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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%.

Matrices, GEN2 2023 VCAA 9

The circus is held at five different locations, \(E, F, G, H\) and \(I\).

The table below shows the total revenue for the ticket sales, rounded to the nearest hundred dollars, for the last 20 performances held at each of the five locations.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Location} \rule[-1ex]{0pt}{0pt} & E & F & G & H & I \\
\hline
\rule{0pt}{2.5ex} \textbf{Ticket Sales} \rule[-1ex]{0pt}{0pt} & \$960\ 000 & \$990\ 500 & \$940\ 100 & \$920\ 800 & \$901\ 300 \\
\hline
\end{array}

The ticket sales information is presented in matrix \(R\) below.

\(R=\begin{bmatrix}
960\ 000 & 990\ 500 & 940\ 100 & 920\ 800 & 901\ 300
\end{bmatrix}\)

  1. Complete the matrix equation below that calculates the average ticket sales per performance at each of the five locations.  (1 mark)

\(\begin {bmatrix}\rule{2cm}{0.25mm} \end {bmatrix}\times R = \begin {bmatrix}\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} \end {bmatrix}\)

The circus would like to increase its total revenue from the ticket sales from all five locations.

The circus will use the following matrix calculation to target the next 20 performances.

\( [t] \times R \times \begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{bmatrix}\)

  1. Determine the value of \(t\) if the circus would like to increase its revenue from ticket sales by 25%.  (1 mark)

The circus moves from one location to the next each month. It rotates through each of the five locations, before starting the cycle again.

The following matrix displays the movement between the five locations.

\begin{aligned}
& \quad \ \ \ this \ month\\
& \ \ \ E \ \ \ F \ \ \ G \ \ \ H \ \ \ I \\
& \begin{bmatrix}
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0
\end{bmatrix} \begin{array}{ll}
E & \\
F\\
G & \ \ next \ month \\
H & \\
I
\end{array}\\
&
\end{aligned}

  1. The circus started in town \(I\).
  2. What is the order in which the circus will visit the five towns?  (1 mark)

Show Answers Only

a.    \(\Big{[} \dfrac{1}{20} \Big{]} \times R = [48\ 000\ \ \ 49\ 525\ \ \ 47\ 005\ \ \ 46\ 040\ \ \ 45\ 065] \)

b.    \(t=1.25\)

c.    \(I\ H\ E\ G\ F\)

Show Worked Solution

a.    \(\Big{[} \dfrac{1}{20} \Big{]} \times R = [48\ 000\ \ \ 49\ 525\ \ \ 47\ 005\ \ \ 46\ 040\ \ \ 45\ 065] \)

 
b.    \(t=1.25\)

 
c.    \(I\ H\ E\ G\ F\)

\(\text{Process: look for a 1 in “this month” column}\ I \)

\(\Rightarrow\ \text{corresponds to “next month” letter}\ H\)

\(\text{Repeat by then looking for a 1 in “this month” column}\ H \)

\(\Rightarrow\ \text{corresponds to “next month” letter}\ E\ \text{etc…}\)

♦♦♦ Mean mark (a) 25%
♦ Mean mark (b) 47%.

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%.

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`

MATRICES, FUR1 2019 VCAA 4 MC

Stella completed a multiple-choice test that had 10 questions.

Each question had five possible answers, `A, B, C, D` and `E`.

For question number one, Stella chose the answer `E`.

Stella chose each of the nine remaining answers, in order, by following the transition matrix, `T`, below
 

`{:(qquad qquad qquad quad text(this question)),(qquad qquad quad \ A quad\ B quad C quad D quad E),(T = [(0, 0, 1, 0, 0), (0, 0, 0, 1, 0), (0, 0, 0, 0, 1), (1, 0, 0, 0, 0), (0, 1, 0, 0, 0)]{:(A),(B),(C),(D),(E):} qquad text(next question)):}`
 

What answer did Stella choose for question number six?

  1. `A`
  2. `B`
  3. `C`
  4. `D`
  5. `E`
Show Answers Only

`E`

Show Worked Solution

`text(1st question) -> E`

`text(2nd question) -> C`

`text(3rd question) -> A`

`text(then)\ \ D -> B -> E`
 

`=>  E`

MATRICES, FUR1 2018 VCAA 8 MC

A public library organised 500 of its members into five categories according to the number of books each member borrows each month.

These categories are

J = no books borrowed per month
K = one book borrowed per month
L = two books borrowed per month
M = three books borrowed per month
N = four or more books borrowed per month

The transition matrix, `T`, below shows how the number of books borrowed per month by the members is expected to change from month to month.
 

`{:(),(),(T=):}{:(qquadqquadqquad\ text(this month)),((qquadJ,quadK,quadL,quadM,quadN)),([(0.1,0.2,0.2,0,0),(0.5,0.2,0.3,0.1,0),(0.3,0.3,0.4,0.1,0.2),(0.1,0.2,0.1,0.6,0.3),(0,0.1,0,0.2,0.5)]):}{:(),(),({:(J),(K),(L),(M),(N):}):}{:(),(),(text(next month)):}`

 
In the long term, which category is expected to have approximately 96 members each month?

  1. `J`
  2. `K`
  3. `L`
  4. `M`
  5. `N`
Show Answers Only

`B`

Show Worked Solution

`text(Any initial member split by category will)`

♦ Mean mark 46%.

`text(result in the same long term expectations.)`

`text(Starting with 100 in each category:)`
 

`T^50[(100),(100),(100),(100),(100)] = [(49),(96),(124),(151),(80)]`
 

`:.\ text(Category)\ K\ text(is expected to have 96 members.)`

`=> B`