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Matrices, GEN2 2025 VCAA 12

The early learning centre contains three rooms, Nursery \((N)\), Toddler \((T)\) and Pre-kinder \((P)\) .

From one year to the next, children can move between rooms, stay in the same room, or may leave \((L)\) the centre. The following transition matrix, \(M\), shows the expected proportion of children who will move between categories or stay in the same category from one year to the next.

\begin{aligned}
& \quad \quad \quad  \quad \quad \textit{this year} \\
& \quad \quad \ \ \ \ N \quad \quad \ \  T \quad \quad P \quad \  L \\
M&=\begin{bmatrix}
0.25 & 0 & 0 & 0 \\
0.625 & 0.25 & 0 & 0 \\
0 & 0.625 & 0.1 & 0 \\
0.125 & 0.125 & 0.9 & 1
\end{bmatrix}\begin{array}{l}
N\\
T \\
P \\
P
\end{array}\quad \textit{next year}
\end{aligned}

  1. The number of children expected to be in each of the four categories, from one year to the next, can be calculated using the matrix recurrence relation
    1. \(S_{n+1}=M S_n\)
  2. where \(S_n\) represents the expected number of children in each of the four categories at the start of year \(n\).
  3. The state matrix \(S_{2024}\), shown below, gives the number of children in each category at the start of 2024.
      1. \begin{align*}S_{2024}=\left[\begin{array}{c}4 \\15 \\15 \\27\end{array}\right]\begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
  4. Find \(S_{2023}\).   (1 mark)

  5. From the start of 2025, new children commenced in the early learning centre at the start of each year.
  6. A new matrix recurrence relation for determining the expected number of children in each of the four categories from one year to the next is
      1. \begin{align*}
        S_{n+1}=M S_n+B
        \end{align*}
  7. where
      1. \begin{align*}B=\left[\begin{array}{c}12 \\5 \\10 \\0\end{array}\right] \begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
  8. gives the number of new children enrolled in each room of the early learning centre at the start of each year.
  9. Given the state matrix
    1. \begin{align*}S_{2024}=\left[\begin{array}{c}4 \\15 \\15 \\27\end{array}\right]\begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
  10. find the expected total number of children to be enrolled in the early learning centre at the start of 2026. Round your answer to the nearest whole number.   (1 mark)

Show Answers Only

a.   

\(S_{2023}=\begin{bmatrix}16 \\ 20 \\ 25 \\ 0\end{bmatrix}\)

b.    \(\text{Children enrolled (2026) =50}\)

Show Worked Solution

a.    \(S_{n+1}=M S_n \ \Rightarrow \ S_n=M^{-1} S_{n+1}\)

\(S_{2023}=\begin{bmatrix}0.25 & 0 & 0 & 0 \\ 0.625 & 0.25 & 0 & 0 \\ 0 & 0.625 & 0.1 & 0 \\ 0.125 & 0.125 & 0.9 & 1\end{bmatrix}^{-1}\begin{bmatrix}4 \\ 15 \\ 15 \\ 27\end{bmatrix}=\begin{bmatrix}16 \\ 20 \\ 25 \\ 0\end{bmatrix}\)

♦ Mean mark (a) 53%.

b.

\(S_{2025}=M S_{2024} + B=\begin{bmatrix}13 \\ 11.25 \\ 20.87 \\ 42.87\end{bmatrix}\)
 

\(S_{2026}=M S_{2025} + B=\begin{bmatrix}15.25 \\ 15.93 \\ 19.11 \\ 64.69\end{bmatrix}\)
 

\(\text{Children enrolled (2026)\(=15.25+15.93+19.11=50\) (nearest whole)}\)

♦♦♦ Mean mark (b) 13%.

Matrices, GEN2 2024 VCAA 12

When the construction company established the construction site at the beginning of 2023, it employed 390 staff to work on the site.

The staff comprised 330 construction workers \((C)\), 50 foremen \((F)\) and 10 managers \((M)\).

At the beginning of each year, staff can choose to stay in the same job, move to a different job on the site, or leave the site \((L)\) and not return.

The transition diagram below shows the proportion of staff who are expected to change their job at the site each year.
 

This situation can be modelled by the recurrence relation

\(S_{n+1}=T S_n\), where

\(T\) is the transitional matrix, \(S_0=\left[\begin{array}{c}330 \\ 50 \\ 10 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L \end{aligned}\)  and \(n\) is the number of years after 2023.

  1. Calculate the predicted percentage decrease in the number of foremen \((F)\) on the site from 2023 to 2025.   (1 mark)

  2. Determine the total number of staff on the site in the long term.  (1 mark)

To encourage more construction workers \((C)\) to stay, the construction company has given workers an incentive to move into the job of foreman \((F)\).

Matrix \(R\) below shows the ways in which staff are expected to change their jobs from year to year with this new incentive in place.

\begin{aligned}
& \quad \quad \ \ \textit{this year} \\
& \quad  C \quad  \ \ F \quad  \ \  M \quad L\\
R = & \begin{bmatrix}
0.4 & 0.2 & 0 & 0 \\
0.4 & 0.2 & 0.4 & 0 \\
0 & 0.2 & 0.3 & 0 \\
0.2 & 0.4 & 0.3 & 1
\end{bmatrix}\begin{array}{l}
C\\
F\\
M\\
L
\end{array} \quad \textit{next year}
\end{aligned}

The site always requires at least 330 construction workers.

To ensure that this happens, the company hires an additional 190 construction workers \((C)\) at the beginning of 2024 and each year thereafter.

The matrix  \(V_{n+1}\)  will then be given by

\(V_{n+1}=R V_n+Z\), where

\(V_0=\left[\begin{array}{c}330 \\ 50 \\ 10 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L\end{aligned} \quad\quad\quad Z=\left[\begin{array}{c}190 \\ 0 \\ 0 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L\end{aligned} \ \ \) and \(n\) is the number of years after 2023.

  1. How many more staff are there on the site in 2024 than there were in 2023 ?    (1 mark)

  2. Based on this new model, the company has realised that in the long term there will be more than 200 foremen on site.
  3. In which year will the number of foremen first be above 200?   (1 mark)

Show Answers Only

a.    \(14\%\)

b.    \(\text{Zero}\)

c.    \(101\)

d.    \(2027\)

Show Worked Solution

a.    \(\text{Using CAS:}\)

\(\text{Transition matrix}(T):\ \begin{aligned}
& \quad \quad \quad \ \ \ \ \textit{from} \\
& \quad \ \ \ \  C \ \ \ \ \  F \ \ \ \ \ \  M \ \ \ \ \ \  L \ \ \\
\ \ \textit{to}\ \ \ & \begin{array}{l}
C \\
F \\
M \\
L
\end{array}\begin{bmatrix}
0.3 & 0.2 & 0 & 0 \\
0.2 & 0.2 & 0.4 & 0 \\
0 & 0.2 & 0.3 & 0 \\
0.5 & 0.4 & 0.3 & 1
\end{bmatrix}
\end{aligned}\)

♦♦♦ Mean mark (a) 25%.

\(S_{2023}=\begin{bmatrix}
330 \\
50 \\
10 \\
0 \\
\end{bmatrix}\ \ ,\ \  S_{2024}=T\times S_{2023}=\begin{bmatrix}
109 \\
80 \\
13 \\
188 \\
\end{bmatrix}\)

\( S_{2025}=T\times S_{2024}=\begin{bmatrix}
48.7 \\
43 \\
19.9 \\
278.4 \\
\end{bmatrix}\)

 
\(\text{% decrease in foremen}\ (F)=\dfrac{50-43}{50}\times 100\%=14\%\)
 

b.   \(\text{Test for 10 years:}\)

\(\text{Using CAS:}\)

\(S_{2024}=T^{10}\times S_{2023}=\begin{bmatrix}
0.490748 \\
0.734232 \\
0.488858 \\
388.286 \\
\end{bmatrix}\)

 
\(\text{In the long term, there will be zero employees on site.}\)
 

♦♦♦ Mean mark (b) 24%.

c.    \(\text{Using CAS:}\)

\(V_{2024}=R\times V_{2023}+Z=\begin{bmatrix}
332 \\
146 \\
13 \\
89 \\
\end{bmatrix}\)
  

\(\text{Difference in staff from}\ 2023-2024\)

\(=(332+146+13)-390\)

\(=101\ \text{more staff.}\)
 

\(\text{NOTE: 89 not included in calculation as these are the staff}\)

\(\text{who have left the company during the year.}\)
 

♦♦♦ Mean mark (c) 12%.

d.   \(\text{Using CAS:}\)

\(V_{2024}=R\times V_{2023}+Z=\begin{bmatrix}
332 \\
146 \\
13 \\
89 \\
\end{bmatrix}\ \ ,\ \  V_{2025}=R\times V_{2024}+Z=\begin{bmatrix}
352 \\
167.2 \\
33.1 \\
217.7 \\
\end{bmatrix}\ \\\)

♦♦♦ Mean mark (d) 26%.

\(V_{2026}=R\times V_{2025}+Z=\begin{bmatrix}
364.24\\
187.48 \\
43.37 \\
364.91 \\
\end{bmatrix}\ \ ,\ \  V_{2027}=R\times V_{2026}+Z=\begin{bmatrix}
373.192 \\
200.54 \\
50.507 \\
525.761 \\
\end{bmatrix}\ \\\)

 
\(\therefore\ \text{In 2027 the number of foremen will be over 200.}\)

Matrices, GEN1 2024 VCAA 32 MC

A large sporting event is held over a period of four consecutive days: Thursday, Friday, Saturday and Sunday.

People can watch the event at four different sites throughout the city: Botanical Gardens \((G)\), City Square \((C)\), Riverbank \((R)\) or Main Beach \((M)\).

Let \(S_n\) be the state matrix that shows the number of people at each location \(n\) days after Thursday.

The expected number of people at each location can be determined by the matrix recurrence rule

\(S_{n+1}=TS_n+A\)

\begin{aligned}
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \textit{this day} \\
& \quad \quad \quad \quad \quad \quad \quad \ G \quad \ \  C \quad \ \ R \quad \ \  M \\
& \text{where} \quad T=\begin{bmatrix}
0.4 & 0.2 & 0.4 & 0 \\
0.4 & 0.1 & 0.3 & 0.3 \\
0.1 & 0.4 & 0.1 & 0.2 \\
0.1 & 0.3 & 0.2 & 0.5
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array} \text { next day } \quad \text{and}& A=\begin{bmatrix}
300 \\
200\\
100 \\
300
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array}
\end{aligned}

 

\begin{aligned} \text{Given the state matrix}& \quad \quad S_3=\begin{bmatrix}
5620\\
6386\\
4892\\
6902
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array}
\end{aligned}

the number of people watching the event at the Botanical Gardens \((G)\) from Thursday to Sunday has

  1. decreased by 162
  2. decreased by 212
  3. increased by 124
  4. increased by 696
Show Answers Only

\(D\)

Show Worked Solution
\(S_{n+1}\) \(=TS_n+A\)
\(TS_n\) \(=S_{n+1}-A\)
\(S_n\)  \(=T^{-1}\times \left(S_{n+1}-A\right)\)
♦♦♦ Mean mark 33%.

  
\(\text{Using CAS:}\)
  
\(T =\begin{bmatrix}
\ 0.4 & 0.2 & 0.4 & 0 \\
\ 0.4 & 0.1 & 0.3 & 0.3 \\
\ 0.1 & 0.4 & 0.1 & 0.2 \\
\ 0.1 & 0.3 & 0.2 & 0.5 
\end{bmatrix}\ \rightarrow \ \ T^{-1}=
 \begin{bmatrix}
\ -2.6 & 5.4 & 3.4 & -4.6 \\
\ 0.2 & -0.8 & 3.2 & -0.8 \\
\ 5 & -5 & -5 & 5 \\
\ -1.6 & 1.4 & -0.6 & 1.4   
\end{bmatrix}\)

 

\(\text{Sunday} =S_3=\begin{bmatrix}
\ 5620 \\
\ 6386 \\
\ 4892  \\
\ 6902
\end{bmatrix} \)

\(\text{Saturday} =S_2=T^{-1}\times \left(S_{3}-A\right)=\begin{bmatrix}
\ 5496 \\
\ 6168 \\
\ 4720  \\
\ 6516
\end{bmatrix} \)

\(\text{Friday} =S_1=T^{-1}\times \left(S_{2}-A\right)=\begin{bmatrix}
\ 5832 \\
\ 6076 \\
\ 4120  \\
\ 5972
\end{bmatrix} \)

\(\text{Thursday} =S_0=T^{-1}\times \left(S_{1}-A\right)=\begin{bmatrix}
\ 4924 \\
\ 4732 \\
\ 6540  \\
\ 4904
\end{bmatrix} \)

  
\(5620-4924=696\)

\(\therefore\ \text{Botanical Gardens attendance has increased by 696 people.}\)

\(\Rightarrow D\)

MATRICES, FUR2 2021 VCAA 3

A market research study of shoppers showed that the buying preferences for the three olive oils, Carmani (`C`), Linelli (`L`) and Ohana (`O`), change from month to month according to the transition matrix  `T` below.

`qquadqquadqquadqquad \ text(this month)`

`T = {:(qquad\ C quadquadqquad \ L quadquad \ O ),([(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:(C),(L),(O):} qquad text(next month)):}`
 

The initial state matrix `S_0` below shows the number of shoppers who bought each brand of olive oil in July 2021.

`S_0 = {:[(3200),(2000),(2800)]{:(C),(L),(O):} :}`

Let `S_n` represent the state matrix describing the number of shoppers buying each brand `n` months after July 2021.

  1. How many of these 8000 shoppers bought a different brand of olive oil in August 2021 from the brand bought in July 2021?   (1 mark)

  2. Using the rule `S_(n+1) = T xx S_(n)`, complete the matrix `S_1` below.   (1 mark)

`S_1 = {:[(3060),(text{_____}),(text{_____})]{:(C),(L),(O):} :}`

  1. Consider the shoppers who were expected to buy Carmani olive oil in August 2021.
  2. What percentage of these shoppers also bought Carmani olive oil in July 2021?
  3. Round your answer to the nearest percentage.   (1 mark)

  4. Write a calculation that shows Ohana olive oil is the brand bought by 50% of these shoppers in the long run.   (1 mark)

  5. Further research suggests more shoppers will buy olive oil in the coming months.
  6. A rule to model this situation is  `R_(n+1) = T xx R_n + B`, where `R_n` represents the state matrix describing the number of shoppers `n` months after July 2021.

`qquadqquadqquadqquad \ text(this month)`
    `T = {:(qquad\ C quadquadqquad \ L quadquad \ O ),([(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:(C),(L),(O):} ):} qquad text(next month) \ , \ B = {:[(200),(100),(k)]{:(C),(L),(O):} :}, \ R_0 = {:[(3200),(2000),(2800)]{:(C),(L),(O):} :}`

  1. `k` represents the extra number of shoppers expected to buy Ohana olive oil each month.
  2. If  `R_2 = {:[(3333),(2025),(3642)]:}`, what is the value of `k`?   (1 mark)

Show Answers Only
  1. `1160`
  2. `L = 1900 \ , \ O = 3040`
  3. `89text(%)`
  4. `text(See Worked Solutions)`
  5. `200`
Show Worked Solution

a.    `text{Consider matrix} \ T`

`text{Carmani – 15% bought new brand}`

`text{Linelli – 20%, Ohana – 10%}`

`:.  text{Shoppers}` `= 0.15 xx 3200 xx + 0.2 xx 2000 + 0.1 xx 2800`
  `= 1160`

 
b.    `S_1 = [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:[(3200),(2000),(2800)]:} = {:[(3060),(1900),(3040)]:}`
 

`:. \ L = 1900 \ , \ O = 3040`
 

c.    `text{Carmani purchasers in August)} \ = 3060 \ text{(see part b)}`

`text{Carmani purchasers in July} = 3200`

`text{Carmani purchasers in both July and August}`

`= 0.85 xx 3200`

`= 2720`
 

`:.\ text{% of August purchasers who bought in July}`

`= 2720/3060 xx 100`

`= 88.88 …`

`=89text(%)`
 

d.    `S = T^50 xx S_0 = {:[(2400),(1600),(4000)]:}`

`text{Total shoppers} = 8000`

`:.\ text{Ohana purchasers (in long run)}`

`= 4000/8000 xx 100`

`= 50text(%)`
 

e.     `R_1` `= [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)] [(3200),(2000),(2800)] + [(200),(180),(k)]` 
  `R_2` `= [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)] [(3260),(2000),(k+3040)] + [(200),(100),(k)]`
    `= [(3171 + 0.05 (k + 3040)),(text{not required}),(text{not required})]`

 
`text{Equating matrices, solve for} \ k:`

`3171 + 0.05 (k + 3040) = 3333`

`:. k = 200`

MATRICES, FUR1 2021 VCAA 7 MC

The matrix  `S_{n + 1}`  is determined from the matrix  `S_n`  using the recurrence relation  `S_{n + 1} = T xx S_n - C`, where

`T= [(0.6,0.1,0.3),(0.3,0.8,0.2),(0.1,0.1,0.5)] , qquad S_0 = [(21),(51),(31)], qquad S_1 = [(24.0),(54.3),(20.7)] `

and  `C` is a column matrix. 

Matrix  `S_2`  is equal to

A.  `[(23.04),(55.78),(16.18)]` B.  `[(25.34),(56.28),(17.38)]`  
     
C.  `[(26.04),(54.78),(18.18)]` D.  `[(28.34),(55.28),(19.38)]`  
     
E.  `[(29.04),(53.78),(20.18)]`    
Show Answers Only

`A`

Show Worked Solution

`S_1 = TS_0 – C`

♦ Mean mark 40%.
`[(24.0),(54.3),(20.7)]` `= [(0.6,0.1,0.3),(0.3,0.8,0.2),(0.1,0.1,0.5)] [(21),(51),(31)] – C`
`[(24.0),(54.3),(20.7)]` `= [(27),(53.3),(22.7)] – C`
`C` `= [(27),(53.3),(22.7)] – [(24.0),(54.3),(20.7)] = [(3),(-1),(2)]`

 
`S_2 = TS_1 – C`

`S_2` `= [(0.6,0.1,0.3),(0.3,0.8,0.2),(0.1,0.1,0.5)] [(24.0),(54.3),(20.7)] – [(3),(-1),(2)]`
  `= [(23.04),(55.78),(16.18)]`

`=> A`

MATRICES, FUR2 2020 VCAA 4

A second market research project also suggested that if the Westmall shopping centre were sold, each of the three centres (Westmall, Grandmall and Eastmall) would continue to have regular shoppers but would attract and lose shoppers on a weekly basis.

Let `R_n` be the state matrix that shows the expected number of shoppers at each of the three centres `n` weeks after Westmall is sold.

A matrix recurrence relation that generates values of `R_n` is

`R_(n+1) = TR_n + B`

`{:(quad qquad qquad qquad qquad qquad qquad qquad text(this week)),(qquad qquad qquad qquad qquad qquad quad \ W qquad quad G qquad quad \ E),(text(where)\ T = [(quad 0.78, 0.13, 0.10),(quad 0.12, 0.82, 0.10),(quad 0.10, 0.05, 0.80)]{:(W),(G),(E):}\ text(next week,) qquad qquad  B = [(-400), (700), (500)]{:(W),(G),(E):}):}`
 

The matrix `R_2` is the state matrix that shows the expected number of shoppers at each of the three centres in the second week after Westmall is sold

`R_2 = [(239\ 060), (250\ 840), (192\ 900)]{:(W),(G),(E):}`

  1. Determine the expected number of shoppers at Westmall in the third week after it is sold.   (1 mark)

  2. Determine the expected number of shoppers at Westmall in the first week after it is sold.   (1 mark)

Show Answers Only
  1. `237\ 966`
  2. `241\ 000`
Show Worked Solution

♦ Mean mark part (a) 50%.
a.   `R_3` `= TR_2 + B`
    `= [(0.78, 0.13, 0.1),(0.12, 0.82, 0.1),(0.10, 0.05, 0.8)][(239\ 060),(250\ 840),(192\ 900)]+[(-400),(700),(500)] = [(237\ 966),(254\ 366),(191\ 268)]`

 
`:. text(Expected Westmall shoppers) = 237\ 966`
 

♦♦♦ Mean mark part (b) 20%.
b.   `R_2` `= TR_1 + B`
  `R_1` `= T^(-1)[R_2-B]`
    `= [(241\ 000), (246\ 000), (195\ 000)]`

 
`:. text(Expected Westmall shoppers) = 241\ 000`

MATRICES, FUR2-NHT 2019 VCAA 4

After 5.00 pm, tourists will start to arrive in Gillen and they will stay overnight.

As a result, the number of people in Gillen will increase and the television viewing habits of the tourists will also be monitored.

Assume that 50 tourists arrive every hour.

It is expected that 80% of arriving tourists will watch only `C_2` during the hour that they arrive.

The remaining 20% of arriving tourists will not watch television during the hour that they arrive.

Let `W_m` be the state matrix that shows the number of people in each category `m` hours after 5.00 pm on this day.

The recurrence relation that models the change in the television viewing habits of this increasing number of people in Gillen `m` hours after 5.00 pm on this day is shown below.

`W_(m + 1) = TW_m + V` 

where

`{:(quad qquad qquad qquadqquadqquadquadtext(this hour)),(qquadqquadqquad quad \ C_1 qquad quad C_2 qquad \ C_3 quad \ NoTV),(T = [(quad 0.50, 0.05, 0.10, 0.20 quad),(quad 0.10, 0.60, 0.20, 0.20 quad),(quad 0.25, 0.10, 0.50, 0.10 quad),(quad 0.15, 0.25, 0.20, 0.50 quad)]{:(C_1),(C_2),(C_3),(NoTV):}\ text(next hour,) qquad and qquad  W_0 = [(400), (600), (300),(700)]):}`
 

  1. Write down matrix `V`.   (1 mark)

  2. How many people in Gillen are expected to watch `C_2` at 7.00 pm on this day?   (2 marks)

Show Answers Only
  1.  `V=[(0),(40),(0),(10)]`
  2.  `666`
Show Worked Solution

a.   `V=[(0),(40),(0),(10)]`

b.    `W_1` `=TW_0+V`
    `=[(quad 0.50, 0.05, 0.10, 0.20 quad),(quad 0.10, 0.60, 0.20, 0.20 quad),(quad 0.25, 0.10, 0.50, 0.10 quad),(quad 0.15, 0.25, 0.20, 0.50 quad)] [(400),(600),(300),(700)]+[(0),(40),(0),(10)]=[(400),(640),(380),(630)]`
     
  `W_2` `=TW_1+V=[(396),(666),(417),(621)]`

 
`:.\ text(666 people are expected to watch)\ C_2\ text(at 7 pm.)`

MATRICES, FUR2 2019 VCAA 3

On Sunday, matrix `V` is used when calculating the expected number of visitors at each location every hour after 10 am. It is assumed that the park will be at its capacity of 2000 visitors for all of Sunday.

Let `L_0` be the state matrix that shows the number of visitors at each location at 10 am on Sunday.

The number of visitors expected at each location at 11 am on Sunday can be determined by the matrix product

 
`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad text(this hour)),(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad qquad qquad  \ A qquad quad F qquad \  G \ quad quad W),({:V xx L_0 qquad text(where) qquad L_0 = [(500), (600), (500), (400)]{:(A),(F),(G),(W):}, qquad text(and):} qquad V = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`
 

  1. Safety restrictions require that all four locations have a maximum of 600 visitors.
  2. Which location is expected to have more than 600 visitors at 11 am on Sunday?   (1 mark) 
  3. Whenever more than 600 visitors are expected to be at a location on Sunday, the first 600 visitors can stay at that location and all others will be moved directly to Ground World `(G)`.
  4. State matrix `R_n` contains the number of visitors at each location `n` hours after 10 am on Sunday, after the safety restrictions have been enforced.
  5. Matrix `R_1` can be determined from the matrix recurrence relation
  6. `qquad qquad qquad R_0 = [(500),(600),(500),(400)]{:(A),(F),(G),(W):}, qquad qquad R_1 = V xx R_0 + B_1`
  7. where matrix `B_1` shows the required movement of visitors at 11 am.
    1. Determine the matrix `B_1`.   (1 mark)

    2. State matrix `R_2` can be determined from the new matrix rule
    3. `qquad qquad R_2 = VR_1 + B_2`
    4. where matrix `B_2` shows the required movement of visitors at 12 noon.
    5. Determine the state matrix `R_2`.   (1 mark)

Show Answers Only
  1. `text(Location)\ A`
  2.  i. `B_1 = [(-210),(0),(210),(0)]`
    ii. `R_2 = [(600),(288),(512),(600)]`
Show Worked Solution
a.   `text{By inspection (higher decimal values in row 1)}`
  `=>\ text(test location)\ A`
  `text(Visitors at)\ A\ text{(11 am)}` `= 0.3 xx 500 + 0.4 xx 600 + 0.6 xx 500 + 0.3 xx 400`
    `= 810`

 
`text(Location)\ A\ text(will have more than 600 visitors.)`

 

b.i.   `VR_0` `= [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(500),(600),(500),(400)]=[(810),(300),(310),(580)]`
     
  `R_1` `= V xx R_0 + B_1`
    `= [(810),(300),(310),(580)] + [(-210),(0),(210),(0)]`
  `:. B_1` `= [(-210),(0),(210),(0)]`

 

b.ii.   `R_2` `= [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(600),(300),(520),(580)] + B_2`
    `= [(786),(288),(282),(644)] + [(-186),(0),(230),(-44)]= [(600),(288),(512),(600)]`

MATRICES, FUR1 2018 VCAA 7 MC

A study of the antelope population in a wildlife park has shown that antelope regularly move between three locations, east (`E`), north (`N`) and west (`W`).

Let  `A_n` be the state matrix that shows the population of antelope in each location `n` months after the study began.

The expected population of antelope in each location can be determined by the matrix recurrence rule
 

`A_(n + 1) = T A_n - D`
 

where

`{:(),(),(T=):}{:(qquadquadtext(this month)),((qquadE,quadN,quadW)),([(0.4,0.2,0.2),(0.3,0.6,0.3),(0.3,0.2,0.5)]):}{:(),(),({:(E),(N),(W):}):}{:(),(),(text(next month)):}`

and

`D = [(50),(50),(50)]{:(E),(N),(W):}`
 

The state matrix, `A_3`, below shows the population of antelope three months after the study began.
 

`A_3 = [(1616),(2800),(2134)]{:(E),(N),(W):}`
 

The number of antelope in the west (`W`) location two months after the study began, as found in the state matrix `A_2`, is closest to

  1. 2060
  2. 2130
  3. 2200
  4. 2240
  5. 2270
Show Answers Only

`E`

Show Worked Solution

`A_(n + 1) = TA_n – D\ \ (text(given))`

♦♦ Mean mark 28%.

`A_3` `= TA_2 – D`
`TA_2` `= [(1616),(2800),(2134)] + [(50),(50),(50)] = [(1666),(2850),(2184)]`
`A_2` `= [(0.4,0.2,0.2),(0.3,0.6,0.3),(0.3,0.2,0.5)]^(−1)[(1666),(2850),(2184)] = [(1630),(2800),(2270)]`

 
`:.\ text(Antelope population in the west = 2270)`

`=> E`

MATRICES, FUR2 2018 VCAA 4

Beginning in the year 2021, a new company takes over the maintenance of the 2700 km highway with a new contract.

Each year sections of highway must be graded `(G)`, resurfaced `(R)` or sealed `(S)`.

The remaining highway will need no maintenance `(N)` that year.

Let  `M_n` be the state matrix that shows the highway maintenance schedule of this company for the `n`th year after 2020.

The maintenance schedule for 2020 is shown in matrix  `M_0` below.

For these 2700 km of highway, the matrix recurrence relation shown below can be used to determine the number of kilometres of this highway that will require each type of maintenance from year to year.
 
`qquad qquad M_(n + 1) = TM_n + B`

where

  `{:(\ \ qquad qquad qquad qquad quad text(this year)),(qquad qquad quad quad G qquad quad R qquad quad S quad quad \ N):}`  
`M_0 = [(500),(400),(300),(1500)]{:(G),(R),(S),(N):}text (,)` `T = [(0.2,0.1,0.0,0.2),(0.1,0.1,0.0,0.2),(0.2,0.1,0.2,0.1),(0.5, 0.7,0.8,0.5)]{:(G),(R),(S),(N):} \ text(next year,)` `B = [(k),(20),(10),(-60)]`

 

  1. Write down the value of `k` in matrix `B`.   (1 mark)

  2. How many kilometres of highway are expected to be graded `(G)` in the year 2022?   (1 mark)

Show Answers Only
  1. `k =30`
  2. `457\ text(km)`
Show Worked Solution

a.    `text(S) text(ince the length of highway is constant:)`

♦ Mean mark 39%.

`k + 20 + 10-60` `= 0`
`k` `= 30`

 

♦♦ Mean mark 23%.

b.    `M_1 (2021)` `= TM_0 + B= [(470),(410),(360),(1460)]`
     
  `M_2 (2022)` `= TM_1 + B= [(457),(400),(363),(1480)]`

 
`:. 457\ text(km are expected to be graded in 2022.)`

MATRICES, FUR1 2017 VCAA 7 MC

At a fish farm:

    • young fish (`Y`) may eventually grow into juveniles (`J`) or they may die (`D`)
    •  juveniles (`J`) may eventually grow into adults (`A`) or they may die (`D`)
    • adults (`A`) eventually die (`D`).

The initial state of this population, `F_0`, is shown below.
 

`F_0 = [(50\ 000),(10\ 000),(7000),(0)]{:(Y),(J),(A),(D):}`

 

Every month, fish are either sold or bought so that the number of young, juvenile and adult fish in the farm remains constant.

The population of fish in the fish farm after `n` months, `F_n`, can be determined by the recurrence rule
 

`F_(n + 1) = [(0.65,0,0,0),(0.25,0.75,0,0),(0,0.20,0.95,0),(0.10,0.05,0.05,1)]\ F_n + B`
 

where `B` is a column matrix that shows the number of young, juvenile and adult fish bought or sold each month and the number of dead fish that are removed.

Each month, the fish farm will

  1. sell 1650 adult fish.
  2. buy 1750 adult fish.
  3. sell 17 500 young fish.
  4. buy 50 000 young fish.
  5. buy 10 000 juvenile fish.
Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince the number of each type of fish remains constant,)`

`F_0 = F_1 = F_n = F_(n + 1)`

 

`[(50\ 000),(10\ 000),(7000),(0)]` `= [(0.65,0,0,0),(0.25,0.75,0,0),(0,0.20,0.95,0),(0.10,0.05,0.05,1)][(50\ 000),(10\ 000),(7000),(0)] + B`
`:. B` `= [(50\ 000),(10\ 000),(7000),(0)] – [(32\ 500),(20\ 000),(8650),(5850)]`
  `= [(17\ 500),(−10\ 000),(−1650),(−5850)]{:(Y),(J),(A),(D):}`

 
`=> A`

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

  5. Consider the customers who were expected to choose air travel in 2015.
  6. What percentage of these customers had also chosen air travel in 2014?
  7. 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):}`

  1. The company intends to increase the number of customers in the study in 2017 and in 2018.
  2. 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.
    2. Explain this number in the context of selecting customers for the studies in 2017 and 2018.   (1 mark)

    3. Determine the number of customers who are expected to choose sea travel in 2018.
    4. 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(%)`
  5.  
    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.
    5. When does this first occur?   (1 mark)

    6. 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.
  4. Assuming 30% of adults lay eggs each week, the population matrix after one week, `S_1`, is now given by
  5. `qquad S_1 = TS_0 + BS_0`
  6. 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)

    2. This pattern continues. The population matrix after `n` weeks, `S_n`, is given by
    3. `qquad qquad qquad S_n = TS_(n - 1) + BS_(n - 1)`
    4. Determine the number of live eggs in this insect population after two weeks.  (1 mark)

Show Answers Only
  1. `700`
  2. `50text(%)`
  3.  
    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.)`

MATRICES, FUR2 2009 VCAA 4

A 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_(n + 1) = [(0.85,0.25),(0.15,0.75)] xx L_n-[(5),(7)]`

where `L_n` 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

`L_1 = [(95),(97)]{:(text(dancing)),(text(singing)):}`

  1. Calculate the number of students who are expected to attend the extra singing rehearsals in week 3.   (1 mark)

  2. 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)

Show Answers Only
  1. `text(68 students)`
  2. `12`
Show Worked Solution

a.   `text(Using matrix equation,)`

`L_2` `= T xx L_1-[(5),(7)]`
  `= [(0.85,0.25),(0.15,0.75)][(95),(97)]-[(5),(7)]= [(100),(80)]`
`L_3` `= [(0.85,0.25),(0.15,0.75)][(100),(80)]-[(5),(7)]= [(100),(68)]`

 
`:.\ text(68 students are expected to attend singing in week 3.)`
 

b.    `L_4` `= [(0.85,0.25),(0.15,0.75)][(100),(68)]-[(5),(7)]= [(97),(59)]`

 
`:.\ text(Students expected not to return)`

`= (100 + 68)-(97 + 59)`

`= 12`

MATRICES, FUR2 2013 VCAA 2

10 000 trout eggs, 1000 baby trout and 800 adult trout are placed in a pond to establish a trout population.

In establishing this population

    • eggs (`E`) may die (`D`) or they may live and eventually become baby trout (`B`)
    • baby trout (`B`) may die (`D`) or they may live and eventually become adult trout (`A`)
    • adult trout (`A`) may die (`D`) or they may live for a period of time but will eventually die.

From year to year, this situation can be represented by the transition matrix `T`, where
 

`{:(qquadqquadqquadqquadqquadtext(this year)),((qquadqquadqquadE,quad\ B,quad\ A,\ D)),(T = [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)]):}{:(),(),(E),(B),(A),(D):}{:(),(),(qquadtext(next year)):}`
 

  1. Use the information in the transition matrix `T` to
    1. determine the number of eggs in this population that die in the first year.   (1 mark)

    2. complete the transition diagram below, showing the relevant percentages.   (2 marks)

       

          Matrices, FUR2 2013 VCAA 2_a

The initial state matrix for this trout population, `S_0`, can be written as
 

`S_0 = [(10\ 000),(1000),(800),(0)]{:(E),(B),(A),(D):}`
 

Let `S_n` represent the state matrix describing the trout population after `n` years.

  1. Using the rule  `S_n = T S_(n-1)`, determine each of the following.

     

    1. `S_1`   (1 mark)

    2. the number of adult trout predicted to be in the population after four years   (1 mark)

  2. The transition matrix `T` predicts that, in the long term, all of the eggs, baby trout and adult trout will die.
    1. How many years will it take for all of the adult trout to die (that is, when the number of adult trout in the population is first predicted to be less than one)?   (1 mark)

    2. What is the largest number of adult trout that is predicted to be in the pond in any one year?   (1 mark)

  3. Determine the number of eggs, baby trout and adult trout that, if added to or removed from the pond at the end of each year, will ensure that the number of eggs, baby trout and adult trout in the population remains constant from year to year.   (2 marks)

The rule  `S_n = T S_(n – 1)`  that was used to describe the development of the trout in this pond does not take into account new eggs added to the population when the adult trout begin to breed.

  1. To take breeding into account, assume that 50% of the adult trout lay 500 eggs each year.
  2. The matrix describing the population after one year, `S_1`, is now given by the new rule
  3. `S_1 = T S_0 + 500\ M\ S_0`
  4. where      `T=[(0,0,0,0),(0.40,0,0,0),(0,0.25,0.50,0),(0.60,0.75,0.50,1.0)], M=[(0,0,0.50,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]\ text(and)\ S_0=[(10\ 000),(1000),(800),(0)]`
    1. Use this new rule to determine `S_1`.   (1 mark)

  5. This pattern continues so that the matrix describing the population after `n` years, `S_n`, is given by the rule
  6.       `S_n = T\ S_(n-1) + 500\ M\ S_(n-1)`
     

    1. Use this rule to determine the number of eggs in the population after two years   (2 marks)

Show Answers Only

a.i.   `6000`

a.ii.  `text(See Worked Solutions)`

b.i.   `S_1= [(0),(4000),(650),(7150)]`

b.ii. `S_4= [(0),(0),(331.25),(11\ 468.75)]`

c.i.   `text{13 years}`

c.ii.  `1325`

d.    `text(Add 10 000 eggs, remove 3000 baby trout and add 150)`

`text(150 adult trout to keep the population constant.)`

e.i.   `S_1= [(200\ 000),(4000),(650),(7150)]`

e.ii.   `text(See Worked Solutions)`

Show Worked Solution

a.i.   `text(60% of eggs die in 1st year,)`

`:.\ text(Eggs that die in year 1)`

`= 0.60 xx 10\ 000`

`= 6000`
 

MARKER’S COMMENT: A 100% cycle drawn at `D` was a common omission. Do not draw loops and edges of 0%!
a.ii.   

Matrices-FUR2-2013-VCAA-2_a Answer
b.i.    `S_1` `= TS_0`
    `= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)][(10\ 000),(1000),(800),(0)]= [(0),(4000),(650),(7150)]`

 

b.ii.    `S_4` `= T^4S_0`
    `= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)]^4[(10\ 000),(1000),(800),(0)]= [(0),(0),(331.25),(11\ 468.75)]`

 

`:. 331\ text(trout is the predicted population after 4 years.)`

 

c.i.    `S_12 = T^12S_0 = [(0),(0),(1.29),(11\ 791)]`

`S_13 = T^13S_0 = [(0),(0),(0.65),(11\ 799)]`
 

`:.\ text{It will take 13 years (when the trout population drops below 1).}`
 

c.ii.    `S_1 = TS_0 = [(0),(4000),(650),(7150)]`

`text(After 1 year, 650 adult trout.)`

`text(Similarly,)`

`S_2 = T^2S_0 = [(0),(0),(1325),(10\ 475)]`

`S_3 = T^3S_0 = [(0),(0),(662.5),(11\ 137.5)]`

`S_4 = T^4S_0 = [(0),(0),(331),(11\ 469)]`
 

`:.\ text(Largest number of adult trout = 1325.)`
 

d.    `S_0-S_1 = [(10\ 000),(1000),(800),(0)]-[(0),(4000),(650),(7150)] = [(10\ 000),(−3000),(150),(−7150)]`

 

`:.\ text(Add 10 000 eggs, remove 3000 baby trout and add 150 adult)`

`text(trout to keep the population constant.)`

 

e.i.    `S_1` `= TS_0 + 500MS_0`
    `= [(0),(4000),(650),(7150)] + 500 xx [(0,0,0.5,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(10\ 000),(1000),(800),(0)]`
    `= [(0),(4000),(650),(7150)] + 500[(400),(0),(0),(0)]`
    `= [(200\ 000),(4000),(650),(7150)]`

 

e.ii.    `S_2` `= TS_1 + 500MS_1`
   

`= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)][(200\ 000),(4000),(650),(7150)]`

       `+ 500 xx [(0,0,0.5,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(200\ 000),(4000),(650),(7150)]`
    `= [(162\ 500),(80\ 000),(1325),(130\ 475)]`

MATRICES, FUR1 2013 VCAA 8 MC

The matrix  `S_(n+1)`  is determined from the matrix  `S_n`  using the rule  `S_(n+1) = TS_n - C,` where  `T, S_0`  and  `C`  are defined as follows.
 

`T = [(0.5, 0.6), (0.5,0.4)], \ S_0 = [(100), (250)] quad text(and) quad C = [(20), (20)]`
 

Given this information, the matrix  `S_2` equals

A.   `[(100), (250)]`

B.   `[(148), (122)]`

C.   `[(170), (140)]`

D.   `[(180), (130)]`

E.   `[(190), (160)]`

Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 41%.
`S_1` `= TS_0 – C`
  `= [(0.5,0.6),(0.5,0.4)][(100),(250)] – [(20),(20)]`
  `= [(180),(130)]`

 

`S_2` `= TS_1 – C`
  `= [(0.5,0.6),(0.5,0.4)][(180),(130)] – [(20),(20)]`
  `= [(148),(122)]`

 
`rArr B`