, If A is a square matrix, then (A T ) T exists and (A T ) T = A. 19.True or False, If A is an invertible square matrix, then (A T ) -1 exists and (A T ) -1 = (A -1 ) T. Possible points this page = 17. POINTS EARNED THIS PAGE =
MATH 251 EXAM I Fall 2005 Prof. Moseley Page 3 On the back of the previous sheet, solve the x 1 + x 2 + x 3 - x 4 = 1 system of linear algebraic equations Be sure to write you answer in the correct form. Circle the correct answer x 1 + 2x 2 + x 3 = 0 from the possibilities below x 3 + x 4 = 0 x 2-2x 3 + x 4 = 1 20. (3 pts.) x 1 =.A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. DE. 21. (3 pts.) x 2 =.A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. DE. 22. (3 pts.) x 3 =.A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. DE. 23. (3 pts.) x 4 =.A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. DE. Possible answers this page. A.1, B.2, C. 3, D. 4, E. 5, AB. 6, AC. 7, AD. 8, AE. 9, BC.10, BD.1, BE.2, CD.3, CE.4, DE.5, ABC.6, ABD.7, ABE 8, BCD.9, BCE.10, CDE. None of the above. Possible points this page = 12. POINTS EARNED THIS PAGE =
MATH 251 EXAM I Fall 2005 Prof. Moseley Page 4 ( 5 pts.) True or false. Solution of Linear Algebraic Equations having possibly complex coefficients. Assume A is an m n matrix of possibly complex numbers, that is an n 1 column vector of (possibly complex) unknowns, and that b A x x is an m 1 (possibly complex valued) column vector. Now consider. (*) mxn nx1 b mx1 Under these hypotheses, determine which of the following is true and which is false. It true, circle True. It false, circle False. If I can not read your answer, it is wrong. b 0 24. True or False, If, then (*) always has an infinite number of solutions. 25. True or False, The vector equation (*) always has exactly one solution. 26. True or False, If A is square (n=m) and nonsingular, then (*) always has a unique solution. 27. True or False, The equation (*) can be considered as a mapping problem from one vector space to another. A 1 i i 1 28. True or False, If then (*) has a unique solution. Total points this page = 5. TOTAL POINTS EARNED THIS PAGE
MATH 251 FINAL EXAM Fall 2005 Prof. Moseley Page 5 1 i x 1 You are to solve A x b where A,, and. Be sure you write your 2x2 2x1 2x1 i 1 x y b i answer according to the directions given in class (attendance is mandatory) for these kinds of problems. A b 1 i 1 0 0 0 U c U c 29. (4 pts.) If is reduced to using Gauss elimination, then =. 1 i 0 0 0 0 1 i 1 0 0 1 0 0 0 0 0 0 1 i 1 0 0 i A., B., C., D., E., AB. None of the above are possible. 30. ( 4 pts.) The general solution of can be written as. 1 A x b 2x2 2x1 2x1 1i i i A. No Solution, B. x, C., D., E., 0 x y 1 x y 0 1 x 1 1 i i1 1i x y 0 1 x y 1 0 x y 0 1 AB., AC., AD., BC. None of the above correctly describes the solution or set of solutions. Total points this page = 8. TOTAL POINTS EARNED THIS PAGE
MATH 251 EXAM I Fall 2005 Prof. Moseley Page 6 41 9 a b Let A = and A 1 =. Compute the inverse of A. Be sure to explain clearly 9 2 c d and completely your method. Circle the correct values of a, b, c, and d from the possiblities below: 31. (2 pts.) a =.A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. DE. 32. (2 pts.) b =.A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. DE. 33. (2 pts.) c =.A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. DE. 34. (2 pts.) d =.A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. DE. Possible answers this page. A.1, B.2, C. 3, D. 4, E. 5, AB. 8, AC. 9, AD. 10, AE. 11, BC. 20, BD. 21, BE. 22, CD. 25, CE. 26, DE. 30, ABC. 40, ABD. 41, ABE. 42, BCD. 45, BCE. 55. CDE. None of the above. Possible points this page = 8. POINTS EARNED THIS PAGE =
MATH 251 EXAM II Fall 2005 Prof. Moseley Page 7 {v,v,...,v } 35. ( 2 pts.) Let S = 1 2 n V where V is a vector space. Choose the completion of the following definition of what it means for S to be linearly independent. {v,v,...,v } Definition. The set S = 1 2 n V where V is a vector space is linearly independent if A. The vector equation c has an infinite number of solutions. 1v 1+c2v 2+...+cnv n = 0 B. The vector equation c has a solution other than the trivial solution. 1v 1+c2v 2+...+cnv n = 0 C. The vector equation c has only the trivial solution c 1 = c 2 = = c n = 0. 1v 1+c2v 2+...+cnv n = 0 D. The vector equation c has at least two solutions. 1v 1+c2v 2+...+cnv n = 0 E. The vector equation c has no solution. 1v 1+c2v 2+...+cnv n = 0 AB. The associated matrix is nonsingular. AC. The associated matrix is singular On the back of the previous sheet, determine Directly Using the Definition (DUD) if the following sets of vectors are linearly independent. As explained in class, circle the appropriate answer that gives an appropriate method to prove that your results are correct (Attendance is mandatory). Be careful. If you get them backwards, your grade is zero. 36. (4 pts.) Let S =.{[2, 4, 8] T, [3, 6, 11] T }. Circle the correct answer A. S is linearly independent as c 1 [2, 4, 8] T + c 2 [3, 6, 11] T = [0,0,0] implies c 1 = 0 and c 2 = 0. B. S is linearly independent as 3[2, 4, 8] T + (2) [3, 6, 11] T = [0,0,0]. C. S is linearly dependent as c 1 [2, 4, 8] T + c 2 [3, 6, 11] T = [0,0,0] implies c 1 = 0 and c 2 = 0. D. S is linearly dependent as 3[2, 4, 8] T + (2) [3, 6, 11] T = [0,0,0]. E. S is neither linearly independent or linearly dependent as the definition does not apply. 37. (4 pts.) Let S = {[2, 2, 6] T, [3, 3, 9] T }. Circle the correct answer A. S is linearly independent as c 1 [2, 2, 6] T + c 2 [3, 3, 9] T = [0,0,0] implies c 1 = 0 and c 2 = 0. B. S is linearly independent as 3[2, 2, 6] T + (2) [3, 3, 9] T = [0,0,0]. C. S is linearly dependent as c 1 [2, 2, 6] T + c 2 [3, 3, 9] T = [0,0,0] implies c 1 = 0 and c 2 = 0. D. S is linearly dependent as 3[2, 2, 6] T + (2) [3, 3, 9] T = [0,0,0]. E. S is neither linearly independent or linearly dependent as the definition does not apply. Total points this page = 10. TOTAL POINTS EARNED THIS PAGE
MATH 251 EXAM I Fall 2005 Prof. Moseley Page 8 a x + b y = 3 c x + d y = 2 Use Cramer's rule to solve the system of linear algebraic equations given above. Assume a d b c. Solutions using a method other than Cramer's rule will receive very little credit. Using Cramer's Rule, we have 38. (4 pts.) x =. A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. 39.(4 pts.) y =. A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD.CE. Possible answers. 3 b 3 a b 3 a 3 2 d 2 c d 2 c 2 2 d 2 c d 2 c 2 A., B., C., D., E., AB., AC., AD., ad bc ad bc ad bc ad bc bc ad bc ad bc ad bc ad 3 2 3 2 b d a c b d a b 3 2 3 2 c d AE., BC., BD., BE., CD., CE. None of the above ad bc ad bc ad bc ad bc ad bc 3 b a b 3 a Possible points this page = 8. POINTS EARNED THIS PAGE = b 3 a 3
MATH 251 EXAM I Fall 2005 Prof. Moseley Page 9 1 0 3 0 Let A = On the back of the previous sheet, compute the determinant of A. 0 2 1 4 3 0 9 1 40. ( 3 pts.) The first step of the Laplace Expansion in terms of the first column yields det(a) =. 41. (3 pts.) The first step in using Gauss Elimination to find det(a) yields det(a) =.: 42. (4 pts.) The numerical value of det(a) is det(a) =. 0 3 0 0 3 0 0 3 1 0 2 0 3 0 A. (1) 1 0 2 (3) 1 0 2, B. (1) 1 0 2 (3) 2 1 4, C. (1) 2 1 4 (1) 1 0 2, 2 1 4 2 1 4 2 1 4 0 9 1 0 9 1 2 1 4 1 0 2 0 3 0 1 0 2 0 3 0 0 3 0 0 1 2 D. (3) 2 1 4 (1) 1 0 2, E. (1) 2 1 4 (3) 1 0 2, AB. (1) 1 0 2 ( 3) 0 2 4, 0 9 1 2 1 4 0 9 1 2 1 4 2 1 4 3 0 1 0 3 0 0 1 2 0 3 0 0 1 2 1 0 2 0 1 2 AC. (1) 1 0 2 (1) 0 2 4, AD. (3) 1 0 2 ( 3) 0 2 4, AE. (1) 2 1 4 ( 3) 0 2 4, 2 1 4 3 0 1 2 1 4 3 0 1 0 9 1 3 0 1 1 0 3 0 1 0 3 0 1 0 3 0 1 0 3 0 1 0 3 0 BC., BD., BE., CD., CE., 0 2 1 4 0 2 1 4 0 0 2 4 0 2 1 4 0 2 1 4 0 0 9 1 0 1 0 1 3 0 9 1 0 0 0 0 0 0 0 1 1 0 3 0 1 0 3 0 1 0 3 0 1 0 3 0 DE., ABC., ABD., ABE., ACD.1, ACE.2, 0 2 1 4 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 2 0 0 0 1 0 0 0 2 0 0 0 0 ADE. 3, BCD. 4, BCE. 5, BDE. 8, CDE. 9, ABCD. 0, ABCE. 1, ABDE. 2, ACDE.3 BCDE. None of the above. Possible points this page = 10. POINTS EARNED THIS PAGE =
MATH 251 EXAM I Fall 2005 Prof. Moseley Page 10 PRINT NAME ( ) SS No. Let a and b be the vectors, a = <2,-1,1> = (2,1,1) = [2,1,1] T = 2 î ĵ + ˆk and b = <0,1,3> = (0,1,3) = [0,1,3] T = ĵ + 3 ˆk. 43. (3 pts.) Then the dot product is a b = ( a, b ) = a, b. A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD. CE. DE. ABC. ABD. ABE. BCD. BCE, CDE, ABCD. 44. (5 pts.) The cross product is a b.. A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. CD. CE. DE. ABC. ABD. ABE. BCD. BCE, CDE, ABCD. Possible answers this page. A. 1, B.2, C.3, D.4, E.5 AB.1 AC.2 AD.3 AE.4 BC. 3i ˆ 2j ˆ k ˆ, BD. 3i ˆ 2j ˆ kˆ, BE. 3i ˆ 2j ˆ k ˆ,CD. 3i ˆ 2j ˆ kˆ, CE. 3i ˆ 2j ˆ kˆ, DE. 3i ˆ 2j ˆ kˆ, ABC. 3i ˆ 2j ˆ kˆ, ABD. 3i ˆ 2j ˆ kˆ, ABE. 3i ˆ 2j ˆ k ˆ, BCD. 3i ˆ 2j ˆ kˆ, BCE. 3i ˆ 2j ˆ kˆ, CDE. 3i ˆ 2j ˆ kˆ, ABCD. None of the above. Possible points this page = 8. POINTS EARNED THIS PAGE =