University of Ottawa Department of Mathematics and Statistics MAT B: Mathematical Methods II Instructor: Hadi Salmasian Final Exam Solutions April 7 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 5 through 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 (zero) for the exam. By signing below, you acknowledge that you have ensured that you are complying with statement (h) above: Signature: Good luck! Please do not write in the table below. Question 5 6 7 8 9 Total Maximum 7 5 8 6 6 Grade
Part A: Answer Only Questions For Questions, only your final answer will be considered for marks. answers in the spaces provided. MAT B Final Exam If applicable, write your final. [ points] Consider the matrices A =, B =, I = Compute I + A T B. Answer: I + A T B = [ ] + [ ] = [ ]. [ points] Let z = i + and w = i. Write the complex number and b are real numbers. Answer: [ ]. z w z w = ( ) + (i i) = ( )( i) = + i ( + i)( i) = 5 + 5 i. in the form a + bi where a. [ points] Let A, B, and C be matrices such that det A =, det B =, and det(c) = 6. Calculate det( AB T CA ). Answer: det( AB T CA ) = 8. Page of
MAT B Final Exam. [ points] Determine all values of t R such that the linear system { y y + y = y 6y + ty = is consistent. Answer: t 5. [ points] Let A and B be two m n matrices such that n > m. Each of the statements below is either always true or always false. For each statement below, write T if the statement is always true, and write F if the statement is always false. You will receive.5 points for each correct answer, lose.5 points for each incorrect answer, and receive zero points for an answer left blank. You cannot receive a negative score on this question. F Col(A T ) is a subspace of R m. T Nul(B) {} T det(ab T ) = det(ba T ) F rank(ab T ) + dim Nul (AB T ) = n. + i 6. [ points] Let A =. Write down the eigenvalues of A and their multiplicities. i + i i You should write the eigenvalues in the form a + bi, where a, b R. Answer: Note that i = + i. Therefore + i has multiplicity and has multiplicity. 7. [ point] Suppose that A is a 6 6 matrix which has non-pivot columns. Write down the dimension of the subspace W = { v R 6 : the equation Ax = v has a solution x R 6}. Answer: W is equal to Col A. Therefore the dimension of W is equal to the number of pivot columns, i.e., 6 =. Page of
MAT B Final Exam 8. [ 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.5 points for each correct answer, lose.5 points for each incorrect answer, and receive zero points for an answer left blank. You cannot receive a negative score on this question. Y x x x + x x = N x x + y x y x y x, y Y The line in R that passes through the points (,, ) and (,, ). x x x Y y y = y z z z 9. [ point] Suppose that A and B are 7 matrices, and suppose that C = A + B. Let x be a vector in R 7 such that Ax = and Bx = 6. Calculate Cx. Solution: Cx = Ax + Bx = 6. [ points] Let A be an n n matrix, and consider the sentence A is invertible. ( ) For each statement below, write Y if it is equivalent to the sentence ( ), and write N if it is not equivalent to ( ). You will receive.5 points for each correct answer, lose.5 points for each incorrect answer, and receive zero points for an answer left blank. You cannot receive a negative score on this question. Y det(a). N The homogeneous linear system Ax = is consistent. Y Every column vector in R n can be expressed as a linear combination of the columns of A. Y The columns of A form a basis for R n. N The reduced echelon form of A has non-pivot columns. N rank A =. Page of
Solutions MAT B Final Exam. [ points] Consider the traffic flow described by the following diagram. The letters A through C label intersections. The arrows indicate the direction of flow (all roads are one-way) and their labels indicate flow in cars per minute. x A x B x x C x 5 Write down a linear system describing the traffic flow, i.e., all constraints on the variables x i, i =,..., 5. (Do not solve the linear system.) Solution: A : x = x + B : x = x + x C : x + x = x 5 + Total : x = 6 + x 5. [ points] Let A be an 8 8 matrix such that the characteristic equation of A is λ 8 λ + λ =. Also recall that I 8 denotes the 8 8 identity matrix. For each statement below, write T if the statement is true, and write F if the statement is false. You will receive.5 points for each correct answer, lose.5 points for each incorrect answer, and receive zero points for an answer left blank. You cannot receive a negative score on this question. F A is invertible. T There exists a nonzero vector x R 8 such that Ax = x. T The eigenvalue λ = of A has multiplicity. F The homogeneous linear system (A I 8 )x = has a nontrivial solution. Page 5 of
MAT B Final Exam. [ points] Write down det(a ), where 5 A = 7. Answer: 8.. [ points] Let A and B be n n invertible matrices. Solve the matrix equation AB T XB + BA T = for the matrix X. Answer: X = (B T ) A BA T B. Page 6 of
MAT B Final Exam Part B: Long Answer Questions For Questions 5, 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. 5. (a) [ 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 + 9x = x + x + 5x = 8x x = x + 8x Solution: We reduce the augmented matrix of the system to the REF: 9 5 R R+R 9 5 8 8 9 R R R 5 R R R 5 It follows that the system is consistent (since the rightmost column is not pivot). Moreover, x, x are basic, x, x are free. The general solution is x = x + x + x = free x = 5x x = free The vector parametric form of the solution is x x + x + x x = x 5x = x + x x x 5 + (b) [ point] Using your solution to part (a), write down the general solution of the following linear system in vector parametric form. x 9 5 x x 8 8 = Hint: Compare the augemented matrices of the two systems. Solution: The second system is the homogeneous version of the previous system. Therefore its general solution is x x x = x + x 5 x x Page 7 of
MAT B Final Exam 6. [ points] Calculate the determinant of the following matrix using the method of cofactor expansion. M = 7 6. Solution: Expanding with respect to the second column, and then with respect to the second row of the matrix, we obtain ( 7 6 = ( ) = ( ) ) + () () = (5 + ( )) =. Page 8 of
7. Consider the matrix B =. MAT B Final Exam (a) [ points] By finding the roots of the characteristic polynomial, show that the eigenvalues of B are and. Solution: Starting with an expansion with respect to the third row, we have λ det(b λi) = λ λ = ( λ) λ λ = ( λ)(λ λ ) = (λ+) (λ ). Thus, the eigenvalues are and. (b) [ points] For each of the eigenvalues of B found in part (a), find a basis of the corresponding eigenspace. There is extra space for this question on the next page. Solution: For the eigenvalue λ =, we have: [ ] B ( )I = R R + R R R R R R R R Thus, the eigenspace is given by x = x x = x x = x x and a basis of this eigenspace is. For the eigenvalue λ =, we have: [ ] B I = R R + R R R + R R R R Thus, the eigenspace is given by x x x = x = x = x, x R, x and a basis of this eigenspace is.. Page 9 of
MAT B Final Exam (c) [ point] Is B a diagonalizable matrix? Justify your answer. Solution: B is not diagonalizable because there does not exist a basis consisting of the eigenvectors of B. Page of
MAT B Final Exam 8. A European country has two private jet rental companies: AeroNocia and GloboWings. In September, of the country s businessmen flew with AeroNocia, while of the businessmen flew with GloboWings. After three months of tough marketing campaign between the two companies, of AeroNocia s customers switched to GloboWings, while of GloboWings customers switched to AeroNocia. (a) [ point] Write down the migration matrix M and the initial state vector x for this problem. Solution: [ ] [ ] M =, x = Another possible migration matrix (if one reverses the order of the companies) is M = [ ] x = [ ]. (b) [ point] What is the market share of each of the companies at the end of the three-month marketing campaign? [ 9 ] 6 Solution: x = M x = 7 6 (c) [ points] Suppose that the same marketing campaign continues for several more three-month time periods, and its effect on customer migration remains the same. Then, in the long run, what is the predicted market share of each company? Put down your final answer in the following blank spots and write your detailed solution below. In the long run, the market share of AeroNocia will be will be., and the market share of GloboWings Solution: To find the steady-state vector, we must find an eigenvector of eigenvalue. We row reduce: [ ] [ ] [ ] [ ] M I = R R +R R R The general solution is thus [ ] x = x. We choose the value of the free variable so that the sum of the entries of x is equal to one: ( + ) x = = x =. Thus the steady-state vector is [ ] x =. Since M is regular stochastic, the long term behaviour is given by the steady-state vector. Thus, in the long term, The maket share of AeroNocia will be, and the market share of GloboWings will be. Page of
MAT B Final Exam 9. [ points] Consider the matrix Find a basis for Nul A. A = 8. 8 5 5 Solution: We row reduce to find an echelon form of A. R R R 5 R 8 R R 5 R R R 5 8 5 5 R R +R 5 Denoting the variables x,..., x 5 as usual, we realize that x, x, x are basic and x, x 5 are free. It follows that the general solution is x = x 5x 5 x = x x 5 x = x 5 and therefore the vector parametric form of the solution is x 5 and the basis for Nul A is x x x x 5 = x + x 5 5, Page of
MAT B Final Exam. [ points] Let Find the inverse of A. Solution: R R R A = 6 We row reduce the super-augmented matrix: R R R R R + R 6 R R+R Thus A is invertible, and its inverse is R R +R 6 A = 6. 6 Page of
MAT B Final Exam. [ points] Determine the dimension of Span{v, v, v, v }, where v =, v =, v =, v =. Solution: Note that Span{v, v, v, v } is equal to Col A where A = [v v v v ]. We now obtain an echelon form of A: We form the matrix whose columns are the given vectors and row reduce: R R +R 6 R R + R R R R R R R R R 6 R 6 Clearly the pivot columns are columns,, and. Therefore the latter columns form a basis for Col A. It follows that the dimension of Span{v, v, v, v } is equal to. Page of