vcaa extended response questions · module 6: matrices – question 4 – continued question 3 the...

REVISION for MATRICES

OUTCOME 2013

VCAA EXTENDED RESPONSE

QUESTIONS

Module 6: Matrices

Module 6: Matrices

Question 1I B L E

E

birds

eat eat

eat

lizards insects

I000

B101

L100

E =IBL

a. i. B L E

ii. E

I B L F Z

Z

I000_

B101_

L100_

F____

Z =

IBLF

eateat

eat eat

eateat

birds

insects

frogs

lizards

b. Z

35

Module 6: MatricesTURN OVER

Question 2

I B L FN

I100 000

B400

L1000

F800N = [ ]

D

I0.995

000

B0

0.0500

L00

0.0250

F000

0.30

IBLF

dead after sprayingD =

alive before spraying

a. K = ND

K =

b. K

c. M = KF

0111

F =

M =

d. M

Module 6: Matrices – Question 3

Question 3

W0 J A

W0 =3264

JA

a.

J A

W1 = BW0

B

B =0

0.252

0.5JA

J A

b. W1

W1 =

37

Module 6: Matrices – Question 3TURN OVER

J An

Wn = BWn – 1

juvenilesadults

130

120

110

100

90

80

70

60

50

40

30

20

10

00 1 2 3 4 5 6 7 8 9 10

year

numberof ducks

c. i.

ii.

END OF QUESTION AND ANSWER BOOK

B

Wn = PWn – 1

P =0

0.251

0.5JA

J A

W0 =3264

JA

d.

e. P

2010 FURMATH EXAM 2 28

Module 6: Matrices – Question 2 – continued

Module 6: Matrices

Question 1In a game of basketball, a successful shot for goal scores one point, two points, or three points, depending on the position from which the shot is thrown.G is a column matrix that lists the number of points scored for each type of successful shot.

G123

In one game, Oscar was successful with • 4 one-point shots for goal• 8 two-point shots for goal• 2 three-point shots for goal.a. Write a row matrix, N, that shows the number of each type of successful shot for goal that Oscar had in

that game.

N = [ ]1 mark

b. Matrix P is found by multiplying matrix N with matrix G so that P = N × G Evaluate matrix P.

1 mark

c. In this context, what does the information in matrix P provide?

1 mark

Question 2The 300 players in Oscar’s league are involved in a training program. In week one, 90 players are doing heavy training (H), 150 players are doing moderate training (M) and 60 players are doing light training (L).The state matrix, S1, shows the number of players who are undertaking each type of training in the first week.

SH

M

L1

90150

60

29 2010 FURMATH EXAM 2

Module 6: Matrices – continuedTURN OVER

The percentage of players that remain in the same training program, or change their training program from week to week, is shown in the transition diagram below.

50%

H

L

10%30% 20%

30%

50%

10%

40%60%

M

a. What information does the 20% in the diagram above provide?

1 mark

The information in the transition diagram above can also be written as the transition matrix T.

this weekH M L

TH

M

L

next week

0.5 0.1 0.10.2 0.6 0.50.3 0.3 0.4

b. Determine how many players will be doing heavy training in week two.

1 mark

c. Determine how many fewer players will be doing moderate training in week three than in week one.

1 mark

d. Show that, after seven weeks, the number of players (correct to the nearest whole number) who are involved in each type of training will not change.

1 mark



Question 3The basketball coach has written three linear equations which can be used to predict the number of points, p, rebounds, r, and assists, a, that Oscar will have in his next game.

The equations are p + r + a = 332p – r + 3a = 40 p + 2r + a = 43

a. These equations can be written equivalently in matrix form. Complete the missing information below.

pra

334043

1 mark

This matrix equation can be solved in the following way.

pra x

7 1 41 0 1

1 3

334043

b. Determine the value of x shown in the matrix equation above.

1 mark

c. How many rebounds is Oscar predicted to have in his next game?

1 mark

Question 4The Dinosaurs (D) and the Scorpions (S) are two basketball teams that play in different leagues in the same city.The matrix A1 is the attendance matrix for the first game. This matrix shows the number of people who attended the first Dinosaur game and the number of people who attended the first Scorpion game.

ADS1

20001000

The number of people expected to attend the second game for each team can be determined using the matrix equation

A2 = GA1

where G is the matrix

this gameD S

GDS

next game1 2 0 30 2 0 7. .. .



a. i. Determine A2 , the attendance matrix for the second game.

ii. Every person who attends either the second Dinosaur game or the second Scorpion game will be given a free cap.

How many caps, in total, are expected to be given away?

1 + 1 = 2 marks

Assume that the attendance matrices for successive games can be determined as follows.

A3 = GA2A4 = GA3and so on such that An+1 = GAn

b. Determine the attendance matrix (with the elements written correct to the nearest whole number) for game 10.

1 mark

c. Describe the way in which the number of people attending the Dinosaurs’ games is expected to change over the next 80 or so games.

1 mark

The attendance at the first Dinosaur game was 2000 people and the attendance at the first Scorpion game was 1000 people.Suppose, instead, that 2000 people attend the first Dinosaur game, and 1800 people attend the first Scorpion game.d. Describe the way in which the number of people attending the Dinosaurs’ games is expected to change

over the next 80 or so games.

1 mark

Total 15 marks



Module 6: Matrices

Question 1Three types of cheese, Cheddar (C), Gouda (G) and Blue (B), will be bought for a school function.The cost matrix P lists the prices of these cheeses, in dollars, at two stores, Foodway and Safeworth.

PFoodwaySafeworth

6 80 5 30 6 207 30 4 90 6 15. . .. . .

a. What is the order of matrix P?

1 mark

The number of packets of each type of cheese needed is listed in the quantity matrix Q.

QCGB

8113

b. i. Evaluate the matrix W = PQ.

ii. At which store will the total cost of the cheese be lower?

1 + 1 = 2 marks



Question 2Tickets for the function are sold at the school office, the function hall and online.Different prices are charged for students, teachers and parents.Table 1 shows the number of tickets sold at each place and the total value of sales.

Table 1

School office

Function hall

Online

Student tickets 283 35 84

Teacher tickets 28 4 3

Parent tickets 5 2 7

Total sales $8712 $1143 $2609

For this function

• student tickets cost $x• teacher tickets cost $y• parent tickets cost $z.

a. Use the information in Table 1 to complete the following matrix equation by inserting the missing values in the shaded boxes.

x

y

z

283

84

28

4

3

5

7

8712

1143

2609

=

1 mark

b. Use the matrix equation to find the cost of a teacher ticket to the school function.

2 marks



Question 3In 2009, the school entered a Rock Eisteddfod competition.When rehearsals commenced in February, all students were asked whether they thought the school would make the state finals. The students’ responses, ‘yes’, ‘no’ or ‘undecided’ are shown in the initial state matrix S0.

Syesnoundecided

0

160120220

a. How many students attend this school?

1 mark

Each week some students are expected to change their responses. The changes in their responses from one week to the next are modelled by the transition matrix T shown below.

T0 85 0 35 0 600 10 0 40 0 300 05 0 25 0 10

. . .

. . .

. . .

yes no undecidedresponse this week

yes

no

undecided

response next week

The following diagram can also be used to display the information represented in the transition matrix T.

yes 85%

5%

undecidedno

10%

35% 60%

30%10%40%

b. i. Complete the diagram above by writing the missing percentage in the shaded box.

ii. Of the students who respond ‘yes’ one week, what percentage are expected to respond ‘undecided’ the next week when asked whether they think the school will make the state finals?

iii. In total, how many students are not expected to have changed their response at the end of the first week?

1 + 1 + 2 = 4 marks



c. Evaluate the product S1 = T S0 , where S1 is the state matrix at the end of the first week.

1 mark

d. How many students are expected to respond ‘yes’ at the end of the third week when asked whether they think the school will make the state finals?

1 mark

Question 4A series of extra rehearsals commenced in April. Each week participants could choose extra dancing rehearsals or extra singing rehearsals.A matrix equation used to determine the number of students expected to attend these extra rehearsals is given by

L Ln n10 85 0 250 15 0 75

57

. .

. .

where Ln is the column matrix that lists the number of students attending in week n.The attendance matrix for the first week of extra rehearsals is given by

L19597

dancingsinging

a. Calculate the number of students who are expected to attend the extra singing rehearsals in week 3.

1 mark

b. Of the students who attended extra rehearsals in week 3, how many are not expected to return for any extra rehearsals in week 4?

1 mark

Total 15 marks


Module 6: Matrices – Question 1 – continuedTURN OVER

Module 6: Matrices

Question 1Two subjects, Biology and Chemistry, are offered in the fi rst year of a university science course.The matrix N lists the number of students enrolled in each subject.

N =⎡

⎣⎢

⎤

⎦⎥

460360

BiologyChemistry

The matrix P lists the proportion of these students expected to be awarded an A, B, C, D or E grade in each subject.

A B C D EP = [0.05 0.125 0.175 0.45 0.20]

a. Write down the order of matrix P.

1 mark

b. Let the matrix R = NP. i. Evaluate the matrix R.

ii. Explain what the matrix element R24 represents.

1 + 1 = 2 marks


Module 6: Matrices – continued

c. Students enrolled in Biology have to pay a laboratory fee of $110, while students enrolled in Chemistry pay a laboratory fee of $95.

i. Write down a clearly labelled row matrix, called F, that lists these fees.

ii. Show a matrix calculation that will give the total laboratory fees, L, paid in dollars by the students enrolled in Biology and Chemistry. Find this amount.

1 + 1 = 2 marks



Question 2The following transition matrix, T, is used to help predict class attendance of History students at the university on a lecture-by-lecture basis.

this lecture

attend not attend

Tattendnot attend

=⎡

⎣⎢

⎤

⎦⎥

0 90 0 200 10 0 80. .. .

next lecture

S1 is the attendance matrix for the fi rst History lecture.

Sattendnot attend1

54036

=⎡

⎣⎢

⎤

⎦⎥

S1 indicates that 540 History students attended the fi rst lecture and 36 History students did not attend the fi rst lecture. a. Use T and S1 to

i. determine S2 the attendance matrix for the second lecture

ii. predict the number of History students attending the fi fth lecture.

1 + 1 = 2 marks

b. Write down a matrix equation for Sn in terms of T, n and S1.

1 mark

The History lecture can be transferred to a smaller lecture theatre when the number of students predicted to attend falls below 400.c. For which lecture can this fi rst be done?

1 mark

d. In the long term, how many History students are predicted to attend lectures?

1 mark


Module 6: Matrices – continued

Question 3The bookshop manager at the university has developed a matrix formula for determining the number of Mathematics and Physics textbooks he should order each year. For 2009, the starting point for the formula is the column matrix S2008. This lists the number of Mathematics and Physics textbooks sold in 2008.

SMathematicsPhysics2008

456350

=⎡

⎣⎢

⎤

⎦⎥

O2009 is a column matrix listing the number of Mathematics and Physics textbooks to be ordered for 2009. O2009 is given by the matrix formula

O2009 = A S2008 + B where A =⎡

⎣⎢

⎤

⎦⎥

0 75 00 0 68.

. and B =

⎡

⎣⎢

⎤

⎦⎥

1812

a. Determine O2009

1 mark

The matrix formula above only allows the manager to predict the number of books he should order one year ahead. A new matrix formula enables him to determine the number of books to be ordered two or more years ahead.The new matrix formula is

On + 1 = C On – D

where On is a column matrix listing the number of Mathematics and Physics textbooks to be ordered for year n.

Here, C =⎡

⎣⎢

⎤

⎦⎥

0 8 00 0 8.

. and D =

⎡

⎣⎢

⎤

⎦⎥

4038

The number of books ordered in 2008 was given by

OMathematicsPhysics2008

500360

=⎡

⎣⎢

⎤

⎦⎥

b. Use the new matrix formula to determine the number of Mathematics textbooks the bookshop manager should order in 2010.

2 marks



Question 4By the end of each academic year, students at the university will have either passed, failed or deferred the year.Experience has shown that

• 88% of students who pass this year will also pass next year• 10% of students who pass this year will fail next year• 2% of students who pass this year will defer next year

• 52% of students who fail this year will pass next year• 44% of students who fail this year will fail next year• 4% of students who fail this year will defer next year

• 65% of students who defer this year will pass next year• 10 % of students who defer this year will fail next year• 25% of students who defer this year will defer next year.

Twelve hundred and thirty students began a business degree in 2007.By the end of the 2007 academic year, 880 students had passed, 230 had failed, while 120 had deferred the year.No students have dropped out of the business degree permanently.

Use this information to predict the number of business students who will defer the 2009 academic year.

2 marks

Total 15 marks


Module 6: Matrices � Question 1 � continuedTURN OVER

Module 6: Matrices

Question 1The table below displays the energy content and amounts of fat, carbohydrate and protein contained in a serve of four foods: bread, margarine, peanut butter and honey.

Food Energy content (kilojoules/serve)

Fat (grams/serve)

Carbohydrate (grams/serve)

Protein (grams/serve)

Bread 531 1.2 20.1 4.2

Margarine 41 6.7 0.4 0.6

Peanut butter 534 10.7 3.5 4.6

Honey 212 0 12.5 0.1

a. Write down a 2 × 3 matrix that displays the fat, carbohydrate and protein content (in columns) of bread and margarine.

1 mark

b. A and B are two matrices deÞ ned as follows.

A = [2 2 1 1] B =

53141

534212

i. Evaluate the matrix product AB.

ii. Determine the order of matrix product BA.

Matrix A displays the number of servings of the four foods: bread, margarine, peanut butter and honey, needed to make a peanut butter and honey sandwich.

Matrix B displays the energy content per serving of the four foods: bread, margarine, peanut butter and honey.

iii. Explain the information that the matrix product AB provides.

1 + 1 + 1 = 3 marks


Module 6: Matrices � Question 2 � continued

c. The number of serves of bread (b), margarine (m), peanut butter (p) and honey (h) that contain, in total, 53 grams of fat, 101.5 grams of carbohydrate, 28.5 grams of protein and 3568 kilojoules of energy can be determined by solving the matrix equation

Solve the matrix equation to Þ nd the values b, m, p and h.

2 marks

Question 2To study the life-and-death cycle of an insect population, a number of insect eggs (E), juvenile insects (J) and adult insects (A) are placed in a closed environment.The initial state of this population can be described by the column matrix

A row has been included in the state matrix to allow for insects and eggs that die (D).a. What is the total number of insects in the population (including eggs) at the beginning of the study?

1 mark

In this population� eggs may die, or they may live and grow into juveniles� juveniles may die, or they may live and grow into adults� adults will live a period of time but they will eventually die.In this population, the adult insects have been sterilised so that no new eggs are produced. In these circumstances, the life-and-death cycle of the insects can be modelled by the transition matrix

b. What proportion of eggs turn into juveniles each week?

1 mark

1 2 6 7 10 7 020 1 0 4 3 5 12 54 2 0 6 4 6 0 1531 41 534 212

. . .. . . .. . . .

=

bmph

53101 5

28 53568

.

.

S

EJAD

0

400200100

0

=

this weekE J A D

T =

0 4 0 0 00 5 0 4 0 00 0 5 0 8 0

0 1 0 1 0 2 1

.

. .. .

. . .

EJAD

next week


Module 6: Matrices � Question 2 � continuedTURN OVER

c. i. Evaluate the matrix product S1 = T S0

ii. Write down the number of live juveniles in the population after one week.

iii. Determine the number of live juveniles in the population after four weeks. Write your answer correct to the nearest whole number.

iv. After a number of weeks there will be no live eggs (less than one) left in the population. When does this Þ rst occur?

v. Write down the exact steady-state matrix for this population.

1 + 1 + 1 + 1 + 1 = 5 marks

S T S

EJAD

1 0= =

S

EJAD

steady state =



d. If the study is repeated with unsterilised adult insects, eggs will be laid and potentially grow into adults. Assuming 30% of adults lay eggs each week, the population matrix after one week, S1, is now given by

S1 = T S0 + B S0

i. Determine S1

This pattern continues. The population matrix after n weeks, Sn, is given by

Sn = T Sn �1 + B S n�1

ii. Determine the number of live eggs in this insect population after two weeks.

1 + 1 = 2 marks

Total 15 marks

where andB S=

=

0 0 0 3 00 0 0 00 0 0 00 0 0 0

400200100

0

0

.

EJAD

S

E

J

A

D

1 =

vcaa extended response questions · module 6: matrices – question 4 – continued question 3 the...

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