Add. dressing Chapter 3 Answers 1

Chapter 3 Answers

Lesson 3.11. A 2 × 5 matrix: any matrix with 2 rows and 5 columns.

A row matrix with two elements: any matrix with one row of two elements.A square matrix: any matrix in which the number of rows equals the number ofcolumns.

2. 16 elements; 7 elements; mn elements3.

Jacket Shirt PantsShop 1 Shop 2 Shop 3

25 75 7530 50 5020 40 35

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4. a. A2 1 = $1.09, A1 2

= $10.86, A3 2 = $3.89

b. A2 1

represents the cost of drinks at Vin’s.A

1 2 represents the cost of pizza at Toni’s.

A3 2

represents the cost of salad at Toni’s.5. S

3 represents the cost of pizza at Sal’s.

6. a.Cals g Fat g Carbo mg Chol

N =

CerealBananaMilkToast

165 3 33 0 120 0 26 0 120 5 11 15 125 6 14 18

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

b. N2 3 = 26, N3 2

= 5, N4 2

= 6c. N

2 3 represents the grams of carbohydrates in the banana.

N3 2

represents the grams of fat in the milk.N

4 2 represents the grams of fat in the toast.

7.Vin's Toni's Sal's

PizzaSalad

$10.10 $10.86 $10.65$3.69 $3.89 $3.85

⎡

⎣⎢

⎤

⎦⎥

8.Vin's Toni's Sal's

Add. topping $1.15 $1.10 $1.25$0.00 $0.45 $0.50⎡

⎣⎢

⎤

⎦⎥

Chapter 3 Answers 2

9.

A + B =10.10 + 1.15 10.86 + 1.10 10.65 + 1.25

3.69 + 0.00 3.89 + 0.45 3.85 + 0.50⎡

⎣⎢

⎤

⎦⎥

Vin's Toni's Sal's

C = PizzaSalad

$11.25 $11.96 $11.90$3.69 $4.34 $4.35

⎡

⎣⎢

⎤

⎦⎥

10. a. A2 1

= $3.69, B2 1 = $0.00, C

2 1 = $3.69

b. A2 1

represents the cost of a salad at Vin’s.B

2 1 represents the cost of an additional salad dressing at Vin’s.

C2 1

represents the cost of a salad with a choice of two dressings at Vin’s.

11. a.9 −4

−4 7⎡

⎣⎢

⎤

⎦⎥

b. It is not possible to add these matrices because their orders are not the same.c.

0 −56 03 40 −2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

d. It is not possible to add these matrices because their orders are not the same.12.

At Bats Runs Hits HRs RBIs AvgJ. Reyes R. BraunM. Kemp

105 35–199

–15

–1–41

3 4

–73

4 8–16

13 1–57

–0.050–0.013–0.021

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

A decrease in statistics is shown with a negative sign.13. a. Matrix representing increases in times:

Men Women100-meter race200-meter race400-meter race

–0.2–0.48–0.98

–0.48–1.31–2.72

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

b. The greatest decrease in times is in the 400-meter race for women.The smallest decrease is in the 100-meter race for men.

Chapter 3 Answers 3

14. a. Answers vary. Possible answer: Yes. In matrix addition we add thecomponents of the matrices that are in corresponding positions. These valuesare real numbers and since addition of real numbers is both commutative andassociative, it follows that addition of matrices is also commutative andassociative.

b. Commutative property:

A + B = B + A =5 11 6⎡

⎣⎢

⎤

⎦⎥

Associative property:

A + (B + C) =4 –23 1⎡

⎣⎢

⎤

⎦⎥ +

3 7–1 4⎡

⎣⎢

⎤

⎦⎥ =

7 52 5⎡

⎣⎢

⎤

⎦⎥

(A + B) + C =5 11 6⎡

⎣⎢

⎤

⎦⎥ +

2 41 –1⎡

⎣⎢

⎤

⎦⎥ =

7 52 5⎡

⎣⎢

⎤

⎦⎥

15. Answers vary. Possible answer: No, subtraction of matrices is neithercommutative nor associative. This follows since the commutative and associativeproperties do not hold for subtraction of real numbers.To verify that the commutative property does not hold for the given matrices Aand B:

A − B =3 –55 –4⎡

⎣⎢

⎤

⎦⎥ B − A =

–3 5–5 4⎡

⎣⎢

⎤

⎦⎥

To verify that the associative property does not hold for the given matrices Aand B:

A − (B − C) =4 –23 1⎡

⎣⎢

⎤

⎦⎥ −

–1 –1–3 6⎡

⎣⎢

⎤

⎦⎥ =

5 –16 –5⎡

⎣⎢

⎤

⎦⎥

(A − B) − C =3 –55 –4⎡

⎣⎢

⎤

⎦⎥ −

2 41 –1⎡

⎣⎢

⎤

⎦⎥ =

1 –94 –3⎡

⎣⎢

⎤

⎦⎥

16. a.

A +O =4 –23 1⎡

⎣⎢

⎤

⎦⎥ +

0 00 0⎡

⎣⎢

⎤

⎦⎥ =

4 –23 1⎡

⎣⎢

⎤

⎦⎥ = A

O + A =0 00 0⎡

⎣⎢

⎤

⎦⎥ +

4 –23 1⎡

⎣⎢

⎤

⎦⎥ =

4 –23 1⎡

⎣⎢

⎤

⎦⎥ = A

A − A =4 –23 1⎡

⎣⎢

⎤

⎦⎥ −

4 –23 1⎡

⎣⎢

⎤

⎦⎥ =

0 00 0⎡

⎣⎢

⎤

⎦⎥ = O

Chapter 3 Answers 4

b.

A + (–A) =4 –23 1⎡

⎣⎢

⎤

⎦⎥ +

–4 2–3 –1⎡

⎣⎢

⎤

⎦⎥ =

0 00 0⎡

⎣⎢

⎤

⎦⎥ = O

(–A) + A =–4 2–3 –1⎡

⎣⎢

⎤

⎦⎥ +

4 –23 1⎡

⎣⎢

⎤

⎦⎥ =

0 00 0⎡

⎣⎢

⎤

⎦⎥ = O

17. a. They are all zeros.b. Yes, A1 1 = A1 1, A1 2 = A2 1, A1 3 = A3 1, A1 4 = A4 1, A1 5 = A5 1, A2 2 = A2 2, A2 3 = A3 2, etc.c. Sample matrix:

1 2 32 4 63 6 7

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

d. No. If the matrix is symmetric then the number of rows must equal thenumber of columns. In a matrix that is symmetric the entry Cij will be thesame as the entry Cji.

18. a. The 1s along the diagonal represent the relationship of a variable with itself.(Answers may vary.)

b. Answers vary. The values are symmetric because they represent therelationship between variables. The relationship between ACT math andcollege GPA, for example, must be the same as the relationship betweencollege GPA and ACT math. These values are in row 1 column 6 and in row 6column 1.

c. ACT composited. ACT math

Lesson 3.21. a. T represents the cost of four pizzas with additional toppings and four salads

with a choice of two dressings from each of the three pizza houses.b. $47.60c. T

1 2represents the cost of four pizzas at Toni’s. T

2 1 represents the cost four

salads at Vin’s.2. a and b

e p n b

J =PearlJade

16 8 12 1040 20 24 18⎡

⎣⎢

⎤

⎦⎥

c. 24 jade necklacesd. J

2 1 represents the number of jade earrings that the jeweler expects to sell in

June. J1 2

represents the number of pearl pins that the jeweler expects to sell inJune.

Chapter 3 Answers 5

3.% %

V = Vitamin AVitamin CVitamin D

252510

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3V = Vitamin AVitamin CVitamin D

757530

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4. a.Undergrad Grad

ResidentNonresident

$75.00 $99.25 $204.00 $245.25⎡

⎣⎢

⎤

⎦⎥

b. Undergrad Grad

ResidentNonresident

$80.25 $106.20$218.28 $262.42

⎡

⎣⎢

⎤

⎦⎥

c.Undergrad Grad

ResidentNonresident

$5.25 $6.95$14.28 $17.17

⎡

⎣⎢

⎤

⎦⎥

5. a. QP is defined. Order of QP: 1 × 2.b. QP is not defined because the number of entries in the row matrix Q does not

equal the number of rows in P.c. QP is defined. Order of QP: 1 × 4.d. QP is not defined because the number of entries in the row matrix Q does not

equal the number of rows in P.e. QP is defined. Order of QP: 1 × 1.

6.

Loans BondsRate 0.065 0.072 [ ]

MA NE CALoansBonds

230 440 680780 860 940⎡

⎣⎢

⎤

⎦⎥

MA NE CA= 71.11 90.52 111.88[ ]

7. a. Two possible answers: 2 × 6 4 × 6 5 × 5

Q = no. [ 10 4 2 ]Cost

P =2 × 64 × 65 × 5

$3.00$8.50$9.50

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Chapter 3 Answers 6

or 2 × 6 4 × 6 5 × 5

P = Cost [$3.00 $8.50 $9.50]no.

Q =2 × 64 × 65 × 5

1042

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

.

b. No. Answers vary. A possible answer is that the row and column matricesmay be written in the two different ways shown in part a.

c.

[10 4 2] 3.008.509.50

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= [10(3.00) + 4(8.50) + 2(9.50)]

= [30.00 + 34.00 + 19.00] = [83.00] or $83.00 8. a.

3A =9 24 −36 0 12⎡

⎣⎢

⎤

⎦⎥

b. BA = [14 16 14]c. BC = [20]d.

–2C =–4–8⎡

⎣⎢

⎤

⎦⎥

9.

CD CU BondRate 0.073 0.065 0.075[ ]

Amt InvestCDCUBonds

$10,00017,00012,000

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= $2,735

10. a. The transpose of a row matrix is a column matrix.b. The transpose of a column matrix is a row matrix.c.

p j

MT =

epnb

8 204 106 125 9

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

Chapter 3 Answers 7

11. a.e p n b

Hours 2 1 2.5 1.5[ ]b.

p j

Hours 2 1 2.5 1.5[ ]epnb

8 204 106 125 9

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 42.5 93.5[ ]

c. p j

Hours [42.5 93.5]d. It takes the jeweler 42.5 hours to make the pearl jewelry and 93.5 hours to

make the jade jewelry.12.

trn car pl tk$ 40 35 80 45[ ]

N S ETrainCarPlaneTruck

10 8 126 5 43 2 24 3 2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

N S E= dollars 1,030 790 870[ ]

13. a.Calendars

C =

MonTuesWedThursFri

1015203050

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

b. N = [1 1 1 1 1]

N × C = [1 1 1 1 1]

1015203050

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

= [125]

Chapter 3 Answers 8

c. A = [1/5 1/5 1/5 1/5 1/5] or A = [0.2 0.2 0.2 0.2 0.2]

A × C = [0.2 0.2 0.2 0.2 0.2]

1015203050

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

= [25]

Lesson 3.31. a.

AB =0 5 5

–14 8 15⎡

⎣⎢

⎤

⎦⎥

b. BA is not defined because the number of columns of A is not equal to thenumber of rows of B.

c.

BC =1 2–7 6⎡

⎣⎢

⎤

⎦⎥

d.

CB =–6 7 106 –7 –10

–16 12 20

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2. a. Mike Liz Kate

MikeLizKate

$261,000 $0 $250$235,000 $0 $215$255,000 $0 $325

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

b. The entries in row 1 represent the value to Mike of the items that he, Liz, andKate received. Row 2 represents the values to Liz and row 3 the values toKate.

3. a. Matrix Q: Burger Special Potato Fries ShakeRosaMax

0 1 0 1 11 0 1 0 1

⎡

⎣⎢

⎤

⎦⎥

.

Chapter 3 Answers 9

b. Matrix C:Cal Fat Chol

BurgerSpecialPotatoFriesShake

450 40 50570 48 90500 45 25300 30 0400 22 50

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

.

c. The dimensions of Q are 2 by 5, and those of C are 5 by 3.d. The dimensions of the product QC are 2 by 3.

e. The dimensions of C can be described as Foods by Contents, and thedimensions of Q times C can be described as Persons by Contents.

f.Cal Fat Chol

R =RosaMax

1,270 100 1401,350 107 125⎡

⎣⎢

⎤

⎦⎥

g. R1 2 represents the total grams of fat in Rosa's food.R2 1 represents the total number of calories in Max's food.R2 3 represents the mg of cholesterol in Max's food.

4. a. The number of columns of A must equal the number of rows of B.b. The number of columns of A must equal the number of rows of B, and the

number of columns of B must equal the number of rows of A.

Chapter 3 Answers 10

c. Answers vary. A possible answer follows.

A =2 1 34 5 6⎡

⎣⎢

⎤

⎦⎥ B =

2 13 24 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

AB =2 1 34 5 6⎡

⎣⎢

⎤

⎦⎥

2 13 24 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

19 1947 44⎡

⎣⎢

⎤

⎦⎥

BA =2 13 24 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2 1 34 5 6⎡

⎣⎢

⎤

⎦⎥ =

8 7 1214 13 2128 29 42

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

AB ≠ BA. The dimensions of AB are not equal to the dimensions of BA.d. Students may use trial and error to find a solution. Another method is to

examine the algebra involved as follows:

Suppose we let A =3 16 5⎡

⎣⎢

⎤

⎦⎥ and B =

a bc d⎡

⎣⎢

⎤

⎦⎥ . Then if AB = BA, we have

3 16 5⎡

⎣⎢

⎤

⎦⎥

a bc d⎡

⎣⎢

⎤

⎦⎥ =

a bc d⎡

⎣⎢

⎤

⎦⎥

3 16 5⎡

⎣⎢

⎤

⎦⎥ or

3a + c 3b + d6a + 5c 6b + 5d⎡

⎣⎢

⎤

⎦⎥ =

3a + 6b a + 5b3c + 6d c + 5d⎡

⎣⎢

⎤

⎦⎥ .

Since these two matrices are equal, we have a system of 4 equations in fourunknowns.

3a + c = 3a + 6b3b + d = a + 5d6a + 5c = 3c + 6d6b + 5d = c + 5d

Notice that the first and fourth equations areequivalent, leaving three equations in fourunknowns. There is an infinite number ofsolutions for the system. If we let a = 2 and b = 1,then c = 6 and d = 4 will be one solution.

Then AB = BA =3 16 5⎡

⎣⎢

⎤

⎦⎥

2 16 4⎡

⎣⎢

⎤

⎦⎥ =

2 16 4⎡

⎣⎢

⎤

⎦⎥

3 16 5⎡

⎣⎢

⎤

⎦⎥ =

12 742 26⎡

⎣⎢

⎤

⎦⎥ .

Students will also discover in Exercise 8 that if A and B are inverses of eachother then AB = BA.

5. Students should show work for IA = AI = A for any 3 × 3 matrix A they choose.A possible answer follows:

Let A =6 2 13 5 09 8 7

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

. Then AI =6 2 13 5 09 8 7

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 0 00 1 00 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

6 2 13 5 09 8 7

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

and IA =1 0 00 1 00 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

6 2 13 5 09 8 7

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

6 2 13 5 09 8 7

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

.

Chapter 3 Answers 11

6. a. Yes. A(BC) = (AB)Cb. A possible answer follows.

A BC( ) = 1 0 10 1 1⎡

⎣⎢

⎤

⎦⎥

3 12 2

–1 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 –1 02 1 1⎡

⎣⎢

⎤

⎦⎥

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟=

1 0 10 1 1⎡

⎣⎢

⎤

⎦⎥

5 –2 16 0 21 2 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

6 0 27 2 3⎡

⎣⎢

⎤

⎦⎥

AB( )C =1 0 10 1 1⎡

⎣⎢

⎤

⎦⎥

3 12 2

–1 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

1 –1 02 1 1⎡

⎣⎢

⎤

⎦⎥ =

2 21 3⎡

⎣⎢

⎤

⎦⎥

1 –1 02 1 1⎡

⎣⎢

⎤

⎦⎥ =

6 0 27 2 3⎡

⎣⎢

⎤

⎦⎥

c. Yes. Reasons will vary. A possible answer follows: The matrix multiplicationdepends on the associative properties for multiplication and addition of realnumbers.

7. Answers vary. One possibility for matrices A and B follows:

A =1 23 4⎡

⎣⎢

⎤

⎦⎥ B =

1 02 3⎡

⎣⎢

⎤

⎦⎥

A + B( ) A – B( ) = 2 25 7⎡

⎣⎢

⎤

⎦⎥

0 21 1⎡

⎣⎢

⎤

⎦⎥ =

2 67 17⎡

⎣⎢

⎤

⎦⎥

A2 – B2 =1 23 4⎡

⎣⎢

⎤

⎦⎥

1 23 4⎡

⎣⎢

⎤

⎦⎥ –

1 02 3⎡

⎣⎢

⎤

⎦⎥

1 02 3⎡

⎣⎢

⎤

⎦⎥ =

7 1015 22⎡

⎣⎢

⎤

⎦⎥ –

1 08 9⎡

⎣⎢

⎤

⎦⎥ =

6 107 13⎡

⎣⎢

⎤

⎦⎥ .

8. a. Students show work for AB = BA = I.

AB =2 31 2⎡

⎣⎢

⎤

⎦⎥

2 –3–1 2⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥

BA =2 –3

–1 2⎡

⎣⎢

⎤

⎦⎥

2 31 2⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥

b. Suppose matrix C does have an inverse. Then there exists a matrix D whereCD = I. Suppose

D =x zy w⎡

⎣⎢

⎤

⎦⎥.

Then 2 43 6⎡

⎣⎢

⎤

⎦⎥

x zy w⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥.

Performing the matrix multiplication gives

2x + 4y 2z + 4w3x + 6y 3z + 6w⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥.

Since the two matrices are equal, we have a system of four equations in fourunknowns.

Chapter 3 Answers 12

2x + 4y = 12z + 4w = 03x + 6y = 03z + 6w = 1

Attempting to solve the system for x and yshows that the first and third equations areinconsistent and that there is no solution.Since there is no solution for x and y, matrixD does not have an inverse.

Another way to show that the system does not have a solution is to rewritethe equations in slope-intercept form and to use the graphing calculator.

The lines are parallel since theirslopes are equal. It follows that thereis no solution for the system ofequations since the lines do notintersect.

9. The diagram below shows the polygons plotted for parts a through h.

b.

T1P =0 –6 –8 –20 2 6 4⎡

⎣⎢

⎤

⎦⎥

c. Polygon A'B'C'D' is the reflection of polygon ABCD in the y-axis.

Chapter 3 Answers 13

d.

T2P =0 –6 –8 –20 –2 –6 –4⎡

⎣⎢

⎤

⎦⎥

e. Polygon A"B"C"D" is the reflection of polygon A'B'C'D' in the x-axis.f.

T2T1–1 00 –1

⎡

⎣⎢

⎤

⎦⎥ = R

RP =0 –6 –8 –20 –2 –6 –4⎡

⎣⎢

⎤

⎦⎥

The effect of R on P is to rotate ABCD 180 degrees about the origin.g.

T3 =–1 00 1

⎡

⎣⎢

⎤

⎦⎥

T3P" =0 6 8 20 –2 –6 –4⎡

⎣⎢

⎤

⎦⎥

h.

T4 =–1 00 –1

⎡

⎣⎢

⎤

⎦⎥ = T3T2

T4P ' =0 6 8 20 –2 –6 –4⎡

⎣⎢

⎤

⎦⎥

10. a. T: Bag by Work W: Place by WorkP: Place by Bag D: Bag by Month

b. In this exercise we want the result to be a Work-by-Month matrix. We canobtain this result by multiplying a Work-by-Bag matrix times a Bag-by-Month matrix. We do this by multiplying the transpose of T times D.

May June

T TD =CuttingStitchingFinishing

880 1,1801,400 1,880

880 1,180

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

c. We want a Cost-by-Place matrix. We achieve this by multiplying a matrixrepresenting the hours worked on each bag times a matrix representing thewages paid per hour in each place. This translates into multiplying a Bag-by-Work matrix times a Work-by-Place matrix. We do this by multiplying Ttimes the transpose of W.

Chapter 3 Answers 14

NY NE CA

TW T =HandbagStandardRoomy

11.70 11.00 13.1615.05 14.15 16.9318.40 17.30 20.70

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

d.

May 600 800 400[ ] HSR

11.0014.1517.30

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= May $24,840[ ]

e. In this exercise, we want a matrix that represents hours of Work by Place. Wecan achieve this by multiplying a Work-by-Bag matrix times a Bag-by-Placematrix. One way to do this is to multiply the transpose of T times thetranspose of P.

NY NE CA

T TPT =CuttingStitchingFinishing

23.5 24.7 20.038.0 39.0 31.623.5 24.7 20.0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

11. a. The figure below shows the result of rotating ABCD through 30°, 60°, 90°,180°, and –90°.

b.0.866 –0.50.5 0.866⎡

⎣⎢

⎤

⎦⎥

0 6 8 20 2 6 4⎡

⎣⎢

⎤

⎦⎥ =

0 4.196 3.928 –0.2680 4.732 9.196 4.464⎡

⎣⎢

⎤

⎦⎥

c. The polygon A'B'C'D' is the result of rotating polygon ABCD 30°counterclockwise. For 180°, use the transformation matrix:

cos180° –sin180°sin180° cos180°

⎡

⎣⎢

⎤

⎦⎥

–1 00 –1

⎡

⎣⎢

⎤

⎦⎥

0 6 8 20 2 6 4⎡

⎣⎢

⎤

⎦⎥ =

0 –6 –8 –20 –2 –6 –4⎡

⎣⎢

⎤

⎦⎥ .

Chapter 3 Answers 15

d. For 60°, use the transformation matrix

cos60° –sin60°sin60° cos60°

⎡

⎣⎢

⎤

⎦⎥

0.5 –0.8660.866 0.5⎡

⎣⎢

⎤

⎦⎥

0 6 8 20 2 6 4⎡

⎣⎢

⎤

⎦⎥ =

0 1.268 –1.196 –2.4640 6.196 9.928 3.732⎡

⎣⎢

⎤

⎦⎥ .

Yes, this transformation matrix has the same effect as applying T1 twice.

T1( )2 = 0.5 –0.8660.866 0.5⎡

⎣⎢

⎤

⎦⎥

e.

90°: 0 –11 0⎡

⎣⎢

⎤

⎦⎥

0 6 8 20 2 6 4⎡

⎣⎢

⎤

⎦⎥ =

0 –2 –6 –40 6 8 2⎡

⎣⎢

⎤

⎦⎥

–90°:0 –11 0⎡

⎣⎢

⎤

⎦⎥

0 6 8 20 2 6 4⎡

⎣⎢

⎤

⎦⎥ =

0 2 6 40 –6 –8 –2⎡

⎣⎢

⎤

⎦⎥

0 –11 0⎡

⎣⎢

⎤

⎦⎥

0 1–1 0⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥,

0 1–1 0⎡

⎣⎢

⎤

⎦⎥

0 –11 0⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥

These two matrices are inverses of each other. The computations for 60° and–60° are as follows.

0.5 –0.8660.866 0.5⎡

⎣⎢

⎤

⎦⎥

0.5 0.866–0.866 0.5⎡

⎣⎢

⎤

⎦⎥ =

0.5 0.866–0.866 0.5⎡

⎣⎢

⎤

⎦⎥

0.5 –0.8660.866 0.5⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥

12. a.

A =1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥A2 =

1 2 30 1 20 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥A3 =

1 3 60 1 30 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥A4 =

1 4 100 1 40 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

b. Answers vary. Students may notice, for example, that in any power the entryin the first row third column is the sum of the entries in the first row in theprevious power. They should also be able to explain why any statements thatthey make are true.

c.

A5 =1 5 150 1 50 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥A6 =

1 6 210 1 60 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥A7 =

1 7 280 1 70 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Chapter 3 Answers 16

A8 =1 8 360 1 80 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥A9 =

1 9 450 1 90 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

d. A conjecture that can be proved with mathematical induction is that for anynatural number n, the nth power of the matrix has the form

An =1 n 1+ 2 + 3+ 4 + ...+ n0 1 n0 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

.

This is clearly true for n = 1.

A1 =1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Assume true for n = k.

Ak =1 k 1+ 2 + 3+ 4 + ...+ k0 1 k0 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Now, we need to show the statement is true for n = k + 1, that is, that

Ak+1 =1 k +1 1+ 2 + 3+ 4 + ...+ k +10 1 k +10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥.

But, Ak+1 = AAk =1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 k 1+ 2 + 3+ 4 + ...+ k0 1 k0 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=1 k +1 1+ 2 + 3+ 4 + ...+ k + k +10 1 k +10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=1 k +1 1+ 2 + 3+ 4 + ...+ k +10 1 k +10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

.

Therefore the conjecture is true for all natural numbers n.

Note: Students may also know that ∑k=1

nk = 1

2n n +1( ) .

So, another conjecture that they could prove using mathematical induction isthat

An =

1 n 12n n +1( )

0 1 n0 0 1

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

.

Chapter 3 Answers 17

13. a.

Let A =1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥, B =

2 2 20 2 20 0 2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥, C =

3 3 30 3 30 0 3

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥, and D =

4 4 40 4 40 0 4

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥.

Then B2 =4 8 120 4 80 0 4

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 4

1 2 30 1 20 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 4

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2

= 4A2 = 22A2 .

B3 =8 24 480 8 240 0 8

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 8

1 3 60 1 30 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 8

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3

= 8A3 = 23A3

B4 =16 64 1600 16 640 0 16

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 16

1 4 100 1 40 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 16

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4

= 16A4 = 24A4

C 2 =9 18 270 9 180 0 9

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 9

1 2 30 1 20 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 9

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2

= 9A2 = 32A2

C 3 =27 81 1620 27 810 0 27

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 27

1 3 60 1 30 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 27

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3

= 27A3 = 33A3

C 4 =81 324 8100 81 3240 0 81

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 81

1 4 100 1 40 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 81

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4

= 81A4 = 34A4

D2 =16 32 480 16 320 0 16

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 16

1 2 30 1 20 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 16

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2

= 16A2 = 42A2

D3 =64 192 3840 64 1920 0 64

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 64

1 3 60 1 30 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 64

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3

= 64A3 = 43A3

D4 =256 1,024 2,5600 256 1,0240 0 256

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 256

1 4 100 1 40 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 256

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4

= 256A4 = 44A4

b and c. Conjecture: For any natural numbers m and n

Chapter 3 Answers 18

m m m0 m m0 0 m

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

n

= mn

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

n

.

This is clearly true for n = 1.

m m m0 m m0 0 m

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= m

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Assume true for n = k.

m m m0 m m0 0 m

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k

= mk

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k

Now, we need to show that this is true for n = k + 1, that is,

m m m0 m m0 0 m

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k+1

= mk+1

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k+1

.

But m m m0 m m0 0 m

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k+1

=m m m0 m m0 0 m

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

m m m0 m m0 0 m

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k

= m1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥mk

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k

= mk+1

1 1 10 1 10 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k+1

.

Therefore, the conjecture is true for all natural numbers m and n.

Lesson 3.41. a. 18.97 newborn rats

b.

(16.6)(0.6) = 9.96 move up to the 3–6 age group.(9)(0.9) = 8.1 move up to the 6–9 age group.(8.1)(0.9) = 7.29 move up to the 9–12 age group.(11.7)(0.8) = 9.36 move up to the 12–15 age group.(4)(0.6) = 2.4 move up to the 15–18 age group.

c. Distribution after 6 months: 18.97, 9.96, 8.1, 7.29, 9.36, 2.4; Total 56 rats.

Chapter 3 Answers 19

d. Distribution after 9 months: 18.32, 11.38, 8.96, 7.29, 5.83, 5.62; Total 57 rats.Distribution after 12 months: 18.02, 10.99, 10.24, 8.06, 5.83, 3.5; Total 57 rats.

e. Answers may vary. Possible answer: The population continues to grow. Therate of growth seems to have slowed.

f. Answers may vary. Possible answer: The population growth may continue toslow or even become constant.

2. a. 50(0) + 30(0.8) + 24(1.7) + 24(1.7) + 12(0.8) + 8(0.4) = 118.4 newborn deerb.

(50)(0.6) = 30.0 move up to the 2–4 age group.(30)(0.8) = 24.0 move up to the 4–6 age group.(24)(0.9) = 21.6 move up to the 6–8 age group.(24)(0.9) = 21.6 move up to the 8–10 age group.(12)(0.7) = 8.4 move up to the 10–12 age group.(8)(0) = 0 No deer lives beyond 12 months

c. The product is the number of newborn deer after 1 cycle.3.

a. 50 30 24 24 12 8[ ]

0.600000

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

b. 50 30 24 24 12 8[ ]

00.80000

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

c. 50 30 24 24 12 8[ ]

000.9000

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

Chapter 3 Answers 20

d. 50 30 24 24 12 8[ ]

0000.900

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

e. 50 30 24 24 12 8[ ]

00000.70

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

4. a. Distribution after 3 months: 0, 21, 0, 0, 0, 0; Total 21 rats.b. Distribution after 3 months: 11, 3, 4.5, 4.5, 4, 3; Total 30 rats.

5. a. After one cycle: 0 15 0 0 0 0[ ] ; Total 15 deer.

b. After two cycles: 12 0 12 0 0 0[ ] ; Total 24 deer.

c. After three cycles: 20.4 7.2 0 10.8 0 0[ ] ; Total 39 deer.

d. After four cycles: 24.12 12.24 5.76 0 9.72 0[ ] ; Total 52 deer.

Lesson 3.51. a. To use the graphing calculator, store PO = [15 9 13 5 0 0] as matrix [A]

and store the Leslie matrix for the Rattus norvegicus as matrix [B].P5 = P0L5 = [A] × [B]^5 = [19.47 10.81 9.89 9.22 6.45 3.50]

b. Let [C] be a 6-by-1 column matrix containing all 1s. Then the total populationafter 15 months (5 cycles) T = [A] × [B]^5 × [C] = [59.35] orT = [19.47 10.81 9.89 9.22 6.45 3.50] × [C] = [ 59.35 ] = 59.35 rats.

c. P7 = [A] × [B]^7 = [20.47 12.11 10.51 8.76 7.12 4.43]

Note: Stop and record the distribution and then press [×] [C] [ENTER] to findthe total. The calculator will show:

T = 63.41 rats.

Chapter 3 Answers 21

2. Use trial and error to find the number of cycles for the female population toreach 250 members. The number of years equals the number of cycles dividedby 4 since there are three months per cycle.

Cycles Population Yearsa. 61 253.2 15.25b. 69 250.9 17.25c. 76 252.9 19.00d. 41 249.6 10.25

3. a.Cycle Total Population Growth Rateoriginal 42.001 49.40 17.6%2 56.08 13.5%3 57.40 2.35%4 56.65 –1.31%5 59.35 4.77%6 61.76 4.06%

b. The population appears to decline then increase again.c. P2 5 = 108.488, P2 6 = 111.789, P2 7 = 115.191. The growth rate in each case is

3.04%.4. a. P2 5 = 86.054, P2 6 = 88.674, P2 7 = 91.372

P2 5 = 67.098, P2 6 = 69.142, P2 7 = 71.247P2 5 = 54.823, P2 6 = 56.491, P2 7 = 58.210P2 5 = 154.512, P2 6 = 159.215, P2 7 = 164.060The long-term growth rate of the total population is 3.04% in each case.

b. The initial population does not affect the long-term growth rate.5. a.

The Leslie matrix L =

0.0 0.6 0.0 0.0 0.0 0.00.8 0.0 0.8 0.0 0.0 0.01.7 0.0 0.0 0.9 0.0 0.01.7 0.0 0.0 0.0 0.9 0.00.8 0.0 0.0 0.0 0.0 0.70.4 0.0 0.0 0.0 0.0 0.0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

.

b. P2 9 = 371,147.427, P3 0 = 485,321.624. The long-term growth rate is 30.76%.

Chapter 3 Answers 22

c. P7 = 1,013.15, P8 = 1,327.88The female herd size reaches 1,250 between 7 and 8 cycles or about 15 yearssince one cycle equals 2 years.

d. Approximately 30.76% of the animals would need to be removed from theherd each year.

6. a.

The Leslie matrix L =0.0 0.5 0.00.0 0.0 0.76.0 0.0 0.0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

b. P0 = [0 0 5]c. P1 6 = [1,225 0 0]. There will be 1,225 female bugs living in the basement

after 16 weeks.Note: To use the iteration function on the graphing calculator:

Press [A] [×] [B] and press[ENTER].Press [×] [B] and press [ENTER].Continue to press [ENTER] whileyou count the cycles.

7. a.

1 Week 2 Weeks 3 WeeksP0 0 0 5P2 2 5,403.3 0 0P2 3 0 2,701.6 0P2 4 0 0 1,891.1P2 5 11,346.9 0 0P2 6 0 5,673.4 0P2 7 0 0 3,971.4P2 8 23,828.4 0 0P2 9 0 11,914.2 0P3 0 0 0 8,339.9

Answers vary. Students may observe that only bugs of the same age groupare living during each cycle or that the pattern is determined by the growthrate for the previous age group. The number of bugs decreases for 2 cyclesand then increases dramatically. About one third of any population of

Chapter 3 Answers 23

newborns survives to 3 weeks. The population of newborns doubles everyfourth week.

b. The population change in each case is 210%.8. a.

1 Week 2 Weeks 3 Weeks Total % GrowthP0 4 4 4P2 2 4,322.6 360.2 504.3 5,187.1P2 3 3,025.8 2,161.3 252.2 5,439.3 4.9%P2 4 1,512.9 1,512.9 1,512.9 4,538.7 –16.6%P2 5 9,077.5 756.5 1,059.0 10,893.0 140%P2 6 6,354.2 4,538.7 529.5 11,422.4 4.9%P2 7 3,177.1 3,177.1 3,177.1 9,531.3 –16.6%P2 8 19,062.7 1,588.6 2,224.0 22,875.3 140%P2 9 13,343.9 9,531.4 1,112.0 23,987.3 4.9%P3 0 6,672.0 6.672.0 6,672.0 20,016.0 –16.6%

Note: The percentage growth is for the total population for each cycle.Answers vary. Students may observe that the number of bugs in each agegroup is the same every third cycle. They may also notice that the totalpopulation growth falls into a pattern of 2.400, 1.0486, and 0.8344 times theprevious total population.

b. The growth is 110% in each case.9. The statement is true for k = 1 since P1 = P0L by definition. Assume that the

statement is true for k = n, that is, that Pn= P0Ln is true.Then it is necessary to show that Pn+1 = P0Ln+1.But, Pn+1 = Pn(L) = P0Ln(L) = P0Ln+1.It follows by mathematical induction that Pk = P0Lk for all natural numbers k.

Review1. Important points covered in this chapter.

• Matrix terminology and definitions including dimensions of matrices, rowand column matrices, square matrix, identity matrix, inverse of a matrix,triangular matrices, and transpose of a matrix

• How to find the sum and difference of two matrices• How to find scalar multiples of matrices• How to find the product of two matrices• How to use mathematical induction to prove conjectures involving matrices• How to use matrices to model real-world situations

Chapter 3 Answers 24

• How to use powers of matrices to model population growth• How to use the Leslie matrix model for population growth• How to make predictions using powers of matrices

2. a. 6 elementsb. C1 2 = –2; C2 1 = –1

3. a.

A + C =6 –23 72 –1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

b. It is not possible to subtract these matrices because their orders are not thesame.

c.

A + C( ) – D =5 11 9–1 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

d.

2A + D =5 –310 121 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4. a. Mex Chips Salsa DrinksL = 35 6 6 12 [ ]

b. L2 = number of bags of chips orderedL4 = number of six-packs of drinks

c.

Mex Chips Salsa DrinksCost = Number 35 6 6 12 [ ] ×

MexChipsSalsaDrinks

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

$4.50$1.97$2.10$2.89

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= [35($4.50) + 6($1.97) + 6($2.10) + 12($2.89)] = [$216.60]5. a.

Lodging Food Rec

C =

CrystalSpringsBearBeaver

$13.00 $20.00 $5.00$12.50 $19.50 $7.50$20.00 $18.00 $0.00$40.00 $0.00 $0.00

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

Chapter 3 Answers 25

b. C2 2 = $19.50C4 3 = $0.00

c. C1 3 = cost for recreation at Crystal LodgeC3 1 = cost for lodging at Bear Lodge

6. a.System Cartridge Case

Z-MartBase

$39.50 $24.50 $8.50$49.90 $29.95 $12.50⎡

⎣⎢

⎤

⎦⎥

b.System Cartridge Case

Z-MartBase

$35.55 $22.05 $7.65$39.92 $23.96 $10.00⎡

⎣⎢

⎤

⎦⎥

c.System Cartridge Case

Z-MartBase

$3.95 $2.45 $0.85$9.98 $5.99 $2.50⎡

⎣⎢

⎤

⎦⎥

d.System Cartridge Case

4$35.55 $22.05 $7.65$39.92 $23.96 $10.00⎡

⎣⎢

⎤

⎦⎥ =

Z-MartBase

$142.20 $88.20 $30.60$159.68 $95.84 $40.00⎡

⎣⎢

⎤

⎦⎥

7. a. 3 × 2

b. 4 × 3c. Not possible. The number of columns of Q does not equal the number of

rows of S.d. 4 × 3

8. a. Plate Large SmallNo. [ 5 3 7 ]

b. Ebony Walnut Rose MaplePlateLargeSmall

100 800 600 400 200 1,200 1,000 80050 500 450 400

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

c. Ebony Walnut Rose Maple[ 1,450 11,000 9,150 7,200 ]

Chapter 3 Answers 26

d.

Plate Large SmallNo. 5 3 7 [ ] ×

WeeksPlateLargeSmall

3 42

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Weeks= No. 15 12 14[ ]

Total time = 15 + 12 + 14 = 41 weeks9.

Tennis Golf SoccerReturn 0.082 0.065 0.075[ ] ×

DollarsTennisGolfSoccer

50,000100,00075,000

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= $16,225[ ]

10. Jazz Symp Orch$ [300.00 335.00 373.50]

11. a.

AB =8 –2 423 38 –13–1 19 –11

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

b.

BA =2 –129 33⎡

⎣⎢

⎤

⎦⎥

c. CA is not defined. The number of columns of C does not equal the number ofrows of A.

d. DA + E = [12 10]12.

AT =4 52 16 3

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

13. a.

M =1 11 1⎡

⎣⎢

⎤

⎦⎥, M 2 =

2 22 2⎡

⎣⎢

⎤

⎦⎥, M 3 =

4 44 4⎡

⎣⎢

⎤

⎦⎥, M 4 =

8 88 8⎡

⎣⎢

⎤

⎦⎥

b.

M 5 =16 1616 16⎡

⎣⎢

⎤

⎦⎥

c.

M n =2n–1 2n–1

2n–1 2n–1⎡

⎣⎢

⎤

⎦⎥ where n is a natural number.

Chapter 3 Answers 27

d. The conjecture is true for n = 1 since

M 1 =21–1 21–1

21–1 21–1⎡

⎣⎢

⎤

⎦⎥ =

20 20

20 20⎡

⎣⎢

⎤

⎦⎥ =

1 11 1⎡

⎣⎢

⎤

⎦⎥ .

Assume true for n = any natural number k, that is,

M k =2k –1 2k –1

2k –1 2k –1⎡

⎣⎢

⎤

⎦⎥ .

To complete the proof, we need to show that the conjecture is true forn = k + 1. That is, we must show that

M k+1 =2k 2k

2k 2k⎡

⎣⎢

⎤

⎦⎥ .

M k+1 = MM k =1 11 1⎡

⎣⎢

⎤

⎦⎥

2k –1 2k –1

2k –1 2k –1

⎡

⎣⎢

⎤

⎦⎥ =

2k –1 + 2k –1 2k –1 + 2k –1

2k –1 + 2k –1 2k –1 + 2k –1

⎡

⎣⎢

⎤

⎦⎥

=2 ⋅2k –1 2 ⋅2k –1

2 ⋅2k –1 2 ⋅2k –1⎡

⎣⎢

⎤

⎦⎥ =

2k 2k

2k 2k⎡

⎣⎢

⎤

⎦⎥

Therefore the conjecture is true for all natural numbers n.e.

a.

M =1 02 3⎡

⎣⎢

⎤

⎦⎥, M 2 =

1 08 9⎡

⎣⎢

⎤

⎦⎥, M 3 =

1 026 27⎡

⎣⎢

⎤

⎦⎥, M 4 =

1 080 81⎡

⎣⎢

⎤

⎦⎥

b.

M 5 =1 0

242 243⎡

⎣⎢

⎤

⎦⎥

c.

M n =1 0

3n –1 3n⎡

⎣⎢

⎤

⎦⎥ where n is a natural number.

d. The conjecture is true for n = 1 since

M 1 =1 0

31 – 1 31⎡

⎣⎢

⎤

⎦⎥ =

1 02 3⎡

⎣⎢

⎤

⎦⎥ .

Assume true for n = any natural number k, that is,

M k =1 0

3k –1 3k⎡

⎣⎢

⎤

⎦⎥ .

To complete the proof, we need to show the conjecture is true forn = k + 1. That is, we must show that

Chapter 3 Answers 28

M k+1 =1 0

3k+1 – 1 3k+1⎡

⎣⎢

⎤

⎦⎥ .

M k+1 = MM k =1 02 3⎡

⎣⎢

⎤

⎦⎥

1 03k – 1 3k⎡

⎣⎢

⎤

⎦⎥ =

1 02 + 3 ⋅ 3k – 3 3 ⋅ 3k⎡

⎣⎢

⎤

⎦⎥ =

1 03k+1 – 1 3k+1⎡

⎣⎢

⎤

⎦⎥

Therefore the conjecture is true for all natural numbers n.14. If a square matrix A has an inverse A– 1, then the product AA– 1 = the identity

matrix I, where I is a square matrix with ones along the diagonal and zeroselsewhere.

15. a. Yes. AB = BA = I.b. Yes. AB = BA = I.c. No. These are not square matrices. AB ≠ BA ≠ I.

16. a. 10th 11th 12th Male Female Male FemaleABC

0.2 0.35 0.40.3 0.3 0.250.5 0.35 0.35

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥×

10th11th12th

235 225205 215175 190

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

ABC

189 196176 180251 254

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

b. Male FemaleABC

189 196176 180251 254

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥×

11⎡

⎣⎢

⎤

⎦⎥ =

ABC

385356505

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Note: There may be a slight difference in the totals matrix due to rounding.17. a. 24 months

b.

L =

0.0 0.6 0.0 0.0 0.0 0.00.5 0.0 0.8 0.0 0.0 0.01.1 0.0 0.0 0.9 0.0 0.00.9 0.0 0.0 0.0 0.8 0.00.4 0.0 0.0 0.0 0.0 0.60 0.0 0.0 0.0 0.0 0.0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

c. 10%d. After the 17th cycle, or about 35 years