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Matrices, GEN1 2024 VCAA 29 MC

A tennis team consists of five players: Quinn, Rosie, Siobhan, Trinh and Ursula.

When the team competes, players compete in the order of first, then second, then third, then fourth.

The fifth player has a bye (does not compete).

On week 1 of the competition, the players competed in the following order.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad\textbf{First} \quad\rule[-1ex]{0pt}{0pt}& \quad \textbf{Second} \quad& \quad\textbf{Third} \quad& \quad\textbf{Fourth}\quad & \quad\textbf{Bye} \quad\\
\hline
\rule{0pt}{2.5ex} \text{Quinn} \rule[-1ex]{0pt}{0pt}& \text {Rosie} & \text {Siobhan} & \text { Trinh } & \text {Ursula} \\
\hline
\end{array}

This information can be represented by matrix \(G_1\), shown below.

\(G_1=\begin{bmatrix} Q & R & S & T & U \end{bmatrix}\)

Let \(G_n\) be the order of play in week \(n\).

The playing order changes each week and can be determined by the rule  \(G_{n+1}=G_n \times P\)

\(\text{where}\quad \\P=\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0  \\
0 & 0 & 0 & 1 & 0 \end{bmatrix}\)

Which player has a bye in week 4 ?

  1. Quinn
  2. Rosie
  3. Siobhan
  4. Trinh
Show Answers Only

\(C\)

Show Worked Solution

\(G_1=\begin{bmatrix} Q & R & S & T & U \end{bmatrix}\)

\(G_2=\begin{bmatrix} S & Q & T & U & R \end{bmatrix}\)

\(G_3=\begin{bmatrix} T & S & U & R & Q \end{bmatrix}\)

\(G_4=\begin{bmatrix} U & T & R & Q & S \end{bmatrix}\)

\(\text{5th player has a bye.}\)

\(\therefore\ \text{Siobhan does not compete in week 4}\)

\(\Rightarrow C\)

♦ Mean mark 49%.

Filed Under: Transition Matrices - Regular Tagged With: Band 5, smc-618-30-State Matrix in discrete period, smc-618-50-Permutation, smc-618-63-5x5 Matrix

MATRICES, FUR1 2022 VCAA 7 MC

Matrix `K` is a permutation matrix.  

`K = [(0,0,1,0,0),(0,1,0,0,0),(0,0,0,1,0),(0,0,0,0,1),(1,0,0,0,0)]`

Matrix `M` is a column matrix that is multiplied once by matrix `K` to obtain matrix `P`.

When matrix `M` is multiplied by matrix `K`, the element `m_31` moves to element

  1. `p_11`
  2. `p_21`
  3. `p_31`
  4. `p_41`
  5. `p_51`
Show Answers Only

`A`

Show Worked Solution

`text{The matrix product}\ KM\ text{is column matrix}\ P.`

`text{Row 1 of matrix}\ K\ text{moves the 3rd row of matrix}\ M\ text{to the}`

`text{first row of matrix}\ P.`

`=>A`

♦ Mean mark 42%.

Filed Under: Transition Matrices - Regular Tagged With: Band 5, smc-618-50-Permutation, smc-618-63-5x5 Matrix

MATRICES, FUR1 2021 VCAA 4 MC

Ramon and Norma are names that contain the same letters but in different order.

The permutation matrix that can change  `[(R),(A),(M),(O),(N)]` into `[(N),(O),(R),(M),(A)]`  is

A.  `[(0,0,0,0,1),(1,0,0,0,0),(0,0,1,0,0),(0,1,0,0,0),(0,0,0,1,0)]` B.  `[(0,0,0,0,1),(1,0,0,0,0),(0,0,1,0,0),(0,1,0,0,0),(0,0,0,1,0)]`  
     
C.  `[(1,0,0,0,0),(0,0,0,1,0),(0,0,0,0,1),(0,0,1,0,0),(0,1,0,0,0)]` D.  `[(0,0,0,0,1),(0,0,0,1,0),(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0)]`  
     
E.  `[(0,0,0,0,1),(0,0,0,1,0),(1,0,0,0,0),(0,0,1,0,0),(0,1,0,0,0)]`    
Show Answers Only

`E`

Show Worked Solution

`[(0,0,0,0,1),(0,0,0,1,0),(1,0,0,0,0),(0,0,1,0,0),(0,1,0,0,0)] [(R),(A),(M),(O),(N)] = [(N),(O),(R),(M),(A)]`
 

`=> E`

Filed Under: Matrix Applications Tagged With: Band 4, smc-618-50-Permutation, smc-618-63-5x5 Matrix

MATRICES, FUR2 2019 VCAA 2

The theme park has four locations, Air World `(A)`, Food World `(F)`, Ground World `(G)` and Water World `(W)`.

The number of visitors at each of the four locations is counted every hour.

By 10 am on Saturday the park had reached its capacity of 2000 visitors and could take no more visitors.

The park stayed at capacity until the end of the day

The state matrix, `S_0`, below, shows the number of visitors at each location at 10 am on Saturday.
 

`S_0 = [(600), (600), (400), (400)] {:(A),(F),(G),(W):}`
 

  1. What percentage of the park’s visitors were at Water World `(W)` at 10 am on Saturday?   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

Let `S_n` be the state matrix that shows the number of visitors expected at each location `n` hours after 10 am on Saturday.

The number of visitors expected at each location `n` hours after 10 am on Saturday can be determined by the matrix recurrence relation below.
 

`{:(qquad qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad text(  this hour)),(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad qquad qquad quad A qquad quad F qquad \  G \ quad quad W),({:S_0 = [(600), (600), (400), (400)], qquad S_(n+1) = T xx S_n quad quad qquad text(where):}\ T = [(0.1,0.2,0.1,0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`
 

  1. Complete the state matrix, `S_1`, below to show the number of visitors expected at each location at 11 am on Saturday.   (1 mark)

    --- 0 WORK AREA LINES (style=lined) ---

 
`S_1 = [(\ text{______}\ ), (\ text{______}\ ), (300),(\ text{______}\ )]{:(A),(F),(G),(W):}`
 

  1. Of the 300 visitors expected at Ground World `(G)` at 11 am, what percentage was at either Air World `(A)` or Food World `(F)` at 10 am?   (1 mark)

    --- 3 WORK AREA LINES (style=lined) ---

  2. The proportion of visitors moving from one location to another each hour on Sunday is different from Saturday.

     

    Matrix `V`, below, shows the proportion of visitors moving from one location to another each hour after 10 am on Sunday.

     

    `qquad qquad {:(qquadqquadqquadqquadqquadtext(this hour)),(qquad qquad qquad \ A qquad quad F qquad \  G \ quad quad W),(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)):}`

     

     
    Matrix `V` is similar to matrix `T` but has the first two rows of matrix `T` interchanged.

     

  3. The matrix product that will generate matrix `V` from matrix `T` is
  4. `qquad qquad V = M xx T`
  5. where matrix `M` is a binary matrix.
  6. Write down matrix `M`.   (1 mark)

    --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `20%`
  2. `S_1 = [(300 ), (780), (300),(620)]{:(A),(F),(G),(W):}`
  3. `60%`
  4. `M = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)]`
Show Worked Solution

a.   `text(Total visitors)\ =2000`

`text{Percentage}\  (W)` `= 400/2000= 20%`

 

b.   `S_1` `= TS_0`
    `= [(0.1, 0.2, 0.1, 0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(600),(600),(400),(400)]=[(300),(780),(300),(620)]`

 

c.  `text(Visitors from)\ A to G= 0.1 xx 600 = 60`

`text(Visitors from)\ F to G= 0.2 xx 600 = 120`

`:.\ text(Percentage of)\ G\ text(at 11 am)= (60 + 120)/300= 60%`
 

d.   `M = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)]`
 

 
`text(Check:)`

`M xx T = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)][(0.1, 0.2, 0.1, 0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)] = [(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)]`

Filed Under: Transition Matrices - Regular Tagged With: Band 3, Band 4, Band 5, page-break-before-question, smc-618-30-State Matrix in discrete period, smc-618-50-Permutation, smc-618-62-4x4 Matrix

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