Test Bank for Linear Algebra and its Applications, 6th

Test Bank for Linear Algebra and its Applications, 6th Edition Exam Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the system of equations. 1) x 1) 1 - x2 + 3x3 = -8 2x1 + x3 = 0 x1 + 5x2 + x3 = 40 A) (8, 8, 0) B) (0, -8, -8) C) (-8, 0, 0) D) (0, 8, 0) 2) x 2) 1 + 3x2 + 2x3 = 11 4x2 + 9x3 = -12 x3 = -4 A) (1, -4, 6) B) (-4, 1, 6) C) (-4, 6, 1) D) (1, 6, -4) 3) x 3) 1 - x2 + 8x3 = -107 6x1 + x3 = 17 3x2 - 5x3 = 89 A) (5, -8, -13) B) (5, 8, -13) C) (-5, -8, 13) D) (-5, 8, 13) 4) 4x 4) 1 - x2 + 3x3 = 12 2x1 + 9x3 = -5 x1 + 4x2 + 6x3 = -32 A) (2, 7, -1) B) (2, 7, 1) C) (2, -7, 1) D) (2, -7, -1) 5) x 5) 1 + x2 + x3 = 6 x1 - x3 = -2 x2 + 3x3 = 11 A) (0, 1, 2) B) No solution C) (1, 2, 3) D) (-1, 2, -3) 6) x 6) 1 + x2 + x3 = 7 x1 - x2 + 2x3 = 7 5x1 + x2 + x3 = 11 A) (4, 1, 2) B) (4, 2, 1) C) (1, 4, 2) D) (1, 2, 4) 1 7) x 7) 1 - x2 + x3 = 8 x1 + x2 + x3 = 6 x1 + x2 - x3 = -12 A) (2, -1, -9) B) (2, -1, 9) C) (-2, -1, 9) D) (-2, -1, -9) 8) 5x 8) 1 + 2x2 + x3 = -11 2x1 - 3x2 - x3 = 17 7x1 + x2 + 2x3 = -4 A) (3, 0, -4) B) (0, 6, -1) C) (-3, 0, 4) D) (0, -6, 1) 9) 7x 9) 1 + 7x2 + x3 = 1 x1 + 8x2 + 8x3 = 8 9x1 + x2 + 9x3 = 9 A) (0, 1, 0) B) (0, 0, 1) C) (1, -1, 1) D) (-1, 1, 1) 10) 2x 10) 1 + x2 = 0 x1 - 3x2 + x3 = 0 3x1 + x2 - x3 = 0 A) (0, 1, 0) B) (0, 0, 0) C) No solution D) (1, 0, 0) Determine whether the system is consistent. 11) x 11) 1 + x2 + x3 = 7 x1 - x2 + 2x3 = 7 5x1 + x2 + x3 = 11 A) No B) Yes 12) 5x 12) 1 + 2x2 + x3 = -11 2x1 - 3x2 - x3 = 17 7x1 + x2 + 2x3 = -4 A) No B) Yes 13) 4x 13) 1 - x2 + 3x3 = 12 2x1 + 9x3 = -5 x1 + 4x2 + 6x3 = -32 A) Yes B) No 14) 2x 14) 1 + x2 = 0 x1 - 3x2 + x3 = 0 3x1 + x2 - x3 = 0 A) Yes B) No 2 15) x 15) 1 + x2 + x3 = 6 x1 - x3 = -2 x2 + 3x3 = 11 A) No B) Yes 16) x 16) 1 - x2 + 3x3 = -11 -4x1 + 4x2 - 12x3 = -2 x1 + 3x2 + x3 = -17 A) Yes B) No 17) x 17) 1 + x2 + x3 = 7 x1 - x2 + 2x3 = 7 2x1 + 3x3 = 15 A) No B) Yes 18) x 18) 1 + 3x2 + 2x3 = 11 4x2 + 9x3 = -12 x1 + 7x2 + 11x3 = -11 A) No B) Yes 19) 5x 19) 1 + 2x2 + x3 = -11 2x1 - 3x2 - x3 = 17 7x1 - x2 = 12 A) Yes B) No 20) 5x 20) 2 + x4 = -21 x1 + x2 + 4x3 - x4 = 4 5x1 + x3 + 4x4 = 12 x1 + x2 + 6x3 = 5 A) No B) Yes Determine whether the matrix is in echelon form, reduced echelon form, or neither. 21) 1 3 5 -7 0 1 -4 -4 0 0 1 6 21) A) Reduced echelon form B) Echelon form C) Neither 3 22) 1 4 5 -7 0 1 -4 -5 0 6 1 4 22) A) Reduced echelon form B) Echelon form C) Neither 23) 1 4 5 -7 3 1 -4 -6 0 4 1 3 23) A) Reduced echelon form B) Echelon form C) Neither 24) 1 0 0 -7 1 1 0 1 0 3 1 1 24) A) Reduced echelon form B) Neither C) Echelon form 25) 1 4 1 -7 0 1 -4 7 0 0 0 0 25) A) Reduced echelon form B) Neither C) Echelon form 26) 1 0 5 -4 0 1 -5 -2 0 0 0 0 0 0 0 0 26) A) Neither B) Reduced echelon form C) Echelon form 27) 1 -5 -5 -5 0 0 -2 3 0 0 0 -3 0 0 0 0 27) A) Neither B) Echelon form C) Reduced echelon form 4 Use the row reduction algorithm to transform the matrix into echelon form or reduced echelon form as indicated. 28) Find the echelon form of the given matrix. 28) 1 4 -2 3 -3 -11 9 -5 -2 4 -3 4 A) 1 4 -2 3 0 1 3 4 0 12 -7 10 B) 1 4 -2 3 0 1 3 4 0 0 -43 -38 C) 1 4 -2 3 0 1 3 4 0 0 -19 -2 D) 1 4 -2 3 0 1 3 4 0 0 -43 0 29) Find the reduced echelon form of the given matrix. 29) 1 4 -5 1 2 2 5 -4 -1 4 -3 -9 9 2 2 A) 1 0 3 0 14 0 1 -2 0 -4 0 0 0 1 4 B) 1 4 -5 1 2 0 1 -2 1 0 0 0 0 1 4 C) 1 0 0 0 14 0 1 0 0 -4 0 0 0 1 4 D) 1 4 -5 0 -2 0 1 -2 0 -4 0 0 0 1 4 The augmented matrix is given for a system of equations. If the system is consistent, find the general solution. Otherwise state that there is no solution. 30) 1 -5 -1 0 0 3 30) A) x1 = -1 + 5x2 x2 = 3 x3 is free B) No solution C) x1 = -1 + 5x2 x2 is free D) (-1, 3) 31) 1 2 -3 -9 0 1 4 5 0 0 0 1 31) A) x1 = -19 + 11x3 x2 = 5 - 4x3 x3 = 1 B) x1 = -9 - 2x2 + 3x3 x2 is free x3 is free C) No solution D) x1 = -19 + 11x3 x2 = 5 - 4x3 x3 is free 5 32) 1 2 -3 6 0 1 4 -7 0 0 0 0 32) A) x1 = 6 - 2x2 + 3x3 x2 is free x3 is free B) x1 = 20 + 11x3 x2 = -7 - 4x3 x3 is free C) x1 = 20 + 11x3 x2 = -7 - 4x3 x3 = 0 D) x1 = 6 -2x2 + 3x3 x2 = -7 - 4x3 x3 is free 33) 1 0 6 2 0 1 -2 -3 0 0 0 0 33) A) x1 = 2 - 6x3 x2 = -3 + 2x3 x3 is free B) x1 = 2 - 6x3 x2 is free x3 = 3 2 + 1 2 x2 C) x1 = 2 - 6x3 x2 = -3 + 2x3 x3 = 0 D) No solution 34) 1 4 -2 -3 1 0 0 1 4 -4 -1 -4 -1 -9 11 34) A) x1 = - 4x2 +2x3 + 3 x4 + 1 x2 is free x3 = -4 - 4x4 x4 is free B) x1 = -7 - 4x2 - 5x4 x2 is free x3 = -4 - 4x4 x4 is free C) x1 = -7 - 4x2 - 5x3 x2 = -4 - 4x3 x3 is free D) x1 = -7 - 4x2 - 5x4 x2 is free x3 = -4 - 4x4 x4 = 0 6 35) 1 6 3 -1 2 6 0 0 0 -4 3 4 0 0 0 0 -2 8 35) A) No solution B) x1 = -6x2 - 3x3 + 10 x2 is free x3 = -4 x4 = 3 4 x5 - 1 x5 = -4 C) x1 = -6x2 - 3x3 + 10 x2 is free x3 is free x4 = -4 x5 = -4 D) x1 = -6x2 - 3x3 + x4 - 2x5 + 6 x2 is free x3 is free x4 = 3 4 x5 - 1 x5 = -4 Find the indicated vector. 36) Let u = -6 4 , v = 2 1 . Find u + v. 36) A) -4 5 B) -8 3 C) -2 3 D) -5 6 37) Let u = 9 -8 , v = 7 2 . Find u - v. 37) A) 7 -15 B) 17 5 C) 2 -10 D) 16 -6 38) Let u = -5 5 , v = 1 6 . Find v - u. 38) A) 10 5 B) -4 11 C) 11 -4 D) 6 1 39) Let u = 3 2 . Find 4u. 39) A) 12 -8 B) -12 -8 C) 12 8 D) -12 8 7 40) Let u = 8 -6 . Find 3u. 40) A) 24 -18 B) -24 18 C) 24 18 D) -24 -18 41) Let u = -6 -7 . Find -2u. 41) A) -12 -14 B) -12 14 C) 12 14 D) 12 -14 42) Let u = -7 5 . Find -5u. 42) A) -35 -25 B) -35 25 C) 35 25 D) 35 -25 43) Let u = 2 1 , v = 7 -5 . Find -5u + 2v. 43) A) -15 4 B) -24 5 C) -45 -8 D) 4 -15 Display the indicated vector(s) on an xy-graph. 44) Let u = -3 5 and v = -2 2 . Display the vectors u, v, and u + v on the same axes. 44) A) B) 8 C) D) 45) Let u = 5 -4 Display the vector 2u using the given axes. 45) A) B) 9 C) D) Solve the problem. 46) Let a1 = 6 1 -1 , a2 = -5 1 6 , and b = -39 -1 22 . Determine whether b can be written as a linear combination of a1 and a2. In other words, determine whether weights x1 and x2 exist, such that x1 a1 + x2 a2 = b. Determine the weights x1 and x2 if possible. 46) A) x1 = -4, x2 = 4 B) No solution C) x1 = -4, x2 = 3 D) x1 = -3, x2 = 2 47) Let a1 = 1 2 -3 , a2 = -3 -4 1 , a3 = 2 1 6 , and b = -3 6 -1 . Determine whether b can be written as a linear combination of a1, a2, and a3. In other words, determine whether weights x1, x2, and x3 exist, such that x1 a1 + x2 a2 + x3 a3 = b. Determine the weights x1, x2, and x3 if possible. 47) A) x1 = -2, x2 = -1, x3 = 6 B) x1 = 2, x2 = 1, x3 = - 1 C) No solution D) x1 = -5, x2 = 0, x3 = 1 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 48) A company manufactures two products. For $1.00 worth of product A, the company 48) spends $0.50 on materials, $0.20 on labor, and $0.10 on overhead. For $1.00 worth of product B, the company spends $0.40 on materials, $0.20 on labor, and $0.10 on overhead. Let a = 0.50 0.20 0.10 and b = 0.40 0.20 0.10 . Then a and b represent the "costs per dollar of income" for the two products. Evaluate 500a + 100b and give an economic interpretation of the result. 10 49) A company manufactures two products. For $1.00 worth of product A, the company 49) spends $0.45 on materials, $0.20 on labor, and $0.10 on overhead. For $1.00 worth of product B, the company spends $0.40 on materials, $0.20 on labor, and $0.10 on overhead. Let a = 0.45 0.20 0.10 and b = 0.40 0.20 0.10 . Then a and b represent the "costs per dollar of income" for the two products. Suppose the company manufactures x1 dollars worth of product A and x2 dollars worth of product B and that its total costs for materials are $205, its total costs for labor are $100, and its total costs for overhead are $50. Determine x1 and x2, the dollars worth of each product produced. Include a vector equation as part of your solution. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Compute the product or state that it is undefined. 50) [-7 2 7] 9 0 -3 50) A) [435] B) [-84] C) [-63 0 -21] D) -63 0 -21 51) -1 1 -2 -8 2 -6 5 9 9 51) A) -76 -14 B) -14 -76 C) [-14 -76] D) -1 1 -2 -8 2 -6 5 9 9 52) 52) 3 -7 8 -4 -4 1 -4 5 A) -12 -35 -32 -20 16 5 B) -47 -52 21 C) -12 40 28 -20 -4 1 D) Undefined 11 53) 53) -4 8 1 5 2 8 1 -5 7 A) 4 -30 70 B) Undefined C) 5 39 D) -4 8 -5 -25 14 56 Write the system as a vector equation or matrix equation as indicated. 54) Write the following system as a vector equation involving a linear combination of vectors. 54) 5x1 - 2x2 - x3 = 2 4x1 + 3x3 = -1 A) x1 5 4 + x2 -2 1 + x3 1 3 = 2 -1 B) 5 x1 x2 x3 - 2 x1 x2 x3 - x1 x2 x3 = 2 -1 0 C) x1 5 2 -1 + x2 4 0 3 = 2 -1 0 D) x1 5 4 + x2 -2 0 + x3 -1 3 = 2 -1 55) Write the following system as a matrix equation involving the product of a matrix and a vector on 55) the left side and a vector on the right side. 3x1 + x2 - 2x3 = -4 2x1 - 2x2 = 1 A) x1 x2 x3 2 -2 0 3 1 -2 = -4 1 B) 3 1 -2 2 -2 0 x1 x2 x3 = -4 1 C) 3 1 -2 2 2 1 x1 x2 x3 = -4 1 D) 3 2 1 -2 -2 0 x1 x2 = -4 1 Solve the problem. 56) Let A = 1 -3 2 -2 5 -1 3 -4 -7 and b = b1 b2 b3 . Determine if the equation Ax = b is consistent for all possible b1, b2, b3. If the equation is not consistent for all possible b1, b2, b3, give a description of the set of all b for which the equation is consistent (i.e., a condition which must be satisfied by b1, b2, b3). 56) A) Equation is consistent for all b1, b2, b3 satisfying 7b1 + 5b2 + b3 = 0. B) Equation is consistent for all b1, b2, b3 satisfying 2b1 + b2 = 0. C) Equation is consistent for all possible b1, b2, b3. D) Equation is consistent for all b1, b2, b3 satisfying -3b1 + b3 = 0. 12 57) Let A = 1 -3 2 -2 5 -1 3 -3 -12 and b = b1 b2 b3 . Determine if the equation Ax = b is consistent for all possible b1, b2, b3. If the equation is not consistent for all possible b1, b2, b3, give a description of the set of all b for which the equation is consistent (i.e., a condition which must be satisfied by b1, b2, b3 ). 57) A) Equation is consistent for all b1, b2, b3 satisfying -3b1 + b3 = 0. B) Equation is consistent for all b1, b2, b3 satisfying 9b1 + 6b2 + b3 = 0. C) Equation is consistent for all b1, b2, b3 satisfying -b1 + b2 + b3 = 0. D) Equation is consistent for all possible b1, b2, b3. 58) Find the general solution of the simple homogeneous "system" below, which consists of a single 58) linear equation. Give your answer as a linear combination of vectors. Let x2 and x3 be free variables. -2x1 - 8x2 + 16x3 = 0 A) x1 x2 x3 = x2 -4 1 0 + x3 8 0 1 (with x2, x3 free) B) x1 x2 x3 = x2 4 1 0 + x3 -8 0 1 (with x2, x3 free) C) x1 x2 x3 = -4 x1 x2 x3 - 8 x1 x2 x3 (with x2, x3 free) D) x1 x2 x3 = x2 -4 0 1 + x3 8 1 0 (with x2, x3 free) 13 59) Find the general solution of the homogeneous system below. Give your answer as a vector. 59) x1 + 2x2 - 3x3 = 0 4x1 + 7x2 - 9x3 = 0 -x1 - 4x2 + 9x3 = 0 A) x1 x2 x3 = x3 -3 3 0 B) x1 x2 x3 = x3 3 -3 1 C) x1 x2 x3 = -3 3 1 D) x1 x2 x3 = x3 -3 3 1 60) Describe all solutions of Ax = b, where 60) A = 2 -5 3 -2 6 -5 -4 7 0 and b = -5 2 19 . Describe the general solution in parametric vector form. A) x1 x2 x3 = 7/2 2 1 + x3 -10 -3 0 B) x1 x2 x3 = -5 -3 0 + x3 -1 2 1 C) x1 x2 x3 = -10 -3 0 + x3 7/2 2 1 D) x1 x2 x3 = -10 -3 0 + x3 7/2 2 0 14 61) Suppose an economy consists of three sectors: Energy (E), Manufacturing (M), and Agriculture 61) (A). Sector E sells 70% of its output to M and 30% to A. Sector M sells 30% of its output to E, 50% to A, and retains the rest. Sector A sells 15% of its output to E, 30% to M, and retains the rest. Denote the prices (dollar values) of the total annual outputs of the Energy, Manufacturing, and Agriculture sectors by pe, pm, and pa, respectively. If possible, find equilibrium prices that make each sector's income match its expenditures. Find the general solution as a vector, with pa free. A) pe pm pa = 0.308 pa 0.716 pa pa B) pe pm pa = 0.356 pa 0.686 pa pa C) pe pm pa = 0.607 pa 0.481 pa pa D) pe pm pa = 0.465 pa 0.593 pa pa 62) The network in the figure shows the traffic flow (in vehicles per hour) over several one-way 62) streets in the downtown area of a certain city during a typical lunch time. Determine the general flow pattern for the network. In other words, find the general solution of the system of equations that describes the flow. In your general solution let x4 be free. A) x1 = 600 - x4 x2 = 400 + x4 x3 = 300 - x4 x4 is free x5 = 300 B) x1 = 500 + x4 x2 = 400 - x4 x3 = 300 - x4 x4 is free x5 = 200 C) x1 = 600 - x4 x2 = 400 - x4 x3 = 300 + x4 x4 is free x5 = 300 D) x1 = 600 + x5 x2 = 400 - x5 x3 = 300 - x5 x4 = 300 x5 is free 15 63) Let v1 = 1 -3 -4 , v2 = -3 8 4 , v3 = 2 -2 6 . Determine if the set {v1, v2, v3} is linearly independent. 63) A) Yes B) No 64) Determine if the columns of the matrix A = -2 1 4 4 0 -4 2 4 6 are linearly independent. 64) A) Yes B) No 65) For what values of h are the given vectors linearly independent? 65) -1 1 6 , -4 4 h A) Vectors are linearly independent for h J 24 B) Vectors are linearly dependent for all h C) Vectors are linearly independent for all h D) Vectors are linearly independent for h = 24 66) For what values of h are the given vectors linearly dependent? 66) -1 4 6 , 5 2 -3 , 5 -3 5 , -20 12 h A) Vectors are linearly independent for all h B) Vectors are linearly dependent for all h C) Vectors are linearly dependent for h J -20 D) Vectors are linearly dependent for h = -20 67) Let A = 2 8 -2 3 -5 -3 and u = 2 -1 1 . Define a transformation T: [3 -> [2 by T(x) = Ax. Find T(u), the image of u under the transformation T. 67) A) -6 8 B) 4 -8 -2 6 5 -3 C) 10 -3 -5 D) 16 5 16 68) Let T: [2 -> [2 be a linear transformation that maps u = -3 6 into -15 6 and maps v = 6 -6 into 24 -12 . Use the fact that T is linear to find the image of 3u + v. 68) A) -3 12 B) 9 -6 C) 27 -18 D) -21 6 69) Let A = 1 -3 0 -4 0 2 2 -5 -3 and b = 2 6 0 . Define a transformation T: [3 -> [3 by T(x) = Ax. If possible, find a vector x whose image under T is b. Otherwise, state that b is not in the range of the transformation T. 69) A) -1 -1 0 B) b is not in the range of the transformation T. C) -1 -1 1 D) -1 1 -1 17 70) Let A = 1 -3 2 -3 4 -1 2 -5 3 and b = -5 2 -4 . Define a transformation T: [3 -> [3 by T(x) = Ax. If possible, find a vector x whose image under T is b. Otherwise, state that b is not in the range of the transformation T. 70) A) b is not in the range of the transformation T. B) 2 2 0 C) -10 -5 -5 D) 4 0 -4 Describe geometrically the effect of the transformation T. 71) Let A = 1 0 2 1 . Define a transformation T by T(x) = Ax. 71) A) Horizontal shear B) Projection onto x2-axis C) Vertical shear D) Projection onto x1-axis 72) Let A = 0 0 0 0 1 0 0 0 1 . Define a transformation T by T(x) = Ax. 72) A) Vertical shear B) Horizontal shear C) Projection onto the x2-axis D) Projection onto the x2x3-plane 18 Solve the problem. 73) The columns of I3 = 1 0 0 0 1 0 0 0 1 are e1 = 1 0 0 , e2 = 0 1 0 , e3 = 0 0 1 . Suppose that T is a linear transformation from [3 into [2 such that T( e1) = 3 -2 , T( e2) = 2 0 , and T( e3) = -4 1 . Find a formula for the image of an arbitrary x = x1 x2 x3 in [3. 73) A) T x1 x2 x3 = 3x1 + 2x2- 4x3 -2x1 + x3 B) T x1 x2 x3 = 3x1 + 2x2 - 4x3 2x1 -2x1 + x3 C) T x1 x2 x3 = 3x1- 2x2 2x1 D) T x1 x2 x3 = 3x1- 2x2 2x1 4x2 + x3 Find the standard matrix of the linear transformation T. 74) T: [2 -> [2 rotates points (about the origin) through 7 4 Δ radians (with counterclockwise rotation for a positive angle). 74) A) - 2 2 - 2 2 - 2 2 2 2 B) 1 1 -1 1 C) 3 3 3 3 - 3 3 3 3 D) 2 2 2 2 - 2 2 2 2 75) T: [2 -> [2 first performs a vertical shear that maps e1 into e1 + 3e2, but leaves the vector e2 unchanged, then reflects the result through the horizontal x1-axis. 75) A) 1 0 -3 -1 B) -1 -3 0 1 C) 1 3 0 -1 D) -1 0 3 -1 19 Determine whether the linear transformation T is one-to-one and whether it maps as specified. 76) Let T be the linear transformation whose standard matrix is 76) A = 1 -2 3 -1 3 -4 -5 5 -6 . Determine whether the linear transformation T is one-to-one and whether it maps [3 onto [3. A) Not one-to-one; not onto [3 B) One-to-one; not onto [3 C) Not one-to-one; onto [3 D) One-to-one; onto [3 77) T(x 77) 1, x2, x3) = (-2x2 - 2x3, -2x1 + 9x2 + 5x3, -x1 - 2x3, 3x2 + 3x3) Determine whether the linear transformation T is one-to-one and whether it maps [3 onto [4. A) One-to-one; onto [4 B) Not one-to-one; not onto [4 C) Not one-to-one; onto [4 D) One-to-one; not onto [4 Solve the problem. 78) The table shows the amount (in g) of protein, carbohydrate, and fat supplied by one unit (100 g) of 78) three different foods. Food 1 Food 2 Food 3 Protein 15 35 25 Carbohydrate 45 30 50 Fat 6 4 1 Betty would like to prepare a meal using some combination of these three foods. She would like the meal to contain 15 g of protein, 25 g of carbohydrate, and 3 g of fat. How many units of each food should she use so that the meal will contain the desired amounts of protein, carbohydrate, and fat? Round to 3 decimal places. A) 0.280 units of Food 1, 0.192 units of Food 2, 0.164 units of Food 3 B) 0.360 units of Food 1, 0.204 units of Food 2, 0.055 units of Food 3 C) 0.302 units of Food 1, 0.238 units of Food 2, 0.085 units of Food 3 D) 0.326 units of Food 1, 0.247 units of Food 2, 0.059 units of Food 3 79) The population of a city in 2000 was 400,000 while the population of the suburbs of that city in 79) 2000 was 800,000. Suppose that demographic studies show that each year about 5% of the city's population moves to the suburbs (and 95% stays in the city), while 4% of the suburban population moves to the city (and 96% remains in the suburbs). Compute the population of the city and of the suburbs in the year 2002. For simplicity, ignore other influences on the population such as births, deaths, and migration into and out of the city/suburban region. A) City: 361,000 Suburbs: 737,280 B) City: 361,000 Suburbs: 839,000 C) City: 422,920 Suburbs: 777,080 D) City: 412,000 Suburbs: 788,000 20 Answer Key Testname: UNTITLED1 1) D 2) D 3) B 4) D 5) C 6) D 7) C 8) D 9) B 10) B 11) B 12) B 13) A 14) A 15) B 16) B 17) A 18) A 19) B 20) B 21) B 22) C 23) C 24) B 25) C 26) B 27) B 28) B 29) A 30) B 31) C 32) B 33) A 34) B 35) C 36) A 37) C 38) D 39) C 40) A 41) C 42) D 21 Answer Key Testname: UNTITLED1 43) D 44) D 45) B 46) C 47) C 48) 500a + 100b = 290 120 60 500a + 100b lists the various costs for producing $500 worth of product A and $100 worth of product B, namely $290 for materials, $120 for labor, and $60 for overhead. 49) x1a + x2b = 205 100 50 or x1 0.45 0.20 0.10 + x2 0.40 0.20 0.10 = 205 100 50 x1 = 100, x2 = 400 50) B 51) B 52) B 53) B 54) D 55) B 56) C 57) B 58) A 59) D 60) C 61) B 62) A 63) A 64) B 65) A 66) B 67) A 68) D 69) C 70) A 71) C 72) D 22 Answer Key Testname: UNTITLED1 73) A 74) D 75) A 76) D 77) B 78) D 79) C 23 Exam Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Perform the matrix operation. 1) Let A = -3 5 0 2 . Find 5A. 1) A) -15 5 0 2 B) -15 25 0 2 C) 2 10 5 7 D) -15 25 0 10 2) Let B = -1 4 4 -3 . Find -3B. 2) A) 3 4 4 -3 B) -3 12 12 -9 C) 3 -12 -12 9 D) -3 2 2 -5 3) Let C = 2 -2 10 . Find (1/2) C. 3) A) 2 -1 10 B) 1 -1 5 C) 1 -2 10 D) 4 -4 20 4) Let A = 2 3 2 6 and B = 0 4 -1 6 . Find 3A + B. 4) A) 6 21 3 36 B) 6 13 1 12 C) 6 7 5 12 D) 6 13 5 24 5) Let C = 1 -3 2 and D = -1 3 -2 . Find C - 4D. 5) A) -5 15 -10 B) 5 -6 4 C) 5 -15 10 D) -3 9 -6 6) Let A = -3 2 and B = 1 0 . Find 2A + 3B. 6) A) 0 2 B) -5 4 C) -3 4 D) -6 4 1 7) Let A = 1 2 7 3 -9 -8 and B = 1 7 8 -7 -8 2 . Find A + B. 7) A) 2 9 15 -4 -17 -6 B) 2 9 -15 3 -17 6 C) 2 3 15 -4 -17 -6 D) 0 -5 -1 -1 -1 -11 8) Let A = -2 2 -8 -8 and B = 2 5 -8 4 . Find A - B. 8) A) 0 3 -16 12 B) 0 -3 0 -12 C) -4 -3 0 -12 D) 4 -3 -16 -4 9) Let A = -6 3 9 -4 and B = 0 0 0 0 . Find A + B. 9) A) Undefined B) 6 -3 -9 4 C) -6 3 9 -4 D) 0 0 0 0 Find the matrix product AB, if it is defined. 10) A = -1 3 4 2 , B = -2 0 -1 1 . 10) A) 3 -1 2 -10 B) 2 0 -4 2 C) 2 -6 -3 -1 D) -1 3 -10 2 11) A = 0 -1 4 2 , B = -2 0 -1 1 . 11) A) 0 2 -4 2 B) -8 -4 4 3 C) -1 1 -6 -10 D) 1 -1 -10 2 12) A = 3 -1 3 0 , B = 0 -1 2 6 . 12) A) -3 0 24 -2 B) -2 -9 0 -3 C) 0 1 6 0 D) -9 -2 -3 0 2 13) A = -1 3 5 6 , B = 0 -2 5 1 -3 2 . 13) A) 3 6 -7 -28 1 37 B) 3 -7 1 6 -28 37 C) AB is undefined. D) 0 -6 15 5 -18 12 14) A = 3 -2 1 0 4 -2 , B = 5 0 -2 1 . 14) A) AB is undefined. B) 15 -10 5 -6 8 -4 C) 15 0 0 4 D) 15 -6 -10 8 5 -4 15) A = 0 -2 3 3 , B = -1 3 2 0 -2 1 . 15) A) AB is undefined. B) 0 4 -2 -3 3 9 C) 0 -3 4 3 -2 9 D) 0 -6 -8 0 -6 3 16) A = 1 3 -2 4 0 4 , B = 3 0 -2 1 0 4 . 16) A) AB is undefined. B) -5 -3 16 12 C) -3 -5 12 16 D) 3 -6 0 0 0 16 3 17) A = 1 0 0 3 , B = 5 2 -1 2 -1 2 . 17) A) 5 0 0 0 -3 6 B) 5 2 -1 6 -3 6 C) 6 -3 6 5 2 -1 D) AB is undefined. The sizes of two matrices A and B are given. Find the sizes of the product AB and the product BA, if the products are defined.18) A is 3 × 3, B is 3 × 3. 18) A) AB is 3 × 6, BA is 3 × 6. B) AB is 1 × 1, BA is 1 × 1. C) AB is 6 × 3, BA is 6 × 3. D) AB is 3 × 3, BA is 3 × 3. 19) A is 2 × 1, B is 1 × 1. 19) A) AB is 2 × 1, BA is undefined. B) AB is undefined, BA is 1× 2. C) AB is 1 × 2, BA is 1 × 1. D) AB is 2 × 2, BA is 1 × 1. 20) A is 3 × 1, B is 1 × 3. 20) A) AB is 3 × 3, BA is undefined. B) AB is 1 × 1, BA is 3 × 3. C) AB is 3 × 3, BA is 1 × 1. D) AB is undefined, BA is 1 × 1. 21) A is 4 × 3, B is 4 × 3. 21) A) AB is 4 × 3, BA is 3 × 4. B) AB is 3 × 4, BA is 4 × 3. C) AB is undefined, BA is undefined. D) AB is 4 × 4, BA is 3 × 3. Find the transpose of the matrix. 22) 2 4 -4 0 -7 7 22) A) -7 7 -4 0 2 4 B) 4 0 7 2 -4 -7 C) 4 2 0 -4 7 -7 D) 2 -4 -7 4 0 7 4 23) 9 8 9 8 0 -7 0 -7 23) A) 8 9 8 9 -7 0 -7 0 B) 0 9 -7 8 0 9 -7 8 C) 9 0 8 -7 9 0 8 -7 D) 0 -7 0 -7 9 8 9 8 Decide whether or not the matrices are inverses of each other. 24) 5 3 3 2 and 2 -3 -3 5 24) A) No B) Yes 25) 10 1 -1 0 and 0 1 -1 10 25) A) No B) Yes 26) -2 4 4 -4 and 1 2 1 4 1 2 1 4 26) A) Yes B) No 27) -5 1 -7 1 and 1 2 - 1 2 7 2 - 5 2 27) A) Yes B) No 28) 6 -5 -3 5 and 1 3 1 3 1 5 2 5 28) A) No B) Yes 29) 9 4 4 4 and -0.2 0.2 0.2 -0.45 29) A) Yes B) No 5 30) 9 -2 7 -2 and 0.5 0.5 - 7 4 - 9 4 30) A) Yes B) No 31) -5 -1 6 0 and 0 1 6 -1 5 6 31) A) No B) Yes 32) 2 -1 0 -1 1 -2 1 0 -1 and 1 -1 2 -3 -2 4 -1 1 1 32) A) Yes B) No Find the inverse of the matrix, if it exists. 33) A = 4 1 -5 1 33) A) 5 9 4 9 1 9 - 1 9 B) 1 9 - 1 9 5 9 4 9 C) 4 9 - 1 9 5 9 1 9 D) 1 9 1 9 - 5 9 4 9 34) A = 0 3 -4 -6 34) A) 0 - 1 4 1 3 - 1 2 B) 1 3 0 - 1 2 - 1 4 C) - 1 2 1 4 - 1 3 0 D) - 1 2 - 1 4 1 3 0 6 35) A = 4 0 4 -3 35) A) A is not invertible B) - 1 3 0 1 3 1 4 C) 1 4 0 - 1 3 - 1 3 D) 1 4 0 1 3 - 1 3 36) A = -6 -3 4 2 36) A) 1 5 - 3 10 2 5 - 3 5 B) 1 5 3 10 - 2 5 - 3 5 C) - 2 5 - 3 5 1 5 3 10 D) A is not invertible 37) A = 4 -4 0 3 37) A) 0 1 3 1 4 1 3 B) 1 3 1 3 0 1 4 C) 1 4 - 1 3 0 1 3 D) 1 4 1 3 0 1 3 38) A = 4 -6 -6 0 38) A) 0 - 1 6 - 1 6 - 1 9 B) 0 1 6 1 6 - 1 9 C) - 1 6 - 1 9 0 - 1 6 D) - 1 9 - 1 6 - 1 6 0 7 39) 39) 1 0 0 -1 1 0 1 1 1 A) 1 0 0 1 1 0 -2 -1 1 B) -1 0 0 -1 -1 0 -1 -1 -1 C) 1 -1 1 0 1 -1 0 0 1 D) 1 1 1 0 1 1 0 0 1 Solve the system by using the inverse of the coefficient matrix. 40) 2x 40) 1 + 2x2 = 2 3x1 + 7x2 = 15 A) (3, -2) B) No solution C) (-2, 3) D) (-2, -3) 41) 4x 41) 1 + 2x2 = -4 2x1 = -8 A) (-4, -6) B) (-4, 6) C) No solution D) (6, -4) 42) 8x 42) 1 - 9x2 = 3 -16x1 + 18x2 = -3 A) No solution B) 3 8 - 8 9 x2, x2 C) (3, -3) D) (-2, -2) 43) 3x 43) 1 + 9x2 = 3 2x1 - x2 = -5 A) (2, -1) B) (1, -2) C) (-1, 2) D) (-2, 1) 44) 2x 44) 1 - 6x2 = -6 3x1 + 2x2 = 13 A) (2, 3) B) (3, 2) C) (-2, -3) D) (-3, -2) 45) 10x 45) 1 - 4x2 = -6 6x1 - x2 = 2 A) (1, 4) B) (-1, -4) C) (-4, -1) D) (4, 1) 46) 3x 46) 1 - 6x2 = -3 3x1 + 4x2 = -23 A) (2, 5) B) (-5, -2) C) (5, 2) D) (-2, 5) 47) -5x 47) 1 + 3x2 = 8 2x1 - 4x2 = -20 A) (-6, -2) B) (2, 6) C) (6, 2) D) (-2, -6) 8 Find the inverse of the matrix A, if it exists. 48) A = 4 -2 5 4 -1 4 8 -3 9 48) A) A-1 does not exist. B) A-1 = 1 0 3 4 0 1 -1 0 3 4 0 C) A-1 = 1 0 3 4 0 1 -1 0 0 0 D) A-1 = 4 4 8 -2 -1 -3 5 4 9 49) A = 1 1 1 2 1 1 2 2 3 49) A) A-1 = -1 -1 -1 -2 -1 -1 -2 -2 -3 B) A-1 = -1 1 0 4 -1 -1 -2 0 1 C) A-1 does not exist. D) A-1 = 1 1 1 1 2 1 1 1 2 1 2 1 3 50) A = 1 3 2 1 3 3 2 7 8 50) A) A-1 = -3 10 -3 2 -4 1 -1 1 0 B) A-1 = 1 1 3 1 2 1 1 3 1 3 1 2 1 7 1 8 C) A-1 = -1 -3 -2 -1 -3 -3 -2 -7 -8 D) A-1 does not exist. 9 51) A = 1 0 8 1 2 3 2 5 3 51) A) A-1 does not exist. B) A-1 = -1 0 -8 -1 -2 -3 -2 -5 -3 C) A-1 = 9 -40 16 -3 13 -5 -1 5 -2 D) A-1 = 1 1 2 0 2 5 8 3 3 52) A = 6 -3 1 9 -5 2 3 -2 1 52) A) A-1 = 1 6 1 9 1 1 9 - 1 5 1 2 1 3 - 1 2 1 B) A-1 does not exist. C) A-1 = 1 9 2 9 - 3 1 3 6 5 3 2 - 1 2 1 D) A-1 = 6 9 3 -3 -5 -2 1 2 1 53) A = 0 3 3 -1 0 7 0 6 0 53) A) A-1 does not exist. B) A-1 = - 7 3 - 1 - 7 6 - 1 6 0 1 6 1 3 0 0 C) A-1 = 7 3 - 1 - 7 6 0 0 1 6 1 3 0 - 1 6 D) A-1 = 7 3 0 1 3 - 1 0 0 - 7 6 1 6 - 1 6 10 Determine whether the matrix is invertible. 54) 4 3 7 16 54) A) Yes B) No 55) 5 5 -5 6 2 -6 -2 0 2 55) A) Yes B) No Identify the indicated submatrix. 56) A = 0 1 -3 -8 3 -1 0 5 2 5 -5 0 . Find A12. 56) A) 3 B) 1 C) -8 5 D) 2 5 -5 57) A = 7 4 1 -7 0 -1 0 3 -4 3 4 3 . Find A21. 57) A) 3 4 B) -7 C) 1 -1 -4 D) 4 Find the matrix product AB for the partitioned matrices. 58) A = 4 0 1 2 -1 -3 5 3 7 , B = -2 0 8 5 1 6 2 2 4 -1 0 3 58) A) -4 -1 32 23 -17 -3 14 -1 B) -4 -1 0 3 -12 -3 0 -9 28 -7 0 21 C) -4 -1 32 23 -17 -3 14 -1 D) -8 0 32 20 -5 -6 14 8 -) A = 0 I I F , B = W X Y Z 59) A) 0 Z FY FZ B) Y Z W + FY X + FZ C) Y Z W + YF X + ZF D) X W + XF Z Y + ZF 11 Solve the equation Ax = b by using the LU factorization given for A. 60) A = 3 -1 2 -6 4 -5 9 5 6 , b = 6 -3 2 A = 1 0 0 -2 1 0 3 4 1 3 -1 2 0 2 -1 0 0 4 60) A) x = 22 -7 15 B) x = 49 -38 32 C) x = 10 -2 -13 D) x = 25 -58 51 61) A = 1 2 4 3 -1 -3 -1 -4 2 1 19 3 1 5 -9 7 , b = 2 0 6 2 A = 1 0 0 0 -1 1 0 0 2 3 1 0 1 -3 -2 1 1 2 4 3 0 -1 3 -1 0 0 2 0 0 0 0 1 61) A) x = 28 -6 -2 -2 B) x = 2 -2 10 -8 C) x = 32 -16 124 -38 D) x = 32 16 12 -8 Find an LU factorization of the matrix A. 62) A = 3 -1 -24 13 62) A) A = 1 0 -8 1 3 1 0 -5 B) A = 1 0 8 1 -3 -1 0 -5 C) A = 1 0 3 1 -8 -1 0 5 D) A = 1 0 -8 1 3 -1 0 5 63) A = 5 4 3 10 11 1 10 -1 20 63) A) A = 1 0 0 2 1 0 2 -3 1 3 4 3 0 -3 5 0 0 1 B) A = 1 0 0 2 1 0 2 -3 1 5 4 3 0 3 -5 0 0 -1 C) A = 1 0 0 10 1 0 10 -1 1 5 4 3 0 3 -5 0 0 -1 D) A = 1 0 0 10 1 0 10 -1 1 5 4 3 0 11 1 0 0 20 12 Determine the production vector x that will satisfy demand in an economy with the given consumption matrix C and final demand vector d. Round production levels to the nearest whole number. 64) C = .4 .3 .1 .6 , d = 54 72 64) A) x = 43 49 B) x = 5 23 C) x = 51 5 D) x = 206 231 65) C = .2 .1 .1 .3 .2 .3 .4 .1 .3 , d = 212 322 297 65) A) x = 108 105 91 B) x = 104 217 206 C) x = 723 973 -297 D) x = 480 892 826 Solve the problem. 66) Compute the matrix of the transformation that performs the shear transformation x  Ax for 66) A = 1 0.18 0 1 and then scales all x-coordinates by a factor of 0.60. A) 1.6 0.18 0 2 B) 0.60 0.18 0 1 C) 1 0.18 0 0.60 D) 0.60 0.108 0 1 67) Compute the matrix of the transformation that performs the shear transformation x  Ax for 67) A = 1 0.23 0 1 and then scales all y-coordinates by a factor of 0.60. A) 0.60 0.138 0 1 B) 2 0.23 0 1.6 C) 1 0.23 0 0.60 D) 1 0.138 0 0.60 Find the 3 × 3 matrix that produces the described transformation, using homogeneous coordinates. 68) (x, y)  (x + 3, y + 7) 68) A) 1 0 3 0 1 7 0 0 0 B) 1 0 7 0 1 3 0 0 1 C) 3 0 0 0 7 0 0 0 1 D) 1 0 3 0 1 7 0 0 1 69) Reflect through the x-axis 69) A) -1 0 0 0 -1 0 0 0 1 B) 1 0 0 0 1 0 0 0 1 C) -1 0 0 0 1 0 0 0 1 D) 1 0 0 0 -1 0 0 0 1 13 Find the 3 × 3 matrix that produces the described composite 2D transformation, using homogeneous coordinates. 70) Rotate points through 45° and then scale the x-coordinate by 0.6 and the y-coordinate by 0.4. 70) A) 0.3 2 0.3 2 0 -0.2 2 0.2 2 0 0 0 1 B) 0 -0.6 0 0.4 0 0 0 0 1 C) 0.3 2 -0.3 2 0 0.2 2 0.2 2 0 0 0 1 D) 0.3 -0.2 2 0 0.3 2 0.2 0 0 0 1 71) Translate by (4, 3), and then reflect through the line y = x. 71) A) 0 3 1 4 0 0 0 0 1 B) -1 0 -4 0 -1 -3 0 0 1 C) 0 1 4 1 0 3 0 0 1 D) 0 1 3 1 0 4 0 0 1 Find the 4 × 4 matrix that produces the described transformation, using homogeneous coordinates. 72) Translation by the vector (9, -9, -2) 72) A) 9 0 0 0 0 -9 0 0 0 0 -2 0 0 0 0 1 B) 0 0 0 9 0 0 0 -9 0 0 0 -2 0 0 0 1 C) 1 0 0 -9 0 1 0 9 0 0 1 2 0 0 0 1 D) 1 0 0 9 0 1 0 -9 0 0 1 -2 0 0 0 1 73) Rotation about the y-axis through an angle of 60° 73) A) 0.5 3/2 0 0 - 3/2 0.5 0 0 0 0 1 0 0 0 0 1 B) 3/2 0 0.5 0 0 1 0 0 -0.5 0 3/2 0 0 0 0 1 C) 0.5 0 3/2 0 0 1 0 0 - 3/2 0 0.5 0 0 0 0 1 D) 1 0 0 0 0 0.5 3/2 0 0 - 3/2 0.5 0 0 0 0 1 Determine whether b is in the column space of A. 74) A = 1 2 -3 1 4 -6 -3 -2 5 , b = 6 7 -10 74) A) Yes B) No 75) A = -1 0 2 5 8 -10 -3 -3 6 , b = -5 4 4 75) A) Yes B) No 14 Find a basis for the null space of the matrix. 76) A = 1 0 -3 -2 0 1 7 -4 0 0 0 0 76) A) 1 0 0 , 0 1 0 B) -3 7 1 0 , -2 -4 0 1 C) 3 -7 1 0 , 2 4 0 1 D) 1 0 -3 -2 , 0 1 7 -4 77) A = 1 0 -5 0 -2 0 1 5 0 2 0 0 0 1 1 0 0 0 0 0 77) A) 5 -5 1 0 0 , 2 -2 0 -1 1 B) 1 0 -5 0 -2 , 0 1 5 0 2 C) 1 0 0 0 , 0 1 0 0 , 0 0 1 0 D) -5 5 1 0 0 , -2 2 0 -1 1 Find a basis for the column space of the matrix. 78) B = 1 -2 2 -3 2 -4 7 -2 -3 6 -6 9 78) A) 1 2 -3 , -2 -4 6 B) 1 2 -3 , 2 7 -6 C) 1 0 0 , 0 1 0 D) 2 1 0 0 , 17 3 0 - 4 3 1 15 79) B = 1 0 -4 0 -4 0 1 4 0 5 0 0 0 1 1 0 0 0 0 0 79) A) 1 0 0 0 , 0 1 0 0 , -4 4 0 0 B) 4 -4 1 0 0 , 4 -5 0 -1 1 C) 1 0 0 0 , 0 1 0 0 D) 1 0 0 0 , 0 1 0 0 , 0 0 1 0 The vector x is in a subspace H with a basis Ά = {b1, b2}. Find the Ά-coordinate vector of x. 80) b1 = 1 -2 , b2 = -5 3 , x = 28 -21 80) A) -5 3 B) 3 -5 C) -3 5 D) -4 1 81) b1 = 2 -2 4 , b2 = 6 1 -3 , x = 18 10 -24 81) A) 3 -4 B) -3 4 C) -3 4 0 D) 4 -3 Determine the rank of the matrix. 82) 1 -2 2 -5 2 -4 6 -6 -3 6 -6 15 82) A) 1 B) 2 C) 3 D) 4 83) 1 0 -3 0 4 0 1 -4 0 2 0 0 0 1 1 0 0 0 0 0 83) A) 5 B) 2 C) 3 D) 4 16 Answer Key Testname: UNTITLED2 1) D 2) C 3) B 4) D 5) C 6) C 7) A 8) C 9) C 10) D 11) D 12) B 13) B 14) A 15) B 16) C 17) B 18) D 19) A 20) C 21) C 22) D 23) C 24) B 25) A 26) B 27) A 28) B 29) B 30) B 31) A 32) B 33) B 34) D 35) D 36) D 37) D 38) A 39) A 40) C 41) B 42) A 17 Answer Key Testname: UNTITLED2 43) D 44) B 45) A 46) B 47) B 48) A 49) B 50) A 51) C 52) B 53) C 54) A 55) B 56) C 57) A 58) A 59) B 60) C 61) A 62) D 63) B 64) D 65) D 66) D 67) C 68) D 69) D 70) C 71) D 72) D 73) C 74) A 75) B 76) C 77) A 78) B 79) D 80) B 81) B 82) B 83) C 18 Exam Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Compute the determinant of the matrix by cofactor expansion. 1) 7 8 8 2 9 2 9 8 9 1) A) -353 B) -65 C) 65 D) 1,743 2) 2 1 -1 -2 2 5 -2 1 -3 2) A) -20 B) 0 C) -40 D) 40 3) 2 1 4 4 1 4 2 5 1 3) A) 38 B) -38 C) 142 D) 22 4) 8 -7 4 0 6 3 0 0 -3 4) A) 123 B) -144 C) 144 D) -165 5) 4 0 0 4 -1 0 -1 3 5 5) A) -32 B) -20 C) 8 D) 20 6) -4 2 -2 5 0 -1 2 -2 0 3 0 0 0 -3 1 4 6) A) 0 B) 120 C) -120 D) -30 7) 4 2 -2 5 -4 0 2 2 1 0 0 0 -2 3 7 0 0 0 -1 -5 0 0 0 0 2 7) A) -32 B) -16 C) 0 D) 32 1 8) 4 1 2 -4 -1 -6 8 5 3 8) A) -12 B) 0 C) 48 D) -48 9) 1 2 5 6 12 18 -2 3 -1 9) A) 14 B) -42 C) 84 D) 42 10) 2 1 -3 1 0 5 -1 6 -4 3 5 4 -2 5 1 3 10) A) -1 B) 60 C) 1 D) 0 11) 2 -2 6 4 2 2 7 1 6 -6 15 14 -2 2 -6 1 11) A) 60 B) -120 C) -30 D) -60 Solve using Cramer's rule. 12) 2x 12) 1 + 2x2 = 8 6x1 + x2 = -6 A) (-2, 6) B) (2, -6) C) (-6, -2) D) (6, -2) 13) 3x 13) 1 - 2x2 = 6 3x1 + 2x2 = 42 A) (9, 8) B) (-8, -9) C) (8, 9) D) (-9, 8) 14) 2x 14) 1 + 3x2 = 22 2x1 - 2x2 = 12 A) (-8, -2) B) (-2, 8) C) (2, 8) D) (8, 2) 2 Determine the values of the parameter s for which the system has a unique solution, and describe the solution. 15) 5sx 15) 1 + 4x2 = -3 5x1 + sx2 = 4 A) s J ± 4; x1 = -3s - 16 and x2 = 4s + 3 B) s J ± 5; x1 = -3s - 16 5(s - 5)(s + 5) and x2 = 4s+ 3 5(s - 5)(s + 5) C) s J ± 2; x1 = -3s - 16 5(s - 2)(s + 2) and x2 = 4s+ 3 (s - 2)(s + 2) D) s J 2; x1 = -3s + 16 5(s - 2)(s + 2) and x2 = 4s - 3 (s - 2)(s + 2) 16) sx 16) 1 - 4sx2 = 3 3x1 - 12sx2 = 5 A) s J 1; x1 = 4 12(s - 1)(s + 1) and x2 = 9 - 5s 12(s - 1)(s + 1) B) s J 1; x1 = 14 3(s + 1) and x2 = 9 + 5s 12s(s + 1) C) s J ± 1; x1 = 4 3(s + 1) and x2 = 9 - 5s 12s(s + 1) D) s J 0, 1; x1 = 4 3(s - 1) and x2 = 9 - 5s 12s(s - 1) Calculate the area of the parallelogram with the given vertices. 17) (0, 0), (4, 7), (13, 11), (9, 4) 17) A) 94 B) 46 C) 55 D) 47 18) (-1, -2), (2, 7), (7, 2), (10, 11) 18) A) 60 B) 120 C) 59 D) 66 3 Answer Key Testname: UNTITLED3 1) B 2) C 3) A 4) B 5) B 6) B 7) D 8) C 9) C 10) D 11) B 12) A 13) C 14) D 15) C 16) D 17) D 18) A 4 Exam Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the problem. 1) Determine which of the following sets is a subspace of Pn for an appropriate value of n. 1) A: All polynomials of the form p(t) = a + bt 2, where a and b are in [ B: All polynomials of degree exactly 4, with real coefficients C: All polynomials of degree at most 4, with positive coefficients A) B only B) A only C) A and B D) C only 2) Determine which of the following sets is a vector space. 2) V is the line y = x in the xy-plane: V = x y : y = x W is the union of the first and second quadrants in the xy-plane: W = x y : y L 0 U is the line y = x + 1 in the xy-plane: U = x y : y = x + 1 A) V only B) U and V C) U only D) W only 3) Let H be the set of all polynomials having degree at most 4 and rational coefficients. Determine 3) whether H is a vector space. If it is not a vector space, determine which of the following properties it fails to satisfy. A: Contains zero vector B: Closed under vector addition C: Closed under multiplication by scalars A) H is not a vector space; does not contain zero vector B) H is a vector space. C) H is not a vector space; not closed under multiplication by scalars D) H is not a vector space; not closed under vector addition 4) Let H be the set of all polynomials of the form p(t) = a + bt 2 where a and b are in [ and b > a. Determine whether H is a vector space. If it is not a vector space, determine which of the following properties it fails to satisfy. A: Contains zero vector B: Closed under vector addition C: Closed under multiplication by scalars 4) A) H is not a vector space; not closed under multiplication by scalars and does not contain zero vector B) H is not a vector space; does not contain zero vector C) H is not a vector space; not closed under multiplication by scalars D) H is not a vector space; not closed under vector addition 1 5) Let H be the set of all points of the form (s, s-1). Determine whether H is a vector space. If it is not 5) a vector space, determine which of the following properties it fails to satisfy. A: Contains zero vector B: Closed under vector addition C: Closed under multiplication by scalars A) H is not a vector space; not closed under vector addition B) H is not a vector space; fails to satisfy all three properties C) H is a vector space. D) H is not a vector space; does not contain zero vector 6) Let H be the set of all points in the xy-plane having at least one nonzero coordinate: 6) H = x y : x, y not both zero . Determine whether H is a vector space. If it is not a vector space, determine which of the following properties it fails to satisfy: A: Contains zero vector B: Closed under vector addition C: Closed under multiplication by scalars A) H is not a vector space; fails to satisfy all three properties B) H is not a vector space; does not contain zero vector C) H is not a vector space; not closed under vector addition D) H is not a vector space; does not contain zero vector and not closed under multiplication by scalars If the set W is a vector space, find a set S of vectors that spans it. Otherwise, state that W is not a vector space. 7) W is the set of all vectors of the form a + 6b 5b 6a - b -a , where a and b are arbitrary real numbers. 7) A) 1 5 6 -1 , 6 0 -1 0 B) 1 0 6 0 , 6 0 -1 0 , 0 5 0 -1 C) 1 0 6 -1 , 6 5 -1 0 D) Not a vector space 2 8) W is the set of all vectors of the form a - 2b 5 3a + b -a - b , where a and b are arbitrary real numbers. 8) A) 1 0 3 -1 , -2 5 1 -1 B) Not a vector space C) 1 5 3 -1 , -2 0 1 -1 D) 1 0 3 -1 , -2 0 1 -1 , 0 5 0 0 Solve the problem. 9) Find all values of h such that y will be in the subspace of [3 spanned by v1, v2, v3 if v1 = 1 2 -4 , v2 = 3 4 -8 , v3 = -1 0 0 , and y = 7 6 h . 9) A) all h J -12 B) h = -12 or 0 C) h = -28 D) h = -12 Determine whether the vector u belongs to the null space of the matrix A. 10) u = 2 4 1 , A = -2 3 -8 -3 -1 10 10) A) No B) Yes 11) u = -2 -1 1 , A = -1 -1 -3 -3 -4 -10 3 -2 6 11) A) No B) Yes Find an explicit description of the null space of matrix A by listing vectors that span the null space. 12) A = 1 -2 -2 -2 0 1 1 4 12) A) 2 1 0 0 , 0 -1 1 0 , -6 -4 0 1 B) 2 -1 1 0 , 2 -4 0 1 C) 2 1 0 0 , 2 -1 1 0 , 2 -4 0 1 D) 0 -1 1 0 , -6 -4 0 1 3 13) A = 1 -2 3 -3 -1 -2 5 -5 4 -4 -1 3 -2 1 -5 13) A) -5 -1 1 0 0 , 7 -2 0 1 0 , 12 6 0 0 1 B) -5 -1 1 0 0 , 7 2 0 1 0 , 13 6 0 0 1 C) 1 0 5 -7 -13 , 0 1 1 -2 -6 D) 2 1 0 0 0 , -3 -1 1 0 0 , 3 2 0 1 0 , -1 6 0 0 1 Find a matrix A such that W = Col A. 14) W = 2s - 2t 6t -3s + t : s, t in [ 14) A) 2 6 -3 2 0 1 B) 2 0 -3 -2 6 1 C) 2 -2 0 6 -3 1 D) 2 2 6 0 -3 1 15) W = 2r - t 4r - s + 3t s + 3t r - 5s + t : r, s, t in [ 15) A) 2 -1 4 3 1 3 1 -5 B) 2 0 -1 4 -1 3 0 1 3 1 -5 1 C) 2 4 0 1 0 -1 1 -5 -1 3 3 1 D) 0 2 -1 4 -1 3 0 1 3 1 -5 1 For the given matrix A, find k such that Nul A is a subspace of [k and find m such that Col A is a subspace of [m. 16) A = 1 -2 0 6 -4 5 -1 -3 -3 3 16) A) k = 5, m = 2 B) k = 2, m = 2 C) k = 5, m = 5 D) k = 2, m = 5 4 17) A = 4 0 0 -1 1 -7 2 6 -5 -1 0 3 -3 -4 4 -1 -4 4 17) A) k = 6, m = 3 B) k = 3, m = 6 C) k = 6, m = 6 D) k = 3, m = 3 Determine if the vector u is in the column space of matrix A and whether it is in the null space of A. 18) u = 5 -3 -5 , A = 1 -3 4 -1 0 -5 3 -3 6 18) A) Not in Col A, in Nul A B) Not in Col A, not in Nul A C) In Col A and in Nul A D) In Col A, not in Nul A 19) u = -1 3 1 -2 , A = 1 0 3 -2 -1 -4 3 -3 0 -1 3 6 19) A) Not in Col A, in Nul A B) In Col A, not in Nul A C) Not in Col A, not in Nul A D) In Col A and in Nul A Determine which of the sets of vectors is linearly independent. 20) A: The set p1, p2, p3 where p1(t) = 1, p2(t) = t 2, p3(t) = 4 + 4t B: The set p1, p2, p3 where p1(t) = t, p2(t) = t 2, p3(t) = 4t + 4t 2 C: The set p1, p2, p3 where p1(t) = 1, p2(t) = t 2, p3(t) = 4 + 4t + t 2 20) A) A and C B) C only C) B only D) all of them E) A only 21) A: The set sin t , tan t in C[0, 1] 21) B: The set sin t cos t , cos 2t in C[0, 1] C: The set cos2 t , 1 + cos 2t in C[0, 1] A) A only B) B only C) C only D) A and C E) A and B 5 Determine whether {v1, v2, v3} is a basis for [3. 22) v1 = 1 -3 -5 , v2 = -3 8 -2 , v3 = 2 -2 -1 22) A) Yes B) No 23) v1 = -2 4 -6 , v2 = 1 0 -2 , v3 = 4 -4 2 23) A) Yes B) No Solve the problem. 24) Let v1 = 1 3 -3 , v2 = -3 1 -2 , v3 = 8 4 -2 , and H = Span v1, v2, v3 . Note that v3 = 2v1 - 2v2. Which of the following sets form a basis for the subspace H, i.e., which sets form an efficient spanning set containing no unnecessary vectors? A: v1, v2, v3 B: v1, v2 C: v1, v3 D: v2, v3 24) A) A only B) B only C) B and C D) B, C, and D Find a basis for the column space of the matrix. 25) Find a basis for Col B where 25) B = 1 1 0 -4 0 0 0 0 1 -1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 . A) 1 0 0 0 0 , 0 1 0 0 0 , 0 0 1 0 0 , 0 0 0 1 0 B) 1 0 0 0 0 , 0 0 1 0 0 , 0 0 0 1 0 C) 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 D) 1 0 0 0 0 , 1 0 0 0 0 , 0 1 0 0 0 , -4 -1 0 0 0 , 0 0 1 0 0 , 0 0 0 1 0 6 26) Let A = -1 3 7 2 0 1 -2 -7 -1 3 2 -4 -9 -5 1 3 -6 -11 -9 -1 and B = 1 -3 -7 -2 0 0 1 0 1 3 0 0 5 -3 -5 0 0 0 0 0 . It can be shown that matrix A is row equivalent to matrix B. Find a basis for Col A. 26) A) -1 1 2 3 , 3 -2 -4 -6 , 2 -1 -5 -9 B) 1 0 0 0 , -3 1 0 0 , -7 0 5 0 C) -1 1 2 3 , 3 -2 -4 -6 , 7 -7 -9 -11 , 2 -1 -5 -9 , 0 3 1 -1 D) -1 1 2 3 , 3 -2 -4 -6 , 7 -7 -9 -11 Determine whether the set of vectors is a basis for [3. 27) Given the set of vectors 1 0 0 , 0 1 2 , decide which of the following statements is true: A: Set is linearly independent and spans [3. Set is a basis for [3. B: Set is linearly independent but does not span [3. Set is not a basis for [3. C: Set spans [3 but is not linearly independent. Set is not a basis for [3. D: Set is not linearly independent and does not span [3. Set is not a basis for [3. 27) A) D B) C C) B D) A 28) Given the set of vectors 1 0 0 , 0 1 0 , 0 0 1 , 0 1 1 , decide which of the following statements is true: A: Set is linearly independent and spans [3. Set is a basis for [3. B: Set is linearly independent but does not span [3. Set is not a basis for [3. C: Set spans [3 but is not linearly independent. Set is not a basis for [3. D: Set is not linearly independent and does not span [3. Set is not a basis for [3. 28) A) C B) D C) A D) B Find the vector x determined by the given coordinate vector [x]B and the given basis B. 29) B = -3 1 , 0 1 , [x]B = -6 5 29) A) -3 2 B) 18 -1 C) -15 -1 D) 18 5 7 30) B = 1 -3 2 , -3 8 -4 , 2 -2 -5 , [x]B = 2 4 -1 30) A) -12 28 -7 B) 0 12 7 C) -10 28 3 D) -4 6 5 Find the coordinate vector [x]B of the vector x relative to the given basis B. 31) b1 = 3 4 , b2 = 4 -5 , x = 25 -8 , and B = b1, b2 31) A) 3 5 B) 43 140 C) 3 4 D) 25 -8 32) b1 = 2 3 -3 , b2 = 1 -5 -1 , x = -8 1 11 , and B = b1, b2 32) A) -3 -2 B) -3 -3 C) -15 -29 23 D) -24 -2 -44 Use coordinate vectors to determine whether the given polynomials are linearly dependent in P2. Let B be the standard basis of the space P2 of polynomials, that is, let B = 1, t, t 2 . 33) 1 + 2t, 3 + 6t 2, 1 + 3t + 4t 2 33) A) Linearly independent B) Linearly dependent 34) 1 + 2t + t 2, 3 - 9t 2, 1 + 4t + 5t 2 34) A) Linearly independent B) Linearly dependent Solve the problem. 35) Let H = a + 3b + 2d c + d -3a - 9b + 4c - 2d -c - d : a, b, c, d in [ Find the dimension of the subspace H. 35) A) dim H = 2 B) dim H = 1 C) dim H = 4 D) dim H = 3 Find the dimensions of the null space and the column space of the given matrix. 36) A = 1 -3 2 3 0 -2 -3 4 -4 1 36) A) dim Nul A = 4, dim Col A = 1 B) dim Nul A = 3, dim Col A = 2 C) dim Nul A = 2, dim Col A = 3 D) dim Nul A = 3, dim Col A = 3 8 37) A = 1 -2 3 1 0 5 -4 0 0 1 -6 2 -2 0 0 0 0 0 0 1 3 0 0 0 0 0 0 0 37) A) dim Nul A = 4, dim Col A = 3 B) dim Nul A = 2, dim Col A = 5 C) dim Nul A = 3, dim Col A = 4 D) dim Nul A = 5, dim Col A = 2 Solve the problem. 38) Determine which of the following statements is false. 38) A: The dimension of the vector space P6 of polynomials is 7. B: Any line in [3 is a one-dimensional subspace of [3. C: If a vector space V has a basis B = b1, ......, b4 , then any set in V containing 5 vectors must be linearly dependent. A) B B) A and B C) C D) A 39) Determine which of the following statements is true. 39) A: If V is a 7-dimensional vector space, then any set of exactly 7 elements in V is automatically a basis for V. B: If there exists a set v1, ......, v4 that spans V, then dim V = 4. C: If H is a subspace of a finite-dimensional vector space V, then dim H K dim V. A) A B) A and C C) B D) C Assume that the matrix A is row equivalent to B. Find a basis for the row space of the matrix A. 40) A = 1 3 -4 0 1 2 4 -5 -4 2 1 -5 0 -3 2 -3 -1 8 3 -4 , B = 1 3 -4 0 1 0 -2 3 -4 0 0 0 -8 13 1 0 0 0 0 0 40) A) {(1, 0, 0, 0), (3, -2, 0, 0), (-4, 3, -8, 0)} B) {(1, 3, -4, 0, 1), (0, -2, 3, -4, 0), (0, 0, -8, 13, 1), (0, 0, 0, 0, 0)} C) {(1, 3, -4, 0, 1), (2, 4, -5, -4), 2, (1, -5, 0, -3, 2), (-3, -1, 8, 3, -4)} D) {(1, 3, -4, 0, 1), (0, -2, 3, -4, 0), (0, 0, -8, 13, 1)} Solve the problem. 41) If the null space of a 7 × 6 matrix is 5-dimensional, find Rank A, Dim Row A, and Dim Col A. 41) A) Rank A = 1, Dim Row A = 1, Dim Col A = 5 B) Rank A = 1, Dim Row A = 5, Dim Col A = 5 C) Rank A = 2, Dim Row A = 2, Dim Col A = 2 D) Rank A = 1, Dim Row A = 1, Dim Col A = 1 42) If A is a 5 × 9 matrix, what is the smallest possible dimension of Nul A? 42) A) 5 B) 4 C) 9 D) 0 9 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 43) A mathematician has found 5 solutions to a homogeneous system of 26 equations in 30 43) variables. The 5 solutions are linearly independent and all other solutions can be constructed by adding together appropriate multiples of these 5 solutions. Will the system necessarily have a solution for every possible choice of constants on the right side of the equation? Explain. 44) Suppose a nonhomogeneous system of 11 linear equations in 14 unknowns has a solution 44) for all possible constants on the right side of the equation. Is it possible to find 4 nonzero solutions of the associated homogeneous system that are linearly independent? Explain. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the new coordinate vector for the vector x after performing the specified change of basis. 45) Consider two bases B = b 45) 1, b2 and C = c1, c2 for a vector space V such that b1 = c1 - 4c2 and b2 = 5c1 + 2c2. Suppose x = b1 + 3b2. That is, suppose [x]B = 1 3 . Find [x]C. A) 16 2 B) -11 11 C) 8 -10 D) 15 2 46) Consider two bases B = b 46) 1, b2, b3 and C = c1, c2, c3 for a vector space V such that b1 = c1 + 4c3, b2 = c1 + 6c2 - c3, and b3 = 5c1 - c2. Suppose x = b1 + 3b2 + b3. That is, suppose [x]B = 1 3 1 . Find [x]C. A) 9 21 -3 B) 9 17 7 C) 5 18 2 D) 9 17 1 Find the specified change-of-coordinates matrix. 47) Let B = b1, b2 and C = c1, c2 be bases for [2, where b1 = -1 5 , b2 = -2 4 , c1 = 1 3 , c2 = -4 -10 . Find the change-of-coordinates matrix from B to C. 47) A) -1 -2 4 5 B) 5 3 - 6 - 4 3 5 C) 15 18 4 5 D) 1 -4 3 -10 10 48) Consider two bases B = b 48) 1, b2 and C = c1, c2 for a vector space V such that b1 = c1 - 6c2 and b2 = 4c1 - 3c2. Find the change-of-coordinates matrix from B to C. A) 1 4 -6 -3 B) 1 -6 4 -3 C) 0 4 -6 -3 D) 1 4 6 3 Find the indicated sums of the signals in the Table 49) Find the indicated sum of the signals. 49) Λ+Α A) (..., 0, 2, 0, 2, 0, 0, 2, ...) B) (..., 1, 1, 1, 2, 2, 2, 2, ...) C) (..., -1, 1, -1, 2, -1, 1, -1, ...) D) (..., 0, 0, 0, 2, 1, 1, 1, ...) 50) Find the indicated sum of the signals. 50) řΈ+F A) (..., 2, -1, 1, 3, 1, 1, 2, ..) B) (..., 0, 0, 0, 3, 1, 1, 1, ...) C) (..., 2, -1, 1, 3, 4, 4, 5, ..) D) (..., -1, 1, -1, 4, -1, 1, -1, ...) 51) Which signal(s) are in the kernal of I-S 2? 51) A) Έ B) Έ and Λ C) Λ and ΅ D) ΅ 52) Find a nonzero signal in the kernal of T({x})={xk-xk 52) -1-xk-2} A) ΅ B) F C) Α D) ȱΛ 11 53) Construct a linear time invariant transformation that has the signal {x}= 4 5 k in its kernel. 53) A) I4 5 S B) I+ 5 4 S C) I+ 4 5 S D) I5 4 S 54) Let W= {xk} | xk= r if k is a multiple of 2 -r if k is not a multiple of 2 where r can be any real number. A typical signal in W looks like (..., -r, r, -r, r, -r, r, -r, ...). Is W is a subspace of S? Explain. 54) A) Yes. For each signal {xk} in W, there exists a signal {-xk} in W, such that {xk}+{-xk}={0}. B) No. The zero vector is not in W. C) Yes. The zero vector is in W, the sum of any two vectors in W is also in W, and any scalar multiple of a vector in W is also in W. D) Yes. The sum of any two vectors in W is also in W and any scalar multiple of a vector in W is also in W. 55) Let W= {xk} | xk= 0 if k<0 r+k if kL0 where r can be any real number. A typical signal in W looks like (..., 0, 0, 0, r, r+1, r+2, r+3, ...). Is W is a subspace of S? Explain. 55) A) No. The zero vector is not in W. B) Yes. For each signal {xk} in W, there exists a signal {-xk} in W, such that {xk}+{-xk}={0}. C) Yes. The zero vector is in W, the sum of any two vectors in W is also in W, and any scalar multiple of a vector in W is also in W. D) Yes. The sum of any two vectors in W is also in W and any scalar multiple of a vector in W is also in W. 56) Let W= {xk} | xk= r if k is a multiple of 2 2r if k is not a multiple of 2 where r can be any real number. A typical signal in W looks like (..., 2r, r, 2r, r, 2r, r, 2r, ...). Find a basis for the subspace W. What is the dimension of W? 56) A) Basis: {Λ+S -1ǻΈ)} Dimension: Q B) Basis: {3Λ+S(΅)} Dimension: 1 C) Basis: {Α+Ř΅} Dimension: Q D) Basis: {3Λ+΅} Dimension: 1 12 57) Let W= {xk} | xk= -rk if k<0 rk if kL0 where each rk can be any real number. A typical signal in W looks like (..., -r3, -r2, -r1, r0, r1 r2, r3, ...). Describe an infinite linearly independent subset of the subspace W. Does this establish that W is infinite dimensional? 57) A) {S nǻΈ)+S -nǻΈ) | n is a nonnegative integer}. An infinite set of vectors can span a finite dimensional subspace. Therefore, it is unknown whether W is infinite dimensional. B) {S nǻΈ) | n is an integer}. An infinite set of vectors can span a finite dimensional subspace. Therefore, it is unknown whether W is infinite dimensional. C) {S nǻΈ)-S -nǻΈ) | n is a nonnegative integer}. An infinite linearly independent set of vectors spans an infinite dimensional vector space. If that vector space is a subspace of W, then W must also be infinite dimensional. D) {S nǻΈ) | n is a nonnegative integer}. An infinite linearly independent set of vectors spans an infinite dimensional vector space. If that vector space is a subspace of W, then W must also be infinite dimensional. 58) Verify that the signals are solutions of the accompanying difference equation. 58) 2 k, (-3)k; yk+2+yk+1-6yk=0 A) 2 k+2+2 k+1-6·2 k=0 2 k(2 2+2 1-6)=0 2 k(0)=0 2 k+2+2 k+1-6·2 k=0 2 0+2+2 0+1-6·2 0=0 2 2+2 1-6=0 0=0 B) (-3)k+2+(-3)k+1-6·(-3)k=0 (-3)k((-3)2+(-3)1-6)=0 (-3)k(0)=0 (-3)k+2+(-3)k+1-6·(-3)k=0 (-3)0+2+(-3)0+1-6·(-3)0=0 (-3)2+(-3)1-6=0 0=0 C) 2 k+2+2 k+1-6·2 k=0 2 k(2 2+2 1-6)=0 2 k(0)=0 (-3)k+2+(-3)k+1-6·(-3)k=0 (-3)k((-3)2+(-3)1-6)=0 (-3)k(0)=0 D) 2 k+2+2 k+1-6·2 k=0 2 0+2+2 0+1-6·2 0=0 2 2+2 1-6=0 0=0 (-3)k+2+(-3)k+1-6·(-3)k=0 (-3)0+2+(-3)0+1-6·(-3)0=0 (-3)2+(-3)1-6=0 0=0 13 59) Assume the signals listed are solutions of the given difference equation. Determine if the signals 59) form a basis for the the solution space of the equation. (-1)k, 1 k, (-5)k; yk+3+5yk+2-yk+1-5yk=0 A) No. The signals are all linearly independent but do not span the solution space. B) Yes. The signals are all linearly independent and span the solution space. C) No. The signals are neither linearly independent, nor span the solution space. D) Yes. The signals are linearly dependent and span the solution space. 60) Assume the signals listed are solutions of the given difference equation. Determine if the signals 60) form a basis for the the solution space of the equation. (-2)k, (-3)k; yk+3+2yk+2-9yk+1-18yk=0 A) No. The signals are neither linearly independent, nor span the solution space. B) Yes. The signals are linearly dependent and span the solution space. C) Yes. The signals are all linearly independent and span the solution space. D) No. The signals are all linearly independent but do not span the solution space. 61) Find a basis for the solution space of the given difference equation. 61) yk+2-yk+1+ 6 25 yk=0 A) 3 k and 5 k B) 6 k and 1 25 k C) 2 k and 3 k D) 2 5 k and 3 5 k 62) Find a basis for the solution space of the given difference equation. 62) yk+2-16yk=0 A) 2 k and (-8)k B) (-2)k and 8 k C) 2 k and 8 k D) 4 k and (-4)k 63) The Pell Sequence can be viewed as the sequence of numbers where each number is the sum of 63) twice the previous number and the number before that. It can be described by the homogeneous difference equation yk+2-2yk+1-yk=0 with initial conditions y0=0 and y1=1. Find the general solution of the Pell sequence. A) yk= 1 2 2 1+ 2 k - 1 2 2 1- 2 k B) yk= 1+ 2 2 2 k - 1- 2 2 2 k C) yk= 1+ 2 k - 1 - 2 k D) yk= 2+ 2 4 1+ 2 k + 2- 2 4 1- 2 k 14 64) The Pell Sequence can be viewed as the sequence of numbers where each number is the sum of 64) twice the previous number and the number before that. It can be described by the homogeneous difference equation yk+2-2yk+1-yk=0 with initial conditions y0=0 and y1=1. Find the solution to the difference equation if the initial conditions are changed to y0=1 and y1=2. A) yk= 2+ 2 4 1+ 2 k + 2- 2 4 1- 2 k B) yk= 1+ 2 2 2 k - 1- 2 2 2 k C) yk= 1 2 2 1+ 2 k - 1 2 2 1- 2 k D) yk= 1+ 2 k - 1- 2 k 65) A simple model if the national economy can be described by the difference equation 65) Yk+2-a(1+b)Yk+1+abYk=1. Here Yk is the total national income during year k, a is a constant less than 1, called the marginal propensity to consume, and b is a positive constant of adjustment that describes how changes in consumer spending affect the annual rate of private investment. Find the general solution when a=0.75 and b=0.3. [Hint: First find a particular solution of the form Yk=T, where T is a constant called the equilibrium level o f national income.] A) Yk=c1(0.7)k+c2(0.275)k+1 B) Yk=c1(0.6)k+c2(0.375)k+4 C) Yk=c1(0.6)k+c2(0.375)k D) Yk=c1(0.7)k+c2(0.275)k+5 66) When a signal is produced from a sequence of measurements made on a process ( a chemical 66) reaction, a flow of heat through a tube, a moving robot arm, etc.), the signal usually contains random noise produced by measurement errors. A standard method of preprocessing the data to reduce the noise is to smooth or filter the data. One simple filter is a moving average that replaces each yk by its average with the two adjacent values. 1 3 yk+1+ 1 3 yk+ 1 3 yk-1=zk, for k=1, 2, ... Suppose a signal yk, for k=0, ..., 10, is 10, 7, 4, 1, 7, 10, 7, 4, 1, 1, 7. Use the filter to compute z1, ..., z9. A) 7, 4, 4, 6, 8, 7, 3, 2, 3 B) 8.5, 5.5, 2.5, 4, 8.5, 5.5, 2.5, 1, 4 C) 7, 5, 4, 6, 7, 8, 4, 5, 5 D) 8.5, 6, 4, 2.5, 2, 2.5, 7, 8.5, 10 15 67) Show that the given signal is a solution of the difference equation. Then find the general solution 67) of the difference equation. yk=k 2; yk+2+4yk+1-5yk=12k+8 A) (k+2)2+4(k+1)2-5k 2=12k+8 (0+2)2+4(0+1)2-5*0 2=12*0+8 8=8 yk=k 2+c1(-5)k+c2 B) (k+2)2+4(k+1)2-5k 2=12k+8 (k 2+4k+4)+(4k 2+8k+4)-5k 2=12k+8 12k+8=12k+8 yk=c1(-5)k+c2 C) (k+2)2+4(k+1)2-5k 2=12k+8 (0+2)2+4(0+1)2-5*0 2=12*0+8 8=8 yk=c1(-5)k+c2 D) (k+2)2+4(k+1)2-5k 2=12k+8 (k 2+4k+4)+(4k 2+8k+4)-5k 2=12k+8 12k+8=12k+8 yk=k 2+c1(-5)k+c2 68) Show that the given signal is a solution of the difference equation. Then find the general solution 68) of the difference equation. yk=2+k; yk+2+3yk+1+2yk=17+6k A) (4+k)+3(3+k)+2(2+k)=17+6k 4+k+9+3k+4+2k=17+6k 17+6k=17+6k yk=2 k+c1(-1)k+c2(-2)k B) (4+k)+3(3+k)+2(2+k)=17+6k 4+k+9+3k+4+2k=17+6k 17+6k=17+6k yk=2+k+c1(-1)k+c2(-2)k C) (4+k)+3(3+k)+2(2+k)=17+6k 4+k+9+3k+4+2k=17+6k 17+6k=17+6k yk=c1(-1)k+c2(-2)k D) (2+k)+3(3+k)+2(2+k)=15+6k 2+k+9+3k+4+2k=15+6k 15+6k=15+6k yk=c1(-1)k+c2(-2)k 69) Is the following difference equation of order 3? Explain. 69) yk+3-25yk+1=0 A) No. There are only two y-terms; therefore, the equation is of order 3. B) Yes. The equation has three terms, including the constant 0. C) No. Since the equation holds for all k, replacing k+1 with k shows that the equation is of order 2. D) Yes. An nth order homogenous difference equation is given by a0yk+n+...+ anyn= zk. 16 70) Write the difference equation as a first-order system, xk 70) +1=Axk, for all k. yk+3+6yk+2-2yk=0 A) xk+1=Ax, where A= 0 1 0 0 0 1 -6 0 2 and xk= yk+1 yk+2 yk+3 B) xk+1=Ax, where A= 0 1 0 0 0 1 6 0 -2 and xk= yk yk+1 yk+2 C) xk+1=Ax, where A= 0 1 0 0 0 1 2 0 -6 and xk= yk+1 yk+2 yk+3 D) xk+1=Ax, where A= 0 1 0 0 0 1 2 0 -6 and xk= yk yk+1 yk+2 17 Answer Key Testname: UNTITLED4 1) B 2) A 3) C 4) A 5) B 6) A 7) C 8) B 9) D 10) B 11) A 12) D 13) B 14) C 15) B 16) D 17) A 18) D 19) C 20) A 21) E 22) A 23) B 24) D 25) A 26) D 27) C 28) A 29) B 30) A 31) C 32) A 33) A 34) B 35) A 36) B 37) A 38) A 39) D 40) D 41) D 42) B 18 Answer Key Testname: UNTITLED4 43) No. Let A be the 26 × 30 coefficient matrix of the system. The 5 solutions are linearly independent and span Nul A, so dim Nul A = 5. By the Rank Theorem, dim Col A = 30 - 5 = 25. Since 25 < 26, Col A does not span [26. So not every nonhomogeneous equation Ax = b has a solution. 44) No. Since every nonhomogeneous equation Ax = b has a solution, Col A spans [11. So Dim Col A = 11. By the Rank Theorem, dim Nul A = 14 - 11 = 3. So the associated homogeneous system does not have more than 3 linearly independent solutions. 45) A 46) D 47) C 48) A 49) B 50) A 51) C 52) B 53) A 54) C 55) A 56) B 57) C 58) C 59) B 60) D 61) D 62) D 63) A 64) A 65) B 66) A 67) D 68) B 69) C 70) D 19 Exam Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue. 1) A = -24 -5 150 31 , Ώ = 6 1) A) 6 1 B) -6 1 C) 1 6 D) 1 -6 2) A = -18 -5 60 17 , Ώ = 2 2) A) 1 -4 B) 1 17 C) -4 1 D) 1 0 For the given matrix A, find a basis for the corresponding eigenspace for the given eigenvalue. 3) A = 1 - 5 - 5 - 5 1 5 5 - 5 - 9 , Ώ = -4 3) A) 1 0 -1 , 0 1 1 B) 1 0 -1 C) 0 1 -1 D) 1 1 0 , 1 0 1 4) A = 4 0 0 6 1 0 14 5 -4 , Ώ = 4 4) A) 1 2 0 , 1 0 -3 B) 1 -2 -3 C) 1 2 3 D) 1 2 0 , 1 0 3 Find the eigenvalues of the given matrix. 5) -4 -1 6 1 5) A) -2, -1 B) -2, 1 C) -2 D) 1 6) -26 -16 56 34 6) A) 6 B) 2, 6 C) -2 D) -2, -6 1 Find the characteristic equation of the given matrix. 7) A = 7 8 4 2 0 5 -8 3 0 0 9 9 0 0 0 7 7) A) (7 - Ώ) 2(5 - Ώ)(9 - Ώ) = 0 B) (7 - Ώ)(8 - Ώ)(4 - Ώ)(2 - Ώ) = 0 C) (7 - Ώ)(5 - Ώ)(9 - Ώ) = 0 D) (7 - Ώ)(9 - Ώ)(3 - Ώ)(2 - Ώ) = 0 8) A = -2 5 2 6 0 4 -5 1 0 0 -3 2 0 0 0 3 8) A) (-2 - Ώ)(5 - Ώ)(2 - Ώ)(6 - Ώ) = 0 B) (-2 - Ώ)(4 - Ώ)(-3 - Ώ)(3 - Ώ) = 0 C) (3 - Ώ)(2 - Ώ)(1 - Ώ)(6 - Ώ) = 0 D) (6 - Ώ)(1 - Ώ)(2 - Ώ)(3 - Ώ) = 0 The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities. 9) Ώ 5 - 7Ώ 4 - 18Ώ 3 9) A) 0 (multiplicity 3), -9 (multiplicity 1), 2 (multiplicity 1) B) 0 (multiplicity 1), -2 (multiplicity 1), 9 (multiplicity 1) C) 0 (multiplicity 3), -2 (multiplicity 1), 9 (multiplicity 1) D) 0 (multiplicity 1), -9 (multiplicity 1), 2 (multiplicity 1) 10) Ώ 5 - 11Ώ 4 - 45Ώ 3 + 567Ώ 2 10) A) 0 (multiplicity 2), -9 (multiplicity 2), -7 (multiplicity 1) B) 0 (multiplicity 1), 9 (multiplicity 3), -7 (multiplicity 1) C) 0 (multiplicity 2), 9 (multiplicity 2), -7 (multiplicity 1) D) 0 (multiplicity 2), -9 (multiplicity 2), 7 (multiplicity 1) Find a formula for Ak, given that A = PDP-1, where P and D are given below. 11) A = -2 10 -8 16 , P = 5 1 4 1 , D = 6 0 0 8 11) A) 5 · 6 k - 4 · 8 k 5 · 8 k + 5 · 6 k 4 · 6 k + 4 · 8 k 5 · 8 k - 4 · 6 k B) 5 · 6 k + 4 · 8 k 5 · 8 k + 5 · 6 k 4 · 6 k + 4 · 8 k 5 · 8 k + 4 · 6 k C) 6 k 0 0 8 k D) 5 · 6 k - 4 · 8 k 5 · 8 k - 5 · 6 k 4 · 6 k - 4 · 8 k 5 · 8 k - 4 · 6 k 2 Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that A= PDP-1. 12) A = -11 3 -9 0 -5 0 6 -3 4 12) A) P = 1 0 -1 5 3 0 1 1 1 , D = -5 0 -2 0 -5 0 0 -5 -2 B) P = 1 0 -1 5 3 0 1 1 1 , D = -5 0 0 0 -5 0 0 0 -2 C) P = 1 0 -1 0 3 0 1 1 1 , D = -5 0 0 0 1 0 0 0 -2 D) P = 1 5 -1 5 3 0 1 3 1 , D = -5 1 0 0 -5 0 0 0 -2 13) A = 4 0 0 1 4 0 0 0 4 13) A) P = 1 0 0 4 4 0 0 1 1 , D = 4 1 0 0 4 0 0 0 4 B) P = 1 4 1 0 4 1 -1 0 1 , D = 4 0 0 0 4 0 0 0 4 C) P = 1 0 -1 4 4 0 1 1 1 , D = 4 0 1 1 4 1 0 0 4 D) Not diagonalizable 14) A = -4 0 0 0 0 -4 0 0 1 -4 4 0 -1 2 0 4 14) A) P = 8 16 0 0 -4 -4 0 0 1 0 1 0 0 1 0 1 , D = 4 0 0 0 0 4 0 0 0 0 -4 0 0 0 0 -4 B) Not diagonalizable C) P = 8 -4 1 0 16 -4 0 0 0 0 1 0 0 0 0 1 , D = 4 0 0 0 0 4 0 0 0 0 -4 0 0 0 0 -4 D) P = 8 16 0 0 4 4 0 0 1 0 1 0 0 1 0 1 , D = -4 0 0 0 0 -4 0 0 0 0 4 0 0 0 0 4 3 15) A = 8 0 0 0 0 8 0 0 - 12 3 2 12 0 0 0 8 15) A) P = 2 0 -2 1 0 2 1 0 1 0 0 1 0 0 1 0 , D = 8 0 0 0 0 8 0 0 0 0 8 0 0 0 0 2 B) P = 2 0 1 0 0 2 0 0 -2 1 0 1 1 0 1 0 , D = 8 0 0 0 0 8 0 0 0 0 8 0 0 0 0 2 C) P = 4 -2 1 0 8 -2 0 0 1 0 1 1 0 1 1 0 , D = 8 0 0 0 0 8 0 0 0 0 2 0 0 0 0 2 D) Not diagonalizable Find the matrix of the linear transformation T: V  W relative to B and C. 16) Suppose B = {b 16)