Linear Algebra and Its Applications, 6th edition David C Lay Test Bank

Linear Algebra and Its Applications, 6th edition By David C. Lay

Email: richard@qwconsultancy.com

Exam Name___________________________________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Solve the system of equations.

1) x1 - x2 + 3x3 = -8 2x1

1)

+ x3 = 0

x1 + 5x2 + x3 = 40

A) (8, 8, 0)

B) (0, -8, -8)

C) (-8, 0, 0)

D) (0, 8, 0)

2) x1 + 3x2 + 2x3 = 11

2)

4x2 + 9x3 = -12 x3 = -4

A) (1, -4, 6)

B) (-4, 1, 6)

C) (-4, 6, 1)

D) (1, 6, -4)

3) x1 - x2 + 8x3 = -107 6x1

3)

+ x3 = 17 3x2 - 5x3 = 89

A) (5, -8, -13)

B) (5, 8, -13)

C) (-5, -8, 13)

D) (-5, 8, 13)

4) 4x1 - x2 + 3x3 = 12 2x1

4)

+ 9x3 = -5

x1 + 4x2 + 6x3 = -32

A) (2, 7, -1)

B) (2, 7, 1)

C) (2, -7, 1)

D) (2, -7, -1)

5) x1 + x2 + x3 = 6 x1

5)

- x3 = -2 x2 + 3x3 = 11

A) (0, 1, 2)

B) No solution

C) (1, 2, 3)

D) (-1, 2, -3)

6) x1 + x2 + x3 = 7

6)

x1 - x2 + 2x3 = 7 5x1 + x2 + x3 = 11

A) (4, 1, 2)

B) (4, 2, 1)

C) (1, 4, 2)

1

D) (1, 2, 4)

7) x1 - x2 + x3 = 8

7)

x1 + x 2 + x 3 = 6 x1 + x2 - x3 = -12

A) (2, -1, -9)

B) (2, -1, 9)

C) (-2, -1, 9)

D) (-2, -1, -9)

8) 5x1 + 2x2 + x3 = -11

8)

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) 7x1 + 7x2 + x3 = 1

9)

x1 + 8x2 + 8x3 = 8 9x1 + x2 + 9x3 = 9

A) (0, 1, 0) 10) 2x1 + x2

B) (0, 0, 1)

C) (1, -1, 1)

D) (-1, 1, 1)

=0

10)

x1 - 3x2 + x3 = 0 3x1 + x2 - x3 = 0

A) (0, 1, 0)

B) (0, 0, 0)

C) No solution

Determine whether the system is consistent.

11) x1 + x2 + x3 = 7

D) (1, 0, 0)

11)

x1 - x2 + 2x3 = 7 5x1 + x2 + x3 = 11

A) No

B) Yes

12) 5x1 + 2x2 + x3 = -11

12)

2x1 - 3x2 - x3 = 17 7x1 + x2 + 2x3 = -4

A) No

B) Yes

13) 4x1 - x2 + 3x3 = 12 2x1

13)

+ 9x3 = -5

x1 + 4x2 + 6x3 = -32

A) Yes 14) 2x1 + x2

B) No =0

14)

x1 - 3x2 + x3 = 0 3x1 + x2 - x3 = 0

A) Yes

B) No

2

15) x1 + x2 + x3 = 6 x1

15)

- x3 = -2 x2 + 3x3 = 11

A) No 16)

B) Yes

x1 - x2 + 3x3 = -11

16)

-4x1 + 4x2 - 12x3 = -2 x1 + 3x2

+ x3 = -17

A) Yes

B) No

17) x1 + x2 + x3 = 7

17)

x1 - x2 + 2x3 = 7 2x1

+ 3x3 = 15

A) No

B) Yes

18) x1 + 3x2 + 2x3 = 11

18)

4x2 + 9x3 = -12 x1 + 7x2 + 11x3 = -11

A) No

B) Yes

19) 5x1 + 2x2 + x3 = -11

19)

2x1 - 3x2 - x3 = 17 7x1 - x2

= 12

A) Yes

B) No

5x2

20)

+ x4 = -21

20)

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.

1 3 5 -7 21) 0 1 -4 -4 0 0 1 6

A) Reduced echelon form

B) Echelon form

3

C) Neither

21)

1 4

5 -7

0 6

1

22) 0 1 -4 -5

22)

4

A) Reduced echelon form 1 4

5 -7

0 4

1

B) Echelon form

C) Neither

23) 3 1 -4 -6

23)

3

A) Reduced echelon form

B) Echelon form

C) Neither

1 0 0 -7 1 0 3 1 1

24) 1 1 0

A) Reduced echelon form 1 4

24) B) Neither

C) Echelon form

1 -7 7 0 0

25) 0 1 -4 0 0

A) Reduced echelon form

26)

1 0 0 0

B) Neither

C) Echelon form

0 5 -4 1 -5 -2 0 0 0 0 0 0

A) Neither

27)

25)

26)

B) Reduced echelon form

C) Echelon form

1 -5 -5 -5 0 0 -2 3 0 0 0 -3 0 0 0 0

A) Neither

27)

B) Echelon form

4

C) Reduced echelon form

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)

B) 1 4 -2 3 0 1 3 4 0 12 -7 10

1 4 -2 3 0 1 3 4 0 0 -43 -38

C)

D) 1 4 -2 3 0 1 3 4 0 0 -19 -2

1 4 -2 3 0 1 3 4 0 0 -43 0

29) Find the reduced echelon form of the given matrix.

1 4 -5 1 2 5 -4 -1 -3 -9 9 2

29)

2 4 2

A)

B) 1 0 3 0 14 0 1 -2 0 -4 0 0 0 1 4

C)

1 0 0

4 -5 1 2 1 -2 1 0 0 0 1 4

1 0 0

4 -5 1 -2 0 0

D) 1 0 0

0 1 0

0 0 0

0 14 0 -4 1 4

0 -2 0 -4 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 30) 0 0 3

A) x1 = -1 + 5x2

B) No solution

C) x1 = -1 + 5x2

x2 = 3

D) (-1, 3)

x2 is free

x3 is free

1 2 -3 -9 4 5 0 0 0 1

31) 0 1

31)

A) x1 = -19 + 11x3

B) x1 = -9 - 2x2 + 3x3

x2 = 5 - 4x3

x2 is free

x3 = 1

x3 is free

C) No solution

D) x1 = -19 + 11x3 x2 = 5 - 4x3 x3 is free

5

1 2 -3 6 4 -7 0 0 0 0

32) 0 1

32)

A) x1 = 6 - 2x2 + 3x3

B) x1 = 20 + 11x3

x2 is free

x2 = -7 - 4x3

x3 is free

x3 is free

C) x1 = 20 + 11x3

D) x1 = 6 -2x2 + 3x3

x2 = -7 - 4x3

x2 = -7 - 4x3

x3 = 0

x3 is free

1 0

6

2

0 0

0

0

33) 0 1 -2 -3

A) x1 = 2 - 6x3

33)

x2 = -3 + 2x3 x3 is free

34)

C) x1 = 2 - 6x3

B) x1 = 2 - 6x3

x2 = -3 + 2x3

x2 is free x3 =

D) No solution

3 1 + x 2 2 2

x3 = 0

1 4 -2 -3 1 0 0 1 4 -4 -1 -4 -1 -9 11

34) B) x1 = -7 - 4x2 - 5x4

A) x1 = - 4x2 +2x3 + 3 x4 + 1 x2 is free

x2 is free

x3 = -4 - 4x4

x3 = -4 - 4x4

x4 is free

x4 is free

D) x1 = -7 - 4x2 - 5x4

C) x1 = -7 - 4x2 - 5x3 x2 = -4 - 4x3

x2 is free

x3 is free

x3 = -4 - 4x4 x4 = 0

6

1 6 3 -1

2 6

35) 0 0 0 -4 3 4 0 0 0

35)

0 -2 8

A) No solution

B) x1 = -6x2 - 3x3 + 10 x2 is free x3 = -4 x4 =

3 x -1 4 5

x5 = -4

D) x1 = -6x2 - 3x3 + x4 - 2x5 + 6

C) x1 = -6x2 - 3x3 + 10 x2 is free

x2 is free

x3 is free

x3 is free

x4 = -4

x4 =

x5 = -4

3 x -1 4 5

x5 = -4

Find the indicated vector.

36) Let u = -6 , v = 2 . Find u + v.

4 1

A)

B) -4 5

37) Let u =

36) C)

-8 3

D) -2 3

-5 6

9 , v = 7 . Find u - v.

2 -8

A)

37)

B) 7 -15

C) 17 5

D) 2 -10

16 -6

38) Let u = -5 , v = 1 . Find v - u.

5

38)

6

A)

B) 10 5

C)

D) 11 -4

-4 11

6 1

39) Let u = 3 . Find 4u.

39)

2

A)

B) 12 -8

C)

D) 12 8

-12 -8

7

-12 8

40) Let u =

8 . Find 3u.

-6

A)

40) B)

24 -18

C)

D) 24 18

-24 18

-24 -18

41) Let u = -6 . Find -2u.

41)

-7

A)

B) -12 -14

C)

D) 12 14

-12 14

12 -14

42) Let u = -7 . Find -5u.

42)

5

A)

B) -35 -25

43) Let u = 2 , v = 1

A)

C)

35 -25

7 . Find -5u + 2v.

-5

43)

B) -15 4

D) 35 25

-35 25

C) -24 5

D) -45 -8

Display the indicated vector(s) on an xy-graph.

44) Let u = -3 and v = -2 . Display the vectors u, v, and u + v on the same axes.

5 2

A)

B)

8

4 -15

44)

C)

45) Let u =

D)

5 -4

Display the vector 2u using the given axes.

A)

B)

9

45)

C)

D)

Solve the problem.

6 -5 -39 1 , a 2 = 1 , and b = -1 .

6 22 -1 Determine whether b can be written as a linear combination of a 1 and a 2 . In other words,

46) Let a 1 =

46)

determine whether weights x1 and x2 exist, such that x1 a 1 + x2 a 2 = b. Determine the weights x1 and x2 if possible.

A) x1 = -4, x2 = 4 1

B) No solution

-3

C) x1 = -4, x2 = 3

D) x1 = -3, x2 = 2

2

-3 6 .

1 6 -3 -1 Determine whether b can be written as a linear combination of a 1 , a 2 , and a 3 . In other words,

47) Let a 1 = 2 , a 2 = -4 , a 3 = 1 , and b =

47)

determine whether weights x1 , x2 , and x3 exist, such that x1 a 1 + x2 a 2 + x3 a 3 = b. Determine the weights x1 , x2 , and x3 if possible.

A) x1 = -2, x2 = -1, x3 = 6 C) No solution

B) x1 = 2, x2 = 1, x3 = - 1 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

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 0.50 0.40 a = 0.20 and b = 0.20 .

0.10 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

48)

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 0.45 0.40 a = 0.20 and b = 0.20 .

0.10 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.

9 50) [-7 2 7] 0 -3

A)

50)

B) [435]

51) -1 1 -2 -8 2 -6

C) [-84]

D) [-63 0 -21]

-63 0 -21

5 9 9

A)

51) B)

-76 -14

C)

D) [-14 -76]

-14 -76

-1 1 -2 -8 2 -6 59 9

52)

52) 3 -7 8 -4 -4 1

-4 5

A)

B) -12 -35 -32 -20 16 5

C) -12 40 28 -20 -4 1

-47 -52 21

11

D) Undefined

53)

53) -4 8 1 5 2 8

1 -5 7

B) Undefined

A) 4 -30 70

C)

D) 8 -4 -5 -25 14 56

5 39

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.

5x1 - 2x2 - x3 = 2 4x1 +

54)

3x3 = -1 x1

A) x1 5 + x2 -2 + x3 1 = 2 4 1 3 -1 5

4

x3

2

3

x1

2

B) 5 x2 - 2 x2 - x2 = -1 x3

x3

0

D) x1 5 + x2 -2 + x3 -1 =

C) x1 2 + x2 0 = -1 -1

x1

4

0

0

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

A)

C)

=1

x1 x2 x3 2 -2

0

3 1 -2

= -4 1

x1 3 1 -2 x = -4 2 2 2 1 1 x3

B)

x1 3 1 -2 x = -4 2 2 -2 0 1 x3

D)

3 1 -2

2 -2 0

x1 x2

= -4 1

Solve the problem.

b1 1 -3 2 56) Let A = -2 5 -1 and b = b2 .

3 -4 -7 b3

56)

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 ).

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

b1 1 -3 2 57) Let A = -2 5 -1 and b = b2 .

3 -3 -12 b3

57)

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 ).

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 linear equation. Give your answer as a linear combination of vectors. Let x2 and x3 be free variables.

-2x1 - 8x2 + 16x3 = 0

A) x1

8 -4 1 + x3 0 0 1

(with x2 , x3 free)

4 -8 x 2 = x2 1 + x3 0 x3 0 1

(with x2 , x3 free)

x 2 = x2 x3

B) x1

C) x1

x1

x1

x2 = -4 x2 - 8 x2 x3 x3 x3

(with x2 , x3 free)

D) x1 x 2 = x2 x3

8 -4 + x 0 3 1 1 0

(with x2 , x3 free)

13

58)

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)

B) x1

x1

3 x2 = x3 -3 x3 1

-3 x 2 = x3 3 x3 0

C)

D) x1

x1

-3 x2 = 3 x3 1

x 2 = x3 x3

-3 3 1

60) Describe all solutions of Ax = b, where

60)

2 -5 3 -5 A = -2 6 -5 and b = 2 .

-4 7 0 19

Describe the general solution in parametric vector form.

A)

B) x1

x1

7/2 -10 x2 = 2 + x3 -3 x3 1 0

-5 -1 x2 = -3 + x3 2 x3 0 1

C)

D) x1

x1

7/2 -10 x2 = -3 + x3 2 x3 0 1

x2 = x3

14

7/2 -10 + x -3 3 2 0 0

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)

B) pe

0.308 pa

pe

pm = 0.716 pa pa pa

0.356 pa

pm = 0.686 pa pa pa

C)

D) pe

0.607 pa

pe

pm = 0.481 pa pa pa

0.465 pa

pm = 0.593 pa pa pa

62) The network in the figure shows the traffic flow (in vehicles per hour) over several one-way

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

B) x1 = 500 + x4

C) x1 = 600 - x4

D) x1 = 600 + x5

x2 = 400 + x4

x2 = 400 - x4

x2 = 400 - x4

x2 = 400 - x5

x3 = 300 - x4

x3 = 300 - x4

x3 = 300 + x4

x3 = 300 - x5

x4 is free

x4 is free

x4 is free

x4 = 300

x5 = 300

x5 = 200

x5 = 300

x5 is free

15

62)

1

63) Let v1 = -3 , v2 = -4

2 -3 8 , v3 = -2 .

4 6

63)

Determine if the set {v1 , v2 , v3 } is linearly independent.

A) Yes

B) No

64) Determine if the columns of the matrix A = A) Yes

-2 1 4 4 0 -4 are linearly independent.

2 4 6 B) No

64)

65) For what values of h are the given vectors linearly independent?

65)

-1 -4 1 , 4 6 h A) Vectors are linearly independent for h 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)

5 5 -1 -20 4 , 2 , -3 , 12 6 -3 5 h A) Vectors are linearly independent for all h

B) Vectors are linearly dependent for all h C) Vectors are linearly dependent for h -20 D) Vectors are linearly dependent for h = -20 2 8 -2 and u = -1 .

3 -5 -3 1 Define a transformation T: 3 -> 2 by T(x) = Ax. Find T(u), the image of u under the

67) Let A = 2

67)

transformation T.

A)

B) -6 8

C) 4 -8 -2 6 5 -3

D) 10 -3 -5

16

16 5

68) Let T: 2 -> 2 be a linear transformation that maps u = -3 into -15 and maps v = 24 .

-12 Use the fact that T is linear to find the image of 3u + v.

A)

B) -3 12

6

C) 9 -6

6 into -6

68)

D) 27 -18

1 -3 0 2 0 2 and b = 6 .

2 -5 -3 0 Define a transformation T: 3 -> 3 by T(x) = Ax.

6

-21 6

69) Let A = -4

69)

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.

A) -1 -1 0

B) b is not in the range of the transformation T.

C) -1 -1 1

D) -1 1 -1

17

1 -3

2

-5 2 .

-4 2 -5 3 3 Define a transformation T: -> 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) Let A = -3 4 -1 and b =

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. A) Horizontal shear

71) B) Projection onto x2 -axis D) Projection onto x1 -axis

C) Vertical shear 0

0

0

72) Let A = 0 1 0 .

72)

0 0 1 Define a transformation T by T(x) = Ax. A) Vertical shear

B) Horizontal shear D) Projection onto the x2 x3-plane

C) Projection onto the x2 -axis

18

Solve the problem.

1

0

0

1

0

0

0

0

1

0

0

1

73) The columns of I3 = 0 1 0 are e1 = 0 , e 2 = 1 , e 3 = 0 .

73)

Suppose that T is a linear transformation from 3 into 2 such that T( e 1 ) =

3 , T( e ) = 2 -2

2 , and T( e ) = -4 .

3 0 1 x1

Find a formula for the image of an arbitrary x = x2 in 3 . x3

A)

B) x1

x1

3x1 + 2x2 - 4x3

T x2 = -2x1 x3

3x1 + 2x2 - 4x3

T x2 = 2x1 -2x1 + x3 x3

+ x3

C)

D) x1

x1

3x1 - 2x2

T x2 = 2x1 x3

3x1 - 2x2

T x2 = 2x1 x3 4x2 + x3

Find the standard matrix of the linear transformation T.

74) T: 2 -> 2 rotates points (about the origin) through 7 4

radians (with counterclockwise rotation

74)

for a positive angle).

A)

B) -

2 2 2 2

-

2 2

C)

D)

1 1 -1 1

2 2

-

3 3

3 3

3 3

3 3

-

2 2

2 2

2 2

2 2

75) T: 2 -> 2 first performs a vertical shear that maps e 1 into e 1 + 3e2 , but leaves the vector e 2 unchanged, then reflects the result through the horizontal x1 -axis.

A)

B) 1 0 -3 -1

C)

D) 1 3 0 -1

-1 -3 0 1

19

-1 0 3 -1

75)

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 1 -2 3 A = -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

76)

D) One-to-one; onto 3

77) T(x1 , x2 , x3 ) = (-2x2 - 2x3 , -2x1 + 9x2 + 5x3 , -x1 - 2x3, 3x2 + 3x3 )

77)

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 three different foods.

Food 1 Food 2 Protein 15 35 Carbohydrate 45 30 Fat 6 4

78)

Food 3 25 50 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

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: B) City: 361,000 361,000 Suburbs: 737,280 Suburbs: 839,000 C) City: D) City: 422,920 412,000 Suburbs: 777,080 Suburbs: 788,000

20

79)

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 290

48) 500a + 100b = 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.

205 49) x1 a + x2 b = 100 50 or 0.45 x1 0.20 0.10

0.40 + x2 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 . Find 5A. 02

A)

1) B)

-15 5 02

C) -15 25 0 2

D) 2 10 5 7

2) Let B = -1 4 4 -3 . Find -3B. A) 3 4 4 -3 C) 3 -12 -12 9

-15 25 0 10

2) B) -3 12 12 -9 D) -3 2 2 -5

2

3) Let C = -2 . Find (1/2) C.

3)

10

A)

B) 2 -1 10

C) 1 -1 5

4) Let A = 2 3 and B = 26

04 -1 6

A)

4)

B)

1 2

A)

C) 6 13 1 12

-1 3 -2

D) 6 7 5 12

6 13 5 24

. Find C - 4D.

5)

B) -5 15 -10

4 -4 20

. Find 3A + B.

6 21 3 36

5) Let C = -3 and D =

D) 1 -2 10

C) 5 -6 4

D) 5 -15 10

6) Let A = -3 2 and B = 1 0 . Find 2A + 3B. A) 0 2 B) -5 4

-3 9 -6

6) C) -3 4

1

D) -6 4

7) Let A =

1 2 7 3 -9 -8

1 7 8 -7 . Find A + B. -8 2

and B =

A)

7)

B) 2 9 15 -4 -17 -6

C) 29 -15 3 -17 6

D) 2 3 15 -4 -17 -6

0 -5 -1 -1 -1 -11

8) Let A = -2 2 and B = 2 5 . Find A - B.

8)

-8 4

-8 -8

A)

B) 0 3 -16 12

C) 0 -3 0 -12

D) 4 -3 -16 -4

-4 -3 0 -12

9) Let A = -6 3 and B = 0 0 . Find A + B. 9 -4 A) Undefined

9)

00

B)

C) 6 -3 -9 4

D) -6 3 9 -4

00 00

Find the matrix product AB, if it is defined.

10) A = -1 3 , B = -2 0 .

42 -1 1

A)

10)

B) 3 -1 2 -10

C) 20 -4 2

D) 2 -6 -3 -1

-1 3 -10 2

11) A = 0 -1 , B = -2 0 .

4

2

11)

-1 1

A)

B) 02 -4 2

C)

D) 1 -1 -6 -10

-8 -4 4 3

1 -1 -10 2

12) A = 3 -1 , B = 0 -1 .

3

0

A) -3 0 24 -2

2

12)

6

C) 0 1

B) -2 -9 0 -3

60

2

D) -9 -2 -3

0

13) A = -1 3 , B = 0 -2 5 .

56

13)

1 -3 2

A)

B) 3 6 -7 -28 1 37

3 -7 1 6 -28 37

C) AB is undefined.

D) 0 -6 15 5 -18 12

14) A = 3 -2 1 , B =

50 .

0 4 -2 -2 1 A) AB is undefined.

B)

C)

D)

14) 15 -10 5 8 -4 -6

15 0 0 4

15 -6 -10 8 5 -4

15) A = 0 -2 , B = -1 3 2 .

3

3

15)

0 -2 1

A) AB is undefined.

B)

C)

D)

0 4 -2 -3 3 9

0 -3 4 3 -2 9

0 -6 -8 0 -6 3 30

16) A = 1 3 -2 , B = -2 1 .

4 0

4

16)

04

A) AB is undefined.

B) -5 -3 16 12

C)

D) 3 -6 0 0 0 16

-3 -5 12 16

3

17) A = 1 0 , B = 5 2 -1 .

03

17)

2 -1 2

A)

B) 5 0 0 0 -3 6

5 2 -1 6 -3 6 D) AB is undefined.

C) 6 -3 6 5 2 -1

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.

A) AB is 2 × 1, BA is undefined.

C) AB is 1 × 2, BA is 1 × 1.

19) B) AB is undefined, BA is 1× 2.

D) AB is 2 × 2, BA is 1 × 1.

20) A is 3 × 1, B is 1 × 3.

A) AB is 3 × 3, BA is undefined.

C) AB is 3 × 3, BA is 1 × 1.

20) B) AB is 1 × 1, BA is 3 × 3.

D) AB is undefined, BA is 1 × 1.

21) A is 4 × 3, B is 4 × 3.

A) AB is 4 × 3, BA is 3 × 4.

C) AB is undefined, BA is undefined.

21) B) AB is 3 × 4, BA is 4 × 3.

D) AB is 4 × 4, BA is 3 × 3.

Find the transpose of the matrix.

2 4 22) -4 0 -7 7

A)

22) B)

-7 7 -4 0 2 4

C) 4 0 7 2 -4 -7

D) 4 2 0 -4 7 -7

4

2 -4 -7 4 0 7

23) 9

8 9 8 0 -7 0 -7

23)

A)

B) 8 9 8 9 -7 0 -7 0

0 -7 0 -7

C)

9 8 9 8

D) 0 -7 0 -7 9 8 9 8

9 0 8 -7 9 0 8 -7

Decide whether or not the matrices are inverses of each other.

24) 5 3 and 2 -3 32 -3 5 A) No

25) 10 1 and -1 0

24) B) Yes

0 1 -1 10

25)

A) No

B) Yes 1 1 2 4

26) -2 4 and 1 1 4 -4

26)

2 4

A) Yes

B) No 1 1 2 - 2

27) -5 1 and 7 -7 1

27)

5 2 2

A) Yes

28)

6 -5 -3 5

and

B) No 1 1 3 3

28)

1 2 5 5

A) No

B) Yes

29) 9 4 and -0.2 44

0.2 0.2 -0.45

29)

A) Yes

B) No

5

30)

9 -2 7 -2

0.5 0.5 and - 7 - 9 4 4

30)

A) Yes

B) No

31) -5 -1 and 6 0

0

1 6

-1

5 6

31)

A) No

B) Yes

2 -1 0

1 -1 2

1 0 -1 A) Yes

-1 1 1

32) -1 1 -2 and -3 -2 4

32) B) No

Find the inverse of the matrix, if it exists.

33) A = 4 1 -5 1

A)

B) 5 9

4 9

1 1 9 9

34) A =

33) C)

D)

1 1 9 9

4 1 9 9

5 9

5 9

4 9

1 1 9 9

1 9

-

5 4 9 9

0 3 -4 -6

34)

A)

B) 0 -

C)

1 4

1 1 3 2

1 3 -

D)

0

-

1 1 2 4

1 1 2 4

-

1 3

6

0

-

1 1 2 4 1 3

0

35) A = 4 0

35)

4 -3 A) A is not invertible

B) -

1 3

0

1 1 3 4

C)

D) 1 4 -

1 4

0

1 1 3 3

0

1 1 3 3

36) A = -6 -3 4

36)

2

A)

B) 1 3 5 10

1 5

2 3 5 5

-

2 3 5 5

D) A is not invertible

C) 2 3 5 5 1 5

3 10

3 10

37) A = 4 -4 0

A)

38) A =

37)

3

B)

C)

1 0 3

1 1 3 3

1 1 4 3

0

D) 1 1 4 3

1 4

0

1 1 4 3

1 3

1 3

0

4 -6 -6 0

A)

38) B)

-

C)

1 06

1 6

0

1 1 6 9

1 1 6 9

D) 1 1 - 6 9

-

1 1 9 6

1 6

-

1 6

0-

7

0

39)

39) 100 -1 1 0 111

A)

B) 1 0 0 1 1 0 -2 -1 1

C) -1 0 0 -1 -1 0 -1 -1 -1

D) 1 -1 1 0 1 -1 0 0 1

111 011 001

Solve the system by using the inverse of the coefficient matrix.

40) 2x1 + 2x2 = 2

40)

3x1 + 7x2 = 15

A) (3, -2)

B) No solution

C) (-2, 3)

D) (-2, -3)

41) 4x1 + 2x2 = -4 2x1

41)

= -8

A) (-4, -6)

B) (-4, 6)

C) No solution

D) (6, -4)

42) 8x1 - 9x2 = 3

42)

-16x1 + 18x2 = -3

A) No solution

B) 3 - 8 x2 , x2 8

C) (3, -3)

9

D) (-2, -2)

43) 3x1 + 9x2 = 3

43)

2x1 - x2 = -5

A) (2, -1)

B) (1, -2)

C) (-1, 2)

D) (-2, 1)

44) 2x1 - 6x2 = -6

44)

3x1 + 2x2 = 13

A) (2, 3)

B) (3, 2)

C) (-2, -3)

D) (-3, -2)

45) 10x1 - 4x2 = -6

45)

6x1 - x2 = 2

A) (1, 4)

B) (-1, -4)

C) (-4, -1)

D) (4, 1)

46) 3x1 - 6x2 = -3

46)

3x1 + 4x2 = -23

A) (2, 5)

B) (-5, -2)

C) (5, 2)

D) (-2, 5)

47) -5x1 + 3x2 = 8

47)

2x1 - 4x2 = -20

A) (-6, -2)

B) (2, 6)

C) (6, 2)

8

D) (-2, -6)

Find the inverse of the matrix A, if it exists.

4 -2 5 48) A = 4 -1 4 8 -3 9

48) 1

B) A-1 = 0 1 -1

A) A-1 does not exist.

0

C) A-1 =

1 0

3 4

0

3 4

3 4

0

4

4

8

5

4

9

D) A-1 = -2 -1 -3

0 1 -1 0 0 0

1 1 1

49) A = 2 1 1

49)

2 2 3 -1 -1 -1

A) A-1 = -2 -1 -1

B) A-1 =

-2 -2 -3

C) A-1 does not exist.

D) A-1 =

-1 1 0 4 -1 -1 -2 0 1 1 1 2

1 1

1 2

1 1 2 3

1 1

1 3 2

50) A = 1 3 3

50)

2 7 8

A) A-1 =

-3 10 -3 2 -4 1 -1 1 0

-1 -3 -2

C) A-1 = -1 -3 -3

1

1 3

1 2

B) A-1 = 1

1 3

1 3

1 2

1 7

1 8

D) A-1 does not exist.

-2 -7 -8

9

1 0 8

51) A = 1 2 3

51)

2 5 3

-1

0 -8

A) A-1 does not exist.

B) A-1 = -1 -2 -3

9 -40 16 C) A-1 = -3 13 -5 5 -2 -1

-2 -5 -3 1 1 2 D) A-1 = 0 2 5 8 3 3

6 -3 1

52) A = 9 -5 2 3 -2 1

52) 1 6

1 9

1

1

1

1

1 1 3 2

1

A) A-1 = 9 - 5 2

B) A-1 does not exist.

1 9

2 -3 9

1

6 5

3

1 2

1

C) A-1 = 3 2 -

6

9

3

1

2

1

D) A-1 = -3 -5 -2

0 3 3

53) A = -1 0 7

53)

0 6 0 -

1

A) A-1 does not exist.

7 7 -1 3 6

C) A-1 = 0 1 3

0 0 -

7 7 -1 3 6

1 6

B) A-1 = - 6

0

1 6

1 3

0

0

7 3

0

D) A-1 = - 1 0 7 6

1 6

10

1 3

0 1 1 6 6

Determine whether the matrix is invertible.

54) 4 3 7 16 A) Yes

55)

54) B) No

5 5 -5 6 2 -6 -2 0 2 A) Yes

55) B) No

Identify the indicated submatrix.

0 1 -3 -8 56) A = 3 -1 0 5 . Find A12.

2 5 -5 0

C) -8

B) 1

A) 3 7 -7 57) A = 0 3

56)

5

4 1 0 -1 . Find A .

21 3 -4 4 3

A) 3 4

D) 2 5 -5

57) 1

B) -7

C) -1

D) 4

-4

Find the matrix product AB for the partitioned matrices.

4 0 1 -2 0 8 5 58) A = 2 -1 -3 , B = 1 6 2 2 5 3 7 4 -1 0 3

58)

A)

B) -4 -1 32 23 -17 -3 14 -1 21 11 46 52

-4 -1 0 3 -12 -3 0 -9 28 -7 0 21

C)

D) -4 -1 32 23 -17 -3 14 -1 21 11 46 52

-8 0 32 20 -5 -6 14 8 -7 18 46 31

59) A = 0 I , B = W X I F

59)

Y Z

A)

B) 0 Z FY FZ

Y Z W + FY X + FZ

C)

D) Y Z W + YF X + ZF

X W + XF Z Y + ZF

11

Solve the equation Ax = b by using the LU factorization given for A. 3 -1 2 6 60) A = -6 4 -5 , b = -3 9 5 6 2 100 A = -2 1 0 341

60)

3 -1 2 0 2 -1 0 0 4

22

49

A) x = -7

10

B) x = -38

15

32

1 2 4 3 61) A = -1 -3 -1 -4 , b = 2 1 19 3 1 5 -9 7 1 0 00 A = -1 1 0 0 2 3 10 1 -3 -2 1

25

C) x = -2

D) x = -58

51

-13

2 0 6 2

61)

1 24 3 0 -1 3 -1 0 02 0 0 00 1

28 A) x = -6 -2 -2

2 B) x = -2 10 -8

32 C) x = -16 124 -38

32 D) x = 16 12 -8

Find an LU factorization of the matrix A. 3 -1 62) A = -24 13

A) A = 1 0 -8 1 C) A = 1 0 3 1 5

62)

3 1 0 -5 -8 -1 0 5

B) A = 1 0

8 1 D) A = 1 0 -8 1

-3 -1 0 -5 3 -1 0 5

4

3 1 10 -1 20

63) A = 10 11

1

0 0

A) A = 2 1 0 2 -3 1 1

0 0 1 0 10 -1 1

C) A = 10

63) 3 4 3 0 -3 5 0 0 1

B) A = 2 1 0

1

0 0

5 4 3 0 3 -5 0 0 -1

D) A = 10

2 -3 1 1

0 0 1 0 10 -1 1

12

5 4 3 0 3 -5 0 0 -1 5 4 3 0 11 1 0 0 20

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 , d = 54 64) .1 .6 72

A) x = 43

B) x =

49

.2 .1 .1

212

.4 .1 .3

297

5 23

C) x = 51 5

D) x = 206 231

65) C = .3 .2 .3 , d = 322 108

A) x = 105 91

65) 104

B) x = 217

C) x =

206

723 973 -297

480

D) x = 892 826

Solve the problem.

66) Compute the matrix of the transformation that performs the shear transformation x A = 1 0.18 and then scales all x-coordinates by a factor of 0.60.

0 1

A)

B) 1.6 0.18 0 2

C) 0.60 0.18 0 1

Ax for

66)

D) 1 0.18 0 0.60

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 and then scales all y-coordinates by a factor of 0.60.

0 1

A)

B) 0.60 0.138 0 1

C) 2 0.23 0 1.6

D) 1 0.23 0 0.60

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)

A)

B) 103 017 000

C) 107 013 001

-1 0 0 0 -1 0 0 01

D) 300 070 001

69) Reflect through the x-axis A) B)

68)

103 017 001

69) C) 100 010 001

D) -1 0 0 010 001

13

1 00 0 -1 0 0 01

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)

B) 0.3 2 -0.2 2 0

0.3 2 0.2 2 0

0 0 1

C)

0 -0.6 0.4 0 0 0

0 0 1

0.3 0.3 2 0

-0.2 2 0.2 0

D) 0.3 2 0.2 2 0

-0.3 2 0.2 2 0

0 0 1

0 0 1

71) Translate by (4, 3), and then reflect through the line y = x.

A) B) C) 031 400 001

-1 0 -4 0 -1 -3 0 0 1

71) D) 014 103 001

013 104 001

Find the 4 × 4 matrix that produces the described transformation, using homogeneous coordinates.

72) Translation by the vector (9, -9, -2)

A)

B) 9 0 00 0 -9 0 0 0 0 -2 0 0 0 01

C) 000 9 0 0 0 -9 0 0 0 -2 000 1

0.5 - 3/2 0 0

3/2 0.5 0 0

0 0 1 0

0 0 0 1

0.5 0 - 3/2 0

0 1 0 0

3/2 0 0.5 0

0 0 0 1

D) 1 0 0 -9 010 9 001 2 000 1

73) Rotation about the y-axis through an angle of 60° A)

72)

100 9 0 1 0 -9 0 0 1 -2 000 1

73) B)

C)

3/2 0 -0.5 0

0 1 0 0

0.5 0 3/2 0

0 0 0 1

1 0 0 0

0 0.5 3/2 0

0 3/2 0.5 0

0 0 0 1

D)

Determine whether b is in the column space of A. 1 2 -3 6 74) A = 1 4 -6 , b = 7 -3 -2 5 -10 A) Yes -1 0 2

75) A = 5 8 -10 , b = -3 -3 6 A) Yes

74) B) No

-5 4 4

75) B) No

14

Find a basis for the null space of the matrix.

1 0 -3 -2 76) A = 0 1 7 -4 00 0 0

A)

76)

B) 1 0 0 , 1 0 0

C)

D) 3 2 -7 , 4 1 0 0 1

-3 -2 7 , -4 1 0 0 1

1 0 0 , 1 -3 7 -2 -4

1 0 -5 0 -2 77) A = 0 1 5 0 2 00 01 1 00 00 0

A)

77)

B) 5 2 -5 -2 1 , 0 0 -1 0 1

C)

D) 1 0 0 0 , 1 , 0 0 0 1 0 0 0

1 0 0 1 -5 , 5 0 0 -2 2

-5 -2 5 2 1 , 0 0 -1 0 1

Find a basis for the column space of the matrix.

1 -2 2 -3 78) B = 2 -4 7 -2 -3 6 -6 9

A)

78)

B) 1 -2 2 , -4 -3 6

C) 1 2 2 , 7 -3 -6

D) 1 0 0 , 1 0 0

17 3

2 1 , 0 0 4 0 3 1

15

1 0 -4 0 -4 79) B = 0 1 4 0 5 00 0 1 1 00 0 0 0

79)

A)

B) 1 0 -4 0 , 1 , 4 0 0 0 0 0 0

4 4 -4 -5 1 , 0 0 -1 0 1

C)

D) 1 0 0 , 1 0 0 0 0

1 0 0 0 , 1 , 0 0 0 1 0 0 0

The vector x is in a subspace H with a basis

80) b1 = 1 , b2 = -5 , x = 3

-2

A)

= {b1 , b2 }. Find the -coordinate vector of x.

28 -21

80)

B)

C) 3 -5

-5 3 2

81) b1 = -2 , b2 = 4

D) -3 5

-4 1

6 18 1 , x = 10 -3 -24

A)

81)

B) 3 -4

C)

D) -3 4 0

-3 4

4 -3

Determine the rank of the matrix.

1 -2 2 -5 82) 2 -4 6 -6 -3 6 -6 15

A) 1

83)

1 0 -3 0 0 1 -4 0 00 0 1 00 0 0

A) 5

82) B) 2

C) 3

D) 4

4 2 1 0

83)

B) 2

C) 3

16

D) 4

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.

7 8 8 1) 2 9 2 9 8 9 A) -353 B) -65 2 1 -1 5 -2 1 -3 A) -20

1) C) 65

D) 1,743

2) -2 2

2) B) 0

C) -40

D) 40

2 1 4

3) 4 1 4 2 5 1 A) 38 8 -7 4 6 3 0 0 -3 A) 123

3) C) 142

B) -38

D) 22

4) 0

5)

6)

4 0 0 4 -1 0 -1 3 5 A) -32

C) 144

B) -144

D) -165

5) C) 8

B) -20

D) 20

-4 2 -2 5 0 -1 2 -2 0 3 0 0 0 -3 1 4

A) 0

7)

4)

4 0 0 0 0

6)

B) 120

C) -120

D) -30

2 -2 5 -4 2 2 1 0 0 -2 3 7 0 0 -1 -5 0 0 0 2

A) -32

7)

C) 0

B) -16

1

D) 32

4

1

2

8

5

3

8) -4 -1 -6

B) 0

A) -12

9)

D) -48

2 0 -4 -2

9) C) 84

B) -42

D) 42

1 -3 1 5 -1 6 3 5 4 5 1 3

2 -2 2 2 6 -6 -2 2

10)

B) 60

A) -1

11)

C) 48

1 2 5 6 12 18 -2 3 -1

A) 14

10)

8)

C) 1

D) 0

6 4 7 1 15 14 -6 1

A) 60

11)

B) -120

C) -30

D) -60

Solve using Cramer's rule.

12) 2x1 + 2x2 = 8

12)

6x1 + x2 = -6

A) (-2, 6)

B) (2, -6)

C) (-6, -2)

D) (6, -2)

13) 3x1 - 2x2 = 6

13)

3x1 + 2x2 = 42

A) (9, 8)

B) (-8, -9)

C) (8, 9)

D) (-9, 8)

14) 2x1 + 3x2 = 22

14)

2x1 - 2x2 = 12

A) (-8, -2)

B) (-2, 8)

C) (2, 8)

2

D) (8, 2)

Determine the values of the parameter s for which the system has a unique solution, and describe the solution.

15) 5sx1 + 4x2 = -3 15) 5x1 + sx2 = 4

A) s ± 4; x1 = -3s - 16 and x2 = 4s + 3 B) s ± 5; x1 =

4s+ 3 -3s - 16 and x2 = 5(s - 5)(s + 5) 5(s - 5)(s + 5)

C) s ± 2; x1 =

4s+ 3 -3s - 16 and x2 = 5(s - 2)(s + 2) (s - 2)(s + 2)

D) s 2; x1 = -3s + 16 and x2 = 4s - 3 5(s - 2)(s + 2) (s - 2)(s + 2) 16) sx1 - 4sx2 = 3

16)

3x1 - 12sx2 = 5

4 9 - 5s A) s 1; x1 = and x2 = 12(s - 1)(s + 1) 12(s - 1)(s + 1)

B) s 1; x1 = 14 and x2 = 9 + 5s 3(s + 1) 12s(s + 1) C) s ± 1; x1 =

4 9 - 5s and x2 = 3(s + 1) 12s(s + 1)

D) s 0, 1; x1 =

4 9 - 5s and x2 = 3(s - 1) 12s(s - 1)

Calculate the area of the parallelogram with the given vertices.

17) (0, 0), (4, 7), (13, 11), (9, 4) A) 94 B) 46 C) 55

18) (-1, -2), (2, 7), (7, 2), (10, 11) A) 60 B) 120

17) D) 47 18)

C) 59

3

D) 66

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.

A: All polynomials of the form p(t) = a + bt2 , 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.

V is the line y = x in the xy-plane: V =

x : y=x y

W is the union of the first and second quadrants in the xy-plane: W = U is the line y = x + 1 in the xy-plane: U =

A) V only

B) U and V

x : y=x+1 y C) U only

1)

2) x :y 0 y

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 + bt2 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 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

4)

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 : x, y not both zero . Determine whether H is a vector space. If it is not a vector space, y 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. a + 6b 7) W is the set of all vectors of the form 5b , where a and b are arbitrary real numbers.

7) 6a - b -a

A)

B) 1 6 5 , 0 6 -1 0 -1

1 6 0 0 , 0 , 5 6 -1 0 0 0 -1 D) Not a vector space

C) 1 6 0 , 5 6 -1 0 -1

2

a - 2b 8) W is the set of all vectors of the form 5 , where a and b are arbitrary real numbers.

3a + b -a - b A) B) Not a vector space 1 -2 0 , 5 3 1 -1 -1

C)

8)

D) 1 -2 5 , 0 3 1 -1 -1

1 -2 0 0 , 0 , 5 3 1 0 -1 -1 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 3 7 = = , v , and y 4 3 0 6 .

0 h -8 A) all h -12 B) h = -12 or 0

1 2 , -4

9)

v2 =

C) h = -28

D) h = -12

Determine whether the vector u belongs to the null space of the matrix A. 2 10) u = 4 , A = -2 3 -8 -3 -1 10 1 A) No B) Yes -2

-1 -1

-3

1

3 -2

6

10)

11) u = -1 , A = -3 -4 -10

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

A)

B) 2 0 -6 1 , -1 , -4 0 1 0 0 0 1

C) 2 2 -1 , -4 1 0 0 1

D) 2 2 2 1 , -1 , -4 0 1 0 0 0 1

3

0 -6 -1 , -4 1 0 0 1

12)

1 -2 3 -3 -1

13) A = -2 5 -5 4 -4

13)

-1 3 -2 1 -5

A)

B) 7 12 -5 6 -1 -2 1 , 0 , 0 0 1 0 0 0 1

-5 7 13 6 -1 2 1 , 0 , 0 0 1 0 0 0 1

C)

D) 1 0 0 1 , 5 1 -7 -2 -13 -6

2 -3 3 -1 1 -1 2 6 , , , 0 1 0 0 0 0 1 0 0 0 0 1

Find a matrix A such that W = Col A. 2s - 2t 14) W = : s, t in 6t -3s + t

A)

B) 2 6 -3 2 0 1

15) W =

14) C) 2 0 -3 -2 6 1

D) 2 -2 0 6 -3 1

22 60 -3 1

2r - t 4r - s + 3t : r, s, t in s + 3t r - 5s + t

15)

A)

B) 2 -1 4 3 1 3 1 -5

2 0 -1 4 -1 3 0 1 3 1 -5 1

C)

D) 2 4 0 -1 -1 3

0 1 1 -5 3 1

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 .

1 -2 0 6 16) A = -4 5 16) -1 -3 -3 3

A) k = 5, m = 2

B) k = 2, m = 2

C) k = 5, m = 5

4

D) k = 2, m = 5

17) A =

4 0 0 -1 1 -7 2 6 -5 -1 0 3 -3 -4 4 -1 -4 4

A) k = 6, m = 3

17)

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. 5 1 -3 4 18) u = -3 , A = -1 0 -5 3 -3 6 -5

A) Not in Col A, in Nul A C) In Col A and in Nul A 1 -1 3 -2 19) u = , A= 1 3 -2 -1

0 -1 -3 3

18)

B) Not in Col A, not in Nul A D) In Col A, not in Nul A

3 -4 0 6

19)

A) Not in Col A, in Nul A C) Not in Col A, not in Nul A

B) 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) = t2 , p3 (t) = 4 + 4t B: The set p1 , p2 , p3

where p1 (t) = t, p2 (t) = t2 , p3 (t) = 4t + 4t2

C: The set p1 , p2 , p3

where p1 (t) = 1, p2 (t) = t2 , p3 (t) = 4 + 4t + t2

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] 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

21)

C) C only

5

D) A and C

E) A and B

Determine whether {v1 , v2 , v3 } is a basis for 3 .

1

22) v1 = -3 , v2 = -5

2 -3 8 , v3 = -2 -2 -1

22)

A) Yes

B) No

1 4 -2 4 , v2 = 0 , v3 = -4 2 -6 -2 A) Yes

23) v1 =

Solve the problem.

23) B) No

1 8 -3 3 , v2 = 1 , v3 = 4 , and H = Span v1 , v2 , v3 .

-3 -2 -2

24) Let v1 =

24)

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

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 1 0 B= 0 0 0

1 0 0 0 0

0 -4 1 -1 0 0 0 0 0 0

0 0 1 0 0

25)

0 0 0 .

1 0

A)

B) 1 0 0 0 0 1 0 0 0 , 0 , 1 , 0 0 0 0 1 0 0 0 0

1 0 0 0 0 0 0 , 1 , 0 0 0 1 0 0 0

C)

D) 1 0 0 0 0 , 1 , 0 , 0 0 0 1 0 0 0 0 1

1 0 0 , 0 0

6

1 0 0 1 0 , 0 , 0 0 0 0

0 0 -4 0 0 -1 0 , 1 , 0 0 0 1 0 0 0

-1 26) Let A = 1 2 3

3 -2 -4 -6

7 -7 -9 -11

2 -1 -5 -9

0 1 -3 3 and B = 0 1 1 0 0 -1 0 0

-7 0 5 0

-2 1 -3 0

0 3 .

-5 0

26)

It can be shown that matrix A is row equivalent to matrix B. Find a basis for Col A.

A)

B) 3 2 -1 1 , -2 , -1 2 -4 -5 3 -6 -9

1 -3 -7 0 , 1 , 0 0 0 5 0 0 0

C)

D) 3 7 2 0 -1 1 , -2 , -7 , -1 , 3 -9 -5 2 -4 1 3 -6 -11 -9 -1

3 7 -1 1 , -2 , -7 -9 2 -4 3 -6 -11

Determine whether the set of vectors is a basis for 3 .

1 0 27) Given the set of vectors 0 , 1 , decide which of the following statements is true: 0 2 A: Set is linearly independent and spans 3 . Set is a basis for 3 .

27)

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 .

A) D B) C C) B D) A

28) Given the set of vectors

1 0 0 0 0 , 1 , 0 , 1 0 0 1 1

, decide which of the following statements is true:

28)

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 .

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 , 0 1 1

A)

, [x]B = -6 5

29) B)

-3 2

C) 18 -1

D) -15 -1

7

18 5

30) B =

1 -3 2 -3 , 8 , -2 2 -4 -5

, [x]B =

A)

2 4 -1

30)

B)

C) 0 12 7

-12 28 -7

D) -10 28 3

-4 6 5

Find the coordinate vector [x]B of the vector x relative to the given basis B.

31) b1 = 3 , b2 = 4

4 , x = 25 , and B = -5 -8

A)

31)

B) 3 5

32) b1 =

b1 , b 2

C) 43 140

2 1 -8 3 , b2 = -5 , x = 1 , and B = 11 -3 -1

A)

3 4

25 -8

b1 , b2

32)

B) -3 -2

D)

C)

D) -15 -29 23

-3 -3

-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, t2 .

33) 1 + 2t, 3 + 6t2, 1 + 3t + 4t2 A) Linearly independent

33) B) Linearly dependent

34) 1 + 2t + t2, 3 - 9t2, 1 + 4t + 5t2 A) Linearly independent Solve the problem.

35) Let H =

34) B) Linearly dependent

a + 3b + 2d c+d : a, b, c, d in -3a - 9b + 4c - 2d -c - d

35)

Find the dimension of the subspace H. 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 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

36)

1 -2 3 1 0 5 -4 0 0 1 2 0 -6 -2 37) A = 0 0 0 0 0 1 3 0 0 0 0 0 0 0 A) dim Nul A = 4, dim Col A = 3

37) B) dim Nul A = 2, dim Col A = 5 D) dim Nul A = 5, dim Col A = 2

C) dim Nul A = 3, dim Col A = 4

Solve the problem.

38) Determine which of the following statements is false.

A: The dimension of the vector space P6 of polynomials is 7.

38)

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 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. 1 3 -4 0 1 1 3 -4 0 1 2 4 2 0 3 0 -5 -4 -2 -4 40) A = ,B= 1 -5 0 -3 2 0 0 -8 13 1 -3 -1 8 3 -4 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. A) Rank A = 1, Dim Row A = 1, Dim Col A = 5

41)

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? A) 5 B) 4 C) 9

9

42) D) 0

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)

44) Suppose a nonhomogeneous system of 11 linear equations in 14 unknowns has a solution

44)

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.

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 = b1, b2 and C = c1 , c2 for a vector space V such that

45)

b1 = c1 - 4c2 and b2 = 5c1 + 2c2 . Suppose x = b1 + 3b2 . That is, suppose [x]B = 1 . Find [x]C. 3

A)

B) 16 2

C)

D) 8 -10

-11 11

15 2

46) Consider two bases B = b1, b2 , b3 and C = c1, c2 , c3 for a vector space V such that

46)

b1 = c1 + 4c3 , b2 = c1 + 6c2 - c3 , and b3 = 5c1 - c2 . Suppose x = b1 + 3b2 + b3 . That is, 1 suppose [x]B = 3 . Find [x]C. 1

A)

B) 9 21 -3

C) 9 17 7

D) 5 18 2

9 17 1

Find the specified change-of-coordinates matrix.

47) Let B = b1 , b2 and C = c1 , c2 be bases for 2 , where

47)

b1 = -1 , b2 = -2 , c1 = 1 , c2 = -4 .

-10 5 4 3 Find the change-of-coordinates matrix from B to C.

A)

B) -1 4

C)

-

4 3

D) 15 4

5 -6 3

-2 5

5

10

18 5

1 -4 3 -10

48) Consider two bases B = b1, b2 and C = c1 , c2 for a vector space V such that

48)

b1 = c1 - 6c2 and b2 = 4c1 - 3c2 . Find the change-of-coordinates matrix from B to C.

A)

B) 1 4 -6 -3

C) 1 -6 4 -3

D) 0 4 -6 -3

14 63

Find the indicated sums of the signals in the Table

49) Find the indicated sum of the signals.

49)

+

A) (..., 0, 2, 0, 2, 0, 0, 2, ...)

C) (..., -1, 1, -1, 2, -1, 1, -1, ...)

B) (..., 1, 1, 1, 2, 2, 2, 2, ...)

D) (..., 0, 0, 0, 2, 1, 1, 1, ...)

50) Find the indicated sum of the signals.

50)

+F

A) (..., 2, -1, 1, 3, 1, 1, 2, ..)

C) (..., 2, -1, 1, 3, 4, 4, 5, ..)

B) (..., 0, 0, 0, 3, 1, 1, 1, ...)

D) (..., -1, 1, -1, 4, -1, 1, -1, ...)

51) Which signal(s) are in the kernal of I-S2 ?

A) B) and

51) C) and

52) Find a nonzero signal in the kernal of T({x})={xk-xk-1 -xk-2 } A) B) F C)

11

D) 52) D)

k 53) Construct a linear time invariant transformation that has the signal {x}= 4 in its kernel.

5

A) I- 4 S 5

B) I+ 5 S

C) I+ 4 S

4

5

53)

D) I- 5 S 4

54) Let W= {xk} | xk= r if k is a multiple of 2

where r can be any real number. A typical signal -r if k is not a multiple of 2 in W looks like (..., -r, r, -r, r, -r, r, -r, ...). Is W is a subspace of S? Explain.

A) Yes. For each signal {xk} in W, there exists a signal {-xk} in W, such that {xk}+{-xk}={0}.

54)

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 where r can be any real number. A typical signal in W looks like r+k if k 0 (..., 0, 0, 0, r, r+1, r+2, r+3, ...). Is W is a subspace of S? Explain.

A) No. The zero vector is not in W.

55)

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

where r can be any real number. A typical signal 2r if k is not a multiple of 2 in W looks like (..., 2r, r, 2r, r, 2r, r, 2r, ...). Find a basis for the subspace W. What is the dimension of W? A) Basis: { +S-1 )} B) Basis: {3 +S( )}

Dimension:

Dimension: 1

C) Basis: { + }

D) Basis: {3 + }

Dimension:

Dimension: 1

12

56)

57) Let W= {xk} | xk=

-rk if k<0 where each rk can be any real number. A typical signal in W looks rk if k 0

57)

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?

A) {Sn )+S-n ) | 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) {Sn ) | 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) {Sn )-S-n ) | 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) {Sn ) | 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.

2 k, (-3)k; yk+2 +yk+1 -6yk=0

A) 2k+2 +2 k+1-6·2k=0

B) (-3)k+2+(-3)k+1 -6·(-3)k=0

2 k+2 +2 k+1 -6·2 k=0 2 0+2 +2 0+1 -6·2 0 =0

(-3)k+2 +(-3)k+1 -6·(-3)k=0 (-3)0+2 +(-3)0+1 -6·(-3)0 =0

2 k(2 2 +2 1 -6)=0 2 k(0)=0

(-3)k((-3)2 +(-3)1 -6)=0 (-3)k(0)=0

2 2 +2 1 -6=0 0=0

(-3)2 +(-3)1 -6=0 0=0

D) 2k+2 +2 k+1-6·2k=0

C) 2k+2 +2 k+1-6·2k=0

2 0+2 +2 0+1 -6·2 0 =0 2 2 +2 1 -6=0

2 k(2 2 +2 1 -6)=0 2 k(0)=0

0=0

(-3)k+2 +(-3)k+1 -6·(-3)k=0 (-3)k((-3)2 +(-3)1 -6)=0

(-3)k+2 +(-3)k+1 -6·(-3)k=0 (-3)0+2 +(-3)0+1 -6·(-3)0 =0

(-3)k(0)=0

(-3)2 +(-3)1 -6=0 0=0

13

58)

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)

6 yk+2 -yk+1 + yk=0 25

A) 3k and 5 k

k B) 6k and 1

D) 2

C) 2k and 3 k

25

5

k

and

3 k 5

62) Find a basis for the solution space of the given difference equation. yk+2 -16yk=0 A) 2k and (-8)k

B) (-2)k and 8 k

62)

C) 2k and 8 k

D) 4k and (-4)k

63) The Pell Sequence can be viewed as the sequence of numbers where each number is the sum of

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 1+ 2 k - 1 1- 2 k

k k B) yk= 1+ 2 - 1- 2

C) yk= 1+ 2 k - 1 - 2 k

D) yk= 2+ 2 1+ 2 k + 2- 2 1- 2 k

2 2

2 2

2 2

4

14

2 2

4

63)

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 1+ 2 k + 2- 2 1- 2 k

k k B) yk= 1+ 2 - 1- 2

C) yk= 1 1+ 2 k - 1 1- 2 k

D) yk= 1+ 2 k - 1- 2 k

4

2 2

4

2 2

2 2

2 2

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

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 1 1 + y + y =z , for k=1, 2, ... y 3 k+1 3 k 3 k-1 k 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 z 1 , ..., z9 .

A) 7, 4, 4, 6, 8, 7, 3, 2, 3 C) 7, 5, 4, 6, 7, 8, 4, 5, 5

B) 8.5, 5.5, 2.5, 4, 8.5, 5.5, 2.5, 1, 4 D) 8.5, 6, 4, 2.5, 2, 2.5, 7, 8.5, 10

15

66)

67) Show that the given signal is a solution of the difference equation. Then find the general solution

67)

of the difference equation.

yk=k2 ; yk+2 +4yk+1 -5yk=12k+8 A) (k+2)2 +4(k+1)2 -5k2 =12k+8

B) (k+2)2 +4(k+1)2 -5k2 =12k+8

yk=k2 +c1 (-5)k+c2 C) (k+2)2 +4(k+1)2 -5k2 =12k+8

yk=c1 (-5)k+c2 D) (k+2)2 +4(k+1)2 -5k2 =12k+8

(0+2)2 +4(0+1)2 -5*0 2 =12*0+8 8=8

(k2 +4k+4)+(4k2 +8k+4)-5k2 =12k+8 12k+8=12k+8

(0+2)2 +4(0+1)2 -5*0 2 =12*0+8 8=8

(k2 +4k+4)+(4k2 +8k+4)-5k2 =12k+8 12k+8=12k+8

yk=c1 (-5)k+c2

yk=k2 +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

B) (4+k)+3(3+k)+2(2+k)=17+6k

yk=2 k+c1 (-1)k+c2 (-2)k

yk=2+k+c1 (-1)k+c2 (-2)k

C) (4+k)+3(3+k)+2(2+k)=17+6k

D) (2+k)+3(3+k)+2(2+k)=15+6k

4+k+9+3k+4+2k=17+6k 17+6k=17+6k

4+k+9+3k+4+2k=17+6k 17+6k=17+6k

4+k+9+3k+4+2k=17+6k 17+6k=17+6k

2+k+9+3k+4+2k=15+6k 15+6k=15+6k

yk=c1 (-1)k+c2 (-2)k

yk=c1 (-1)k+c2 (-2)k

69) Is the following difference equation of order 3? Explain. 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 a 0 yk+n+...+ anyn= z k.

16

69)

70) Write the difference equation as a first-order system, x k+1=Ax k, for all k. yk+3 +6yk+2 -2yk=0

A) x k+1=Ax, where

B) x k+1=Ax, where

yk+1 0 10 A= 0 0 1 and x k= yk+2 yk+3 -6 0 2

yk 010 A= 0 0 1 and x k= yk+1 6 0 -2 yk+2

C) x k+1=Ax, where

D) x k+1=Ax, where

yk+1 010 A= 0 0 1 and x k= yk+2 2 0 -6 yk+3

yk 010 A= 0 0 1 and x k= yk+1 2 0 -6 yk+2

17

70)

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 , = 6 150 31

A)

B) 6 1

C)

2) A = -18 -5 , 60 17

D) 1 6

-6 1

1) 1 -6

=2

A)

2) B)

1 -4

C) 1 17

D) 1 0

-4 1

For the given matrix A, find a basis for the corresponding eigenspace for the given eigenvalue.

1 -5 -5 3) A = - 5 1 5 , = -4 5 -5 -9

A)

B) 1 0 0 , 1 -1 1

4 0

0 0 , 14 5 -4

4) A = 6 1 A)

C) 1 0 -1

D) 0 1 -1

1 1 1 , 0 0 1

=4

4) B)

1 1 2 , 0 0 -3

C) 1 -2 -3

D) 1 2 3

1 1 2 , 0 0 3

Find the eigenvalues of the given matrix.

5) -4 -1 6 1

A) -2, -1

5)

B) -2, 1

C) -2

D) 1

6) -26 -16

6)

56 34

A) 6

3)

B) 2, 6

C) -2

1

D) -2, -6

Find the characteristic equation of the given matrix.

7 8 4 2 7) A = 0 5 -8 3 0 0 9 9 0 0 0 7

7)

A) (7 - )2(5 - )(9 - ) = 0 C) (7 - )(5 - )(9 - ) = 0

B) (7 - )(8 - )(4 - )(2 - ) = 0 D) (7 - )(9 - )(3 - )(2 - ) = 0

-2 5 2 6 8) A = 0 4 -5 1 0 0 -3 2 00 03

8)

A) (-2 - )(5 - )(2 - )(6 - ) = 0 C) (3 - )(2 - )(1 - )(6 - ) = 0

B) (-2 - )(4 - )(-3 - )(3 - ) = 0 D) (6 - )(1 - )(2 - )(3 - ) = 0

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities.

9) 5 - 7 4 - 18 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) 5 - 11 4 - 45 3 + 567 2 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)

10)

Find a formula for Ak, given that A = PDP-1 , where P and D are given below.

11) A = -2 10 , P = 5 1 , D = 6 0 41 0 8 -8 16

A)

11)

B) 5 · 6k - 4 · 8k 4 · 6k + 4 · 8k

5 · 8k + 5 · 6k 5 · 8k - 4 · 6k

C)

5 · 6k + 4 · 8k 4 · 6k + 4 · 8k

5 · 8k + 5 · 6k 5 · 8k + 4 · 6k

5 · 6k - 4 · 8k 4 · 6k - 4 · 8k

5 · 8k - 5 · 6k 5 · 8k - 4 · 6k

D) 6k 0

0

8k

2

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that A= PDP-1 .

-11 3 -9 12) A = 0 -5 0 12) 6 -3 4

A)

B) 1 0 -1 -5 0 -2 P = 5 3 0 , D = 0 -5 0 1 1 1 0 -5 -2

1 0 -1 -5 0 0 P = 5 3 0 , D = 0 -5 0 1 1 1 0 0 -2

C)

D) 1 0 -1 -5 0 0 P= 0 3 0 ,D= 0 1 0 1 1 1 0 0 -2

1 5 -1 -5 1 0 P = 5 3 0 , D = 0 -5 0 1 3 1 0 0 -2

4 0 0

13) A = 1 4 0

13)

0 0 4

A)

B) 1 0 0 4 1 0 P= 4 4 0 ,D= 0 4 0 0 1 1 0 0 4

1 4 1 4 0 0 0 4 1 ,D= 0 4 0 -1 0 1 0 0 4 D) Not diagonalizable P=

C) 1 0 -1 4 0 1 P= 4 4 0 ,D= 1 4 1 1 1 1 0 0 4 -4 0 0 0 14) A = 0 -4 0 0 1 -4 4 0 -1 2 0 4

14) B) Not diagonalizable

A) 8 16 0 0 -4 -4 0 0 , D = P= 1 0 1 0 0 1 0 1

4 0 0 0

0 0 0 4 0 0 0 -4 0 0 0 -4

8 -4 1 0 4 16 0 0 -4 P= ,D= 0 0 0 1 0 0 0 0 0 1 0

0 0 0 4 0 0 0 -4 0 0 0 -4

C)

D) 8 16 0 0 -4 0 0 0 4 4 0 0 P= , D = 0 -4 0 0 1 0 1 0 0 0 4 0 0 1 0 1 0 0 0 4

3

8 0 15) A = - 12 0

0 8 3 0

0 0 0 0 2 12 0 8

15)

A)

B) 2 P= 0 1 0

0 -2 1 2 1 0 ,D= 0 0 1 0 1 0

8 0 0 0

0 8 0 0

0 0 8 0

0 0 0 2

4 -2 1 0 P = 8 -2 0 0 , D = 1 0 1 1 0 1 1 0

8 0 0 0

0 8 0 0

0 0 2 0

0 0 0 2

2 0 1 0 P= 0 2 0 0 ,D= -2 1 0 1 1 0 1 0 D) Not diagonalizable

C)

8 0 0 0

0 8 0 0

0 0 8 0

0 0 0 2

Find the matrix of the linear transformation T: V W relative to B and C. 16) Suppose B = {b1 , b2} is a basis for V and C = {c1, c2 , c3 } is a basis for W. Let T be defined by

16)

T(b1 ) = -4c1 - 7c2 + 4c3 T(b2 ) = -4c1 - 14c2 + 5c3

A)

B) -4 -7 4 0 7 -1

C) -4 -7 4 -4 -14 5

D) -4 0 -7 -7 4 1

-4 -4 -7 -14 4 5

17) Suppose B = {b1 , b2, b3 } is a basis for V and C = {c1 , c2} is a basis for W. Let T be defined by

17)

T(b1 ) = 5c1 + c2 T(b2 ) = -5c1 + 5c2 T(b3 ) = 5c1 + 2c2

A)

B) 6 0 7 1 5 2

C) 5 1 -5 5 5 2

D) 5 6 -5 0 5 7

5 -5 5 1 5 2

Define T: R2 R2 by T(x) = Ax, where A is the matrix defined below. Find the requested basis B for R2 and the corresponding B-matrix for T. 18) Find a basis B for R2 and the B-matrix D for T with the property that D is a diagonal matrix.

18) A = -1 -3 4 6

A)

B) B=

1 , -1 3 4

,D= 2 0 0 3

B=

1 , 3 -1 4

,D= 2 0 0 3

C)

B=

1 , 3 -1 -4

,D= 2 0 0 3

B=

3 , 1 -4 -1

,D= 2 0 0 3

D)

4

19) Find a basis B for R2 and the B-matrix D for T with the property that D is an upper triangular

19)

matrix.

A = -134 -529 36 142

A)

B) B=

23 , 4 6 1

, D = -4 1 0 -4

C)

B=

23 , -4 1 -6

,D= 4 1 0 4

B=

23 , -4 1 -6

,D= 4 1 0 5

D) B=

23 , -6 1 -4

,D= 4 1 0 4

Find the eigenvalues of A, and find a basis for each eigenspace.

20) A = 0.4 3.4 -0.2 0.8 A) 0.6 + 0.8i, 1 + 4i ; 0.6 - 0.8i, 1 - 4i 1 1 1 4i B) -0.6 + 0.8i, ; -0.6 - 0.8i, 1 + 4i 1 1 1 4i 1 4i + C) -0.6 - 0.8i, ; -0.6 + 0.8i, 1 1 D) 0.6 - 0.8i, 1 + 4i ; 0.6 + 0.8i, 1 - 4i 1 1

21) A =

3 2 -2 3

A) 3 - 2i, C) 3 - 2i,

20)

21) 1 ; 3 + 2i, 1 i -i 1 ; 3 + 2i, 1 i -i

B) 3 - 2i, D) 3 - 2i,

1 - 3i -2 1 + 3i -2

; 3 + 2i, ; 3 + 2i,

1 + 3i -2 1 - 3i -2

Determine whether the origin is an attractor, repellor, or a saddle point of the dynamical system x k+1 = Ax k, where A is given below. Determine the direction of greatest attraction or repulsion, appropriately.

22) A = 1 -0.6 0.8 -0.4 A) Attractor; direction of greatest attraction: along the line through 0 and -3 -4

22)

B) Saddle point; direction of greatest attraction: along the line through 0 and -3 , direction of greatest repulsion: along the line through 0 and 1 1

C) Attractor; direction of greatest attraction: along the line through 0 and 1

1 1 D) Repellor; direction of greatest repulsion: along the line through 0 and 1

5

-4

23) A = 2.4

0 0 1.2

23)

A) Attractor; direction of greatest attraction: along the line through 0 and 1 0 B) Attractor; direction of greatest repulsion: along the line through 0 and 0 1 C) Repellor; direction of greatest repulsion: along the line through 0 and 1 0

D) Saddle point; direction of greatest attraction: along the line through 0 and 1 , direction of 0

greatest repulsion: along the line through 0 and 0 1

24) A = -4.5 7.5 -5

24)

8

A) Repellor; direction of greatest repulsion: along the line through 0 and -3 -2

B) Saddle point; direction of greatest attraction: along the line through 0 and 1 , direction of 1

greatest repulsion: along the line through 0 and 3 2

C) Saddle point; direction of greatest attraction: along the line through 0 and -3 , direction of -2

greatest repulsion: along the line through 0 and 1 1

D) Attractor; direction of greatest attraction: along the line through 0 and 1 1

Consider the difference equation x k+1 = Ax k, where A has eigenvalues and corresponding eigenvectors v1 , v2 , and v3 given below. Find the general solution of this difference equation if x 0 is given as below.

2 1 -3 -17 3 , v2 = 1 , v3 = 2 , and x 0 = 17 1 2 -2 -16 k k k A) xk = 4(2) v1 - 5(0.9) v2 + (0.3) v3 B) xk = 4(2)kv1 - 5(0.9)kv2 + 5(0.3)kv3 C) xk = (2)kv1 + (0.9)kv2 + (0.3)kv3 D) xk = (2)kv1 - 5(0.9)kv2 + 5(0.3)kv3

25) 1 = 2, 2 = 0.9, 3 = 0.3, v1 =

Solve the initial value problem.

26) x = Ax, x(0) = -5 , where A = 4 -0.4 15 10 4 3 sin 2t 5 cos 2t + 4t A) x(t) = e -15 cos 2t + 25 sin 2t

25)

26) B) x(t) =

C) x(t) = -3 sin 2t - 5 cos 2t e4t

3 sin 2t + 5 cos 2t e-4t -15 cos 2t + 25 sin 2t

D) x(t) = -3 sin 2t - 5 cos 2t e-4t

15 cos 2t - 25 sin 2t

15 cos 2t - 25 sin 2t

6

Apply the power method to the matrix A below with x 0 = 0 . Stop when k = 5, and determine the dominant eigenvalue 1 and corresponding eigenvector.

27) A = -6 -2 27) 3 -1

A)

B) -4, 2 -3

C) -4, 1 -1

D) -3, 1 -1

-3,

2 -3

2 -28 11

28) A = -4 A)

28) B)

3, 1 4

C) 4, 2 7

D) 4, 1 4

3, 2 7

Use the inverse power method to determine the smallest eigenvalue of the matrix A. 29) Assume that the eigenvalues are roughly 0.8, 4.1, and 17.

1 0 0 A = -1.5 2.5 1.5 -1.5 1.5 5.5 A) 7

B) 1

C) 0.8

29)

D) 4

Solve the problem.

30) Suppose that demographic studies show that each year about 6% of a city's population moves to the suburbs (and 94% stays in the city), while 4% of the suburban population moves to the city (and 96% remains in the suburbs). In the year 2000, 65.7% of the population of the region lived in the city and 34.3% lived in the suburbs. What is the distribution of the population in 2002? For simplicity, ignore other influences on the population such as births, deaths, and migration into and out of the city/suburban region.

A) 60.8% in the city and 39.2% in the suburbs

30)

B) 59.6% in the city and 40.4% in the suburbs C) 61.9% in the city and 38.1% in the suburbs D) 63.1% in the city and 36.9% in the suburbs 31) Suppose that the weather in a certain city is either sunny, cloudy, or rainy on a given day, and

consider the following: 1) If it is sunny today there is a 70% chance it will be sunny tomorrow and a 30% chance that it will be cloudy.

2) If it is cloudy today there is a 40% chance it will be sunny tomorrow, a 40% chance that it will be cloudy, and a 20% chance that it will be rainy.

3) If it is rainy today there is a 40% chance it will be sunny tomorrow, a 30% chance that it will be cloudy, and a 30% chance that it will be rainy.

Suppose the predicted weather for Friday is 45% sunny, 30% cloudy, and 25% rainy. What are the chances that Sunday will be rainy?

A) 9.65% B) 10.65% C) 13.5% D) 14.5%

7

31)

Find the steady-state probability vector for the stochastic matrix P. 32) P = 0.8 0.7 0.2 0.3

A)

B) 7 2

0.9

0.3 0.3

0

0.1 0.5

32)

C)

D)

4 5

2 9

7 9

1 5

7 9

2 9

33) P = 0.1 0.6 0.2 A)

33) B)

18 5 1

C)

D)

19 26

3 4

3 5

3 13

5 24

3 10

1 26

1 24

1 10

34) Determine if P= 0.75 0 is a regular stochastic matrix.

0.25 1 A) Yes, because the entries in each column sum to 1.

34)

B) No, because the entries in each row do not sum to 1.

C) No, because Pk has a zero in the upper right corner for all k>0.

D) Yes, because P2 has all positive entries.

35) Let P be an n×n stochastic matrix. Determine whether the statement below is true or false. Justify the answer.

A steady-state vector of a matrix P has a corresponding eigenvalue of 1.

A) The statement is true. A steady state vector is a probability vector. Therefore, its entries sum to 1. Any vector whose entries sum to 1 has an eigenvalue of 1.

B) The statement is false. A steady state vector is not an eigenvector. Therefore, it cannot have a corresponding eigenvalue.

C) The statement is false. Although a steady state vector is an eigenvector, its corresponding 1 eigenvalue is not 1. A steady state vector q satisfies the equation Pq= q, where = .

2

D) The statement is true. A steady state vector q satisfies the equation Pq= q, where =1.

8

35)

36) Let P be an n×n stochastic matrix. Determine whether the statement below is true or false. Justify

36)

the answer.

If P has a unique steady-state vector, then P is regular.

A) The statement is true. All stochastic matrices are regular, regardless of the number of steady-state vectors..

B) The statement is false. Regular stochastic matrixes have multiple steady-state vectors.

C) The statement is true. If P is an n×n regular stochastic matrix, then P has a unique steady-state vector.

D) The statement is false. If P is an n×n regular stochastic matrix, then P has a unique steady-state vector, but the converse is not true.

5 8

37) Is q= 3 a steady state vector for A= 0.4 1 ? Justify your answer.

8

0.6 0

37)

A) No, because Aq q.

B) No, because q is not a probability vector.

C) No, because A is not a stochastic matrix.

D) Yes, because A is a stochastic matrix, q is a probability vector, and Aq=q.

38) Is q= 0.5 a steady state vector for A= 0.5 0.2 ? Justify your answer.

0.5 0.5 0.8 A) No, because A is not a stochastic matrix.

38)

B) No, because Aq q.

C) No, because q is not a probability vector.

D) Yes, because A is a stochastic matrix, q is a probability vector, and Aq=q.

Solve the problem.

39) Suppose that demographic studies show that each year about 7% of a city's population moves to the suburbs (and 93% stays in the city), while 2% of the suburban population moves to the city (and 98% remains in the suburbs). In the year 2000, 60% of the population of the region lived in the city and 40% lived in the suburbs. What percentage of the population of the region would eventually live in the city if the migration probabilities were to remain constant over many years?

For simplicity, ignore other influences on the population such as births, deaths, and migration into and out of the region.

A) 22.2% B) 50% C) 77.8% D) 36.1%

9

39)

40) Suppose that the weather in a certain city is either sunny, cloudy, or rainy on a given day, and

40)

consider the following: 1) If it is sunny today there is a 70% chance it will be sunny tomorrow and a 30% chance that it will be cloudy.

2) If it is cloudy today there is a 40% chance it will be sunny tomorrow, a 40% chance that it will be cloudy, and a 20% chance that it will be rainy.

3) If it is rainy today there is a 40% chance it will be sunny tomorrow, a 30% chance that it will be cloudy, and a 30% chance that it will be rainy.

In the long run, how likely is it that the weather will be rainy on a given day?

A) 10.6% B) 9.5% C) 11.3%

D) 12.1%

41) Let P be an n×n stochastic matrix. The following argument shows that the equation Px=x has a

nontrivial solution. Justify each assertion below. a. If all the other rows of P-I are added to the bottom row, the result is a row of zeros. b. The rows of P-I are linearly dependent. c. The dimension of the row space of P-I is less than n. d. P-I has a nontrivial null space.

A) a. The entries in a column of P sum to 1. A column in the matrix P-I has the same entries as in P except that one of the entries is increased by 1. Hence each column sum is 2. b. By (a), the bottom row of P-I is double the sum of the other rows. c. By (b) and the Spanning Set Theorem, the bottom row of P-I can be removed and the remaining (n-1) rows will still span the row space. Let A be the matrix obtained from P-I by adding to the bottom row all the other rows. By (a), the row space is spanned by the first (n-1) rows of A. d. By the Rank Theorem and (c), the dimension of the column space of P-I is less than n, and hence the null space is nontrivial.

B) a. The entries in a column of P sum to 1. A column in the matrix P-I has the same entries as in P except that n of the entries are decreased by 1. Hence each column sum is 0. b. By (a), the bottom row of P-I is the negative of the sum of the other rows. c. By (b) and the Spanning Set Theorem, the bottom row of P-I can be removed and the remaining (n-1) rows will still span the row space. Let A be the matrix obtained from P-I by adding to the bottom row all the other rows. By (a), the row space is spanned by the first (n-1) rows of A. d. By the Diagonalization Theorem and (c), the dimension of the column space of P-I is less than n, and hence the null space is nontrivial.

C) a. The entries in a row of P sum to 1. A row in the matrix P-I has the same entries as in P except that n of the entries is decreased by 1. Hence each row sum is 0. b. By (a), the right column of P-I is the negative of the sum of the other columns. c. By (b) and the Spanning Set Theorem, the right column of P-I can be removed and the remaining (n-1) columns will still span the column space. Let A be the matrix obtained from P-I by adding to the right column all the other columns. By (a), the column space is spanned by the first (n-1) columns of A. d. By the Diagonalization Theorem and (c), the dimension of the row space of P-I is less than n, and hence the null space is nontrivial.

10

41)

D) a. The entries in a column of P sum to 1. A column in the matrix P-I has the same entries as

in P except that one of the entries is decreased by 1. Hence each column sum is 0. b. By (a), the bottom row of P-I is the negative of the sum of the other rows. c. By (b) and the Spanning Set Theorem, the bottom row of P-I can be removed and the remaining (n-1) rows will still span the row space. Let A be the matrix obtained from P-I by adding to the bottom row all the other rows. By (a), the row space is spanned by the first (n-1) rows of A. d. By the Rank Theorem and (c), the dimension of the column space of P-I is less than n, and hence the null space is nontrivial.

42) Does every 2×2 stochastic matrix have at least one steady-state vector? Explain.

Note that any 2×2 stochastic matrix can be written in the form A= 1constants between 0 and 1.

, where

and

are

and -1 , with corresponding eigenvalues 1 and 1- - , 1

A) Yes. The eigenvectors of A are respectively. Normalizing

1-

42)

shows that every 2×2 stochastic matrix has at least one

steady-state vector.

B) Yes. The eigenvectors of A are 1-

and 1 , with corresponding eigenvalues + and 1+ 1 - , respectively. Therefore, a 2×2 stochastic matrix will only have a steady-state vector if + =1 or = .

C) Yes. The eigenvectors of A are 1- and 1 , with corresponding eigenvalues 1 and 1- 1 , respectively. Normalizing 1- shows that every 2×2 stochastic matrix has at least one steady-state vector.

and -1 , with corresponding eigenvalues + and 1+ 1 , respectively. Therefore, a 2×2 stochastic matrix will only have a steady-state vector if + =1 or = .

D) No. The eigenvectors of A are

11

Answer Key Testname: UNTITLED5

1) D 2) A 3) D 4) C 5) A 6) B 7) A 8) B 9) C 10) C 11) D 12) B 13) D 14) D 15) B 16) D 17) D 18) B 19) B 20) D 21) C 22) C 23) C 24) B 25) B 26) C 27) B 28) C 29) B 30) A 31) B 32) D 33) C 34) C 35) D 36) D 37) D 38) B 39) A 40) B 41) D 42) A

12

Exam Name___________________________________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Compute the dot product u · v.

1) u = 12 , v = 4 11 -11 A) -121

1) B) 169

C) 48

D) -73

2) u = 6 , v = 12 0

2)

-3

A) 72

B) 75

D) 69

C) -18

3) u = -11 , v = 0 9 A) 54

3)

6

B) 65

C) 43

D) -66

4) u = 1 , v = 3 19 A) 24

4)

1

B) 59

6 -1 5 ,v= 2 3 -5 A) -1

C) 22

D) 16

5) u =

2 -18 0 ,v= 1 -9 -1 A) 0

5) C) 11

B) -11

D) 0

6) u =

6) B) -27

C) -26

D) -45

Find a unit vector in the direction of the given vector.

7) 12 -24

A)

7)

B)

-

C)

1 5

1 5

2 5

2 5

D) 1 5 -

1

2 5

1 3 -

2 3

8)

-28 28 -14

8)

A)

B) -

2 5

C) -

2 5 -

1 5

-

2 9

2 3

2 9

2 3

1 9

1 3

Find the distance between the two vectors.

9) u = (10, -1), v = (1, 10) A) 202 B) 101

10) u = (-3, 6), v = (6, -6) A) 75 11) u = (0, 0, 0) , v = (8, 2, 2) A) 2 3

D) -

2 3 2 3

-

1 3

9) C) 101

D) 202 10)

B) 3

C) 15

D) 225 11)

B) 72

C) 6 2

D) 12

12) u = (0, 0, 0) , v = (-5, 2, -5) A) 3 6 B) 54

C) 2 -2

D) -8

13) u = (-4, 6, -5) , v = (-3, 10, 4) A) 3 34 B) 7 2

C) -14

D) 98

14) u = (14, 17, 15) , v = (3, 6, 4) A) 363 B) 11 3

C) 33

D) 3 131

12)

13)

14)

15) u = (25, 16, 20) , v = (-5, 4, 8) A) 6 33 B) 12 11

15) C) 1,188

Determine whether the set of vectors is orthogonal.

5 -5 5 16) 10 , 0 , -5 5 5 5 A) Yes

D) 54

16) B) No

12 -12 0 , -12 -12 -12 -12 A) Yes -12

17) -24 ,

17) B) No

2

Express the vector x as a linear combination of the u's.

18) u1 = 2 , u2 = 12 , x = 16 6 28 -4 A) x = -4u1 + 2u2 B) x = 4u1 - 2u2 -2

3

18) C) x = 2u1 - 4u2

D) x = -4u1 - 2u2

4 40 -12

-2

19) u1 = 0 , u2 = 5 , u3 = 6 , x = 1 6 -4 A) x = -8u1 + 4u2 + 10u3

19) B) x = 4u1 - 2u2 - 5u3 D) x = 4u1 + 4u2 - 5u3

C) x = -4u1 + 2u2 + 5u3 Find the orthogonal projection of y onto u.

20) y = -10 , u = -4 30 2

A)

20)

B)

C) 4 5

-20 10

-

D) -20 2

-4 2

2 5

21) y = -24 , u = 4 10

21)

20

A)

B) 1 5

C) 16 80

D) 4 20

1 4 5 4

Let W be the subspace spanned by the u's. Write y as the sum of a vector in W and a vector orthogonal to W. 13 2 -1 22) y = 11 , u1 = 2 , u2 = 3 21 4 -1

A)

B) 4 9 y = 36 + -25 -5 26

2 -11 y = 18 + 7 -8 13

C)

D) 2 15 y = 18 + 29 13 34

2 y = 18 + 13

3

11 -7 8

22)

16

1

2

10

-1

2

23) y = 2 , u1 = 0 , u2 = 1

23)

A)

B) y=

15 1 6 + -4 9 1

C)

y=

12 4 0 + 2 -12 4

y=

15 31 6 + 8 9 19

D) 15 -1 y= 6 + 4 -1 9

Find the closest point to y in the subspace W spanned by u1 and u2 .

23

2

9

-1

24) y = 13 , u1 = 2 , u2 =

-1 3 4

A)

24)

B) -8 96 95

23

25) y = 3 , u1 = 7

A)

C) -12 -20 -1

D) 12 20 1

-3 25 26

1 2 0 , u2 = 1 2 -1

25) B)

22 7 6

C)

D) 23 8 9

-22 -7 -6

20 9 16

The given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W. -4 -4 26) Let x1 = 2 , x2 = 12 26) 0 3

A)

B) 6 -4 2 , 12 0 3

C)

D) -4 4 2 , 8 0 3

-4 -12 2 , 16 0 3

4

4 -4 2 , 12 0 -3

0 1 1 1 1 27) Let x1 = , x2 = , x3 = 0 1 -1 -1 -1 1 1

27)

A)

B) 0 3 18 1 , 4 , 4 -1 -4 19 1 -2 13

0 1 1 1 , 1 , 0 -1 -1 1 1 -1 1

C)

D) 0 3 14 1 , 2 , 2 -1 -2 9 1 -4 7

0 1 6 1 , 0 , 0 -1 0 1 1 -2 3

Find a QR factorization of the matrix A. -6 -12 28) A = 3 36 0 -12

28)

A)

B) 2 1 5 6

-15 -60 5 5

1 2 Q= 5 6 ,R= 0

2 1 5 6

0

1 6

Q=

-72 6

C)

-

1 2 -3 -12 5 6 ,R= 0 -12 1 0 6

D) Q=

-60 5

0

-6 -12 3 36 , R = -72 -15 0 -12 6 5

Q=

0 11 29) A = 1 1 0 -1 -1 1 1 -1 1

0

-72 6

29)

A) 0 3 14 Q= 1 2 2 ,R= -1 -2 9 1 -4 7

-6 12 3 24 , R = 0 -12

-15 -60 5 5

3 3

1 3

0

0

11 33

3 33

0

0

30 330

5

B)

Q=

-

0

3 33

14 330

1 3

2 33

2 330

1 3

2 33

9 330

1 3

4 33

7 330

,R=

3 3

1 3

0

0

11 33

3 33

0

0

30 330

C) 0 3 14 Q= 1 2 2 ,R= -1 -2 9 1 -4 7

3 3

0

0

1 3

11 33

0

0-

3 33

30 330

D)

Q=

-

0

3 33

14 330

1 3

2 33

2 330

1 3

2 33

3 1 0 , R = 0 11 -3 9 0 0 30 330

1 3

4 33

7 330

Find a least-squares solution of the inconsistent system Ax = b.

1 2 3 30) A = 3 4 , b = 5 5 9 1

A)

B)

C)

1 2 -

11 7

30) D)

52 21 -

43 9

131 84

-

6

22 9

129 3508 -

33 1754

1 1 31) A = 1 1 1 1

1 1 0 0 0 0

0 0 1 1 0 0

0 0 0 ,b= 0 1 1

7 8 0 2 4 1

A)

31)

B) 5 2 5 +x 4 3 2

-1 1 1 1

C) 5 4 5 +x 4 3 2

0

D) 5 2

-1 1 1 0

5 2

-1 4 + x -1 4 3 1 2 1

0

5 +x 4 7 2

0

-1 1 1 1

0

Given A and b, determine the least-squares error in the least-squares solution of Ax = b.

43 3 32) A = 2 1 , b = 0 32 1 A) 2.362,907,81 B) 60.231,036,7 C) 162.109,188 D) 0.408,248,29

32)

Find the equation y = 0 + 1 x of the least-squares line that best fits the given data points.

33) Data points: (2, 1), (3, 2), (7, 3), (8, 3)

33)

12 1 X= 1 3 ,y= 2 17 3 18 3 A) y = - 139 + 4 x 52 13

B) y = 37 + 4 x 52

C) y = 37 + 5 x 52

13

D) y = - 43 + 4 x

26

52

13

34) Data points: (5, -4), (2, 2), (4, 3), (5, 1) 15 -4 1 2 X= ,y= 2 14 3 15 1 31 1 A) y = + x 6 6

34)

B) y = 31 - 3x

C) y = 12 - 7 x

6

3

D) y = 31 - 7 x 6

6

35) Find the equation y= 0+ 1 x of the least-squares line that best fits the given data points. Round parameters to one decimal place.

(-3,-2), (0,1), (1,2), (2,5) A) y=2.2+1.4x

B) y=1+x

C) y=2.2-1.4x

7

D) y=1.5 +1.3x

35)

36) Use the given data to answer parts (a) and (b).

36)

a. Rewrite the data with new x-coordinates in mean deviation form. Let X be the associated design matrix. Why are the columns of X orthogonal? b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, y= 0 + 1 x*, where x*=x-4.5.

(1,1), (4,2), (6,3), (7,3) A) a. (1,-3.5), (4,-2.5), (6,-1.5), (7,-1.5); the columns of x are orthogonal because the entries in the first column sum to 0.

0

b. 4 -9 18 -33

1

= -9 ; y=x*=y=(x-4.5) -33

B) a. (5.5,1), (8.5,2), (10.5,3), (11.5,3); the columns of x are orthogonal because the entries in the second column sum to 0.

0

b. 4 9 36 88.5

1

= 9 ; y=x*=y=x-4.5 88.5

C) a. (-3.5,1), (-0.5,2), (1.5,3), (2.5,3); the columns of x are orthogonal because the entries in the first column sum to 0. b. 21 0 0 4

0

5 9 5 9 = 7.5 ; y= + x*=y= + (x-4.5) 14 4 14 4 9 1

D) a. (-3.5,1), (-0.5,2), (1.5,3), (2.5,3); the columns of x are orthogonal because the entries in the second column sum to 0. b. 4 0 0 21

0 1

9 5 9 5 = 9 ; y= + x*=y= + (x-4.5) 4 14 4 14 7.5

37) The least-squares line that best fits the given data points is approximately y=1.8+1.2x. If a

machine learns this, what will the machine pick for the value of y when x=2? How closely does this match the data point at x=2 fed into the machine?

(-4,-3), (-1,1), (1,3), (2,4) A) 3.8; lower by 0.2

B) 4; matches exactly D) 4.2; higher by 0.2

C) 3.9; lower by 0.1

8

37)

38) A certain experiment produces the data (0,7.2), (1,4.6), (2,-1.3). Describe the model that produces a

38)

least-squares fit of these points by a function of the form y=Acosx+Bsinx.

1 A , = , = 2 B 3

A) y=

7.2 cos 0 sin 0 + , where y= 4.6 , X= cos sin -1.3 cos 2 sin 2

B) y=

1 7.2 sin 0 cos 0 + , where y= 4.6 , X= sin 1 cos 1 , = cos A , = 2 sin B -1.3 sin 2 cos 2 3

C) y=

1 7.2 00 + , where y= 4.6 , X= 1 1 , = cos A , = 2 sin B -1.3 22 3

D) y=

1 7.2 cos 0 sin 0 A , = 2 + , where y= 4.6 , X= cos 1 sin 1 , = B -1.3 cos 2 sin 2 3

39) According to Kepler's first law, a comet should have an elliptic, parabolic, or hyperbolic orbit

39)

(with gravitational attractions from the planets ignored). In suitable polar coordinates, the position (r, ) of a comet satisfies an equation of the form r= +e(r·cos ) where is a constant and e is the eccentricity of the orbit with 0 e <1 for an ellipse, e=1 for a parabola, and e > 1 for a hyperbola. Suppose observations of a newly discovered comet provide the data below. Determine the type of orbit, and predict where the comet will be when =3.8 (radians), rounding to two decimal places.

r

.86 3.07

1.20 2.08

1.39 1.71

A) ellipse; r=0.85 C) ellipse; r=0.52

1.75 1.25

2.19 0.92

B) hyperbola; r=0.52 D) parabola; r=3.28

40) To measure a car's ability to accelerate from a stopped position, the horizontal position of the car

was measured every second, from t=0 to t=12. The positions (in feet) were: 0, 10.1, 32.3, 65.6, 109.7, 164.5, 229.8, 305.5, 391.3, 487.1, 592.8, 708.1, 832.8. a. Find the least-squares cubic curve y= 0 + 1 t+ 2 t2 + 3 t3 for these data. Round parameters to three decimal places. b. If a machine learned the curve given in part (a), what would it estimate the velocity of the car to be when t=6.5 seconds? Round to one decimal place.

A) a. y=-0.248+4.955t+5.754t2-0.032t3 B) a. y=-0.248-4.955t+5.754t2+0.032t3

b. 266.3 feet per second

b. 73.9 feet per second

C) a. y=-0.248+4.955t+5.754t2-0.032t3 b. 75.7 feet per second

D) a. y=-0.248-4.955t+5.754t2+0.032t3 b. 219.4 feet per second

9

40)

41) Given data for a least-squares problem, (x1 ,y1),...,(xn,yn), the following abbreviations are helpful: x=

n

x2 =

xi ,

i=1

n i=1

2 xi ,

y=

n i=1

yi ,

xy =

n i=1

41)

xi yi

The normal equations for a least-squares line y= 0 + 1 x may be written in the following form: ^

^

n 0+ 1 x = y ^

2 0 x + 1 x = xy ^

^

Use a matrix inverse to solve these normal equations and thereby obtain formulas for 0 and ^

1 that appear in many statistics texts. ^

A) 0= ^

1=

^

C) 0= ^

1=

n xy -( x )( y )

^

B) 0=

2 n x2 -( x ) ( x2 )( y )-( x )( xy )

^

2 n x2 -( x ) ( x2 )( y )-( x )( xy )

1=

^

D) 0=

2 n x2 -( x ) n xy -( x )( y )

^

2 n x2 -( x )

1=

( x2 )( y )-( x )( xy ) 2 x2 -n( x ) n xy -( x )( y ) 2 x2 -n( x ) n xy -( x )( y ) 2 x2 -n( x ) ( x2 )( y )-( x )( xy ) 2 x2 -n( x )

^

42) Assume a design matrix X with two or more columns and a least-squares solution of y= Consider the following numbers. (i) X

^ 2

.

the sum of the squares of the "regression term." Denote this number by SS(R).

^ 2

(ii) y-X the sum of the squares for the error term." Denote this number by SS(E).

2 (iii) y the "total" sum of the squares of the y-values. Denote this number by SS(T).

Every statistics text that discusses regression and the linear model y= + introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume the mean of the y-values is zero. In this case, SS(T) is proportional to what is called the variance of the set of y-values.

The equation SS(T)=SS(R)-SS(E) relates the three terms. Using this equation, the definitions of SS(T), SS(R), and SS(E) above, the transpose property (AB)T=BTAT, and the normal equations XT

=XTy, find an equivalent formula for SS(E). ^ A) SS(E)=yTy- TXTy ^ C) SS(E)=XTy T-yTy

B) SS(E)=yTy-X Ty T ^

D) SS(E)= TXTy-yTy ^

10

42)

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Solve the problem.

43) With the given positive numbers, show that vectors u = (u1, u2 ) and v = (v1 , v2 ) define an

43)

inner product in 2 using the 4 axioms.

Set u, v = 5u1 v1 + 7u2 v2

44) Let t0 , ...., tn be distinct real numbers. For p and q in n, define p, q = p(t0 )q(t0 ) +

44)

p(t1 )q(t1 ) + ... + p(tn )q(tn ).

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Compute the length of the given vector.

45) p(t) = 10t2 and q(t) = t - 1, where t0 = 0, t1 = 1 , t2 = 1

45)

2

A) p = 5 17; q =

5 4

B) p = 0.5 70; q =

5 4

C) p = 10 17; q =

5 4

D) p =

5 ; 4

5 4

2

q

= 10

46) p(t) = 2t + 4 and q(t) = 4t - 5, where t0 = 0, t1 = 1, t2 = 2 A) p = -84; q = 15 B) p = 116; q = 145 C) p = 116; q = 35 D) p = 104; q = 195

46)

Solve the problem.

47) Let V be in 4 , involving evaluation of polynomials at -2, -1, 0, 1, and 2, and view 2 by applying the Gram-Schmidt process to the polynomials 1, t, and t2 .

A) p2(t) = t2 - 5 6

B) p2(t) = t2 + 2

C) p2(t) = t2 - 2

D) p2(t) = t2 + 6 5

48) Let V be in 4 , involving evaluation of polynomials at -5, -3, 0, 3, and 5, and view 2 by applying the Gram-Schmidt process to the polynomials 1, t, and t2 .

A) p2(t) = t2 + 68 B) p2(t) = t2 - 18 C) p2(t) = t2 - 10 5 5

47)

48)

D) p2(t) = t2 - 68 5

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

49) For f, g in C[a, b], set f, g =

b

f(t) g(t) dt . a Show that f, g defines an inner product of C[a, b].

11

49)

50) Let V be the space C[0, 1] and let W be the subspace spanned by the polynomials

50)

p1 (t) = 1, p2 (t) = 2t - 1, and p3 (t) = 12t2 . Use the Gram-Schmidt process to find an orthogonal basis for W.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Find the least-squares line y = 0 + z x that best fits the given data.

51) Given: The data points (-3, 2), (-2, 5), (0, 5), (2, 2), (3, 5).

51)

Suppose the errors in measuring the y-values of the last two data points are greater than for the other points. Weight these data points half as much as the rest of the data.

1 -3 2 1 -2 5 1 X= 1 0 , = ,y= 5 2 1 2 2 1 3 5 A) y = 0.8 + 0.91x

B) y = 4.6 + 0.91x

C) y = 4.2 + 0.28x

D) y = 0.28 + 4.2x

52) Given: The data points (-2, 2), (-1, 5), (0, 5), (1, 2), (2, 5).

52)

Suppose the errors in measuring the y-values of the last two data points are greater than for the other points. Weight these data points twice as much as the rest of the data.

1 -2 2 1 -1 5 1 X= 1 0 , = ,y= 5 2 1 1 2 1 2 5 A) y = 6.7 + 0.71x

B) y = 2.6 + 0.27x

C) y = 3.3 + 0.10x

D) y = 3.3 + 0.36x

Solve the problem.

53) Let C[0, ] have the inner product f, g =

f(t)g(t) dt, and let m and n be unequal positive

0 integers. Prove that cos(mt) and cos(nt) are orthogonal.

A) cos(mt), cos(nt) =

cos(mt)cos(nt)dt 0 1 2

[cos(mt + nt) + cos(mt - nt)]dt 0 1 sin(mt + nt) sin(mt - nt) from [0, ] = + 2 m-n m+n =

= 1.

12

53)

B) cos(mt), cos(nt) =

cos(mt)cos(nt)dt 0 1 2

[cos(mt + nt) - cos(mt - nt)]dt 0 1 sin(mt + nt) sin(mt - nt) from [0, ] = 2 m-n m+ n =

= 0.

C) cos(mt), cos(nt) =

cos(mt)cos(nt)dt 0

=

1 2

[cos(mt + nt) + cos(mt - nt)]dt

0 1 sin(mt + nt) sin(mt - nt) from [0, ] = + 2 m+n m-n = 0.

D) cos(mt), cos(nt) =

cos(mt)cos(nt)dt 0

=

[cos(mt + nt) + cos(mt - nt)]dt

0 sin(mt - nt) sin(mt + nt) from [0, ] = + m+n m-n = 1.

54) Find the nth-order Fourier approximation to the function f(t) = 3t on the interval [0, 2 ].

A) 3 - 6sin(t) - 3sin(2t) - 1sin(3t) - ... - 3 sin(nt) n B)

- cos(t) - cos(2t) - cos(3t) - ... -

6 cos(nt) n

C) 3 - 6cos(t) - 3sin(2t) - 2cos(3t) - ... - 6 cos(nt) n

D) 3 - 6sin(t) - 3sin(2t) - 2sin(3t) - ... - 6 sin(nt) n

13

54)

Answer Key Testname: UNTITLED6

1) D 2) A 3) A 4) C 5) B 6) B 7) C 8) D 9) D 10) C 11) C 12) A 13) B 14) B 15) A 16) A 17) B 18) A 19) C 20) A 21) A 22) D 23) A 24) D 25) A 26) C 27) C 28) A 29) B 30) C 31) A 32) D 33) B 34) D 35) D 36) D 37) D 38) D 39) A 40) C 41) C 42) A

14

Answer Key Testname: UNTITLED6

43) Axiom 1: u, v = 5u1 v1 + 7u2 v2 = 5v1u1 + 7v2 u2 = v, u Axiom 2: If w = (w1 , w2 ), then u + v, w = 5(u1 + v1 )w1 + 7(u2 + v2 )w2 = 5u1 w1 + 7u2 w2 + 5v1 w1 + 7v2 w2 = u, w + v, w Axiom 3: cu, v = 5c(u1 )v1 + 7(cu2 )v2 = c(5u1 v1 + 7u2 v2 ) = c u, v Axiom 4: u, u = 5u1 2 + 7u2 2 0, and 5u1 2 + 7u2 2 = 0 only if u1 = u2 = 0. Also, 0, 0 = 0.

44) Axioms 1-3 are readily checked. For Axiom 4, note that p, p = [p(t0)]2 + [p(t1)]2 +... + [p(tn)]2 0.

Also, 0, 0 = 0.

If p, p = 0, then p must vanish at n + 1 points: t0 , ...., tn . This is possible only if p is the zero polynomial, because the degree of p is less than n + 1.

45) A 46) C 47) C 48) D 49) Answers will vary.

Inner product Axioms 1-3 follow from elementary properties of definite integrals. b Axiom 4: f, f = [f(t)]2 dt 0. a 2 The function [f(t)] is continuous and nonnegative on [a, b]. If the definite integral of [f(t)]2 is zero, then [f(t)]2 must

be identically zero on [a, b]. Thus, f, f = 0 implies that f is the zero function on [a, b].

50) As a function, q3 (t) = 12t2 - 12t + 2. The orthogonal basis for the subspace W is {q1 , q2 , q3 }.

51) C 52) D 53) C 54) D

15

Exam Name___________________________________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Determine whether the matrix is symmetric.

1) 1 -3 -2 0 A) No 2

4

1) B) Yes

3

2) 4 -5 -2

2)

3 -2 0 A) Yes

B) No

Orthogonally diagonalize the matrix, giving an orthogonal matrix P and a diagonal matrix D. 3) 4 2 27

A)

B) P = 1 -2 , D = 9 0 2 1 04

P = 1/ 5 -2/ 5 , D = 8 0 03 2/ 5 1/ 5

C)

D) P = 1 -2 , D = 8 0 2 1 03

P = 1/ 5 -2/ 5 , D = 3 0 08 -2/ 5 1/ 5

4) 12 - 6 -6

3)

4)

7

A)

B) P=

3 2 , D = 16 0 0 3 -2 3

C)

P=

3/ 13 -2/ 13 , D = 3 0 0 16 -2/ 13 3/ 13

P=

3 -2 , D = 16 0 0 3 -2 3

D) P=

3/ 13 2/ 13 , D = 16 0 0 3 -2/ 13 3/ 13

13 10 4

5) 10 12 6

5)

4 65

A)

B) 2/3 -2/3 1/3 1 0 0 P = 2/3 1/3 -2/3 , D = 0 4 0 1/3 2/3 2/3 0 0 25

2 -2 1 1 0 0 P = 2 1 -2 , D = 0 4 0 1 2 2 0 0 25

C)

D) 2 -2 1 25 0 0 P = 2 1 -2 , D = 0 4 0 1 2 2 0 0 1

2/3 -2/3 1/3 25 0 0 P = 2/3 1/3 -2/3 , D = 0 4 0 1/3 2/3 2/3 0 0 1

1

6)

11 7 7 7 11 7 7 7 11

6)

A) 1/ 3 -1/ 2 -1/ 6 4 0 0 P = 1/ 3 1/ 2 -1/ 6 , D = 0 4 0 0 0 25 1/ 3 0 2/ 6

B) P=

1/ 3 1/ 2 1/ 6 25 0 0 1/ 3 -1/ 2 1/ 6 , D = 0 25 0 0 0 4 0 2/ 6 -1/ 3

C) 1/ 3 -1/ 2 -1/ 6 25 0 0 P = 1/ 3 1/ 2 -1/ 6 , D = 0 4 0 0 0 4 1/ 3 0 2/ 6

D) 1 -1 -1 25 0 0 P = 1 1 -1 , D = 0 4 0 1 0 2 0 0 4

Compute the quadratic form x TAx for the given matrix A and vector x. x1 7) A = -8 0 , x = x2 0 7

A) -8 x 21 + 7 x 22

B) -8x1 + 7x2

C) -16x1 + 14x2

7) D) 7 x 21 - 8 x 22

x1 3 7 0 8) A = 7 1 2 , x = x2 0 2 -2 x3

8)

A) 3 x 21 + x 22 - 2 x 23 + 7x1 + 9x2 + 2x3

B) 3 x 21 - x 22 + 2 x 23 - 14x1 x2 - 4x2 x3

C) 3 x 21 + x 22 - 2 x 23

D) 3 x 21 + x 22 - 2 x 23 + 14x1 x2 + 4x2 x3

Find the matrix of the quadratic form.

9) 3 x 21 + 2 x 22 + 6 x 23 - 8x1x2 + x2 x3

9)

A)

B) 300 020 006

3 -8 0 -8 2 1 0 1 6

C)

D) 3 -4 1/2 2 0 -4 1/2 0 6

3 -4 0 -4 2 1/2 0 1/2 6

2

10) 10x1x2 - 6x1x3 + 14x2 x3 A)

10) B)

0 5 -3 5 0 7 -3 7 0

C)

D)

1 5 -3 5 1 7 -3 7 1

5 00 0 -3 0 0 07

0 10 -6 10 0 14 -6 14 0

Make a change of variable, x = Py, that transforms the given quadratic form into a quadratic form with no cross-product term. Give P and the new quadratic form.

11) Q(x) = 4 x 21 + 7 x 22 + 4x1 x2

11)

A)

B) P = 1 -2 ; 8y1 + 3y2 2 1

2 2 P = -1/ 5 -2/ 5 ; -8 y 1 - 3 y 2 2/ 5 1/ 5

C)

D) 2 2 P = 1/ 5 -2/ 5 ; 8 y 1 + 3 y 2 2/ 5 1/ 5

2 2 P = 1 -2 ; 8 y 1 + 3 y 2 2 1

12) Q(x) = 13 x 21 + 12 x 22 + 5 x 23 + 20x1 x2 + 8x1 x3 + 12x2x3 A)

12)

B) -2/3 -2/3 1/3 2 2 2 P = -2/3 1/3 -2/3 ; 5 y 1 + 2 y 2 + y 3 1/3 -2/3 2/3

2 -2 1 2 2 2 P = 2 1 -2 ; y 1 + 4 y 2 + 25 y 3 1 2 2

C)

D) 2/3 -2/3 1/3 2 2 2 P = 2/3 1/3 -2/3 ; 25 y 1 + 4 y 2 + y 3 1/3 2/3 2/3

2 -2 1 2 2 2 P = 2 1 -2 ; 25 y 1 + 4 y 2 + y 3 1 2 2

Find the maximum value of Q(x) subject to the constraint x T x = 1.

13) Q(x) = 3 x 21 + 7 x 22 + 4 x 23 A) 14

13) B) 4

C) 7

D) 3

14) Q(x) = 14 x 21 + 14 x 22 + 18 x 23 + 26x1x2 + 18x1 x3 + 18x2x3 A) 9

B) 36

14)

C) 25

D) 1

Find a unit vector at which the quadratic form x TAx is maximized, subject to the constraint x T x = 1.

15) A = 3 2 2 6

A)

B) 1 2

C) -2/ 5 1/ 5

D) 1/ 5 2/ 5

3

-1/ 5 2/ 5

15)

8 -6

4

4 -2

4

16) A = -6 9 -2

16)

A)

B) 1/3 2/3 2/3

C)

D)

2 -2 1

2/3 -2/3 1/3

2/ 5 -2/ 5 1/ 5

Find the maximum value of Q(x) subject to the constraints x Tx = 1 and x Tu = 0, where u is a unit eigenvector corresponding to the greatest eigenvalue of the matrix of the quadratic form.

17) Q(x) = 2 x 21 + 8 x 22 + 5 x 23 A) 5

17) B) 0

C) 2

D) 8

18) Q(x) = 14 x 21 + 14 x 22 + 18 x 23 + 26x1x2 + 18x1 x3 + 18x2x3 A) 36

B) 16

18)

C) 1

Find the singular values of the matrix.

19) -9 0 0 2 A) -9,2 B) -81, 4

D) 9

19) C) 9, 2

D) 81, 4

20) 1 4 4

20)

4 1 -4 A) 0, 25, 41

B) 5, 41

C) 0, 5, 41

D) 25, 41

Find a singular value decomposition of the matrix A. 21) A = -5 0 0 3

21)

A)

B) A= 1 0 0 1

-5 0 0 3

1 0 0 1

A = -5 0 0 3

5 0 0 3

-1 0 0 1

C)

A = -5 0 0 3

5 0 0 3

1 0 0 1

A = -1 0 0 1

5 0 0 3

10 01

D)

4

22) A =

3 -1 -1 3

22)

A) A = -1 1 1 1

4 0 0 2

-1 1 1 1

B) A = -1/ 2 1/ 2 1/ 2 1/ 2

4 0 0 2

-1/ 2 1/ 2 1/ 2 1/ 2

A = -1/ 2 1/ 2 1/ 2 1/ 2

4 0 0 4

-1/ 2 1/ 2 1/ 2 1/ 2

A = -1 1 1 1

-1 1 1 1

C)

D) 3 -1 -1 3

23) A = 3 0 -3 3 3

23)

3

A) 1/ 3 1/ 3 0 -1/ 2 1/ 6 -2/ 6

A = 0 -1 1 0

3 3 0 0 0 3 2 0

A = 0 -1 1 0

3 3 0 0 3 2

A = 0 -1 1 0

3 3 0 0 0 3 2 0

A = 0 -1 1 0

27 0 0 0 18 0

1/ 3 1/ 2 1/ 6

B) 1 1 1 -1 0 1 1 -2 1

C) 1/ 3 -1/ 2 1/ 6 1/ 3 0 -2/ 6 1/ 3 1/ 2 1/ 6

D) 1/ 3 1/ 3 1/ 3 0 1/ 2 -1/ 2 1/ 6 -2/ 6 1/ 6

Convert the matrix of observations to mean-deviation form, and construct the sample covariance matrix.

8 11 -1 2 24) 9 6 12 9 15 3 3 3

A)

B) 30 -3 -15 -12 18 -12 -3 -3 S= -15 -12 18 9 9 2 -12 -3

S=

C)

-30 12 -12

12 -12 0 -6 0 -36

D) 90 -36 36 S = -36 18 0 36 0 108

30 -12 12 S = -12 6 0 12 0 36

5

24)

25) 24 29 24 29 9 29

25)

9 39 54 24 49 29

A)

B) 55 S = -60 55 -280

C)

S=

300 -275 -275 1400

S=

60 -55

D) S=

50

5 -45 6

5 6

233

-45

-55 280

1 3

Use the given covariance matrix to compute the percentage of the total variance that is contained in the first principal component. Round to one decimal place.

30 -12 12 26) S = -12 6 0 26) 12 0 36 A) 34.4% B) 65.0% C) 79.1% D) 41.7% 60 -55 -55 280 A) 89.7%

27) S =

27) B) 13.8%

C) 86.2%

6

D) 82.4%

Answer Key Testname: UNTITLED7

1) A 2) A 3) B 4) C 5) D 6) C 7) A 8) D 9) D 10) A 11) C 12) C 13) C 14) B 15) C 16) D 17) A 18) D 19) C 20) C 21) D 22) B 23) A 24) D 25) D 26) B 27) C

7

Exam Name___________________________________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Write y as an affine combination of the other points listed.

1) v1 = 5 , v2 = 3 , v3 = 11 , y = 9 3 -3 -2 -2 A) y = 4v1 - 2v2 - v3

1) B) y = 5v1 + 3v2 - 7v3 D) y = v1 - 2v2 + 2v3

C) y = -5v1 + 4v2 + 2v3 1

0

1

6

2

-2

1

20

2) v1 = 1 , v2 = 4 , v3 = -5 , y = -26

2)

A) y = -4v1 - 2v2 + 7v3 C) y = 2v1 + 4v2 - 5v3

B) y = 6v1 + 4v2 - 5v3 D) y = 4v1 - 5v2 + 2v3

Determine if the vector p is in Span S or aff S. 2 3 5 5 1 4 -1 -3 3) Let v1 = , v2 = , v3 = ,p= , and S = {v1 , v2 , v3 }. It can be shown that S is linearly 3 0 5 -1 -2 -2 4 3 independent.

A) p span S and p aff S B) p span S and p aff S

C) p span S and p aff S 2 3 5 1 -1 4) Let v1 = , v2 = , v3 = 4 , p = 3 0 -1 -2 -2 4 independent.

A) p span S and p aff S

3)

D) p span S and p aff S 10 12 , and S = {v , v , v }. It can be shown that S is linearly 1 2 3 -6 -8

4)

B) p span S and p aff S D) p span S and p aff S

C) p span S and p aff S

2 3 5 2 1 4 -1 -3 5) Let v1 = , v2 = , v3 = ,p= , and S = {v1 , v2 , v3 }. It can be shown that S is linearly 3 0 7 -1 -2 -2 4 2 independent.

A) p span S and p aff S B) p span S and p aff S

C) p span S and p aff S

D) p span S and p aff S

1

5)

Provide an appropriate response.

6) Which of the following statements are true?

I: 2v1 + 2v2 - 3v3 is an affine combination of the 3 vectors.

6)

II: The affine hull of two distinct points is a plane.

III: If S = {x}, then aff S = {x}.

IV: If a set of vectors in n is linearly independent, then every vector in n can be written as an affine combination of these vectors.

A) I and III B) I and II

C) I , III, and IV

D) II and IV

7) Suppose {v1, v2, v3} is a basis for 3 . Which of the following statements are true?

7)

I: Span {v2 -v1 , v3 - v1 } is a plane in 3 .

II: Aff {v1 , v2 , v3 } is the plane through v1 , v2 , and v3 .

A) I

B) Both I and II

C) Neither I or II

D) II

6

6 6 3 , v3 = 4 and S = {v1 , v2 , v3 }. Aff S is a plane in 3 . Give its equation.

5 3 -1 A) 13x +14y =17 B) x + y + z = 0

8) v1 = 2 , v2 = C) x = 6

D) 18x + 9y + 7z = 0

9) Pick a set S of four distinct points in 3 such that aff S is the plane 5x1 + x2 - 4x3 = 13.

A) S= C) S=

8)

2 -1 3 0 12 , 4 , 3 , 0 1 -1 0 0 0 1 2 0 9 , 4 , 23 , 13 1 1 0 0

B) S= D) S=

9)

5 10 0 1 1 , 2 , 13 , 12 0 1 -4 -8 0 1 -2 0 17 , 4 , 23 , 13 1 -1 0 0

10) Which of the following statements are true?

10)

I: Suppose f : n

m is a linear transformation and S is an affine subset of n . It follows that the set of images f(S) is an affine subset of m .

II: If A B, then aff A B. A) Both I and II

B) II

C) I

11) [(aff A) (aff B)] (A B)

D) Neither I or II 11)

What property must the set (A B) have if the above statement is true?

A) A B so that (A B) = A.

B) (A B) has to be affine.

C) (A B) must contain collinear points.

D) There is nothing special about (A B). The statement is always true.

2

Determine if the set of points is affinely dependent. If so, construct an affine dependence relation for the points.

12) -12 , -12 , -12 12) -12 -12 -12 A) The set is affinely dependent. 3 -3 + -3 - -12 = 0 0 -3 -3 -12 B) The set is affinely dependent. 3 -12 + -12 - 4 -12 = 0 0 -12 -12 -12 -12 -12 -12 C) The set is affinely dependent. 3 +4 = 0 0 -12 -12 -12 D) The set is affinely independent.

13)

3 0 -12 -6 6 , -12 , 1 , -23 3 12 19 -1 A) The set is affinely independent.

13)

3

0

-6

0

-12

B) The set is affinely dependent. 5 6 - 3 -12 - 1 - -23 = 0

3 12 19 0 -1 3 0 0 -6 -12 C) The set is affinely dependent. 2 6 - 3 -12 - 1 + -23 = 0 3 12 19 0 -1 3 0 0 -6 -12 D) The set is affinely dependent. 2 6 + 3 -12 + 1 - -23 = 0 3 12 19 0 -1

Find the barycentric coordinates of p with respect to the affinely independent set of points that precedes it.

1 2 1 -4 14) -1 , 1 , 2 , p = 4 2 0 -2 -8 -4 1 1 0 A) (0, -4, 5) B) (1, -5, 5) C) (5, -5, 1) D) (1, 0, 0)

15) 1 , 3 , 0 , p = 1

5

6

A) 3 , 1, - 3 2

2

14)

9 2 -

15)

5 2

C) 2, 5 , - 7

B) (-9, 3, 7)

2

2

D) 1, - 3 , 3 2 2

Answer the question.

16) Which of the following statements are true for the set S={v1 , . . . , vk} in n?

I: {v2 -v1 , . . . , vk-v1 } is linearly dependent if and only if S is affinely dependent.

II: If S is affinely independent and a point p in n has all positive barycentric coordinates determined by S, then p is not in aff S. A) Neither I or II B) II C) I D) Both I and II

3

16)

17) Which of the following statements are true for the set S={v1 , . . . , vk} in n?

17)

I: If S is affinely independent, then a point p in n cannot have any barycentric coordinates determined by S that are equal to 0.

II: If S is affinely independent, then a point p in aff S has a unique representation as an affine combination of v1 , ... , vk.

A) II

B) Neither I or II

C) Both I and II

D) I

18) How many points need to be in a set in 6 to guarantee that the set is affinely dependent?

A) 7 B) 6 C) 9 D) 8

18)

19) If {v1 , v2 } in n is affinely dependent, what is known about the 2 points?

A) {v1 , v2 } is linearly independent.

B) v1 = v2 C) v1 = cv2 where c D) {v2 - v1 } is linearly dependent.

19)

20) Consider the affinely independent set S = {v1, v2, v3} in 2 . These points form a triangular region.

20)

If the barycentric coordinates of a point p are all positive (+, +, +), then where is p with respect to the triangle?

A) Outside B) Inside C) Edge D) Vertex

21) Consider the affinely independent set S = {v1, v2, v3} in 2 . These points form a triangular region.

If the barycentric coordinates of a point p are (+, 0, +), then where is p with respect to the triangle?

A) Inside B) Vertex C) Edge

D) Outside

22) Which of the following statements are true if p is a point on the line through a and b?

I: p is an affine combination a and b. ~~ ~ II: a, b, and p are linearly independent. ~~~ III: The determinant of [a b p] is 0.

A) I and III B) II

C) I and II

2 8 8

pbc with respect to the area of abc?

A) 1 · (area of abc) 8

B) 3 · (area of abc)

C) 1 · (area of abc)

D) Not enough information to determine.

8

Sketch the graph of the convex hull of S.

4

22)

D) None are true.

23) Let p be a point in the interior of abc with barycentric coordinates 1 , 3 , 1 . What is the area of

2

21)

23)

24) In 2 , let S =

0 5

x :0 x 4 0

24)

A)

B)

C)

D)

5

25) In 2 , S is the set of points x where y = x2 and x 0

25)

y

A)

B)

C)

D)

6

Determine whether the point p is in the convex hull of S. 26) S = {v1 , v2 , v3 , v4 }

26)

2 0 1 1 -1 v1 = 2 , v2 = 3 , v3 = 8 , v4 = 2 , p = 3 1 4 0 -1 -6 A) p conv S. p = 1 v1 + 1 v2 + 1 v3 2 4 4

B) p conv S. p = 1 v1 + 1 v2 + 1 v3 + 1 v4 2

C) p conv S.

4

8

8

D) p conv S. p = 2v1 + 3v2 - 2v3 - 2v4

27) S = {v1 , v2 , v3 , v4 }

27)

2 1 3 0 0 1 2 -1 v1 = , v2 = , v3 = ,p= , 5 0 4 -1 -3 -4 1 2

A) p conv S.

B) p conv S. p = 3 v1 + 1 v2 + 3 v3

C) p conv S. p = v1 + v2 - v3

D) p conv S. p = 2v1 + 3v2 - 4v3

8

4

8

Use the barycentric coordinates with respect to S to determine if the point p is inside, outside, on a face, or on the edge of conv S which is a tetrahedron.

28) S = {v1 , v2 , v3 , v4 ,} 28) Barycentric coordinates: (3, 2, 0, -4) A) Face B) Inside

C) Edge

D) Outside

29) S = {v1 , v2 , v3 , v4 ,} Barycentric coordinates:

A) Face

29) 1 3 1 1 , , , 4 8 8 4

B) Edge

C) Inside

D) Outside

30) S = {v1 , v2 , v3 , v4 ,} Barycentric coordinates:

A) Inside

30) 1 1 3 , 0, , 2 8 8

B) Edge

C) Face

D) Outside

31) S = {v1 , v2 , v3 , v4 ,}

31)

Barycentric coordinates: 0, 0,

A) Outside

1 7 , 8 8

B) Inside

C) Edge

7

D) Face

Provide an appropriate response.

32) p = 1 v1 + 1 v2 + 1 v3 + 1 v4 and 2v1 + v2 - 2v3 - v4 = 0 2 8 8 4

32)

Use the procedure in the proof of Caratheodory's Theorem to express p as a convex combination of three of the vi's.

A) p = 1 v2 + 3 v3 + 1 v4

B) p = 1 v1 + 3 v2 + 3 v4

C) p = 5 v1 + 3 v2 + 3 v4

D) It cannot be done using only 3 vi's.

8

8

8

2

16

4

16

8

8

33) Which of the following statements are true?

33)

I: If S is a nonempty set, then conv S S. II: If S and T are convex sets, then S T is also convex.

A) Neither I or II B) I C) Both I and II

D) II

34) Let (pos S) be the set of all positive combinations of the points of S. Which of the following

34)

35) Which of the following statements are true?

35)

statements are true?

I: (pos S) (aff S) = conv S II: If a unique linear combination of points is both positive and affine, then it must be convex.

A) Neither I or II B) I C) II D) Both I and II

I: Suppose f: n

m is a linear transformation and S is a convex subset of n . It follows that the set of images f(S) is a convex subset of m .

II: If A B, then conv A B. A) I B) Both I and II

C) Neither I or II

D) II

36) Which of the following statements are true?

36)

I: If A B, then A conv B. II: If A B, then conv A conv B. A) I B) Both I and II

C) Neither I or II

D) II

37) Let p0 , p1 , and p2 be points in n and define f0(t) = (1 - t)p0 + tp1 , f1 (t) = (1 - t)p1 + tp2, and g(t) = (1 - t)f0 (t) + tf1 (t) for 0 t 1.

Find g

1 in terms of the three points.

2

A) g 1 = 9 p0 + 3 p1 + 1 p2

B) g 1 = 3 p0 + 1 p1 - 1 p2

C) g 1 = 1 p0 + 1 p1

D) g 1 = 1 p0 + 1 p1 + 1 p2

2 2

16 2

8

16

2

2

2

8

2 4

2 2

2 4

37)

Let H be the hyperplane through the points. Find a linear functional f and a real number d such that H = [f : d].

38) -1 , 4 38) 3 -2 A) f(x1, x2) = -5x1 + 5x2 , d = 20 B) f(x1, x2) = -x1 + 3x2 , d = 10

C) f(x1, x2) = 5x1 + 5x2 , d = 10 1

3

4

1

5

-2

D) f(x1, x2) = 4x1 - 2x2 , d = -10

39) 1 , -1 , 2

39)

A) f(x1, x2, x3) = x1 + 9x2 + 4x3 , d = 14

B) f(x1, x2, x3) = 8x1 + 2x2 + 4x3, d = 14

C) f(x1, x2, x3) = x1 + x2 + 5 x3 , d = 9

D) f(x1, x2, x3) = -2x1 + 9x2 + x3, d = 8

2

40)

2

1 2 -1 3 0 , 3 , 1 , 2 0 1 2 -1 1 2 1 1 A) f(x1, x2, x3, x4) = x1 - 2x2 - 6x3 + 13x4 , d = 12

40)

B) f(x1, x2, x3, x4) = -5x1 + 2x2 - 6x3 + 5x4 , d = 0 C) f(x1, x2, x3, x4) = 5x1 + 6x2 + 2x3 + 5x4 , d = 10 D) f(x1, x2, x3, x4) = 2x2 - x3 + 5x4, d = 5 Answer the question.

1

3

1

2

41) Let H be the hyperplane through the three points 1 , 3 , side of H as the origin? Justify your answer.

A) No, because n · v > 0.

2 7 5 . Is the point v = 1 on the same 1 -2

41)

B) Yes, because n · v < 3.

D) Yes, because n · v < 9.

C) No, because n · v > 3.

42) Which of the following statements are true?

42)

I: If F1 and F2 are 5 dimensional flats in 7 , then the dimension of F1 F2 could have a dimension of 6.

II: If F1 and F2 are strictly separated, then F1 F2 = .

A) Neither I or II

B) Both I and II

C) II

D) I

43) Which of the following statements are true?

43)

I: A linear transformation from to n is called a linear functional.

II: The convex hull of a closed set is closed.

A) II B) Neither I or II C) Both I and II

9

D) I

44) Which of the following statements are true?

44)

I: The open ball B(p, ) is a convex set.

II: The convex hull of a compact set is compact.

A) Both I and II B) II

C) I

D) Neither I or II

45) Consider the set S of points x in 2 such that y = 1 and x 1 . Are the sets S and conv S both y

x

closed?

A) Yes. They are both closed sets.

2

45)

B) No. S is closed but conv S is not.

D) No. Neither set is closed.

C) No. Conv S is closed but S is not.

46) Let int S be the set of all interior points of S, and let cl S be the closure of S (S the set of all boundary points of S). Which of the following statements are true?

I: If S is convex, then int S is convex.

II: If S is convex, then cl S is convex.

A) Neither I or II B) Both I and II C) II

46)

D) I

47) Which of the following statements are true?

47)

I: If S = {(x, y): x - y = 0 and x 0} and if P is its profile, then conv P = S. II: If S = {(x, y): x - y = 0 and 0 x 5} and if P is its profile, then conv P = S. A) Neither I or II B) Both I and II C) I

D) II

48) A four dimensional simplex S4 has how many 2-faces?

A) 4 B) 10 C) 5

D) 15

49) A five dimensional hypercube C5 has how many 2-faces ?

A) 40 B) 24 C) 80

D) 32

48)

49)

50) Which of the following statements are true?

50)

51) Which of the following statements are true?

51)

I: A polytope is the affine hull of a finite set of points.

II: An extreme point of a polytope P is any point in the convex hull of 2 vertices.

A) Both I and II B) II C) Neither I or II D) I

I: If A and B are convex sets then A + B is convex.

II: A four dimensional polytope always has the same number of vertices and edges.

A) Neither I or II B) Both I and II C) II D) I

Provide an appropriate response.

52) If a Bezier ´ ´ curve is translated, x(t) + b, will the new curve always be a Bezier curve as well?

A) No, it will almost never be a Bezier ´ curve. There are only a few choices that would work.

B) Yes, it will always be a Bezier ´ curve.

C) No, but it will almost always be a Bezier ´ curve with only a few exceptions.

D) No, it will never be a Bezier ´ curve.

10

52)

53) Let x(t) be a Bezier ´ curve and the tangent vector x (t) is computed. What does knowing that x (0) =

53)

3(p1 - p0 ) tell you?

A) The tangent vector points in the direction from p1 to p0 and its length is 3 times the length of p1 - p0 .

B) The tangent vector points in the direction from p0 to p1 and its length is 3.

C) The tangent vector points in the direction from p0 to -p1 and its length is 3 times the length of p1 .

D) The tangent vector points in the direction from p0 to p1 and its length is 3 times the length of p1 - p0 .

54) If 2 Bezier ´ curves are joined at the point p3 what is necessary for G1 geometric continuity?

A) Both tangent vectors at p3 need to point in the same direction.

B) Both tangent vectors at p3 need to be equal.

C) The second derivative at p3 needs to equal 0.

D) p3 just needs to be a control point for both curves.

54)

55) If 2 Bezier ´ curves are joined at the point p3 what is necessary for C1 parametric continuity?

A) p3 just needs to be a control point for both curves.

B) The second derivative at p3 needs to equal 0.

C) Both tangent vectors at p3 need to point in the same direction.

D) Both tangent vectors at p3 need to have the same magnitude and direction.

55)

56) How many control points are needed for a cubic Bezier ´ curve?

A) 2 B) 4 C) 5

56) D) 3

57) A quadratic Bezier ´ curve is determined by 3 control points p0 , p1 , and p2 . The equation is x(t) =

´ (1 - t)2 p0 + 2t(1 - t)p1 + t2 p2 . Construct the quadratic Bezier basis matrix MB for x(t).

1

A) 0 0

2 2 0

2 -2 1

1

B) 2 0

1 -2 1

1

C) 0 0

11

-2 2 0

1 -2 1

1

D) 0 0

0 1 0

1 -1 1

57)

Answer Key Testname: UNTITLED8

1) C 2) D 3) C 4) D 5) A 6) A 7) B 8) C 9) D 10) C 11) B 12) B 13) A 14) B 15) A 16) C 17) A 18) D 19) B 20) B 21) C 22) A 23) C 24) D 25) D 26) B 27) A 28) D 29) C 30) C 31) C 32) C 33) D 34) C 35) A 36) B 37) D 38) C 39) A 40) B 41) B 42) C

12

Answer Key Testname: UNTITLED8

43) B 44) A 45) B 46) B 47) D 48) B 49) C 50) C 51) D 52) B 53) D 54) A 55) D 56) B 57) C

13

Exam Name___________________________________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Write the payoff matrix for the game.

1) Player R has a supply of nickels, dimes, and quarters. Player R chooses one of the coins, and player C must guess which coin R has chosen. If the guess is correct, C takes the coin. If the guess is incorrect, C gives R an amount equal to R's chosen coin.

A)

n d n 5 -5 d -10 10 q -25 -25

q -5 -10 25

B)

n d q n -5 10 25 d 5 -10 25 q 5 10 -25

C)

n d n 5 -10 d -5 10 q -5 -10

q -25 -25 25

D)

n d q n -5 5 5 d 10 -10 10 q 25 25 -25

2) Each player has a supply of nickels, dimes, and quarters. At a given signal, both players display

1)

2)

one coin. If the displayed coins are not the same, then the player showing the higher valued coin gets to keep both. If they are both nickels or dimes, then player R keeps both; but if they are both quarters, then player C keeps both.

A)

n n 5 d 5 q 5

d -5 10 10

q -5 -10 -25

B)

n d n 5 -10 d 10 10 q 25 25

q -25 -25 -25

C)

n d n -5 -10 d 10 -10 q 25 25

q -25 -25 25

D)

n d n -5 -5 d 5 -10 q 5 10

q -5 -10 25

3) Each player has a supply of pennies, nickels, and dimes. At a given signal, both players display

one coin. If the total number of cents N is even, then R pays N cents to C. If N is odd, then C pays N cents to R.

A)

p n p 1 1 n 5 5 d -1 -5

d -10 -10 10

B)

p n p 2 6 n 6 10 d -11 -15

d -11 -15 20

C)

p p -1 n -5 d 1

d 10 10 -10

D)

p p -2 n -6 d 11

d 11 15 -20

n -1 -5 5

1

n -6 -10 15

3)

4) Player R has two cards: a red 2 and a black 9. Player C has three cards: a red 4, a black 8, and a

4)

black 10. They each show one of their cards. If the cards are the same color, C receives the larger of the two numbers. If the cards are of different colors, R receives the sum of the two numbers.

A)

r4 r2 4 b9 -13

b8 -10 9

b10 -12 10

B)

r4 -6 9

b8 8 -17

b10 10 -19

D)

C) r2 b9

r2 b9

r4 -4 13

b8 10 -9

b10 12 -10

r4 b8 b10 -10 r2 6 -8 b9 -9 17 19

5) Players R and C each show 1, 2, or 5 fingers. If the total number N of fingers shown is even, then R

5)

pays N dollars to C. If N is odd, C pays N dollars to R. 1 2 -3 6

2 -3 4 -7

5 6 -7 10

B)

1 2 5

1 1 1 2 -1 5 5

2 -2 2 -2

5 1 -5 5

2 2 -2 2

5 -1 5 -5

D)

1 2 5

1 -1 1 -5

1 1 -2 2 3 5 -6

2 3 -4 7

5 -6 7 -10

A)

C)

6) In the traditional Japanese children's game janken (or "stone, scissors, paper"), at a given signal,

6)

each of two players shows either no fingers (stone), two fingers (scissors), or all five fingers (paper). Stone beats scissors, scissors beats paper, and paper beats stone. In the case of a tie, there is no payoff. In the case of a win, the winner collects 40 yen.

A)

C)

st sc p st 0 -40 40 sc 40 0 -40 p -40 40 0

B)

st sc p st 40 40 -40 sc -40 40 40 p 40 -40 40

D)

st sc p

Find all saddle points for the matrix game.

7) -9 5 4 6 A) Entry a 21

st sc p 0 40 -40 40 0 -40 40 -40 0

st sc p st 0 40 -40 sc -40 0 40 p 40 -40 0

7) B) Entry a 12 and entry a 21 D) No saddle points

C) Entry a 12 8) -6 2

8)

3 1

A) Entry a 22 C) Entry a 11

B) Entry a 12 D) No saddle point

2

9) 6 1 3

9)

2 -5 -7 A) Entry a 12

B) Entry a 12 and entry a23 D) No saddle points

C) Entry a 23 10) 7 4 1

10)

2 5 9 A) Entry a 21

B) Entry a 13 D) No saddle points

C) Entry a 11

11)

5 -1 -6 4 -1 -4 -4 5 -7 A) Entry a 13

11) B) Entry a 23 and entry a 33 D) No saddle points

C) Entry a 23

12)

1 -3 -8 5 5 8 -1 2 3 A) Entry a 21

12) B) Entry a 21 and entry a 22 D) No saddle points

C) Entry a 21 and entry a 13 0

3 -1 4

1

9

13) -3 -3 -5 4

13)

1 5

A) Entry a 31 C) Entry a 31 and entry a 33

B) Entry a 13 and entry a 31 D) No saddle points

Find the expected payoff.

14) Let M be the matrix game having payoff matrix 1 3

-1 3 5 0 -3 3 .

-2 1 0

14)

1 4

1 1 Find E(x, y) when x = 2 and y = 4 .

1 6

A) 13 24

1 2

B) 11

C) 4

8

3

3

D) 29 24

15) Let M be the matrix game having payoff matrix 1 4 1 Find E(x, y) when x = 2

and y =

1 2 .

1 6

C) 1

B) 0

24

15)

1 3

1 4

A) - 1

-4 0 4 2 -1 1 .

-2 1 0

8

D) - 1

12

0 1 3 -1

16) Let M be the matrix game having payoff matrix -2 0 2 -1 .

16)

0 3 0 -1

1 4

1 4 1 Find E(x, y) when x = 2

and y =

1 4

A) 5

0 1 2 1 4

B) 3

16

C) 3

4

8

17) Let M be the matrix game having payoff matrix

Find E(x, y) when x =

0 2 3 1 3

D) 1 2

0 2 -3 -1 2 0 2 0 .

-2 2 0 -2

17)

1 4 and y =

1 4 1 2 0

A) 4 3

B) 1

C) 1

3

2

4

D) 1

Find the value of the strategy.

0 -2 4

18) Let M be the matrix game having payoff matrix -2 5 1 .

-2

18)

1 0

1 3 1 Find (x) when x = 2 .

1 6

A) - 4

B) 11

3

C) - 7

6

6

D) 2

0 -2 3

19) Let M be the matrix game having payoff matrix -3 3 4 .

19)

2 -2 0

1 4 1 Find (x) when x = 4 .

1 2

A) 7

C) 1

B) 0

4

D) - 3

4

20) Let M be the matrix game having payoff matrix

4

0 -2 4 2 5 -2 .

-2 1 0

20)

1 3 1 Find (y) when y = 2 .

1 6

A) 1 3

B) - 1

C) - 1

3

6

5

D) 17 6

1 -1

4

2 -2

0

21) Let M be the matrix game having payoff matrix -3 1 -2 .

21)

1 4 1 Find (y) when y = 4 .

1 2

A) - 3

B) 0

2

C) - 1

22) Let M be the matrix game having payoff matrix

Find (x) when x =

0 2 3 1 3

D) 2

0 3 -1 -1 1 0 2 0 .

-2 1 0 -1

22)

.

B) 4

A) 0

C) 1

3

3

23) Let M be the matrix game having payoff matrix

D) - 1 3

0 3 -1 -1 1 0 1 0 .

-3 3 0 -3

23)

1 4 Find (y) when y =

1 4 .

1 2 0

A) 3 4

B) - 7

C) 0

4

6

D) 1 4

Find the optimal row or column strategy of the matrix game.

24) Let M be the matrix game having payoff matrix 5 -3 .

0 1

24)

Find the optimal row strategy.

^

5 9

A) x = 4 9

^

4 9

^

B) x = 5

8 9

C) x = 1

9

9

^

1 9

D) x = 8 9

25) Let M be the matrix game having payoff matrix 5 -4 .

0

25)

3

Find the optimal column strategy.

^

7 12

A) y = 5

12

^

1 4

^

B) y = 3

C) y = 7

4

26) Let M be the matrix game having payoff matrix

5 12 12

^

3 4

D) y = 1 4

5 5 0 10 .

-3 3 4 2

26)

Find the optimal row strategy.

^

7 12

A) x = 5

12

^

2 3

^

B) x = 1

C) x = 7

3

27) Let M be the matrix game having payoff matrix

5 12 12

^

1 3

D) x = 2 3

2 2 -1 5 .

-3 1 3 4

27)

Find the optimal column strategy.

2 3

2 3

4 9

4 9

^ A) y = 0

^ B) y = 1

^ C) y = 5

^ D) y = 0

0

0 0

0 0

0

1 3

3

9

7

5 9

4 -4 4

28) Let M be the matrix game having payoff matrix -1 6 4 .

28)

0 3

-3 Find the optimal row strategy.

^

7 15

A) x = 8

7 15 ^ B) x = 0

^

0

29) Let M be the matrix game having payoff matrix

^ D) x = 0

C) x = 1

8 15

15

2 3

2 3

1 3

3 0

3 -2 3 -3 1 4 .

-10 0 0

29)

Find the optimal column strategy.

^

4 9

^

A) y = 5

B) y = 2

0

0

9

2 3

1 3

^ C) y = 0

^

D) y = 1

1 3

3

1

1

3 -1

2 3 3 0

4

30) Let M be the matrix game having payoff matrix 1 -2 -1 1 -2 .

2 -1

0

30)

3 -2

Find the optimal row strategy.

1 3

2 3

1 2

^ A) x = 0

^ B) x = 0

^ C) x = 0

2 3

1 3

1 2

8

1 2 ^

1

D) x = 6 1 3

1

1

3 -1

4

31) Let M be the matrix game having payoff matrix 1 -2 -1 1 -2 .

2 -1

0

31)

3 -2

Find the optimal column strategy.

0 1 2 ^

0

^

0 2 3

A) y = 1

B) y = 0

1 6

0

Find the value of the matrix game.

32) Let M be the matrix game having payoff matrix A) 6

11

9

^

3 1 3

0

5 -1 .

-4 1

32)

C) 1

11

D) 1

10

16

4 4 -2 9 .

-3 1 5 3

B) 7

33)

C) 17

9

D) 1

7

6 -1

6

34) Let M be the matrix game having payoff matrix -3 5 1 .

7

B) 9

34)

0 -2

-9

A) 9

C) 19

5

D) 27

5

1

1

10

3 -1

4

35) Let M be the matrix game having payoff matrix 1 -2 -1 1 -2 .

2 -1

A) 1 2

B) 2

0

C) 1

3

4

9

0

D) y = 1

2 3

B) 1

33) Let M be the matrix game having payoff matrix A) 14

^

C) y = 0

1 3

3

0 1 3

0 1 3

35)

3 -2

D) 1 3

Solve the problem.

36) Each player has a supply of nickels, dimes, and quarters. At a given signal, both players display one coin. If the displayed coins are not the same, then the player showing the higher valued coin gets to keep both. If they are both nickels or dimes, then player R keeps both; but if they are both quarters, then player C keeps both. Find the optimal strategy for player C. n d q n 5 -5 -5 d 5 10 -10 q 5 10 -25 0 0 1 1 0 0 ^ ^ ^ ^ 2 A) y = 0 B) y = C) y = 3 D) y = 1 1 2 1 0 2 3

36)

37) Player R has two cards: a red 2 and a black 8. Player C has three cards: a red 3, a black 5, and a

37)

38) Player R has two cards: a red 2 and a black 8. Player C has three cards: a red 3, a black 5, and a

38)

black 10. They each show one of their cards. If the cards are the same color, C receives the larger of the two numbers. If the cards are of different colors, R receives the sum of the two numbers. The payoff matrix is : r3 b5 b10 r2 -3 7 12 b8 11 -8 -10 Find the optimal strategy for player R. 19 2 10 29 3 29 ^ ^ ^ ^ A) x = 10 B) x = 1 C) x = 1 D) x = 19 0 29 3 29

black 10. They each show one of their cards. If the cards are the same color, C receives the larger of the two numbers. If the cards are of different colors, R receives the sum of the two numbers. The payoff matrix is : r3 b5 b10 r2 -3 7 12 b8 11 -8 -10 Find the value of the game.

A) 19 B) 53 29 29

C) 10 29

10

D) 1 2

39) A certain army is engaged in guerrilla warfare. It has two ways of getting supplies to its troops: it

39)

can send a convoy up the river road or it can send a convoy overland through the jungle. On a given day, the guerrillas can watch only one of the two roads. If the convoy goes along the river and the guerrillas are there, the convoy will have to turn back and 7 army soldiers will be lost. If 1 the convoy goes overland and encounters the guerrillas, of the supplies will get through, but 9 3

army soldiers will be lost. Each day a supply convoy travels one of the roads, and if the guerrillas are watching the other road, the convoy gets through with no losses. What is the optimal strategy for the army if it wants to maximize the amount of supplies it gets to its troops?

A) 2 river, 3 land B) 0 river, 1 land 5 5

C) 3 river, 2 land 5

D) 1 river, 1 land

5

2

2

40) A certain army is engaged in guerrilla warfare. It has two ways of getting supplies to its troops: it

40)

can send a convoy up the river road or it can send a convoy overland through the jungle. On a given day, the guerrillas can watch only one of the two roads. If the convoy goes along the river and the guerrillas are there, the convoy will have to turn back and 5 army soldiers will be lost. If 3 the convoy goes overland and encounters the guerrillas, of the supplies will get through, but 7 4

army soldiers will be lost. Each day a supply convoy travels one of the roads, and if the guerrillas are watching the other road, the convoy gets through with no losses. If the army chooses the optimal strategy to maximize the amount of supplies it gets to its troops and the guerrillas choose the optimal strategy to prevent the most supplies from getting through, then what portion of the supplies will get through?

A) 4 B) 4 C) 1 D) 1 5 9 2 5

41) A certain army is engaged in guerrilla warfare. It has two ways of getting supplies to its troops: it can send a convoy up the river road or it can send a convoy overland through the jungle. On a given day, the guerrillas can watch only one of the two roads. If the convoy goes along the river and the guerrillas are there, the convoy will have to turn back and 4 army soldiers will be lost. If 1 the convoy goes overland and encounters the guerrillas, of the supplies will get through, but 7 3

army soldiers will be lost. Each day a supply convoy travels one of the roads, and if the guerrillas are watching the other road, the convoy gets through with no losses. What is the optimal strategy for the army if it wants to minimize its casualties?

A) 7 river, 4 land B) 1 river, 1 land 11 11 2 2

C) 4 river, 7 land 11

D) 1 river, 0 land

11

11

41)

42) A certain army is engaged in guerrilla warfare. It has two ways of getting supplies to its troops: it

42)

can send a convoy up the river road or it can send a convoy overland through the jungle. On a given day, the guerrillas can watch only one of the two roads. If the convoy goes along the river and the guerrillas are there, the convoy will have to turn back and 4 army soldiers will be lost. If 1 the convoy goes overland and encounters the guerrillas, of the supplies will get through, but 9 2

army soldiers will be lost. Each day a supply convoy travels one of the roads, and if the guerrillas are watching the other road, the convoy gets through with no losses. What is the optimal strategy for the guerrillas if they want to inflict maximum losses on the army?

A) 9 river, 4 land B) 4 river, 9 land 13 13 13 13

C) 1 river, 1 land 2

D) 0 river, 1 land

2

Provide an appropriate response.

43) Anne and Michael are playing a game in which each player has a choice of two colors: green or blue. The payoff matrix with Anne as the row player is given below: gr bl green a b blue c d

43)

Using the same payoffs for Anne and Michael, write the matrix that shows the winnings with Michael as the row player.

A) B) gr bl gr bl green a c green d c blue b d blue b a C) D) gr bl gr bl green -a -c green -a -b blue -b -d blue -c -d

44) A matrix game has payoff matrix

44)

a 11 a 12 a 13 a 21 a 22 a 23 in which entry a 21 is a saddle point. What are the optimal strategies for player R and player C?

0

A) x = 0 , y = 0 ^

^

1

^

^

0

^

1

1

1

C) x = 1 , y = 0 ^

1

B) x = 0 , y = 0

^

1 2

0 1 3

^

1

D) x = 1 , y = 3

0

2

12

1 3

45) If the payoff matrix of a matrix game contains a saddle point, what is the optimal strategy for the

45)

row player?

A) Always choose the row with the smallest maximum.

B) Always choose the row with the largest minimum.

C) Always choose the row with the smallest minimum.

D) Always choose the row with the largest maximum.

46) If the payoff matrix of a matrix game contains a saddle point, what is the optimal strategy for the

46)

column player?

A) Always choose the column with the smallest minimum.

B) Always choose the column with the smallest maximum.

C) Always choose the column with the largest maximum.

D) Always choose the column with the largest minimum.

47) In a matrix game with payoff matrix A, how can you find the value (y) of a strategy y to player

47)

C?

A) (y) is the minimum of the inner product of y with each of the rows of A. B) (y) is the maximum of the inner product of y with each of the columns of A. C) (y) is the minimum of the inner product of y with each of the columns of A. D) (y) is the maximum of the inner product of y with each of the rows of A. 48) In a matrix game with payoff matrix A, how can you find the value (x) of a strategy x to row

48)

player R? A) (x) is the minimum of the inner product of x with each of the columns of A.

B) (x) is the maximum of the inner product of x with each of the rows of A. C) (x) is the maximum of the inner product of x with each of the columns of A. D) (x) is the minimum of the inner product of x with each of the rows of A. 49) In certain situations, a matrix game can be reduced to a smaller game by deleting certain rows

and/or columns from the payoff matrix. The optimal strategy for the reduced game will then determine the optimal strategy for the original game. In what circumstances may a row or column be deleted from the payoff matrix?

A) A row may be deleted if it is dominant to some other row and a column may be deleted if it is dominant to some other column.

B) A row may be deleted if it is recessive to some other row and a column may be deleted if it is dominant to some other column.

C) A row may be deleted if it is recessive to some other row and a column may be deleted if it is recessive to some other column.

D) A row may be deleted if it is dominant to some other row and a column may be deleted if it is recessive to some other column.

13

49)

50) The optimal strategy for a 2 × n matrix game can be found as follows: obtain n linear functions by

50)

51) The optimal strategy for a 2 × 5 matrix game can be found as follows: obtain 5 linear functions by

51)

finding the inner product of x(t) = 1 - t with each of the columns of the payoff matrix A. Graph t the n linear functions on a t-z coordinate system. Then (x(t)) is the minimum value of the n linear functions which will be seen on the graph as a polygonal path. The z-coordinate of any point on this path is the minimum of the corresponding z coordinates of points on the n lines. The highest point on the path (x(t)) is M. Suppose that M has coordinates (a, b). What information is given by these coordinates?

A) The optimal strategy for R is a and the value of the game for R is b.

1-a B) The optimal strategy for C is a and the value of the game for C is b.

1-a 1 C) The optimal strategy for R is - a and the optimal strategy for C is 1 - b . a b D) The optimal strategy for R is 1 - a and the value of the game for R is b. a

finding the inner product of x(t) = 1 - t with each of the columns of the payoff matrix A. Graph t the 5 linear functions on a t-z coordinate system. Then (x(t)) is the minimum value of the 5 linear functions which will be seen on the graph as a polygonal path. The z-coordinate of any point on this path is the minimum of the corresponding z coordinates of points on the 5 lines. The highest point on the path (x(t)) is M. Suppose that only the lines corresponding to columns 1 and 4 of ^

matrix A pass through the point M. What can be said about the optimal column strategy y ?

^

0 c2

c1 ^

A) y = c3 where c2 + c3 + c5 = 1

0

B) y = 0 where c1 + c4 = 1

0 c5

c4 0 1 2

0 0 ^ C) y = 0 1 0

^

0

D) y = 0

1 2

0

Determine whether the statement is true or false.

52) If the payoff matrix of a matrix game contains a saddle point, the optimal strategy for each player will be a pure strategy.

A) True B) False

14

52)

53) If the payoff matrix of a matrix game contains a saddle point, the optimal strategy for the row

53)

54) The value C of a matrix game to player C is the maximum of the values of the various possible

54)

player will be to always choose the row with the largest minimum while the optimal strategy for the column player will be to always choose the column with the smallest maximum.

A) True B) False

strategies for C. A) True

B) False

55) In a matrix game, the value (y) of a particular strategy y to player C is equal to the minimum of

55)

56) In a matrix game, if row s is dominant to some other row in payoff matrix A, then row s will not

56)

the inner product of y with each of the rows of the payoff matrix A. A) True B) False

be used in some optimal strategy for row player R. A) True

15

B) False

Solve the problem.

57) Alan wants to invest a total of $16,000 in mutual funds, CDs, and a high yield savings account. He wants to invest no more in mutual funds than half the amount he invests in CDs. He also wants the amount in savings to be at least twice the sum of his CDs and mutual funds. His expected return on mutual funds is 8%, on the CDs is 6%, and on savings is 3%. How much money should Alan invest in each area in order to have the largest return on his investments? Set this up as a linear programming problem in the following form: Maximize cTx subject to Ax b and x 0. Do not find the solution.

A) Let x1 be the amount invested in mutual funds, x2 the amount in CDs, and x3 the amount in savings.

x1 16,000 0.08 1 1 1 Then b = 0 , x = x2 , c = 0.06 , and A = 2 -1 0 0 x3 0.03 2 2 -1

B) Let x1 be the amount invested in mutual funds, x2 the amount in CDs, and x3 the amount in savings.

x1 16,000 0.08 1 2 2 Then b = 0 , x = x2 , c = 0.06 , and A = 1 -1 2 0 x3 0.03 1 0 -1

C) Let x1 be the amount invested in mutual funds, x2 the amount in CDs, and x3 the amount in savings.

x1 0 0.08 1 Then b = 0 , x = x2 , c = 0.06 , and A = 0 0 x3 0.03 2

1 1 2 -1 2 -1

D) Let x1 be the amount invested in mutual funds, x2 the amount in CDs, and x3 the amount in savings.

x1 16,000 0.08 1 1 1 x Then b = , x , c , and A = = = 2 0.5 0.06 2 -1 0 2 x3 0.03 2 2 -1

16

57)

58) Daniel decides to feed his cat a combination of two foods: Max Cat and Mighty Cat. He wants his

cat to receive four nutritional factors each month. The amounts of these factors (a, b, c, and d) contained in one bag of each food are shown in the chart, together with the total amounts needed. a b c d Max Cat 1 3 2 3 Mighty Cat 3 1 2 1 Needed 21 22 23 19 The costs per bag are $43 for Max Cat and $45 for Mighty Cat. How many bags of each food should be blended to meet the nutritional requirements at the lowest cost? Set this up as a linear programming problem in the following form: Minimize cTx subject to Ax b and x 0. Do not find the solution.

A) Let x1 be the number of bags of Max Cat and x2 the number of bags of Mighty Cat. 21 1 3 x1 22 43 Then b = ,x= ,c= , and A = 3 1 x2 23 45 2 2 19 3 1 B) Let x1 be the number of bags of Max Cat and x2 the number of bags of Mighty Cat. 21 1 3 x1 43 22 Then b = ,x= ,c= , and A = 3 1 x2 45 23 2 2 19 3 1 C) Let x1 be the number of bags of Max Cat and x2 the number of bags of Mighty Cat. 0 1 3 21 x1 0 43 Then b = ,x= ,c= , and A = 3 1 22 x2 0 45 2 2 23 0 3 1 19 D) Let x1 be the number of bags of Max Cat and x2 the number of bags of Mighty Cat. 21 x1 Then b = 22 , x = , c = 43 , and A = x2 23 45 19

1 3 2 3 3 1 2 1

17

58)

59) An airline with two types of airplanes, P1 and P2 , has contracted with a tour group to provide transportation for a minimum of 390 first class, 770 tourist class, and 1,440 economy class passengers. For a certain trip, airplane P1 costs $12,000 to operate and can accommodate 20 first class, 51 tourist class, and 100 economy class passengers. Airplane P2 costs $9,000 to operate and can accommodate 18 first class, 30 tourist class, and 44 economy class passengers. How many of each type of airplane should be used in order to minimize the operating cost? Set this up as a linear programming problem in the following form: Minimize cTx subject to Ax b and x 0. Do not find the solution.

A) Let x1 be the number of airplanes of type P1 and x2 the number of airplanes of type P2 0 20 18 390 x1 Then b = 0 , x = , c = 12,000 , and A = 51 30 770 x2 9,000 0 100 44 1,440 B) Let x1 be the number of airplanes of type P1 and x2 the number of airplanes of type P2 390 x1 , c = 12,000 , and A = 20 51 100 770 , x = x2 9,000 18 30 44 1,440 C) Let x1 be the number of airplanes of type P1 and x2 the number of airplanes of type P2 Then b =

390 20 18 x1 , c = 12,000 , and A = 51 30 770 , x = x2 9,000 1,440 100 44 D) Let x1 be the number of airplanes of type P1 and x2 the number of airplanes of type P2 Then b =

x1 Then b = 12,000 , x = ,c= x2 9,000

390 20 18 770 , and A = 51 30 1,440 100 44

18

59)

60) A furniture company makes two different types of lamp stands. Each lamp stand A requires 12

60)

minutes for sanding, 18 minutes for assembly, and 9 minutes for packaging. Each lamp stand B requires 15 minutes for sanding, 23 minutes for assembly, and 8 minutes for packaging. The total number of minutes available each day in each department are as follows: for sanding 4000 minutes, for assembly 8000 minutes, and for packaging 2000 minutes. The profit on each lamp stand A is $20 and the profit on each lamp stand B is $27. How many of each type of lamp stand should the company make per day to maximize their profit? Set this up as a linear programming problem in the following form: Maximize cTx subject to Ax b and x 0. Do not find the solution.

A) Let x1 be the number of lamp stands of type A made per day and x2 the number of lamp stands of type B made per day.

4000 12 15 x1 Then b = 20 , x = , c = 8000 , and A = 18 23 x2 27 2000 9 8 B) Let x1 be the number of lamp stands of type A made per day and x2 the number of lamp stands of type B made per day.

4000 x1 Then b = 8000 , x = , c = 20 , and A = 12 18 9 x2 27 15 23 8 2000 C) Let x1 be the number of lamp stands of type A made per day and x2 the number of lamp stands of type B made per day.

4000 12 15 x1 Then b = 8000 , x = , c = 20 , and A = 18 23 x2 27 2000 9 8 D) Let x1 be the number of lamp stands of type A made per day and x2 the number of lamp stands of type B made per day.

0 12 15 4000 x1 Then b = 0 , x = , c = 20 , and A = 18 23 8000 x2 27 0 9 8 2000

Find vectors b and c and matrix A so that the problem is set up as a canonical linear programming problem of the form: maximize cT x subject to Ax b and x 0. Do not find the solution.

61) Maximize

5x1 + x2 + 6x3

subject to

61)

x1 + 3x2 - 2x3 23 -4x2 + 6x3 30

and x1 0, x2 0, x3 0

5

5

1

6

5

A) b = 23 , c = 1 , and A = 1 3 -2

B) b = 23 , c = 1 , and A = 1 3 -2

30

30

6 0 -4 6 5 1 0 C) b = 23 , c = 1 , and A = 3 -4 30 6 -2 6

5

6

0 -4

6

D) b = 1 , c = 23 , and A = 1 3 -2 6

19

30

0 -4

6

62) Maximize subject to

x1 - 2x2 + 4x3

62)

4x1 + x2 + 4x3 26 3x1

2x3 29

-

and x1 0, x2 0, x3 0

1

1

A) b = 26 , c = -2 , and A = 4 1 4

B) b = -2 , c =

29

3 0 -2 4 1 4 3 C) b = 26 , c = -2 , and A = 1 0 -29 4 4 -2

63) Maximize subject to

4

D) b =

26 , and A = 4 1 4 -29 -3 0 2

1 26 , c = 4 1 4 -2 , and A = -29 -3 0 2 4

x1 - 6x2 + 3x3

63)

4x1 + x2 - 3x3 39 4x1 + 6x2 - x3 = 40

and x1

0, x2 0, x3

0

1

1

4 4 1 6 40 3 -3 -1 39 1 4 1 -3 D) b = 40 , c = -6 , and A = 4 6 -1 3 -40 -4 -6 1

A) b = 39 , c = -6 , and A = 4 1 -3 40

4

3

B) b = 39 , c = -6 , and A =

6 -1

39 1 4 4 -4 40 , c = -6 , and A = 1 6 -6 3 -40 -3 -1 1

C) b =

64) Minimize subject to

4x1 +

x2 - 4x3

64)

x1 + 5x2 - 3x3 28 -3x2 + 4x3 40

and x1 0, x2 0, x3 0

4 -1 -5 3 1 , and A = -40 0 3 -4 -4 -4 C) b = 28 , c = -1 , and A = 1 5 -3 40 0 -3 4 4

4 1 5 -3 1 , and A = 40 0 -3 4 -4 -4 D) b = -28 , c = -1 , and A = -1 -5 3 -40 0 3 -4 4

B) b = 28 , c =

A) b = -28 , c =

65) Minimize

x1 - 4x2 + 4x3

65)

subject to 5x1 + x2 + 6x3

33

3x1

37

and x1

0, x2 0, x3

- 5x3 0

1

-1 5 1 6 4 , and A = 3 0 -5 -4 1 C) b = 33 , c = -4 , and A = 5 1 6 37 3 0 -5 4

A) b =

33 , c = -37

B) b = 33 , c = -4 , and A = 37

5 1 6 -3 0 5

4 -1 D) b = 33 , c = 4 , and A = 5 1 6 -37 -3 0 5 -4

20

66) Minimize subject to

x1 - 5x2 + 4x3 2x1

66)

-

5x3 26

2x1 - x2

= 23

and x1 0, x2 0, x3 0

-26 -1 -2 0 5 23 , c = 5 , and A = 2 -1 0 -23 -4 -2 1 0 1 C) b = -26 , c = -5 , and A = -2 0 5 23 2 -1 0 4

2 0 -5 -26 -1 23 , c = 5 , and A = 2 -1 0 -23 -4 -2 1 0 1 -26 -2 0 5 D) b = 23 , c = -5 , and A = 2 -1 0 4 -23 -2 1 0

A) b =

B) b =

Solve the linear programming problem.

67) Maximize 6x1 + 7x2 subject to

67)

2x1 + 3x2 12 2x1 + x2 8

and x1 0, x2 0

A) maximum = 32 when x1 = 3 and x2 = 2 C) maximum = 39 when x1 = 3 and x2 = 3 68) Minimize subject to

B) maximum = 38 when x1 = 4 and x2 = 2 D) maximum = 28 when x1 = 0 and x2 = 4

6x1 + 8x2

68)

2x1 + 4x2 12 2x1 + x2 8

and x1 0, x2 0

A) Minimum = 86 when x1 = 3 and x2 = 4 3

3

B) Minimum = 36 when x1 = 6 and x2 = 0 C) Minimum = 64 when x1 = 0 and x2 = 8 D) Minimum = 92 when x1 = 10 and x2 = 4 3

69) Maximize subject to

3

3

2x1 + 5x2

69)

3x1 + 2x2 6 -2x1 + 4x2 8

and x1 0, x2 0

A) Maximum = 15 when x1 = 0 and x2 = 3 B) Maximum = 4 when x1 = 2 and x2 = 0 C) Maximum = 27 when x1 = 1 and x2 = 5 2

2

2

D) Maximum = 49 when x1 = 1 and x2 = 9 4

2

4

21

70) Minimize subject to

4x1 + 5x2

70)

2x1 - 4x2 10 2x1 + x2 15

and x1 0, x2 0

A) Minimum = 33 when x1 = 7 and x2 = 1 B) Minimum = 75 when x1 = 0 and x2 = 15 C) Minimum = 30 when x1 = 15 and x2 = 0 2

D) Minimum = 20 when x1 = 5 and x2 = 0 71) Maximize 50x1 + 35x2 subject to

71)

3x1 + x2 24 x1 + x2 16 2x1 + 3x2 30

and x1 0, x2 0

A) Maximum = 400 when x1 = 8 and x2 = 0 B) Maximum = 510 when x1 = 6 and x2 = 6 C) Maximum = 350 when x1 = 0 and x2 = 10 D) Maximum = 620 when x1 = 4 and x2 = 12 72) Minimize subject to

4x1 + 5x2

72)

5x1 + 4x2 20 4x1 + x2 12

x1 + 3x2 9 and x1 0, x2 0

A) Minimum = 36 when x1 = 9 and x2 = 0 B) Minimum = 221 when x1 = 24 and x2 = 25 11

11

11

C) Minimum = 228 when x1 = 27 and x2 = 24 11

11

11

D) Minimum = 212 when x1 = 28 and x2 = 20 11

11

11

22

73) Maximize 2x1 + 4x2 subject to

73)

x1 - 2x2 -4 -2x1 + x2 -6

and x1 0, x2 0

A) Infeasible B) Maximum = 8 when x1 = 0 and x2 = 2 C) Maximum = 88 when x1 = 16 and x2 = 14 3

3

3

D) Unbounded 74) Minimize subject to

8x1 + 3x2

74)

x1 - 2x2 -4 -3x1 + x2 -9

and x1 0, x2 0

A) Unbounded B) Infeasible C) Minimum = 6 when x1 = 0 and x2 = 2 D) Minimum = 239 when x1 = 22 and x2 = 21 5

5

5

75) Maximize 4x1 + 8x2 subject to

75)

x1 - x 2 2 -x1 + 8x2 -4

and x1 0, x2 0

A) Maximum = 32 when x1 = 12 and x2 = - 2 7

7

7

B) Maximum = 8 when x1 = 2 and x2 = 0 C) Unbounded D) Infeasible Solve the problem.

76) Alan wants to invest a total of $18,000 in mutual funds and a certificate of deposit (CD). He wants to invest no more in mutual funds than half the amount he invests in the CD. His expected return on mutual funds is 10% and on the CD is 5%. How much money should Alan invest in each area in order to have the largest return on his investments? What is his maximum one-year return?

A) Maximum one-year return is $900 when he invests $0 in mutual funds and $18,000 in the CD. B) Maximum one-year return is $1,200 when he invests $6,000 in mutual funds and $12,000 in the CD. C) Maximum one-year return is $1,800 when he invests $18,000 in mutual funds and $0 in the CD. D) Maximum one-year return is $1,500 when he invests $12,000 in mutual funds and $6,000 in the CD.

23

76)

77) A store makes two different types of smoothies by blending different fruit juices. Each bottle of

77)

78) Daniel decides to feed his cat a combination of two foods: Max Cat and Mighty Cat. He wants his

78)

Orange Daze smoothie contains 10 fluid ounces of orange juice, 4 fluid ounces of pineapple juice, and 2 fluid ounces of blueberry juice. Each bottle of Pineapple Blue smoothie contains 5 fluid ounces of orange juice, 6 fluid ounces of pineapple juice, and 4 fluid ounces of blueberry juice. The store has 500 fluid ounces of orange juice, 360 fluid ounces of pineapple juice, and 250 fluid ounces of blueberry juice available to put into its smoothies. The store makes a profit of $1.50 on each bottle of Orange Daze and $1 on each bottle of Pineapple Blue. How many bottles of each smoothie should the store make to maximize its profit? What is the maximum profit?

A) Maximum profit is $80 when the store makes 40 bottles of Orange Daze and 20 bottles of Pineapple Blue.

B) Maximum profit is $75 when the store makes 50 bottles of Orange Daze and 0 bottles of Pineapple Blue.

C) Maximum profit is $87.50 when the store makes 35 bottles of Orange Daze and 35 bottles of Pineapple Blue.

D) Maximum profit is $85 when the store makes 30 bottles of Orange Daze and 40 bottles of Pineapple Blue.

cat to receive four nutritional factors each month. The amounts of these factors (a, b, c, and d) contained in one bag of each food are shown in the chart, together with the total amounts needed. a b c d Max Cat 4 3 1 6 Mighty Cat 5 2 2 3 Needed 54 37 20 60 The costs per bag are $40 for Max Cat and $35 for Mighty Cat. How many bags of each food should be blended to meet the nutritional requirements at the lowest cost? What is the minimum cost?

A) Minimum cost = $500 when he blends 20 bags of Max Cat and 20 bags of Mighty Cat. 3 3

B) Minimum cost = $700 when he blends 0 bags of Max Cat and 20 bags of Mighty Cat. C) Minimum cost = $541.25 when he blends 17 bags of Max Cat and 23 bags of Mighty Cat. 2

4

D) Minimum cost = $610 when he blends 3 bags of Max Cat and 14 bags of Mighty Cat. 79) A furniture company makes two different types of lamp stand. Each lamp stand A requires 20

minutes for sanding, 48 minutes for assembly, and 6 minutes for packaging. Each lamp stand B requires 9 minutes for sanding, 32 minutes for assembly, and 8 minutes for packaging. The total number of minutes available each day in each department are as follows: for sanding 3600 minutes, for assembly 9600 minutes, and for packaging 2000 minutes. The profit on each lamp stand A is $30 and the profit on each lamp stand B is $22. How many of each type of lamp stand should the company make per day to maximize their profit? What is the maximum profit?

A) Maximum profit is $6380 when they make 66 of lamp stand A and 200 of lamp stand B.

B) Maximum profit is $6164 when they make 138 of lamp stand A and 92 of lamp stand B. C) Maximum profit is $5400 when they make 180 of lamp stand A and 0 of lamp stand B. D) Maximum profit is $6850 when they make 100 of lamp stand A and 175 of lamp stand B.

24

79)

80) The Acme Class Ring Company designs and sells two types of rings: the VIP and the SST. They

80)

can produce up to 24 rings each day using up to 60 total man-hours of labor. It takes 3 man-hours to make one VIP ring and 2 man-hours to make one SST ring. How many of each type of ring should be made daily to maximize the company's profit, if the profit on a VIP ring is $40 and on an SST ring is $35?

A) 12 VIP and 12 SST B) 18 VIP and 6 SST

C) 16 VIP and 8 SST

D) 14 VIP and 10 SST

81) An airline with two types of airplanes, P1 and P2 , has contracted with a tour group to provide

81)

transportation for a minimum of 400 first class, 750 tourist class, and 1500 economy class passengers. For a certain trip, airplane P1 costs $10,000 to operate and can accommodate 20 first class, 50 tourist class, and 110 economy class passengers. Airplane P2 costs $8500 to operate and can accommodate 18 first class, 30 tourist class, and 44 economy class passengers. How many of each type of airplane should be used in order to minimize the operating cost?

A) 7 P1 planes and 11 P2 planes B) 20 P1 planes and 0 P2 planes

C) 9 P1 planes and 13 P2 planes

D) 11 P1 planes and 7 P2 planes

82) A summer camp wants to hire counselors and aides to fill its staffing needs at minimum cost. The

82)

average monthly salary of a counselor is $2400 and the average monthly salary of an aide is $1100.

The camp can accommodate up to 45 staff members and needs at least 30 to run properly. They must have at least 10 aides, and may have up to 3 aides for every 2 counselors. How many counselors and how many aides should the camp hire to minimize cost?

A) 12 counselors and 18 aides B) 35 counselors and 10 aides

C) 18 counselors and 12 aides

D) 27 counselors and 18 aides

Provide an appropriate response.

83) A linear programming problem has objective function 3x1 + 4x2 and constraints ax1 + bx2 10 cx1 + dx2 20 x1 0, x2 0 where a, b, c, and d are positive numbers.

Does a maximum exist for the objective function? If not, why not? Does a minimum exist? If not, why not?

A) Maximum does not exist as the objective function may be arbitrarily large (unbounded).

Minimum does exist.

B) Neither maximum nor minimum exist as the feasible set is the empty set.

C) Neither maximum nor minimum exist as the problem is unbounded.

D) Maximum does exist. Minimum does not exist as the objective function may be arbitrarily small (unbounded).

25

83)

84) A linear programming problem has objective function 3x1 + 4x2 and constraints

84)

ax1 + bx2 10 cx1 + dx2 20 x1 0, x2 0 where a, b, c, and d are positive numbers.

Does a maximum exist for the objective function? If not, why not? Does a minimum exist? If not, why not?

A) Maximum and minimum both exist.

B) Neither maximum nor minimum exist as the feasible set is the empty set.

C) Maximum exists. Minimum does not exist as the objective function may be arbitrarily small

(unbounded).

D) Maximum does not exist as the objective function may be arbitrarily large (unbounded).

Minimum does exist.

85) A linear programming problem has objective function 4x and constraints

85)

x a -x -b x 0 where a and b are positive numbers and b > a.

Does a maximum exist for the objective function? If not, why not? Does a minimum exist? If not, why not?

A) Maximum exists. Minimum does not exist as the objective function may be arbitrarily small (unbounded).

B) Neither maximum nor minimum exist as the feasible set is the empty set.

C) Maximum does not exist as the objective function may be arbitrarily large (unbounded).

Minimum does exist.

D) Maximum and minimum both exist.

86) A linear programming problem has objective function 6x and constraints

x a -x -b where a and b are positive numbers.

Does a maximum exist for the objective function? If not, why not? Does a minimum exist? If not, why not?

A) Maximum exists. Minimum does not exist as the objective function may be arbitrarily small (unbounded).

B) Maximum and minimum both exist.

C) Maximum does not exist as the objective function may be arbitrarily large (unbounded).

Minimum does exist.

D) Neither maximum nor minimum exist as the feasible set is the empty set.

26

86)

87) Determine whether the statement below is true or false. If it is false, give an explanation.

87)

If the feasible set for a canonical linear programming problem is not empty, there must be at least one optimal solution.

A) False. The feasible set may contain only one point.

B) True C) False. The objective function may be unbounded.

D) False. The objective function may have the same value for more than one extreme point of the feasible set.

88) Determine whether the statement below is true or false. If it is false, give an explanation.

88)

Given a canonical linear programming problem, if f(x) is greater than or equal to f(x) for all vectors x in the feasible set, then x must be an optimal solution.

A) False. The objective function may be unbounded.

B) False. The problem may ask for the objective function to be minimized.

C) False. x itself may not be in the feasible set.

D) True 89) Determine whether the statement below is true or false. If it is false, give an explanation.

89)

Given a canonical linear programming problem with an optimal solution, if the vector x is an extreme point of the feasible set, then x must be an optimal solution.

A) False. x may not be in the feasible set.

B) True C) False. Some extreme point is an optimal solution but not every extreme point is an optimal solution.

D) False. Not every optimal solution is an extreme point.

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

90) The set S of all x that satisfy the inequality ax b is a ray in R1. Show that the set S is

90)

convex.

[Hint: A set S in Rn is convex if, for each p and q in S, the line segment between p and q lies in S. This line segment is the set of points of the form (1 - t)p + tq for 0 t 1. ]

91) A set S in Rn is convex if, for each p and q in S, the line segment between p and q lies in S. [This line segment is the set of points of the form (1 - t)p + tq for 0 t 1. ] Let F be the feasible set of all solutions x of a linear programming problem Ax b with x 0. Assume that F is nonempty. Show that F is a convex set in Rn .

[Hint: Consider points p and q in F and t such that 0 t 1. Show that (1 - t)p + tq is in F.]

27

91)

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Set up the initial simplex tableau for the given linear programming problem.

92) Maximize 24x1 + 23x2 subject to

4x1 + 7x2

37

7x1 + 7x2

27

92)

and x1 0, x2 0

A)

C)

x1

x2

4 7 7 7 24 23 x1 x 2

x3 1 0 0 x3

x4 M 0 0 37 1 0 27 0 1 0 x4 x 5 M

4 7 7 7 -24 -23

1 0 0

0 1 0

B)

D)

1 0 37 1 0 27 0 1 0

93) Maximize

29x1 + 19x2 + 26x3

subject to

8x1 + 11x2 + 4x3

30

8x1

40

+ 2x2

+ 5x3

x1

x2

4 7 7 7 -24 -23 x1 x 2 4 7 7 7 -24 -23

x3 x4 M 1 0 0 0 1 0 0 0 0 x3 x4 M 1 0 0 0 1 0 0 0 1

37 27 0 37 27 0

93)

and x1 0, x2 0, x3 0

A)

C)

x1 x2

x 3 x 4 x5 M

8 11 4 1 0 8 2 5 0 1 -29 -19 -26 0 0 x 1 x 2 x 3 x 4 x5

0 30 0 40 1 0 M

8 8 29

0 30 0 40 1 0

11 2 19

4 5 26

1 0 0

0 1 0

B)

x1

x2

x3

x4

x5

D)

8 8 29 x1

11 2 19 x2

4 5 26 x3

1 0 0 x4

0 30 1 40 0 0 x5

8 8 -29

11 2 -19

4 5 -26

1 0 0

0 30 1 40 0 0

94) Maximize 16x1 + 27x2 subject to

94)

11x1 + 7x2

31

5x1 + 3x2

29

x1

32

+ 4x2

and x1 0, x2 0

A)

x1

x2

C)

11 7 5 3 1 4 16 27 x1 x 2 11 7 5 3 1 4 -16 -27

x3 x4 x5 1 0 0 0 1 0 0 0 1 0 0 0 x3 x4 x5 1 0 0 0 1 0 0 0 1 0 0 0

M 0 31 0 29 0 32 1 0 M 0 31 0 29 0 32 1 0

B)

x1

D)

11 7 5 3 1 4 -16 -27 x1 x2 5 5 1 16

28

x2

7 3 4 27

x3 x4 x5 M 1 0 0 0 -31 0 1 0 0 -29 0 0 1 0 -32 0 0 0 1 0 x3 x4 x5 1 0 0 31 0 1 0 29 0 0 1 32 0 0 0 0

95) Minimize 5x1 + 4x2 subject to

95)

4x1 - x2

11

4x1 + 5x2

7

and x1 0, x2 0

A)

C)

x1 x2 4 -1 -4 -5 5 4 x1 x2

x3 x 4 M 1 0 0 0 1 0 0 0 1 x3 x 4 M

11 -7 0

4 4 5

1 0 0

11 7 0

-1 5 4

0 1 0

0 0 1

B)

D)

x1

x2

x3 x 4 M 4 1 0 0 11 -1 0 1 0 -7 -4 -5 0 0 1 0 -5 -4 x 1 x 2 x3 x 4 M 4 -1 1 0 0 11 4 5 0 1 0 7 0 0 1 0 -5 -4

Identify the basic feasible solution corresponding to the given simplex tableau.

96) x1 x2 x3 x4 M

96)

9 4 -31

10 1 0 0 42 4 0 1 0 38 0 0 1 0 -35 A) x1 = -31, x2 = -35, x3 = 0, x4 = 0, M = 1

B) x1 = 0, x2 = 0, x3 = 42, x4 = 38, M = 0 C) x1 = 9, x2 = 10, x3 = 1, x4 = 0, M = 0 D) x1 = -31, x2 = -35, x3 = 42, x4 = 38, M = 0 97)

x 1 x 2 x 3 x 4 x5 M 8 6 10 1 0 0 45 45 31 11 0 1 0 31 -15 -20 -29 0 0 1 0 A) x1 = 8, x2 = 6, x3 = 10, x4 = 45, x5 = 31, M = 0

97)

B) x1 = -15, x2 = -20, x3 = -29, x4 = 0, x5 = 0, M = 1 C) x1 = 0, x2 = 0, x3 = 0, x4 = 45, x5 = 31, M = 0 D) x1 = 8, x2 = 6, x3 = 10, x4 = 1, x5 = 0, M = 0 98)

x1

x2

x3 x4 x5 M 2 12 1 0 0 0 50 8 8 0 1 0 0 24 1 11 0 0 1 0 49 -25 -31 0 0 0 1 0 A) x1 = 0, x2 = 0, x3 = 50, x4 = 24, x5 = 49, M = 0

B) x1 = 2, x2 = 12, x3 = 50, x4 = 24, x5 = 49, M = 0 C) x1 = -25, x2 = -31, x3 = 0, x4 = 0, x5 = 0, M = 1 D) x1 = 2, x2 = 12, x3 = 1, x4 = 0, x5 = 0, M = 0

29

98)

99) x1 x2 x3

x4

M

99)

0

1

1 2

-1

0

8

1

0

3 -

4 3

0

10

0

0

4

3

1 130 A) x1 = 0, x2 = 1, x3 = 1 , x4 = -1, M = 0 2

B) x1 = 0, x2 = 0, x3 = 4, x4 = 3, M = 130

C) x1 = 8, x2 = 10, x3 = 0, x4 = 0, M = 130

D) x1 = 10, x2 = 8, x3 = 0, x4 = 0, M = 130

100) x1 x2

x3

x4 M

1 -

5 2

0

2

0

9

0

-2

1

4 3

0

5

0

2

0

3

1 54

101)

100)

A) x1 = 9, x2 = 0, x3 = 5, x4 = 0, M = 54

B) x1 = 1, x2 = - 5 , x3 = 0, x4 = 2, M = 0

C) x1 = 5, x2 = 0, x3 = 9, x4 = 0, M = 54

D) x1 = 0, x2 = 2, x3 = 0, x4 = 3, M = 1

x1 x2 1 0 6

x3

x4

M

3

1

0

9 8

2

101)

-3

1 -

1 3

0

0

-1

0

4

0

1 32

A) x1 = -1, x2 = 0, x3 = 4, x4 = 0, M = 1

B) x1 = 1 , x2 = 0, x3 = 3, x4 = 1, M = 0

C) x1 = 0, x2 = 8, x3 = 0, x4 = 9, M = 32

D) x1 = 0, x2 = 9, x3 = 0, x4 = 8, M = 32

6

For the given simplex tableau, determine which variable should be brought into the solution and which row to use as a pivot.

102) x1 x2 x3 x4 M 102) 12 8 -5

11 1 9 0 0 -9 A) x2 , row 1

0 1 0

0 35 0 46 1 0

B) x1 , row 1

C) x2 , row 2

30

D) x1 , row 2

103)

104)

105)

x1

x2

x3 11 9 1 4 10 0 0 -4 -8 A) x1 , row 2 x1

x2

x3 8 5 1 2 12 0 0 -9 -4 A) x2 , row 1 x1

x3 5 5 1 2 2 0 0 -6 -6 A) x1 , row 2

106) x1

x2

x4 M 0 1 0

103)

0 42 0 38 1 0

B) x2 , row 1

C) x2 , row 2

D) x1 , row 1

x4 M 0 1 0

104)

0 37 0 47 1 0

B) x2 , row 2

C) x1 , row 2

D) x1 , row 1

x4 M 2 3 8

105)

0 22 0 43 1 0

B) x1 , row 1

C) x2 , row 2

x3 x4 M 1 11 0 0 41 -9 0 9 4 1 0 44 0 12 0 1 0 -4 A) x1 , row 2 B) x1 , row 1

D) x2 , row 1

x2

106)

C) x2 , row 1

D) x2 , row 2

107) x1 x2 x3 x4 x5 M 5 3

107)

0

1

2

0

0 18

1 1 2

0

7

0

0

5 3

0

4

1

0 12

-11 0 0 19 A) x1 , row 3

0

1

-

0

5

0

B) x4 , row 2

C) x1 , row 2

31

D) x1 , row 1

A simplex tableau is given. Compute the next tableau.

108) x1 x2 x3 x4 M

108)

11 1 1 0 0 22 12 2 0 1 0 47 -8 -10 0 0 1 0 A) x1 x2 x3 x4 M 0 1 1 0 0 22 -10 0 -2 1 0 3 -30 0 -2 0 1 44 C) x1 x2 x3 x4 M 11 1 1 -10 0 -2 102 0 10

109)

0 1 0

C)

- 12

1

1

6

0

0

- 51

0 -

x1

x2

0 - 12 1 0 0

51

D)

0 22 0 3 1 220

x1 x2 x3 x4 M 3 6 1 0 0 25 2 12 0 1 0 16 -9 -3 0 0 1 0 A) x1 x2 x3 x4 0

B) x1 x2 x3 x4 M

-

3 2

11 10 102 x1 11 10 -30

1 1 0 0 22 0 -2 1 0 3 0 10 10 1 220 x2 x3 x4 M 1 1 0 0 22 0 -2 1 0 3 0 -2 0 1 44

109)

M

B)

x1

x2

x3 x4 M 3 1 0 2

3 2

0

1

0 - 12

1 2

0

8

1

6

0

1 2

0

8

1 24

0

51

0

9 2

1

72

D) x1 x2

x3

x4

0

- 12

-2

-

1

6

0

51

x3 x4 M 1 -1 0 1 0 0 0 8 9 0 1 72 2

32

1

M 3 2

0

1

0

1 2

0

8

9

9 2

1 72

110) x1 x2 x3 x4 M 2 6 -4

A)

1 0 0

0 2 0 10 1 3 0 33 0 12 1 0 x1 x2 x3 x4 M 1

C)

110)

1 2

0

1

0

B) x1 x2 x3 x4 M 5

1

0 - 3 1 -3 0 3 0 - 2 0 16 1 30 x1 x2 x3 x4 M

1 0 0 -3 0 2

0 0 1 -3 0 16

D)

0 5 0 3 1 20

1 2

0

1

0

5

0 - 3 1 -3 0 3 0 2 0 16 1 20 x1 x2 x3 x4 M 1

1 2

0

1

0

5

0 - 3 -5 -3 -6 3 0 2 0 16 1 20

111) x1 x2 x3 x4 M 1 -8 0 4 0 -5

A)

111)

4 0 0 30 5 1 0 16 10 0 1 0 x 1 x 2 x3 x 4 M 1 0

0 8

0

0

14 10 65 4

2 2 5 4

B) x1 x2 x3 x4 M

0 62 0 32

1

0

41 4

5 4

0 50

1 20

0

1

5 4

1 4

0

0

0

65 4

5 4

1 20

C) x1 x2 x3 x4 M 1

0

0

1

0

0

14 2 5 1 4 4 65 4

2

4

D) x1 x2 x3 x4 M

0 62

1

0

0

4

0

1

1 32

0

0

33

14 5 4

2 1 4

65 4

5 4

0 62 0

4

1 20

x1 x2 x3 x4 x5 M

112)

-

112)

1 3

0

1

3

0

0 13

1 2

1

0 12

0

0

1 3

0

0

1

0 10

-9

A)

3

2

0 0 7 0 1 x1 x2 x3 x4

0 x5

0

0

1

0

1

0

0 33 2

-1 3 2

1

0

0

9

3

0

0

0

10

1

C) x1 x2 x3 x4 x5 0

0

1

0

1

0

1 0

0 0

0 0

0 15 2

-1 3 2

9 88

3 27

B) x1 x2 x3 x4 x5

M 0

3

0

0

1

0

17

0

1

0

30 13 1 3

1 0

0 0

0 0

0

M 0

0 33 2

-1 3 2

9 88

3 27

D) x1 x2 x3 x4 x5 3

0

0

1

0 - 13

0

1

0

0 1

1 0

0 0

0 0

30 270

0 33 2

-1 3 2

3 88

1 27

M 0

3

0

17

0 30 1 270 M 0

3

0

17

0 10 1 270

Determine whether the basic feasible solution corresponding to the given tableau is optimal.

113) x1 x2 x3 x4 M 10 7 -3

1 1 0 2 0 1 0 0 -5 A) Optimal

114)

x1

0 10 0 36 1 0

B) Not optimal

x2

x3 x4 M 3 1 1 0 0 8 0 0 -3 1 0 8 13 0 6 0 1 0 A) Not optimal

114)

B) Optimal

115) x1 x2 x3 x4 M 0

44

1

1

5

0

0

50

0

-2 0 1 0 2

113)

115) 4 4

11 1 44 2

A) Optimal

B) Not optimal

34

116) x1 x2 x3 x4 M

116)

2 2 -11

1 0 12 0 14 0 1 7 0 29 0 0 5 1 0 A) Not optimal

B) Optimal

117) x1 x2 x3 x4 x5 M

117)

1 3

0

1

2

0

0 16

1 2

1

0 10

0

0

1 3

0

0

2

1

0 15

0 0 10 A) Optimal

0

1

-

-5

0

1

0

1

0

0

B) Not optimal

118) x1 x2 x3 x4 x5 0

2

M

118)

-1 19 2

-1 3 2

0

2

0

19

0 0 9 0 0 96 A) Optimal

3 30

0 30 1 300

1 0

B) Not optimal

Solve by using the simplex method.

119) Maximize 13x1 + 12x2 subject to and x1

119)

x1

+ 2x2

12

5x1

+ 4x2

30

0, x2 0

A) Maximum = 72 when x1 = 0 and x2 = 6 B) Maximum = 86 when x1 = 2 and x2 = 5 C) Maximum = 110 when x1 = 2 and x2 = 7 D) Maximum = 78 when x1 = 6 and x2 = 0 120) Maximize 6x1 + 7x2 subject to and x1

120)

2x1 + 3x2

12

2x1 + x2

8

0, x2 0

A) Maximum = 32 when x1 = 3 and x2 = 2 C) Maximum = 39 when x1 = 3 and x2 = 3

B) Maximum = 28 when x1 = 0 and x2 = 4 D) Maximum = 38 when x1 = 4 and x2 = 2

35

121) Maximize 50x1 + 35x2 subject to

and x1

121)

3x1

+ x2

24

x1

+ x2

16

2x1 + 3x2

30

0, x2 0

A) Maximum = 350 when x1 = 0 and x2 = 10 B) Maximum = 400 when x1 = 8 and x2 = 0 C) Maximum = 620 when x1 = 4 and x2 = 12 D) Maximum = 510 when x1 = 6 and x2 = 6 122) Minimize

7x1 + 6x2

subject to

3x1 - 4x2

12

2x1 + x2

30

and x1

122)

0, x2 0

A) Minimum = 180 when x1 = 0 and x2 = 30 B) Minimum = 120 when x1 = 42 and x2 = 6 11

C) Minimum = - 120 when x1 = 12 and x2 = 6 D) Minimum = 120 when x1 = 12 and x2 = 6 123) Maximize 2x1 + 5x2 subject to and x1

123)

3x1 + 2x2

6

-2x1 + 4x2

8

0, x2 0

A) Maximum = 27 when x1 = 1 and x2 = 5 2

2

2

B) Maximum = 15 when x1 = 0 and x2 = 3 C) Maximum = 4 when x1 = 2 and x2 = 0 D) Maximum = 49 when x1 = 1 and x2 = 9 4

2

124) Minimize

4x1 + 5x2

subject to

2x1 - 4x2

10

2x1 + x2

15

and x1

4

124)

0, x2 0

A) Minimum = 75 when x1 = 0 and x2 = 15 B) Minimum = 30 when x1 = 15 and x2 = 0 2

C) Minimum = 33 when x1 = 7 and x2 = 1 D) Minimum = 20 when x1 = 5 and x2 = 0

36

125) Maximize 3x1 + 4x2 + 2x3 subject to

x1

x3

16

2x2 + x3

20

+

3x1 + 6x2 and x1

125)

36

0, x2 0, x3 0

A) Maximum = 52 when x1 = 4, x2 = 4, and x3 = 12 B) Maximum = 64 when x1 = 8, x2 = 6, and x3 = 8 C) Maximum = 48 when x1 = 4, x2 = 4, and x3 = 10 D) Maximum = 58 when x1 = 6, x2 = 5, and x3 = 10 126) Minimize 3x1 + 3x2 + 2x3 subject to -2x1

and x1

+

x3

126) -12

2x1 + x2

20

-x1 + x2 + 2x3

24

0, x2 0, x3 0

A) Minimum = 36 when x1 = 6, x2 = 6, and x3 = 0 B) Minimum = 42 when x1 = 6, x2 = 8, and x3 = 0 C) Minimum = 44 when x1 = 4, x2 = 8, and x3 = 4 D) Minimum = 40 when x1 = 6, x2 = 6, and x3 = 2 Use the simplex method to solve the linear programming problem.

127) Alan wants to invest a total of $24,000 in mutual funds and a certificate of deposit (CD). He wants to invest no more in mutual funds than half the amount he invests in the CD. His expected return on mutual funds is 10% and on the CD is 4%. How much money should Alan invest in each area in order to have the largest return on his investments? What is his maximum one-year return?

A) Maximum one-year return is $960 when he invests $0 in mutual funds and $24,000 in the CD. B) Maximum one-year return is $5,760 when he invests $16,000 in mutual funds and $8,000 in the CD. C) Maximum one-year return is $2,400 when he invests $24,000 in mutual funds and $0 in the CD. D) Maximum one-year return is $1,440 when he invests $8,000 in mutual funds and $16,000 in the CD.

37

127)

128) A store makes two different types of smoothies by blending different fruit juices. Each bottle of

128)

129) A furniture company makes two different types of lamp stand. Each lamp stand A requires 16

129)

Orange Daze smoothie contains 10 fluid ounces of orange juice, 4 fluid ounces of pineapple juice, and 2 fluid ounces of blueberry juice. Each bottle of Pineapple Blue smoothie contains 5 fluid ounces of orange juice, 6 fluid ounces of pineapple juice, and 4 fluid ounces of blueberry juice. The store has 500 fluid ounces of orange juice, 360 fluid ounces of pineapple juice, and 250 fluid ounces of blueberry juice available to put into its smoothies. The store makes a profit of $1.50 on each bottle of Orange Daze and $1 on each bottle of Pineapple Blue. How many bottles of each smoothie should the store make to maximize its profit? What is the maximum profit?

A) Maximum profit is $85 when the store makes 30 bottles of Orange Daze and 40 bottles of Pineapple Blue.

B) Maximum profit is $80 when the store makes 40 bottles of Orange Daze and 20 bottles of Pineapple Blue.

C) Maximum profit is $87.50 when the store makes 35 bottles of Orange Daze and 35 bottles of Pineapple Blue.

D) Maximum profit is $75 when the store makes 50 bottles of Orange Daze and 0 bottles of Pineapple Blue.

minutes for sanding, 48 minutes for assembly, and 6 minutes for packaging. Each lamp stand B requires 8 minutes for sanding, 32 minutes for assembly, and 8 minutes for packaging. The total number of minutes available each day in each department are as follows: for sanding 3360 minutes, for assembly 9600 minutes, and for packaging 2000 minutes. The profit on each lamp stand A is $30 and the profit on each lamp stand B is $22. How many of each type of lamp stand should the company make per day to maximize their profit? What is the maximum profit?

A) Maximum profit is $6164 when they make 138 of lamp stand A and 92 of lamp stand B

B) Maximum profit is $6380 when they make 66 of lamp stand A and 200 of lamp stand B C) Maximum profit is $7336 when they make 136 of lamp stand A and 148 of lamp stand B D) Maximum profit is $6000 when they make 200 of lamp stand A and 0 of lamp stand B 130) The Acme Class Ring Company designs and sells two types of rings: the VIP and the SST. They

130)

can produce up to 24 rings each day using up to 60 total man-hours of labor. It takes 3 man-hours to make one VIP ring and 2 man-hours to make one SST ring. How many of each type of ring should be made daily to maximize the company's profit, if the profit on a VIP ring is $40 and on an SST ring is $35?

A) 16 VIP and 8 SST B) 12 VIP and 12 SST

C) 14 VIP and 10 SST

D) 18 VIP and 6 SST

Determine whether the statement is true or false.

131) In the simplex method, a basic solution is infeasible if one of the slack variables is negative.

A) True B) False

132) In the simplex method, a basic variable is a variable which has a coefficient of 1 and which occurs in only one equation.

A) True

B) False

38

131)

132)

133) In the simplex method for a canonical linear programming problem, if all entries in the bottom

133)

134) In the simplex method for a canonical linear programming problem, if a solution is not optimal

134)

row to the left of the vertical line are nonnegative, then the solution is optimal.

A) True B) False

then the variable which should be brought into the solution is the variable xk for which the entry in the bottom row is as negative as possible.

A) True

B) False

135) Consider the simplex method for a canonical linear programming problem in which each entry in

135)

the vector b is positive. If the variable x3 is to be brought into solution then row i will be chosen as the pivot row if the ratio

bi is the largest among all ratios for which a i3 > 0. a i3

A) True

B) False

136) Consider the simplex method for a canonical linear programming problem. If x is an extreme

136)

137) Suppose a system contains three inequalities and three unknowns x1 , x2 , and x3 . If the simplex

137)

point of the feasible set and if the objective function cannot be decreased by moving along any of the edges of the feasible set which join at x, then x is an optimal solution.

A) True B) False

method results in an optimal solution in which the slack variable x6 is zero, this means that at the optimal values of x1 , x2 , and x3 , the third inequality is an equality.

A) True

B) False

138) Suppose a system contains three inequalities and three unknowns x1 , x2 , and x3 . If the simplex

138)

method results in an optimal solution in which the slack variable x4 is greater than zero, this means that at the optimal values of x1 , x2 , and x3 , the first inequality is an equality.

A) True

B) False

139) Suppose that you are using the simplex method to solve a canonical linear programming problem in which each entry in the vector b is positive. If you obtain a tableau which contains just one negative entry in the bottom row and no positive entry a ik above it, then the objective function is unbounded and no optimal solution exists.

A) True

B) False

39

139)

140) You wish to solve the following linear programming problem: Minimize subject to

2x1 + x2 x1 - x2

4

x1 + 2x2

10

140)

and x1 0, x2 0 This can be solved by the simplex method by solving the equivalent problem: Maximize -2x1 - x2 subject to

x1 - x2 x1 + 2x2

4 -10

and x1 0, x2 0

A) True

B) False

141) You wish to solve the following linear programming problem: Minimize

2x1 + x2

subject to

x1 - x2

4

x1 + 2x2

10

141)

and x1 0, x2 0 This can be solved by the simplex method by solving the equivalent problem: Maximize 2x1 + x2 subject to

x1 - x2

4

-x1 - 2x2

-10

and x1 0, x2 0

A) True

B) False

142) You wish to solve the following linear programming problem: Minimize subject to

2x1 + x2 x1 - x2

4

x1 + 2x2

10

142)

and x1 0, x2 0 This can be solved by the simplex method by solving the equivalent problem: Maximize -2x1 - x2 subject to

x1 - x2 -x1 - 2x2

4 -10

and x1 0, x2 0

A) True

B) False

143) In an initial simplex tableau, if the augmented column above the horizontal line contains a negative number, then the objective function is unbounded and no optimal solution exists.

A) True B) False

40

143)

State the dual of the linear programming problem.

144) Maximize 6x1 + 4x2 subject to and x1

x1 + 4x2

13

8x1 + 6x2

33

144)

0, x2 0

A) Minimize subject to

6y1 + 4y2 y1 + 8y2

B) Minimize 13

4y1 + 6y2 0, y2 0

and y1

C) Minimize subject to

33 and y1

13y1 + 33y2 y1 + 4y2

8y1 + 6y2 0, y2 0

and y1

subject to

4y1 + 6y2 0, y2 0

D) Minimize 6

subject to

4 and y1

13y1 + 33y2 y1 + 8y2 6 4

13y1 + 33y2 y1 + 8y2 6

4y1 + 6y2 0, y2 0

4

145) Maximize 3x1 + 5x2 subject to

and x1

145)

4x1 + 5x2

45

2x1 + x2

16

x1 + 4x2

35

0, x2 0

A) Minimize 45y1+ 16y2 + 35y3 subject to and y1

4y1 + 2y2 + y3 5y1 + y2 + 4y3

B) Minimize 3

subject to

2y1 + y2 y1 + 4y2

5

0, y2 0, y3 0 and y1

and y1

45 16 35

0, y2 0

D) Minimize 45y1+ 16y2 + 35y3

C) Minimize 45y1+ 16y2 subject to

3y1 + 5y2 4y1 + 5y2

4y1 + 5y2 2y1 + y2

3

y1 + 4y2 0, y2 0

35

subject to

5

and y1

41

4y1 + 2y2 + y3 5y1 + y2 + 4y3 0, y2 0, y3 0

3 5

146) Maximize subject to

7x1 + 3x2 + 6x3 5x1 + 2x2 4x1

and x1

146) 48

+

x3

18

2x2

+ x3

28

0, x2 0, x3 0

A) Minimize 48y1 + 18y2 + 28y3 subject to

5y1 4y1

+ 2y2 + y2

and y1

B) Minimize 7

2y3 + y3

subject to

3 6

0, y2 0, y3 0

and y1

C) Minimize 48y1 + 18y2 + 28y3 subject to

5y1 2y1

+ 4y2 + y2

and y1

147) Maximize subject to

subject to

3 6

0, y2 0, y3 0

18 28

0, y2 0, y3 0

and y1

5y1 + 4y2 2y1 + y2 0, y2 0, y3 0

7 2y3 + y3

3 6

7x1 + 6x2 + 7x3 2x1 + 3x2 4x1

and x1

48

D) Minimize 48y1 + 18y2 + 28y3 7

2y3 + y3

7y1 + 3y2 + 6y3 5y1 + 4y2 2y1 + 2y3 y2 + y3

147) 38

x3

15

3x1 + x2 + 4x3

49

+

0, x2 0, x3 0

A) Minimize 38y1 + 15y2 + 49y3 subject to

and y1

2y1 + 4y2 + 3y3 3y1 + y3

7

y2 + 4y3 0, y2 0, y3 0

7

C) Minimize subject to

and y1

B) Minimize 38y1 + 15y2 + 49y3 6 and y1

7y1 + 6y2 + 7y3 2y1 + 4y2 + 3y3 3y1 y3 +

38

y2

49

+

4y3

subject to

2y1 4y1

+ 3y2

7

y3 + y2 + 4y3 3y1 0, y2 0, y3 0

6

+

7

D) Minimize 38y1 + 15y2 + 49y3 subject to

15

0, y2 0, y3 0

and y1

42

2y1 3y1

+ 4y2 + 3y3

7

y3 y2 + 4y3 0, y2 0, y3 0

6

+

7

148) Maximize

7x1 + 6x2 + 8x3

subject to and x1

2x1

148)

+

x3

20

4x1 + 5x2

+ x3

43

0, x2 0, x3 0

A) Minimize 7y1 + 6y2 subject to

2y1 y1 0, y2 0

and y1

B) Minimize 20y1 + 43y2

+ 4y2

20

5y2 + y2

43

subject to

8 and y1

2y1 y1 0, y2 0

and y1

y1 0, y2 0

+ 4y2

7

5y2 + y2

6 8

D) Minimize 7y1 + 6y2 + 8y3

C) Minimize 20y1 + 43y2 subject to

2y1

+ 4y2

7

5y2 + y2

6

subject to and y1

8

2y1 4y1

+ + 5y2

y3 + y3

20 43

0, y2 0, y3 0

The primal problem and the final tableau in the solution of the primal problem are given. Use the final tableau to solve the dual problem.

149) Primal problem : 149) Maximize 6x1 + 9x2 subject to and x1

x1 + 2x2

14

5x1 + 4x2

45

0, x2 0

Final tableau in solution to primal problem: x 1 x2 x 3 x 4 M 0

1

5 6

1 6

0

25 6

1

0 -

2 3

1 3

0

17 3

0

0

7 2

1 2

1

143 2

-

A) Minimum = 143 when y1 = 25 and y2 = 17 2

6

6

B) Minimum = 0 when y1 = 0 and y2 = 0 C) Maximum = 143 when y1 = 7 and y2 = 1 2

2

2

D) Minimum = 143 when y1 = 7 and y2 = 1 2

2

2

43

150) Primal problem:

150)

Maximize

6x1 + 7x2

subject to

2x1 + 3x2 12 2x1 + x2 8

and x1 0, x2 0 Final tableau in solution to primal problem: x 1 x 2 x 3 x4 M 0

1

1 1 2 2

0 2

1

0 -

1 4

0 3

3 4

0

0 2 1 1 32 A) Minimum = 32 when y1 = 2 and y2 = 1

B) Minimum = 32 when y1 = 2 and y2 = 3 D) Maximum = 32 when y1 = 2 and y2 = 1

C) Minimum = 0 when y1 = 0 and y2 = 0 151) Primal problem:

151)

Maximize

50x1 + 35x2

subject to

3x1 + x2 24 x1 + x2 16 2x1 + 3x2 30

and x1 0, x2 0 Final tableau in solution to primal problem: x 1 x 2 x3 x4 x 5 M 1

0

3 7

0

-

1 7

0

6

0

0 -

1 7

1

-

2 7

0

4

0

1 -

2 7

0

3 7

0

6

0

0

80 7

0

55 7

1 510

A) Minimum = 510 when y1 = 80 , y2 = 0, and y3 = 55 7

7

B) Minimum = 510 when y1 = 6, y2 = 4, and y3 = 6 C) Minimum = 0 when y1 = 0, y2 = 0 D) Minimum = 510 when y1 = 6, y2 = 6

44

152) Primal problem:

152)

Maximize

3x1 + 4x2 + 2x3

subject to

x1

x3 16

+

2x2 + x3 20 3x1 + 6x2

36

and x1 0, x2 0, x3 0 Final tableau in solution to primal problem: x 1 x 2 x3 x 4 x 5 x 6 M 1

0

0

1 1 2 2

1 6

0

0

1

1 2

1 1 2 6

0

1

0 -

1 4

1 4

1 0 12

0

0

0

3 2

1 2

1 2

0

4

0 12 4

1 52

A) Minimum = 15 when y1 = 3 , y2 = 1 , and y3 = 1 2

2

2

2

B) Minimum = 52 when y1 = 3 , y2 = 1 , and y3 = 1 2

2

2

C) Minimum = 52 when y1 = 4, y2 = 4, and y3 = 12 D) Minimum = 576 when y1 = 4, y2 = 4, and y3 = 12 153) Primal problem:

153)

Maximize

x1 + 4x2 + 3x3

subject to

x1 + 2x2 + x3 24 3x2 + 4x3 30

and x1 0, x2 0, x3 0 Final tableau in solution to primal problem: x 1 x 2 x3 x4 x 5 M 5 3

1 -

2 3

0

1

4 3

0

1 3

0 10

0

2 3

1

2 3

1 44

1

0

0 0

-

4

A) Minimum = 44 when y1 = 4 and y2 = 10 B) Minimum = 44 when y1 = 2 , y2 = 1, and y3 = 2 3

3

C) Minimum = 44 when y1 = 1 and y2 = 2 3

D) Maximum = 44 when y1 = 0, y2 = 0, and y3 = 2 3

45

Solve the minimization problem by using the simplex method to solve the dual problem.

154) Minimize 12x1 + 35x2 subject to

154)

x1 + 5x2 8 2x1 + 4x2 9

and x1 0, x2 0

A) Minimum = 96 when x1 = 8 and x2 = 0 B) Minimum = 56 when x1 = 0 and x2 = 8 5

C) Minimum = 401 when x1 = 13 and x2 = 7 6

6

6

D) Minimum = 401 when x1 = 25 and x2 = 11 6

155) Minimize subject to

6

6

12x1 + 8x2

155)

x1 + x2 3 3x1 + x2 7

and x1 0, x2 0

A) Minimum = 56 when x1 = 0 and x2 = 7 C) Minimum = 32 when x1 = 2 and x2 = 1 156) Minimize subject to

B) Minimum = 36 when x1 = 3 and x2 = 0 D) Minimum = 24 when x1 = 0 and x2 = 3

24x1 + 16x2 + 30x3

156)

3x1 + x2 + 2x3 50 x1 + x2 + 3x3 35

and x1 0, x2 0, x3 0

A) Minimum = 3810 when x1 = 10, x2 = 30 , and x3 = 55 7

7

7

B) Minimum = 696 when x1 = 17, x2 = 18, and x3 = 0 C) Minimum = 498 when x1 = 12, x2 = 0, and x3 = 7 D) Minimum = 510 when x1 = 80 , x2 = 0, and x3 = 55 7

7

46

157) Minimize subject to

16x1 + 20x2 + 36x3 x1

157)

3x3

3

2x2 + 6x3

4

+

x1 + x 2

2

and x1 0, x2 0, x3 0

A) Minimum = 172 when x1 = 2, x2 = 2 , and x3 = 1 3

3

3

B) Minimum = 60 when x1 = 1, x2 = 1, and x3 = 2 3

C) Minimum = 52 when x1 = 3 , x2 = 1 , and x3 = 1 2

2

2

D) Minimum = 152 when x1 = 2, x2 = 1 , and x3 = 1 3

158) Minimize subject to

3

3

24x1 + 30x2

158)

x1

1

2x1 + 3x2

4

x1 + 4x2

3

and x1 0, x2 0

A) Minimum = 44 when x1 = 1 and x2 = 2 3

B) Minimum = 111 when x1 = 2 and x2 = 1 2

4

C) Minimum = 39 when x1 = 1 and x2 = 1 2

D) Minimum = 46 when x1 = 3 and x2 = 1 2

159) Minimize subject to

3

6x1 + 8x2

159)

x1 + 2x2 6 2x1 + x2 8

and x1 0, x2 0

A) Minimum = 86 when x1 = 3 and x2 = 4 3

3

B) Minimum = 36 when x1 = 6 and x2 = 0 C) Minimum = 92 when x1 = 10 and x2 = 4 3

3

3

D) Minimum = 64 when x1 = 0 and x2 = 8

47

160) Minimize subject to

4x1 + 5x2

160)

5x1 + 4x2 20 4x1 + x2 12 x1 + 3x2 9

and x1 0, x2 0

A) Minimum = 221 when x1 = 24 and x2 = 25 11

11

11

B) Minimum = 228 when x1 = 27 and x2 = 24 11

11

11

C) Minimum = 41 when x1 = 2 and x2 = 5 2

2

D) Minimum = 212 when x1 = 28 and x2 = 20 11

11

11

161) Daniel decides to feed his cat a combination of two foods: Max Cat and Mighty Cat. He wants his

161)

cat to receive four nutritional factors each month. The amounts of these factors (a, b, c, and d) contained in one bag of each food are shown in the chart, together with the total amounts needed. a b c d Max Cat 4 2 1 4 Mighty Cat 6 2 2 3 Needed 54 36 20 60 The costs per bag are $40 for Max Cat and $36 for Mighty Cat. How many bags of each food should be blended to meet the nutritional requirements at the lowest cost? What is the minimum cost?

A) Minimum cost = $624 when he blends 12 bags of Max Cat and 4 bags of Mighty Cat.

B) Minimum cost = $720 when he blends 0 bags of Max Cat and 20 bags of Mighty Cat. C) Minimum cost = $672 when he blends 6 bags of Max Cat and 12 bags of Mighty Cat. D) Minimum cost = $712 when he blends 16 bags of Max Cat and 2 bags of Mighty Cat. 162) An airline with two types of airplanes, P1 and P2 , has contracted with a tour group to provide transportation for a minimum of 400 first class, 800 tourist class, and 1500 economy class passengers. For a certain trip, airplane P1 costs $10,000 to operate and can accommodate 20 first class, 50 tourist class, and 100 economy class passengers. Airplane P2 costs $8000 to operate and can accommodate 20 first class, 30 tourist class, and 50 economy class passengers. How many of each type of airplane should be used in order to minimize the operating cost? What is the minimum operating cost?

A) Minimum cost is $178,000 when they use 9 of airplane P1 and 11 of airplane P2 .

B) Minimum cost is $182,000 when they use 11 of airplane P1 and 9 of airplane P2 .

C) Minimum cost is $180,000 when they use 10 of airplane P1 and 10 of airplane P2 .

D) Minimum cost is $186,000 when they use 9 of airplane P1 and 12 of airplane P2 .

48

162)

163) A store makes two different types of smoothies by blending different fruit juices. Each bottle of

163)

Orange Daze smoothie contains 10 fluid ounces of orange juice, 4 fluid ounces of pineapple juice, and 2 fluid ounces of blueberry juice. Each bottle of Pineapple Blue smoothie contains 5 fluid ounces of orange juice, 6 fluid ounces of pineapple juice, and 4 fluid ounces of blueberry juice. The store has 500 fluid ounces of orange juice, 360 fluid ounces of pineapple juice, and 250 fluid ounces of blueberry juice available to put into its smoothies. The store makes a profit of $1.50 on each bottle of Orange Daze and $1 on each bottle of Pineapple Blue. To determine the maximum profit, the simplex method can be used and the final tableau is: x1 x2 x3 x4 x5 M 3 20

1

0

0

1

0

0

1 10

0

0

1 8

-

-

1 10 -

1 8

0

0 30

1 4

0

0 40

3 4

1

0 30

1 16

0

1 85

What is the marginal value of pineapple juice?

A) 1 B) 1 16 8

C) 30

D) 0

164) A store makes two different types of smoothies by blending different fruit juices. Each bottle of

Orange Daze smoothie contains 10 fluid ounces of orange juice, 4 fluid ounces of pineapple juice, and 2 fluid ounces of blueberry juice. Each bottle of Pineapple Blue smoothie contains 5 fluid ounces of orange juice, 6 fluid ounces of pineapple juice, and 4 fluid ounces of blueberry juice. The store has 500 fluid ounces of orange juice, 360 fluid ounces of pineapple juice, and 250 fluid ounces of blueberry juice available to put into its smoothies. The store makes a profit of $1.50 on each bottle of Orange Daze and $1 on each bottle of Pineapple Blue. To determine the maximum profit, the simplex method can be used and the final tableau is: x1 x2 x3 x4 x5 M 3 20

1

0

0

1

0

0

1 10

0

0

1 8

-

-

1 10 -

1 8

0

0 30

1 4

0

0 40

3 4

1

0 30

1 16

0

1 85

What is the marginal value of blueberry juice?

A) 30 B) 1 16

C) 1 8

49

D) 0

164)

165) A store makes two different types of smoothies by blending different fruit juices. Each bottle of

165)

Orange Daze smoothie contains 10 fluid ounces of orange juice, 4 fluid ounces of pineapple juice, and 2 fluid ounces of blueberry juice. Each bottle of Pineapple Blue smoothie contains 5 fluid ounces of orange juice, 6 fluid ounces of pineapple juice, and 4 fluid ounces of blueberry juice. The store has 500 fluid ounces of orange juice, 360 fluid ounces of pineapple juice, and 250 fluid ounces of blueberry juice available to put into its smoothies. The store makes a profit of $1.50 on each bottle of Orange Daze and $1 on each bottle of Pineapple Blue.

What is the marginal value of orange juice?

[Hint: use the simplex method to determine the maximum profit and use the final tableau to determine the marginal value.]

A) 30 B) 0 C) 1 D) 1 16 8

Use linear programming to solve the matrix game with the given payoff matrix.

-3 4 166) A = -2 2 2 -1 3 3 1 10 10 2 ^ ^ ^ ^ 1 0 A) x = B) x = 0 , y = , y = 1 , value = 2 7 7 2 10 10 1 15

2 5

1 9

^ ^ C) x = 0 , y = 1 , value = 2 9

7 45

2 1 -1

167) A = -1

1 4

1 0

167) 3 4

^

2 11

^

4

^

3 11

B) x = 3 , y = 2 , value = 5 11 ^

11

11

3 5

C) x = 3 , y = 2 , value = 1 5 5

2

2

0

^

2 1 , value = 9 2 1 2

3 5

9

4 2 5

1 2

^ ^ D) x = 0 , y = 1 , value = 1

A) x = 3 , y = 1 , value = 1 4 ^

166)

^

3 5

0 3 11 ^

D) x = 2 , y =

5

5

0

50

0 , value = 11 5 2 11

3

1 -2 0 1 3 4 9

5 9

5 9

4 9

5 9

4 9

168) A = 1 2 -1

168)

^ ^ A) x = 0 , y = 0 , value = 7

5 9

6 11

6 11

5 11

4 34

5 34

5 34

4 34

^ ^ B) x = 0 , y = 0 , value = 9

9

^ ^ C) x = 0 , y = 0 , value = 34

4 9

5 11

34

^ ^ D) x = 0 , y = 0 , value = 9

9

34

Solve the problem.

169) Irene wants to invest $36,000 in stocks, bonds, and gold coins. She knows that her rate of return will depend on the economic climate of the country, which is difficult to predict. After some research and analysis, she determines the annual profit in dollars that she would expect per hundred dollars on each type of investment, depending on whether the economy is strong, stable, or weak.

Strong Stocks 3 Bonds 1 Gold -1

169)

Stable Weak -2 1 2 0 1 3

How should Irene invest her money in order to maximize her profit regardless of what the economy does? In other words, consider the problem as a matrix game in which Irene is playing against "fate".

A) Irene should invest $12,000 in stocks, $8000 in bonds, and $16,000 in gold.

B) Irene should invest $14,000 in stocks, $8000 in bonds, and $18,000 in gold.

C) Irene should invest $15,000 in stocks and $21,000 in gold.

D) Irene should invest $16,000 in stocks and $20,000 in gold.

Provide an appropriate response.

170) Given the primal problem P below:

170)

Maximize f(x) = cxT subject to Ax b x 0

where c is a vector in n , b is a vector in n , and A is an m × n matrix, how many constraints and how many variables are in the dual problem?

A) m constraints, m variables B) m constraints, n variables

C) n constraints, m variables

D) n constraints, n variables

51

171) Given the primal and dual problems below:

171)

Primal:

Maximize f(x) = cxT subject to Ax b x 0

where c is a vector in n and A is an m × n matrix Dual:

Minimize g(y) = bTy subject to ATy c y 0

where b is a vector in m If x is an optimal solution to the primal problem and y is an optimal solution to the dual problem, then which of the following statements is (are) true?

A: f(x) = g(y) B: f(x) = f(y) C: g(x) = g(y) D: g(x) = f(y) A) B and C

B) A only

C) A and D

D) All of them.

172) Given the primal linear programming problem below: Maximize f(x) = cxT subject to Ax b x 0 If a slack variable is not in an optimal solution, then what can be said about the marginal value of the item corresponding to its equation?

A) The marginal value is less than zero.

B) The marginal value is greater than zero.

C) The marginal value is zero.

D) There is not enough information to say anything about the marginal value.

52

172)

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

173) A store makes two different types of smoothies by blending different fruit juices. Each

bottle of Orange Daze smoothie contains 10 fluid ounces of orange juice, 4 fluid ounces of pineapple juice, and 2 fluid ounces of blueberry juice. Each bottle of Pineapple Blue smoothie contains 5 fluid ounces of orange juice, 6 fluid ounces of pineapple juice, and 4 fluid ounces of blueberry juice. The store has 500 fluid ounces of orange juice, 360 fluid ounces of pineapple juice, and 250 fluid ounces of blueberry juice available to put into its smoothies. The store makes a profit of $1.50 on each bottle of Orange Daze and $1 on each bottle of Pineapple Blue. To determine the maximum profit, the simplex method can be used and the final tableau is: x1 x2 x3 x4 x5 M 3 20

1

0

0

1

0

0

1 10

0

0

1 8

-

-

1 10 -

1 8

0

0 30

1 4

0

0 40

3 4

1

0 30

1 16

0

1 85

Give an interpretation to the fact that the marginal value of blueberry juice is zero. Refer in your explanation to the value in the final tableau of the slack variable x5 .

53

173)

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

174) A store makes two different types of smoothies by blending different fruit juices. Each bottle of

174)

Orange Daze smoothie contains 10 fluid ounces of orange juice, 4 fluid ounces of pineapple juice, and 2 fluid ounces of blueberry juice. Each bottle of Pineapple Blue smoothie contains 5 fluid ounces of orange juice, 6 fluid ounces of pineapple juice, and 4 fluid ounces of blueberry juice. The store has 500 fluid ounces of orange juice, 360 fluid ounces of pineapple juice, and 250 fluid ounces of blueberry juice available to put into its smoothies. The store makes a profit of $1.50 on each bottle of Orange Daze and $1 on each bottle of Pineapple Blue. To determine the maximum profit, the simplex method can be used and the final tableau is: x1 x2 x3 x4 x5 M 3 20

1

0

0

1

0

0

1 10

0

0

1 8

-

-

1 10 -

1 8

0

0 30

1 4

0

0 40

3 4

1

0 30

1 16

0

1 85

If the store could obtain 10 fluid ounces extra of one of the juices, which type of juice should they get? Why? A) Orange juice, since it has the highest marginal value of 1 .

8

B) It makes no difference which juice they get - the increase in profit would be the same for any of them.

C) Pineapple juice, since it has the highest marginal value of 40.

D) Blueberry juice; in the optimal solution the value of the slack variable x5 corresponding to the blueberry juice constraint is 30, while the values of the other slack variables are zero.

175) A store makes two different types of smoothies by blending different fruit juices. Each bottle of

Orange Daze smoothie contains 10 fluid ounces of orange juice, 4 fluid ounces of pineapple juice, and 2 fluid ounces of blueberry juice. Each bottle of Pineapple Blue smoothie contains 5 fluid ounces of orange juice, 6 fluid ounces of pineapple juice, and 4 fluid ounces of blueberry juice. The store has 500 fluid ounces of orange juice, 360 fluid ounces of pineapple juice, and 250 fluid ounces of blueberry juice available to put into its smoothies. The store makes a profit of $1.50 on each bottle of Orange Daze and $1 on each bottle of Pineapple Blue. To determine the maximum profit, the simplex method can be used and the final tableau is: x1 x2 x3 x4 x5 M 3 20

1

0

0

1

0

0

1 10

0

0

1 8

-

-

1 10 -

1 8

0

0 30

1 4

0

0 40

3 4

1

0 30

1 16

0

1 85

54

175)

Give an interpretation to the number

1 in the bottom row.

8

A) 1 is the marginal value of the orange juice. If the amount of orange juice available were 8

increased by one fluid ounce, the profit would increase by

1 dollars.

8

B) 1 is the marginal value of the pineapple juice. If the amount of pineapple juice available 8

were increased by one fluid ounce, the profit would increase by

1 dollars.

8

C) 1 represents the number of Pineapple Blue smoothies they should make to maximize profit.

8

(In practice this would be rounded to 0).

D) 1 is the value of the slack variable x3 in the optimal solution. The slack variable x3 8 corresponds to the orange juice constraint. Since x3 is greater than zero, this means that in the optimal solution, not all the available orange juice is used. Thus the marginal value of the orange juice is zero.

55

176) You have decided to solve a matrix game by using linear programming. You add 3 to each entry of the payoff matrix A to obtain a matrix B = a b c with only positive entries. You find the d e f optimal strategy for column player C by solving the linear programming problem: Maximize y1 + y2 + y3 subject to ay1 + by2 + cy3 1 dy1 + ey2 + fy3 1 and y1 0, y2 0, y3 0 The initial simplex tableau is y1 y2 y3 y4 y5 M a b c d e f -1 -1 -1

1 0 0

0 1 0

0 0 1

1 1 0

and suppose that the final tableau is y1 y2 y3 g 0 1 j 1 0 m 0 0

y4 y5 M h i 0 q k l 0 r n p 1 v

where g, h, i, j, k, and l are nonzero numbers and m, n, p, q, r, and v are positive numbers.

What is the optimal strategy for player C? 0 n r n+p m 0 ^ ^ ^ ^ A) y = r+q B) y = p C) y = 0 D) y = r q 0 q n+p r+q

56

176)

177) You have decided to solve a matrix game by using linear programming. You add 3 to each entry of the payoff matrix A to obtain a matrix B = a b c with only positive entries. You find the d e f optimal strategy for column player C by solving the linear programming problem: Maximize y1 + y2 + y3 subject to ay1 + by2 + cy3 1 dy1 + ey2 + fy3 1 and y1 0, y2 0, y3 0 The initial simplex tableau is y1 y2 y3 y4 y5 M a b c d e f -1 -1 -1

1 0 0

0 1 0

0 0 1

1 1 0

and suppose that the final tableau is y1 y2 y3 g 0 1 j 1 0 m 0 0

y4 y5 M h i 0 q k l 0 r n p 1 v

where g, h, i, j, k, and l are nonzero numbers and m, n, p, q, r, and v are positive numbers.

What is the optimal strategy for player R? 0 n r n+p 0 ^ ^ ^ ^ A) x = p B) x = r+q C) x = n D) x = n p q p n+p r+q

57

177)

178) You have decided to solve a matrix game by using linear programming. You add 3 to each entry of the payoff matrix A to obtain a matrix B = a b c with only positive entries. You find the d e f optimal strategy for column player C by solving the linear programming problem: Maximize y1 + y2 + y3 subject to ay1 + by2 + cy3 1 dy1 + ey2 + fy3 1 and y1 0, y2 0, y3 0 The initial simplex tableau is y1 y2 y3 y4 y5 M a b c d e f -1 -1 -1

1 0 0

0 1 0

0 0 1

1 1 0

and suppose that the final tableau is y1 y2 y3 g 0 1 j 1 0 m 0 0

y4 y5 M h i 0 q k l 0 r n p 1 v

where g, h, i, j, k, and l are nonzero numbers and m, n, p, q, r, and v are positive numbers.

What is the value of the original game with payoff matrix A? A) v - 3 B) 1 - 3 C) 1 D) v v v

58

178)

Answer Key Testname: UNTITLED9

1) D 2) A 3) D 4) B 5) D 6) D 7) A 8) D 9) A 10) D 11) C 12) B 13) C 14) C 15) A 16) C 17) D 18) A 19) D 20) D 21) D 22) D 23) A 24) D 25) A 26) A 27) D 28) A 29) B 30) B 31) B 32) B 33) D 34) B 35) D 36) A 37) A 38) B 39) A 40) A 41) A 42) A

59

Answer Key Testname: UNTITLED9

43) C 44) B 45) B 46) B 47) D 48) A 49) B 50) D 51) B 52) A 53) A 54) B 55) B 56) B 57) A 58) A 59) C 60) C 61) B 62) D 63) D 64) D 65) D 66) A 67) A 68) D 69) D 70) A 71) B 72) C 73) D 74) D 75) D 76) B 77) D 78) C 79) A 80) A 81) B 82) A 83) A 84) A

60

Answer Key Testname: UNTITLED9

85) B 86) A 87) C 88) C 89) C 90) Take any p and q in S. Then ap b and aq b. Take any scalar t such that 0 t 1. Then

a[(1 - t)p + tq] = (1 - t)ap + taq (1 - t)b + tb = b because (1 - t) and t are both nonnegative and p and q are in S. So the line segment between p and q is in S. Since p and q were any points in S, the set S is convex.

91) Take any p and q in F. Then Ap b, Aq b, p 0, q 0. Take any scalar t such that 0 t 1 and let x = (1 - t)p + tq.

Then Ax = A[(1 - t)p + tq] = (1 - t)Ap + tAq (1 - t)b + tb = b because (1 - t) and t are both nonnegative and p and q are in F. Also x 0 because p and q are 0 and (1 - t) and t are nonnegative. Thus x is in F. So the line segment between p and q is in F. This proves that F is convex.

92) D 93) A 94) C 95) A 96) B 97) C 98) A 99) D 100) A 101) C 102) A 103) C 104) C 105) B 106) D 107) A 108) C 109) B 110) B 111) D 112) B 113) B 114) B 115) A 116) A 117) B 118) A 119) B 120) A 121) D 122) D

61

Answer Key Testname: UNTITLED9

123) D 124) C 125) A 126) B 127) D 128) A 129) B 130) B 131) A 132) A 133) A 134) A 135) B 136) B 137) A 138) B 139) A 140) B 141) B 142) A 143) B 144) D 145) D 146) C 147) D 148) C 149) D 150) A 151) A 152) B 153) C 154) C 155) C 156) D 157) C 158) A 159) C 160) B 161) C 162) C 163) A 164) D

62

Answer Key Testname: UNTITLED9

165) D 166) A 167) C 168) A 169) D 170) C 171) B 172) B 173) The slack variable x5 which corresponds to the blueberry juice constraint has value 30 in the optimal solution. This means that not all the available blueberry juice is used, so there is no increase in profit if the amount of blueberry juice is increased. Hence the marginal value of blueberry juice is zero.

174) A 175) A 176) A 177) A 178) B

63