University of Ottawa Department of Mathematics and Statistics MAT 0A: Mathematical Methods II Instructor: Hadi Salmasian Final Exam Solutions April 08 Surname First Name Student # Seat # Instructions: (a) You have hours to complete this exam. (b) This exam consists of two parts. Questions through are answer only. For these questions, only your final answer will be considered for marks. Questions 4 through 0 are long answer. For these questions, you must show your work and justify your answers to receive full marks. Partial marks may be awarded for making sufficient progress towards a solution. (c) The number of points available for each question is indicated in square brackets. (d) All work to be considered for grading should be written in the space provided. The reverse side of pages is for scrap work. If you find that you need extra space in order to answer a particular question, you should continue on the reverse side of the page and indicate this clearly. Otherwise, the work written on the reverse side of pages will not be considered for marks. (e) Write your student number at the top of each page in the space provided. (f) You may use the second-to-last page of the exam as extra space for solutions. If you do so, indicate clearly on the page of the relevant question that you have continued your solution on the extra page. (g) You may use the last page of the exam as scrap paper. (h) Cellular phones, unauthorized electronic devices or course notes are not allowed during this exam. Phones and devices must be turned off and put away in your bag. Do not keep them in your possession, such as in your pockets. If caught with such a device or document, the following may occur: you will be asked to leave immediately the exam and academic fraud allegations will be filed which may result in you obtaining a 0 (zero) for the exam. By signing below, you acknowledge that you are complying with statement (h) above: Signature: Good luck! Please do not write in the table below. Question 4 5 6 7 8 9 0 Total Maximum 4 4 8 6 4 4 5 58 Grade
MAT 0A Final Exam Part A: Answer Only Questions For Questions, only your final answer will be considered for marks. Write your final answers in the spaces provided.. [ points Consider the matrices Compute B T A. Answer: B T A = A = [ 7 0 0 [, B = [ 4. [ points Suppose that A is a 4 7 matrix and B is a 6 4 matrix. For each statement below, write T if the statement is true, and write F if the statement is false. You will receive 0.5 points for each correct answer, lose 0.5 points for each incorrect answer, and receive zero points for an answer left blank. You cannot receive a negative score on this question. T The columns of B can never form a basis of R 6. F Nul(BA) is a subspace of R 6. T The reduced echelon form of A always has at least one non-pivot column. T rank(ba) = 7 dim Nul (BA). Page of 4
MAT 0A Final Exam. [ points Suppose that A, B, and C are 4 4 matrices such that det(a) =, det(b) = 5, and det( A BC T ) = 5. Calculate det(c). Answer: det(c) = 4. 4. [ point Determine all values of t R such that the columns of the matrix 4 0 5 t 0 5 0 0 0 are linearly dependent. Answer: t = 5. [ points Let z = i and w = i +. Write the complex number and b are real numbers. Answer: 4 4 i w + z in the form a + bi where a Page of 4
6. [ points Let A = 0 0 0 and I = 0 0 0 0. i + i 0 0 Write down the eigenvalues of (A + I) and their multiplicities. MAT 0A Final Exam Answer: 4 with multiplicity ; with multiplicity. 7. [ point Are the vectors 5 0,, 0 linearly independent? Write Yes if these vectors are linearly independent, and write No if they are linearly dependent. Answer: No, they are linearly dependent. 8. [ points Suppose we are given an n n matrix A and a vector b R n. Assume that the linear system Ax = b is inconsistent. For each statement below, write T if it is true, and write F if it is false. You will receive 0.5 points for each correct answer, lose 0.5 points for each incorrect answer, and receive zero points for an answer left blank. You cannot receive a negative score on this question. T Col A is not equal to R n. F A is invertible. T A has an eigenvector with eigenvalue λ = 0. F b = 0 n, where 0 n denotes the zero vector in R n. Page 4 of 4
MAT 0A Final Exam 9. [ points Let a, b, and c be vectors in R 4 such that a b = b + 5c. Justify that a belongs to Span{b, c} by expressing a explicitly as a linear combination of b and c. Solution: a = b + 5 c. 0. [ point Determine the value of the parameter t such that λ = i is an eigenvalue of the matrix [ t A =. i + Answer: From det(a ii) = 0 we obtain the equation t + = 0 i.e., t =.. [ points For each of the following subsets of R, write Y if the set is a subspace of R, and write N if it is not. You will receive 0.5 points for each correct answer, lose 0.5 points for each incorrect answer, and receive zero points for an answer left blank. You cannot receive a negative score on this question. x N x x + x + = 0 Y x x + y 0 + z 7 x, y, z R Y The eigenspace of the matrix A = 0 corresponding to its eigenvalue λ = 6. 0 5 N x y x, y, z R and y = x z Page 5 of 4
Solutions MAT 0A Final Exam. [ points Consider the traffic flow described by the following diagram. The letters A through D label intersections. The arrows indicate the direction of flow (all roads are one-way) and their labels indicate flow in cars per minute. 45 x B x 4 C x 7 A x 5 x 6 x D x 0 0 0 Write down a linear system describing the traffic flow, i.e., all constraints on the variables x i, i =,..., 7. (Do not solve the linear system.) Solution: A : x 5 = x 6 + 0 B : x + x = x 4 + x 5 C : x + x 4 + 45 = x 7 D : x 6 + 0 = x + x + 0 Total : x + 45 + 0 = x 7 + 0 + 0. [ points Let A, B, and C be n n invertible matrices. Solve the matrix equation AX T C B = BC for the matrix X. Answer: X = (C ) T B T (A ) T B T (A ) T. Page 6 of 4
MAT 0A Final Exam Part B: Long Answer Questions For Questions 4 0, you must show your work and justify your answers to receive full marks. Partial marks may be awarded for making sufficient progress towards a solution. 4. [4 points Is the following linear system consistent or inconsistent? If it is consistent, then write down the general solution in vector parametric form. x x x 4 x 5 = x +5x +x 5 = 7 x 4x +x 4 7x 5 = Solution: We reduce the augmented matrix of the system to the REF: 0 0 0 5 0 7 R R R 0 5 0 7 0 4 7 0 0 0 9 0 R R 0 5 0 7 0 0 0 0 0 4 R R +R 0 5 0 7 0 0 0 It follows that the system is consistent (since the rightmost column is not pivot). Moreover, x, x, x 4 are basic, x, x 5 are free. The general solution is x = x + x 5 + 4 x = 5x x 5 7 x = free x 4 = x 5 + x 5 = free The vector parametric form of the solution is x x + x 5 + 4 4 x x x 4 = 5x x 5 7 x x 5 + = x 5 0 + x 5 0 + 7 0 x 5 x 5 0 0 Page 7 of 4
MAT 0A Final Exam 5. [4 points Calculate the determinant of the following matrix using the method of co-factor expansion. 0 5 0 M = 5 7 0 0 0. 9 Solution: Expanding with respect to the rd row, and then with respect to the st row of the matrix, we obtain 0 5 0 5 7 0 0 0 0 0 = () 5 7 9 9 = ()( ) 5 7 = ()( ) = 6. Page 8 of 4
MAT 0A Final Exam 6. (a) [4 points Consider the matrix A = 0 0. Calculate the eigenvalues of A. 4 0 Solution: Starting with an expansion with respect to the second column, we have λ 0 det(a λi) = λ 0 4 0 λ = ( λ) λ 4 λ = ( λ)(λ 4λ 5) = (λ+)(λ 5)(λ+). In fact the roots of λ 4λ 5 are λ = ( 4) ± ( 4) 4( 5) Thus, the eigenvalues of A are and and 5. (b) [ point Is A diagonalizable? You should justify your answer. Solution: Yes because it has distinct eigenvalues. = { 5 Page 9 of 4
MAT 0A Final Exam (c) [ points Let B =. Find a basis for the eigenspace of B corresponding to the eigenvalue 8 λ =. Solution: For the eigenvalue λ =, we have: [ 0 B I 0 = 0 6 0 Thus, the eigenspace is given by and a basis of this eigenspace is x = x x x R R + R R R R x x = x = x + x 0 x 0, 0. 0 0 0 0 0 0 0 0 0 0 Page 0 of 4
MAT 0A Final Exam 7. The city of Quicheton has two culinary arts institutes: ChefAcademie and RoyalToque. In the beginning of January, the two institutes had equal student enrolment levels. During January, ChefAcademie updated its classes. As a result, at the end of January of ChefAcademie s students switched to RoyalToque, 0 while of RoyalToque s students switched to ChefAcademie. 4 (a) [ point Write down the migration matrix M and the initial state vector x 0 for this problem. Solution: [ [.9.5 M =, x 0 =..75 Another possible migration matrix (if one reverses the order of the companies) is M = [.75..5.9 x 0 = [. (b) [ point What fraction of the total number of students will be enrolled at each of the institutes in the beginning of February? Solution: x = M x 0 = [ 40 7 40 (c) [4 points If the same migration trend continues for several months, then in the long run what are the predicted enrolment ratios of each of the institutes? In your final answer, you should clearly indicate the fraction of students for each institute. Solution: reduce: [ M I 0 = [ To find the steady-state vector, we must find an eigenvector of eigenvalue. We row 0 4 0 0 4 0 The general solution is thus R 0R [ R 0R 5 0 5 0 x = x [ 5. [ R R R 5 0 0 0 0 We choose the value of the free variable so that the sum of the entries of x is equal to one: ( ) 5 + x = = x = 7. Thus the steady-state vector is [ 5 7 x =. 7 Since M is regular stochastic, the long term behaviour is given by the steady-state vector. Thus, in the long term, 5 7 of the total number of students will be enrolled in ChefAcademie and of students 7 will be enrolled in RoyalToque. Page of 4
MAT 0A Final Exam 8. [4 points Consider the matrix 0 0 0 A = 5 7 5. 0 4 4 Find a basis for Nul A. Solution: We row reduce to find an echelon form of A. R 0 0 0 R R 0 0 0 0 0 0 R 5 7 5 R + R 0 5 7 R R R 0 5 7 0 4 4 0 0 4 0 0 0 0 R R 0 0 0 R R R R 0 5 7 R + R 0 0 0 0 0 5 7 0 0 0 0 0 0 0 0 0 Denoting the variables x,..., x 5 as usual, we realize that x, x, x 5 are basic and x, x 4 are free. It follows that the general solution is x = 0 x = 5x 7x 4 x = free x 4 = free x 5 = 0 and therefore the vector parametric form of the solution is x 0 0 x x x 4 = x 5 0 + x 7 4 0 0 0 and the basis that we obtain for Nul A is x 5 0 0 5 0, 7 0 0 0 Page of 4
MAT 0A Final Exam 9. [4 points Let Find the inverse of A. Solution: R R 4R 4 A = 0 6 We row reduce the super-augmented matrix: R 4 0 0 R R R 0 0 R R 0 6 0 0 4 0 0 0 0 R ( )R 0 0 4 R R R R R + 4R Thus A is invertible, and its inverse is R R +R 4 0 0 0 0 0 4 0 0 8 0 0 0 0 0 0 0 6 0 0 0 0 0 6 A = 0. 4 0 0 0 0 0 0 Page of 4
MAT 0A Final Exam 0. An economy consists of two sectors: Agriculture and Service. In order to produce one unit, the Agriculture sector consumes units from Agriculture and units from Service. Also, in order to produce one unit, the Service sector consumes units from Agriculture and 6 (a) [ point Write the consumption matrix for this economy. Solution: C = [ 6 units from Service. (b) [ point Write the Leontief Input-Output Model production equation. Solution: The equation is x = C x + d, or (I C) x = d. (c) [ points Determine the production levels needed to satisfy a final demand of units from the Agriculture sector and 6 units from the Service sector. Solution: We have We solve the equation A x = d, where [ [ A = I C =, d = 6 Thus A is invertible and 5 6 det A = 0 8 9 = 0. A = [ 5 6 = [ 5. Therefore, [ 5 [ [ x = A d = 4 =. 6 4 So the Agriculture sector needs to produce 4 units and the Service sector needs to produce 4 units.. Page 4 of 4