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MATRICES, FUR2 2016 VCAA 3

A travel company is studying the choice between air (`A`), land (`L`), sea (`S`) or no (`N`) travel by some of its customers each year.

Matrix `T`, shown below, contains the percentages of customers who are expected to change their choice of travel from year to year.
 

`{:(qquadqquadqquadqquadqquadquadtext(this year)),(qquadqquadqquadquadAqquadqquadLqquadqquadSqquadquadN),(T = [(0.65,0.25,0.25,0.50),(0.15,0.60,0.20,0.15),(0.05,0.10,0.25,0.20),(0.15,0.05,0.30,0.15)]{:(A),(L),(S),(N):}text(next year)):}`
 

Let `S_n` be the matrix that shows the number of customers who choose each type of travel `n` years after 2014.

Matrix `S_0` below shows the number of customers who chose each type of travel in 2014.
 

`S_0 = [(520),(320),(80),(80)]{:(A),(L),(S),(N):}`
 

Matrix `S_1` below shows the number of customers who chose each type of travel in 2015.
 

`S_1 = TS_0 = [(478),(d),(e),(f)]{:(A),(L),(S),(N):}`
 

  1. Find the values missing from matrix `S_1 (d, e, f )`.  (1 mark)
  2. Write a calculation that shows that 478 customers were expected to choose air travel in 2015.  (1 mark)
  3. Consider the customers who chose sea travel in 2014.

     

    How many of these customers were expected to choose sea travel in 2015?  (1 mark)

  4. Consider the customers who were expected to choose air travel in 2015.

     

    What percentage of these customers had also chosen air travel in 2014?

     

    Round your answer to the nearest whole number.  (1 mark)

In 2016, the number of customers studied was increased to 1360.
Matrix `R_2016`, shown below, contains the number of these customers who chose each type of travel in 2016.
 

`R_2016 = [(646),(465),(164),(85)]{:(A),(L),(S),(N):}`
 

The company intends to increase the number of customers in the study in 2017 and in 2018.
The matrix that contains the number of customers who are expected to choose each type of travel in
2017 (`R_2017`) and 2018 (`R_2018`) can be determined using the matrix equations shown below.

 
`R_2017 = TR_2016 + BqquadqquadqquadR_2018 = TR_2017 + B`
 

`{:(qquadqquadqquadqquadqquadquadtext(this year)),(qquadqquadqquadquadAqquadqquadLqquadqquadSqquadquadN),(T = [(0.65,0.25,0.25,0.50),(0.15,0.60,0.20,0.15),(0.05,0.10,0.25,0.20),(0.15,0.05,0.30,0.15)]{:(A),(L),(S),(N):}text(next year)):}qquadqquad{:(),(),(B = [(80),(80),(40),(−80)]{:(A),(L),(S),(N):}):}`

 

    1. The element in the fourth row of matrix `B` is – 80.

       

      Explain this number in the context of selecting customers for the studies in 2017 and 2018.  (1 mark)

    2. Determine the number of customers who are expected to choose sea travel in 2018.

       

      Round your answer to the nearest whole number.  (2 marks)

Show Answers Only
  1. `d=298, \ e=94, \ f=130`
  2. `(0.65 xx 520) + (0.25 xx 320) + (0.25 xx 80) + (0.50 xx 80) = 478`
  3. `20\ text(customers)`
  4. `71text(%)`
    1. `text(80 customers who have no travel in a given)`
      `text(year are removed from the study. This occurs)`
      `text(in both 2017 and 2018.)`
    2. `text(190 customers)`
Show Worked Solution

a.   `d=298, \ e=94, \ f=130`
 

b.   `(0.65 xx 520) + (0.25 xx 320) + (0.25 xx 80) + (0.50 xx 80) = 478`

♦♦ Mean mark part (b) and (c) was 35% and 45% respectively.

 

c.   `text(Sea Travel in 2014 – 80 customers.)`

`text(Of those 80 customers,)`

`text(Sea Travel in 2015)` `= 25text(%) xx 80`
  `= 20\ text(customers)`

 

d.   `text(Expected total for air travel in 2015)`

♦♦♦ Mean mark 17%.

`= 478\ text(customers)`
 

`text(In 2014, 520 customers chose air travel.)`

`text(65% of those chose air travel in 2015)`

`= 65text(%) xx 520`

`= 338\ text(customers)`

`:.\ text(Percentage)` `= 338/478 xx 100`
  `= 70.71…`
  `= 71text(%)`

 

e.i.   `text(80 customers who have no travel in a given)`

♦ Mean mark 14%.
MARKER’S COMMENT: “80 people chose not to travel” was a common answer that received no marks.

`text(year are removed from the study. This occurs)`

`text(in both 2017 and 2018.)`

 

e.ii.   `R_2017` `= TR_2016 + B`
    `= [(0.65,0.25,0.25,0.50),(0.15,0.60,0.20,0.15),(0.05,0.10,0.25,0.20),(0.15,0.05,0.30,0.15)][(646),(465),(164),(85)] + [(80),(80),(40),(−80)]`
    `= [(699.65),(501.45),(176.80),(102.10)]`

 

`R_2018` `= TR_2017 + B`
  `= [(0.65,0.25,0.25,0.50),(0.15,0.60,0.20,0.15),(0.05,0.10,0.25,0.20),(0.15,0.05,0.30,0.15)][(699.65),(501.45),(176.80),(102.10)]`
  `= [(755.39),(536.49),(189.75),(118.38)]`

 

`:. 190\ text(customers are expected to choose)`

`text(sea travel in 2018.)`

MATRICES, FUR2 2007 VCAA 2

To 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
 

`S_0 = [(400),(200),(100),(0)]{:(E),(J),(A),(D):}`
 

A row has been included in the state matrix to allow for insects and eggs that die (`D`).

  1. 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
 

`{:(qquadqquadqquadqquadquadtext(this week)),((qquadqquadqquadE,quad\ J,quadA,\ D)),(T = [(0.4,0,0,0),(0.5,0.4,0,0),(0,0.5,0.8,0),(0.1,0.1,0.2,1)]{:(E),(J),(A),(D):}):}`
 

  1. What proportion of eggs turn into juveniles each week?  (1 mark)
    1. Evaluate the matrix product  `S_1 = TS_0`  (1 mark)
    2. Write down the number of live juveniles in the population after one week.  (1 mark)
    3. Determine the number of live juveniles in the population after four weeks. Write your answer correct to the nearest whole number.  (1 mark)
    4. After a number of weeks there will be no live eggs (less than one) left in the population.
      When does this first occur?  (1 mark)
    5. Write down the exact steady-state matrix for this population.  (1 mark)
  3. 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, `S_1`, is now given by
     
    `qquad S_1 = TS_0 + BS_0`
      
    where   `B = [(0,0,0.3,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]`   and   `S_0 = [(400),(200),(100),(0)]{:(E),(J),(A),(D):}`
     

    1. Determine `S_1`  (1 mark)
       

       

      This pattern continues. The population matrix after `n` weeks, `S_n`, is given by
       
      `qquad qquad qquad S_n = TS_(n - 1) + BS_(n - 1)`
       

    2. Determine the number of live eggs in this insect population after two weeks.  (1 mark)
Show Answers Only
  1. `700`
  2. `50text(%)`
    1. `[(160),(280),(180),(80)]{:(E),(J),(A),(D):}`
    2. `280`
    3. `56`
    4. `text(7th week)`
    5. `[(0),(0),(0),(700)]`
    1. `[(190),(280),(180),(80)]`
    2. `130`
Show Worked Solution

a.   `400 + 200 + 100 + 0 = 700`

 

b.   `50text(%)`

 

c.i.    `S_1` ` = TS_0`
    `= [(0.4,0,0,0),(0.5,0.4,0,0),(0,0.5,0.8,0),(0.1,0.1,0.2,1)][(400),(200),(100),(0)]`
    `= [(160),(280),(180),(80)]{:(E),(J),(A),(D):}`

 

c.ii.   `280`

 

c.iii.    `S_4` ` = T^4S_0`
    `= [(10.24),(56.32),(312.96),(320.48)]{:(E),(J),(A),(D):}\ \ \ text{(by graphics calculator)}`

 

`:. 56\ text(juveniles still alive after 4 weeks.)`

 

c.iv.   `text(Each week, only 40% of eggs remain.)`
 

`text(Find)\ \ n\ \ text(such that)`
`400 xx 0.4^n` `< 1`
`0.4^n` `<1/400`
`n` `> 6.5`

 

`:.\ text(After 7 weeks, no live eggs remain.)`

 

c.v.   `text(Consider)\ \ n\ \ text{large (say}\ \ n = 100 text{)},`

`[(0.4, 0, 0, 0), (0.5, 0.4, 0, 0), (0, 0.5, 0.8, 0), (0.1, 0.1, 0.2, 1)]^100 [(400), (200), (100), (0)] ~~ [(0), (0), (0), (700)]`

 

d.i.   `S_1` `= TS_0 + BS_0`
    `= [(160),(280),(180),(80)] + [(0,0,0.3,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(400),(200),(100),(0)]`
    `= [(190),(280),(180),(80)]`

 

♦♦ Mean mark for part (d) was 30%.
d.ii.   `S_2` `= TS_1 + BS_1`
    `= [(130), (207), (284), (163)]`

 

`:.\ text(There are 130 live egss after 2 weeks.)`