smc-619-70-One/Two Step Dominances

Matrices, GEN1 2024 VCAA 31 MC

The matrix below shows the results of a round-robin chess tournament between five players: \(H, I, J, K\) and \(L\). In each game, there is a winner and a loser.

Two games still need to be played.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \quad \quad \quad \ \textit{loser}\\
&\quad \quad\quad \quad \quad \ \ \ H \quad \quad \ \ \ I \quad \quad \quad J \quad \quad \ \ \ K \quad \quad \quad L \\
& \textit{winner} \quad \begin{array}{ccccc}
H\\
\\
I\\
\\
J\\
\\
K\\
\\
L
\end{array}
\begin {bmatrix}
0 \quad & \quad 1 \quad & \quad 0 \quad & \quad 1 \quad & \quad 0 \\
\\
0 \quad& 0 & \ldots & 1 & \quad \ldots \\
\\
1 \quad& \ldots & 0 & 1 & \quad 0 \\
\\
0 \quad& 0 & 0 & 0 & \quad 1 \\
\\
1 \quad& \ldots & 1 & 0 & \quad 0
\end{bmatrix}\\
&
\end{aligned}

A '1' in the matrix shows that the player named in that row defeated the player named in that column. For example, the 1 in row 4 shows that player \(K\) defeated player \(L\).

A '...' in the matrix shows that the player named in that row has not yet competed against the player in that column.

At the end of the tournament, players will be ranked by calculating the sum of their one-step and two-step dominances.

The player with the highest sum will be ranked first. The player with the second-highest sum will be ranked second, and so on.

Which one of the following is not a potential outcome after the final two games have been played?

Player \(I\) will be ranked first.
Player \(I\) will be ranked fifth.
Player \(J\) will be ranked first.
Player \(J\) will be ranked fifth.

Show Answers Only

\(D\)

Show Worked Solution

\(\text{Games to be played:}\ \ I\ \text{vs}\ J, \ \ I\ \text{vs}\ L\)

\(\text{Let the results matrix be}\ D\)

♦ Mean mark 53%.

\(\text{Case 1: Player}\ I\ \text{wins both games}\)

\(D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 1 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 1 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 1 & 2 & 2 \\
\ 2 & 0 & 2 & 2 & 2 \\
\ 1 & 1 & 0 & 2 & 1 \\
\ 1 & 0 & 1 & 0 & 1 \\
\ 2 & 1 & 1 & 2 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=6\\
\ \ I=8\\
\ \ J=5\\
\ \ K=3\\
\ \ L=6
\end{array}\)

\(\text{Case 2: Player}\ I\ \text{loses to}\ J, \text{defeats}\ L\)

\(D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 0 & 2 & 2 \\
\ 1 & 0 & 1 & 1 & 2 \\
\ 1 & 2 & 0 & 3 & 2 \\
\ 1 & 0 & 1 & 0 & 1 \\
\ 2 & 2 & 1 & 2 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=5\\
\ \ I=5\\
\ \ J=8\\
\ \ K=3\\
\ \ L=7
\end{array}\)

\(\text{Case 3: Player}\ I\ \text{defeats}\ J, \text{loses to}\ L\)

\(D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 0 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 0 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 1 & 2 & 1 \\
\ 1 & 0 & 1 & 2 & 1 \\
\ 1 & 1 & 0 & 2 & 1 \\
\ 1 & 1 & 1 & 0 & 1 \\
\ 2 & 2 & 2 & 3 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=5\\
\ \ I=5\\
\ \ J=5\\
\ \ K=4\\
\ \ L=9
\end{array}\)

\(\text{Case 4: Player}\ I\ \text{loses both games}\)

\(D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 0 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 0 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 0 & 2 & 1 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 2 & 0 & 3 & 1 \\
\ 1 & 1 & 1 & 0 & 1 \\
\ 2 & 3 & 1 & 3 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=4\\
\ \ I=2\\
\ \ J=7\\
\ \ K=4\\
\ \ L=9
\end{array}\)

\(\text{In all cases Player }I\ \text{or}\ K\ \text{will be ranked 5th.}\)

\(\therefore\ \text{Option D is not a potential outcome.}\)

\(\Rightarrow D\)

Matrices, GEN2 2023 VCAA 10

Within the circus, there are different types of employees: directors \((D)\), managers \((M)\), performers \((P)\) and sales staff \((S).\) Customers \((C)\) attend the circus.

Communication between the five groups depends on whether they are customers or employees, and on what type of employee they are.

Matrix \(G\) below shows the communication links between the five groups.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \ \ \textit{receiver}\\
&\quad \quad\quad \quad \quad\quad \quad \quad D \ \ M \ \ P \ \ \ S \ \ \ C \\
& G=\textit{sender} \quad \begin{array}{ccccc}
D\\
M\\
P\\
S\\
C
\end{array}
\begin {bmatrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0
\end{bmatrix}\\
&
\end{aligned}

In this matrix:

- The ' 1 ' in row \(D\), column \(M\) indicates that the directors can communicate directly with the managers.
- The ' 0 ' in row \(P\), column \(D\) indicates that the performers cannot communicate directly with the directors.

A customer wants to make a complaint to a director.
What is the shortest communication sequence that will successfully get this complaint to a director? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Matrix \(H\) below shows the number of two-step communication links between each group. Sixteen elements in this matrix are missing.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \ \ \textit{receiver}\\
&\quad \quad\quad \quad \quad\quad \quad \quad D \quad M \quad P \quad \ S \quad \ C \\
& H=\textit{sender} \quad \begin{array}{ccccc}
D\\
M\\
P\\
S\\
C
\end{array}
\begin {bmatrix} {\displaystyle}
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
0 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
0 & 1 & 0 & 0 & 1
\end{bmatrix}\\
&
\end{aligned}

i. Complete matrix \(H\) above by filling in the missing elements. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
ii. What information do elements \(g_{21}\) and \(h_{21}\) provide about the communication between the circus employees? (1 mark)

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

Show Answers Only

a. \(C\ S\ M\ D\)

b.i. \( \quad \) \(\begin{aligned} & \begin{array}{cccccc}\quad \ \ \ D \ \ M \ \ \ P \ \ \ S \ \ \ C\end{array} \\ &
\begin{array}{c}D \\ M \\P \\ S \\ C\end{array}
\begin{bmatrix}
1 & 2 & 1 & 2 & 2 \\
0 & 3 & 1 & 2 & 2 \\
1 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1
\end{bmatrix}\\ & \end{aligned}\)

b.ii. \(g_{21}\ \text{indicates managers can communicate directly with directors.}\)

\(h_{21}\ \text{indicates managers, however, cannot communicate with}\)

\(\text{directors through another person who speaks directly to a director.}\)

Show Worked Solution

a. \(C\ S\ M\ D\)

b.ii. \(g_{21}\ \text{indicates managers can communicate directly with directors.}\)

\(h_{21}\ \text{indicates managers, however, cannot communicate with}\)

\(\text{directors through another person who speaks directly to a director.}\)

♦♦♦ Mean mark (b)(ii) 25%.

MATRICES, FUR2 2021 VCAA 4

Five staff members in Elena's office played a round-robin video game tournament, where each employee played each of the other employees once. In each game there was a winner and a loser.

A table of their one-step and two-step dominances was prepared to summarise the results.

Consider the results matrix shown below.

A '1' in this matrix shows that the player named in that row defeated the player named in that column.

A '0' in this matrix shows that the player named in that row lost to the player named in that column.

Use all of the information provided to complete the results matrix. (2 marks)

`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad loser),(quadqquadqquadqquadqquadqquad \ \ I qquad\ J qquad \ K qquad\ L qquad M),(wi\n\n\er qquad{:(I),(J),(K),(L),(M):}[(0,…,…,…,…),(…,0,…,…,…),(0,0,0,1,0),(…,…,…,0,…),(…,…,…,…,0)]):}`

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

Show Answers Only

`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad ),(quadqquadqquad \ \ \ I \ \ \ J \ \ K \ \ L \ \ \ M),(qquad{:(I),(J),(K),(L),(M):}[(0,1,1,1,0),(0,0,1,1,1),(0,0,0,1,0),(0,0,0,0,1),(1,0,1,0,0)]):}`

Show Worked Solution

`text{Katie only beat Leslie} \ => \ text{everyone else beat Katie}`

`{:(qquadqquadqquadqquadqquad),(quadqquadqquadqquad I qquad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,…,…,…),(…,0,…,…,…),(0,0,0,1,0),(…,…,…,0,…),(…,…,…,…,0)]):} \ to \ {:(qquadqquadqquadqquadqquadqquadqquadqquadqquad),(quadqquadqquadqquad I qquad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,1,…,…),(…,0,1,…,…),(0,0,0,1,0),(…,…,0,0,…),(…,…,1,…,0)]):}`

`text{Katie only has one 2-step dominances. She only beat Leslie.}`

`=>\ text{Leslie only won 1 game (note she has two 2-step dominances)}`

`=>\ text{Leslie must have beaten Mikki (she is the only person with two 1-step dominances)}`

`to \ {:(qquadqquadqquadqquadqquad),(quadqquadqquadqquad I quad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,1,1,…),(…,0,1,1,…),(0,0,0,1,0),(0,0,0,0,1),(…,…,1,1,0)]):} `

`text{Ike beat either Jolene or Mikki}`

`=> \ text{Jolene has more wins than Mikki and since}`

` qquadqquad text{Ike has the most two-step dominances}`

`=> \ text{Ike beat Jolene and Jolene beat Mikki}`

`to \ {:(qquadqquadqquadqquadqquadqquadqquadqquadqquad),(quadqquadqquad \ \ \ I \ \ J \ \ K \ \ L \ \ M),(qquad{:(I),(J),(K),(L),(M):}[(0,1,1,1,0),(0,0,1,1,1),(0,0,0,1,0),(0,0,0,0,1),(1,0,1,0,0)]):}`

MATRICES, FUR2 2021 VCAA 2

The main computer system in Elena's office has broken down.

The five staff members, Alex (`A`), Brie (`B`), Chai (`C`), Dex (`D`) and Elena (`E`), are having problems sending information to each other.

Matrix `M` below shows the available communication links between the staff members.

`qquadqquadqquadqquadqquadqquadqquadqquadqquad text(receiver)`

`qquadqquadqquadqquadqquadqquadqquad \ \ \ A \ \ B \ \ C \ \ D \ \ E`

`M= \ text{sender} \ \ {:(A),(B),(C),(D),(E):} [(0,1,0,0,1),(0,0,1,1,0),(1,0,0,1,0),(0,1,0,0,0),(0,0,0,1,0)] `

In this matrix:

the '1' in row `A`, column `B` indicates that Alex can send information to Brie
the '0' in row `D`, column `C` indicates that Dex cannot send information to Chai.

Which two staff members can send information directly to each other? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Elena needs to send documents to Chai.
What is the sequence of communication links that will successfully get the information from Elena to Chai? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Matrix `M^2` below is the square of `M` and shows the number of two-step communication links between each pair of staff members.

`qquadqquadqquadqquadqquadqquadqquadqquadqquad text(receiver)`

`qquadqquadqquadqquadqquadqquadqquad \ \ \ A \ \ B \ \ C \ \ D \ \ E`

`M= \ text{sender} \ \ {:(A),(B),(C),(D),(E):} [(0,0,1,2,0),(0,1,0,1,0),(0,1,0,0,0),(0,0,1,1,0),(0,1,0,0,0)] `

Only one pair of individuals has two different two-step communication links.
List each two-step communication link for this pair. (1 mark)

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

Show Answers Only

`text{B and D}`
`text{Elena} to text{Dex} to text{Brie} to text{Chai}`
`text{Alex} to text{Brie} to text{Dex}`
`text{Alex} to text{Elena} to text{Dex}`

Show Worked Solution

a. `text{B (sender) to D (receiver)} => 1`

`text{D (sender) to B (receiver)} => 1`

`:. \ text{B and D can send information to each other}`

b. `text{Elena} to text{Dex} to text{Brie} to text{Chai}`

c. `text{The two 2-step links are from Alex to Dex.}`

`text{These are:}`

`text{Alex} to text{Brie} to text{Dex}`

`text{Alex} to text{Elena} to text{Dex}`

MATRICES, FUR1 2020 VCAA 9 MC

Five competitors, Andy (A), Brie (B), Cleo (C), Della (D) and Eddie (E), participate in a darts tournament.

Each competitor plays each of the other competitors once only, and each match results in a winner and a loser.

The matrix below shows the results of this darts tournament.

There are still two matches that need to be played.

`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad loser),(quadqquadqquadqquadqquadqquad \ A qquad\ B qquad \ C qquad\ D qquad E),(wi\n\n\er qquad{:(A),(B),(C),(D),(E):}[(0,…,0,1,0),(…,0,1,0,1),(1,0,0,…,1),(0,1,…,0,0),(1,0,0,1,0)]):}`

A ‘1’ in the matrix shows that the competitor named in that row defeated the competitor named in that column.

For example, the ‘1’ in row 2, column 3 shows that Brie defeated Cleo.

A ‘…’ in the matrix shows that the competitor named in that row has not yet played the competitor named in that column.

The winner of this darts tournament is the competitor with the highest sum of their one-step and two-step dominances.

Which player, by winning their remaining match, will ensure that they are ranked first by the sum of their one-step and two-step dominances?

Andie
Brie
Cleo
Della
Eddie

Show Answers Only

`B`

Show Worked Solution

`text(Case 1): e_12 = 1,\ e_21 = 0,\ e_34 = 1,\ e_43 = 0`

`[(0,1,0,1,0),(0,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(1,0,0,1,0)]+[(0,1,0,1,0),(0,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(1,0,0,1,0)]^2=[(0,2,1,1,1),(2,0,1,2,2),(2,2,0,3,1),(0,1,1,0,1),(1,2,0,2,0)]{:(5),(7),(8),(3),(5):}`

`text(Case 2): e_12 = 0,\ e_21 = 1,\ e_34 = 1,\ e_43 = 0`

`[(0,0,0,1,0),(1,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(1,0,0,1,0)]+[(0,0,0,1,0),(1,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(1,0,0,1,0)]^2=[(0,1,0,1,0),(3,0,1,3,2),(2,1,0,3,1),(1,1,1,0,1),(1,1,0,2,0)]{:(2),(9),(7),(4),(4):}`

`text(Similarly, totals can be calculated for:)`

`text(Case 3):\ e_34 = 0, e_43 = 1, and e_12=0, e_21=1`

`text(Case 4):\ e_34 = 0, e_43 = 1, and e_12=1, e_21=0`

`text(Only Brie can ensure she is top ranked by winning.)`

`=> B`

MATRICES, FUR1-NHT 2019 VCAA 2 MC

Four teams, blue (`B`), green (`G`), orange (`O`) and pink (`P`), played each other once in a competition.

There were no draws in this competition.

The results of the competition are shown in the matrix below.

`{:(),(),(text(winner)):}{:(qquadqquad\ text(loser)),((qquadquadB,G,O,P)),({: (B), (G), (O), (P):}[(text(−),1,v,1),(0,text(−),1,1),(0,w,text(−),0),(0,0,x,text(−))]):}`

The letters `v`, `w` and `x` each have a value of 0 or 1.

A 1 in the matrix shows that the team named in that row defeated the team named in that column.

A 0 in the matrix shows that the team named in that row was defeated by the team named in that column.

A dash (–) in the matrix shows that no game was played.

The values of `v`, `w` and `x` are

`v = 0, \ w = 1, \ x = 0`
`v = 0, \ w = 1, \ x = 1`
`v = 1, \ w = 0, \ x = 1`
`v = 1, \ w = 1, \ x = 0`
`v = 1, \ w = 1, \ x = 1`

Show Answers Only

`C`

Show Worked Solution

`v = 1\ \ (B\ text(defeated)\ O)`

`w = 0\ \ (O\ text(defeated)\ G)`

`x = 1\ \ (P\ text(defeated)\ O)`

`=> C`

MATRICES, FUR1 2018 VCAA 3 MC

Five people, India (`I`), Jackson (`J`), Krishna (`K`), Leanne (`L`) and Mustafa (`M`), competed in a table tennis tournament.

Each competitor played every other competitor once only.

Each match resulted in a winner and a loser.

The matrix below shows the tournament results.

`{:(),(),(),(),(text(winner)),(),():}{:(),(),(I),(J),(K),(L),(M):}{:(qquadqquadqquadtext(loser)),(qquadIquadJquadKquadLquadM),([(0,1,0,1,0),(0,0,1,0,1),(1,0,0,1,1),(0,1,0,0,0),(0,0,0,1,0)]):}`

A 1 in the matrix shows that the competitor named in that row defeated the competitor named in that column.

For example, the 1 in the fourth row shows that Leanne defeated Jackson.

A 0 in the matrix shows that the competitor named in that row lost to the competitor named in that column.

There is an error in the matrix. The winner of one of the matches has been incorrectly recorded as a 0.

This match was between

India and Mustafa.
India and Krishna.
Krishna and Leanne.
Leanne and Mustafa.
Jackson and Mustafa.

Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince each match has a winner and a loser,)`

`text(If)\ \ x_(ij) = 1 \ => \ x_(ji) = 0`

`text(By trial and error,)`

`text(Consider option)\ A:`

`x_15 = 0\ (text(India loses to Mustafa))`

`x_51 = 0\ (text(Mustafa loses to India))`

`:.\ text(Both cannot be correct.)`

`=> A`

MATRICES, FUR1 2017 VCAA 5 MC

Four teams, `A`, `B`, `C` and `D`, competed in a round-robin competition where each team played each of the other teams once. There were no draws.

The results are shown in the matrix below.

`{:(),(),(text(winner)\ ):}{:(qquadqquadqquad\ text(loser)),(qquadqquadAquadBquadCquadD),({:(A),(B),(C),(D):}[(0,0,f,1),(1,0,0,0),(1,g,0,1),(0,1,0,h)]):}`

A 1 in the matrix shows that the team named in that row defeated the team named in that column.

For example, the 1 in row 2 shows that team `B` defeated team `A`.

In this matrix, the values of `f`, `g` and `h` are

`f = 0,qquadg = 1,qquadh = 0`
`f = 0,qquadg = 1,qquadh = 1`
`f = 1,qquadg = 0,qquadh = 0`
`f = 1,qquadg = 1,qquadh = 0`
`f = 1,qquadg = 1,qquadh = 1`

Show Answers Only

`A`

Show Worked Solution

`text(Consider column 1:)`

`C\ text(defeated)\ A => f = 0`

`text(Consider column 3:)`

`B\ text(lost to)\ C => g = 1`

`h = 0, text(s)text(ince team)\ D\ text(cannot play itself.)`

`=> A`

MATRICES, FUR1 2016 VCAA 8 MC

The matrix below shows the result of each match between four teams, A, B, C and D, in a bowling tournament. Each team played each other team once and there were no draws.

`{:(qquadqquadqquadqquadqquadqquadquadtext(loser)),(qquadqquadqquadqquadqquadquadAquadBquadCquadD),(text(winner)quad{:(A),(B),(C),(D):}[(0,0,1,0),(1,0,0,1),(0,1,0,1),(1,0,0,0)]):}`

In this tournament, each team was given a ranking that was determined by calculating the sum of its one-step and two-step dominances. The team with the highest sum was ranked number one (1). The team with the second-highest sum was ranked number two (2), and so on.

Using this method, team C was ranked number one (1).

Team A would have been ranked number one (1) if the winner of one match had lost instead.

That match was between teams

A and B.
A and D.
B and C.
B and D.
C and D.

Show Answers Only

`A`

Show Worked Solution

`text(Test one and two step dominances after)`

♦ Mean mark 42%.

`text(swapping one result until)\ A\ text(is ranked)`

`text(number one.)`

`text(If we reverse the result and swap)\ A\ text(vs)\ B\ (text(now)\ A\ text(wins,))`

`text(Total dominances:)`

`{:(qquadqquadqquadqquadqquad\ text(lose)),(qquadqquadqquadqquadAquadBquadCquadDqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadtext(total)),(text(win)quad{:(A),(B),(C),(D):}[(0,1,1,0),(0,0,0,1),(0,1,0,1),(1,0,0,0)] + [(0,1,1,0),(0,0,0,1),(0,1,0,1),(1,0,0,0)]^2 = [(0,2,1,2),(1,0,0,1),(1,1,0,2),(1,1,1,0)]{:(5),(2),(4),(3):}):}`

`:.\ text(If the result in)\ A\ text(vs)\ B\ text(was reversed,)`

`A\ text(would have ranked number one.)`

`=> A`

MATRICES*, FUR1 2007 VCAA 8 MC

There are five teams, `A, B, C, D` and `E`, in a volleyball competition. Each team played each other team once in 2007.

The results are summarised in the directed graph below. An arrow from `A` to `E` signifies that `A` defeated `E.`

In 2007, the team that had the highest number of two-step dominances was

A. team `A`

B. team `B`

C. team `C`

D. team `D`

E. team `E`

Show Answers Only

`B`

Show Worked Solution

`text(2-step dominance matrix)`

`{: (\ quad A quad B quad C quad D quad E), ([(0, 0, 1, 2, 1), (3, 0, 0, 2, 1), (0, 3, 0, 0, 0), (0, 0, 1, 0, 1), (0, 0, 1, 0, 0)]{:(A – 4), (B – 6), (C – 3), (D – 2), (E – 1):}\ \ \ \ {:text(“2-step” wins):}):}`

`=> B`

MATRICES*, FUR1 2009 VCAA 9 MC

Five soccer teams played each other once in a tournament. In each game there was a winner and a loser.

A table of one-step and two-step dominances was prepared to summarise the results.

One result in the tournament that must have occurred is that

A. Elephants defeated Bears.

B. Elephants defeated Aardvarks.

C. Aardvarks defeated Donkeys.

D. Donkeys defeated Bears.

E. Bears defeated Chimps.

Show Answers Only

`B`

Show Worked Solution

`text(Consider option)\ A,`

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: The critical step here is to realise that if Team A beats Team B, then Team B’s one-step dominances become Team A’s two-step dominances.

`text(If elephants defeat bears, elephants)`

`text(will have a 2-step dominance over)`

`text(3 other teams.)`

`:.\ text(Incorrect.)`

`text(Consider option)\ B,`

`text(If elephants defeat aardvarks, elephants)`

`text(will have a 2-step dominance over)`

`text(1 other team.)`

`:.\ text(Correct.)`

`=> B`

MATRICES*, FUR1 2011 VCAA 2 MC

The graph below shows the one-step dominances between four farm dogs, Kip, Lab, Max, and Nim.

In this graph, an arrow from Lab to Kip indicates that Lab has a one-step dominance over Kip.

From this graph, it can be concluded that Kip has a two-step dominance over

A. Max only.

B. Nim only.

C. Lab and Nim only.

D. all of the other three dogs.

E. none of the other three dogs.

Show Answers Only

`C`

Show Worked Solution

`=> C`

MATRICES*, FUR1 2013 VCAA 9 MC

Alana, Ben, Ebony, Daniel and Caleb are friends. Each friend has a different age.

The arrows in the graph below show the relative ages of some, but not all, of the friends. For example, the arrow in the graph from Alana to Caleb shows that Alana is older than Caleb.

Using the information in the graph, it can be deduced that the second-oldest person in this group of friends is

A. Alana

B. Ben

C. Caleb

D. Daniel

E. Ebony

Show Answers Only

`B`

Show Worked Solution

`text(Completing the graph,)`

`:.\ text(Oldest to youngest is:)`

`text(Alana, Ben, Daniel, Caleb, Ebony.)`

`=> B`

MATRICES*, FUR1 2013 VCAA 5 MC

Four people, Ash (A), Binh (B), Con (C) and Dan (D), competed in a table tennis tournament.

In this tournament, each competitor played each of the other competitors once.

The results of the tournament are summarised in the directed graph below.

Each arrow shows the winner of a game played in the tournament. For example, the arrow from `C` to `A` shows that Con defeated Ash.

In the tournament, each competitor was given a ranking that was determined by calculating the sum of their one-step and two-step dominances. The competitor with the highest sum is ranked number one (1). The competitor with the second-highest sum was ranked number two (2), and so on.

Using this method, the rankings of the competitors in this tournament were

A. Dan (1), Ash (2), Con (3), Binh (4)

B. Dan (1), Ash (2), Binh (3), Con (4)

C. Con (1), Dan (2), Ash (3), Binh (4)

D. Ash (1), Dan (2), Binh (3), Con (4)

E. Ash (1), Dan (2), Con (3), Binh (4)

Show Answers Only

`E`

Show Worked Solution

`text(One step dominance matrix)`

`{: (quad qquad qquad qquad A quad B quad C quad D), (D_1 = [(0, 1, 0, 1), (0, 0, 1, 0), (1, 0, 0, 0), (0, 1, 1, 0)] {:(A), (B), (C), (D):}\ \ \ text{wins}):}`

`text(Two step dominance matrix)`

`{: (quad qquad qquad qquad A quad B quad C quad D), (D_2 = [(0, 1, 0, 2), (0, 0, 1, 0), (2, 0, 0, 0), (0, 1, 1, 0)] {:(A), (B), (C), (D):}\ \ \ text{wins}):}`

`D_1 + D_2 = [(0, 2, 0, 3), (0, 0, 2, 0), (3, 0, 0, 0), (0, 2, 2, 0)] {:(5), (2), (3), (4):}`

`text(Adding up rows, the ranking)`

`text{(highest to lowest) is:}`

`A, D, C, B`

`=> E`

MATRICES*, FUR2 2006 VCAA 2

The five musicians, George, Harriet, Ian, Josie and Keith, compete in a music trivia game.

Each musician competes once against every other musician.

In each game there is a winner and a loser.

The results are represented in the dominance matrix, Matrix 1, and also in the incomplete directed graph below.

On the directed graph an arrow from Harriet to George shows that Harriet won against George.

Explain why the figures in bold in Matrix 1 are all zero. (1 mark)

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

One of the edges on the directed graph is missing.

Using the information in Matrix 1, draw in the missing edge on the directed graph above and clearly show its direction. (1 mark)

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

The results of each trivia contest (one-step dominances) are summarised as follows.

In order to rank the musicians from first to last in the trivia contest, two-step (two-edge) dominances will be considered.

The following incomplete matrix, Matrix 2, shows two-step dominances.

`{:(qquadqquadqquadtext(Matrix 2)),(qquadqquad{:GquadHquadI\ quadJquad\ K:}),({:(G),(H),(I),(J),(K):}[(0,1,1,2,0),(1,0,1,1,1),(1,0,0,0,0),(0,0,1,0,1),(2,0,1,x,0)]):}`

Explain the two-step dominance that George has over Ian. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Determine the value of the entry `x` in Matrix 2. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Taking into consideration both the one-step and two-step dominances, determine which musician was ranked first and which was ranked last in the trivia contest. (2 marks)

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

Show Answers Only

`text(A musicians does not compete against him/herself.)`
`text(Two step dominance occurs because George is dominant)`

`text(over Keith who is in turn dominant over Ian.)`
`2`
`text{First is Keith (8), last is Ian (2)}`

Show Worked Solution

a. `text(A musicians does not compete against him/herself.)`

b. `text(Josie won against George.)`

c. `text(Two step dominance occurs because George is dominant)`

`text(over Keith who is in turn dominant over Ian.)`

d. `text(Following the edges on network diagram:)`

`text(Keith over Harriet who beats Josie.)`

`text(Keith over Ian who beats Ian.)`

`:. x = 2`

`D_1 + D_2 =`

`[(0,1,2,2,1),(2,0,2,2,1),(1,0,0,1,0),(1,0,1,0,1),(2,1,2,3,0)]{:(G – 6),(H – 7),(I – 2),(J – 3),(K – 8):}`

`text{Summing the rows (above),}`

`:.\ text{First is Keith (8), last is Ian (2).}`

MATRICES*, FUR1 2014 VCAA 4 MC

The directed graph below shows the results of a chess competition between five players: Alex, Ben, Cindi, Donna and Elise.

Each arrow indicates the winner of individual games. For example, the arrow from Alex to Donna indicates that Alex beat Donna in their game.

The sum of their one-step and two-step dominances is calculated to give each player a dominance score. The dominance scores are then used to rank the players.

The ranking of the players in this competition, from highest to lowest dominance score, is

A. Ben, Elise, Donna, Alex, Cindi

B. Ben, Elise, Cindi, Donna, Alex

C. Ben, Elise, Donna, Cindi, Alex

D. Elise, Ben, Donna, Alex, Cindi

E. Elise, Ben, Donna, Cindi, Alex

Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ D_1 = 1 text(-step dominance matrix)`

`{: (quad qquad qquad qquad A quad B\ quad C\ quad D\ quad E), (D_1 = [(0, 1, 1, 0, 1), (0, 0, 0, 1, 0), (0, 1, 0, 1, 1), (1, 0, 0, 0, 1), (0, 1, 0, 0, 0)] {:(A), (B), (C), (D), (E):}\ \ \ \ text(loses)):}`

`text(Let)\ \ D_2 = 2 text(-step dominance matrix)`

`{: (quad qquad qquad qquad A quad B quad C quad D quad E), (D_1 = [(0, 2, 0, 2, 1), (1, 0, 0, 0, 1), (1, 1, 0, 1, 1), (0, 2, 1, 0, 1), (0, 0, 0, 1, 0)] {:(A), (B), (C), (D), (E):}):}`

`{: (D_1 + D_2 = [(0, 3, 1, 2, 2), (1, 0, 0, 1, 1), (1, 2, 0, 2, 2), (1, 2, 1, 0, 2), (0, 1, 0, 1, 0)]), (quad quad qquad qquad qquad qquad qquad 3\ \ \ 8\ \ \ 2\ \ \ \ 6\ \ \ \ 7):}`

`:.\ text(Summing the columns, the High-Low ranking)`

`text(is)\ BEDAC.`

`=> A`

MATRICES*, FUR1 2015 VCAA 8 MC

There are five teams in a table tennis competition.

Every team played one match against every other team, and each match had a winner and a loser.

The results of the matches are summarised in the directed graph below. For example, an arrow from Lions to Eagles indicates that Lions defeated Eagles.

In determining the ranking of these teams, the total of each team’s one-step dominances and two-step dominances will be calculated.

The team with the highest total will be ranked first.

The team with the next highest total will be ranked second, and so on.

The ranking of these five teams from first to last is

A. Lions, Rebels, Dingoes, Eagles, Heavies

B. Lions, Rebels, Eagles, Dingoes, Heavies

C. Rebels, Lions, Dingoes, Eagles, Heavies

D. Rebels, Lions, Eagles, Dingoes, Heavies

E. Eagles, Lions, Rebels, Dingoes, Heavies

Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ D_1 =\ text(1-step dominance matrix)`

`{:(quad qquad qquad qquad L quad E quad\ R quad\ D quad\ H), (D_1 = [(0,0,0,1,0),(1,0,1,0,0),(1,0,0,0,0),(0,1,1,0,0),(1,1,1,1,0)]{:(L),(E),(R),(D),(H):}\ \ \ \ \ \ text(loses)):}`

`text(Let)\ \ D_2 =\ text(2-step dominance matrix)`

`D_2 = [(0,1,1,0,0),(1,0,0,1,0),(0,0,0,1,0),(2,0,1,0,0),(2,1,2,1,0)]`

`{:(\ \ \ \ quad qquad qquad qquad qquad L quad E quad\ R quad\ D quad\ H), (D_1+D_2 = [(0,1,1,1,0),(2,0,1,1,0),(1,0,0,1,0),(2,1,2,0,0),(3,2,3,2,0)]), ({:qquad qquad qquad qquad qquad\ \ \ \ 8 quad\ \ 4 quad\ 7 quad\ 5 quad\ 0:}):}`

`:.\ text(Summing up each column, the ranking from)`

`text(highest to lowest is)\ LRDEH.`

`=> A`

MATRICES*, FUR2 2008 VCAA 4

The children are taken to the zoo where they observe the behaviour of five young male lion cubs. The lion cubs are named Arnold, Barnaby, Cedric, Darcy and Edgar. A dominance hierarchy has emerged within this group of lion cubs. In the directed graph below, the directions of the arrows show which lions are dominant over others.

Name the two pairs of lion cubs who have equal totals of one-step dominances. (2 marks)

--- 2 WORK AREA LINES (style=lined) ---
Over which lion does Cedric have both a one-step dominance and a two-step dominance? (1 mark)

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

In determining the final order of dominance, the number of one-step dominances and two-step dominances are added together.

Complete the table below for the final order of dominance. (1 mark)

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

Over time, the pattern of dominance changes until each lion cub has a one-step dominance over two other lion cubs.

Determine the total number of two-step dominances for this group of five lion cubs. (1 mark)

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

Show Answers Only

`text{Arnold and Edgar (1 each)}`
`text{Barnaby and Cedric (2 each)}`
`text(Edgar)`
`20`

Show Worked Solution

a. `text(One step dominances:)`

`text{Arnold and Edgar (1 each)}`

`text{Barnaby and Cedric (2 each)}`

b. `text(Edgar)`

d. `text(Each lion has a one step dominance over 2 others.)`

`=>\ text(Each lion must have a two step)`

`text(dominance over 2 × 2 = 4 lions)`

`:.\ text(Total 2 step dominances in group)`

`= 5 xx 4`

`= 20`