How many QCAA Specialist Mathematics questions cover Further matrices?

AusGrader has 52 QCAA Specialist Mathematics questions on Further matrices, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics — Further matrices Questions & Answers

Q11

2020

QCAA

Paper 2

5 marks

Q11

Teams A, B, C, D and E participated in a competition with the following results:

A defeated D.
B defeated A, C and E.
C defeated A and E.
D defeated B, C and E.
E defeated A.

To rank the teams at the end of the competition, the organisers constructed a dominance matrix, N, that is partially completed.

Q11a

1 mark

By allocating 1 to represent 'defeated' and 0 to represent either 'was defeated by' or 'no result', complete matrix N.

	A	B	C	D	E
A
B
C
D
E

Rows: Winning teams, Columns: Losing teams

Reveal Answer

\begin{matrix} & \text{A} & \text{B} & \text{C} & \text{D} & \text{E} \\ \text{A} & 0 & 0 & 0 & 1 & 0 \\ \text{B} & 1 & 0 & 1 & 0 & 1 \\ \text{C} & 1 & 0 & 0 & 0 & 1 \\ \text{D} & 0 & 1 & 1 & 0 & 1 \\ \text{E} & 1 & 0 & 0 & 0 & 0 \end{matrix}

Marking Criteria

Descriptor	Marks
correctly completes matrix N	1

Q11b

2 marks

The organisers need to rank the teams into individual places from first to fifth place.
They decide to use the ranking model $\mathbf{N} + \mathbf{N}^2$ to achieve this.

Use the model $\mathbf{N} + \mathbf{N}^2$ to rank the teams.

Reveal Answer

$N + N^2 = \begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 3 & 0 & 1 & 1 & 2 \\ 2 & 0 & 0 & 1 & 1 \\ 3 & 1 & 2 & 0 & 3 \\ 1 & 0 & 0 & 1 & 0 \end{bmatrix}$

Ranking teams using the model $N + N^2$

Team	Model value	Rank position
A	4	3
B	7	2
C	4	3
D	9	1
E	2	5

Marking Criteria

Descriptor	Marks
calculates N + N²	1
ranks the teams to show that teams A and C are tied	1

Q11c

1 mark

Use the result from 11b) to identify a limitation of the organisers' ranking model.

Reveal Answer

The limitation of the ranking model is that it does not provide individual positions from first to fifth.

Marking Criteria

Descriptor	Marks
identifies a limitation of the organiser's model based on the model N + N²	1

Q11d

1 mark

State a mathematical refinement the organisers could consider to overcome the limitation of the ranking model identified in 11c).

Reveal Answer

The ranking model could be improved by including weightings in the calculations (e.g. $N + \frac{1}{2}N^2$ )

Marking Criteria

Descriptor	Marks
correctly describes a suitable mathematical refinement	1

Q7

2023

QCAA

Paper 2

1 mark

Q7

1 mark

Matrix $\mathbf{N}$ represents the results for a competition involving four teams.

\mathbf{N} = \begin{matrix} & \text{Losing teams} \\ \text{Winning teams} & \begin{matrix} \text{P} & \text{Q} & \text{R} & \text{S} \end{matrix} \\ \begin{matrix} \text{P} \\ \text{Q} \\ \text{R} \\ \text{S} \end{matrix} & \begin{bmatrix} 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \end{bmatrix} \end{matrix}

Key: Team P lost to team Q but won against teams R and S.

Using the ranking model $\mathbf{N} + 0.5\mathbf{N}^2$ , the teams that placed first, second and third respectively are

A

P, S and Q.

B

P, S and R.

C

S, P and Q.

D

S, P and R.

Reveal Answer

A

P, S and Q.

Correct Answer

By summing the rows of $\mathbf{N}$ and adding half the row sums of $\mathbf{N}^2$ , the scores are calculated as: P ( $2 + 0.5(3) = 3.5$ ), S ( $2 + 0.5(2) = 3.0$ ), and Q ( $1 + 0.5(2) = 2.0$ ). This results in the ranking 1st: P, 2nd: S, 3rd: Q.

B

P, S and R.

This option incorrectly lists R as third. R's score is $1 + 0.5(1) = 1.5$ , which places it fourth behind Q ( $2.0$ ).

C

S, P and Q.

This option incorrectly ranks S first. P has a higher total score ( $3.5$ ) than S ( $3.0$ ) because P has more two-step wins (indirect victories) represented by the row sums of $\mathbf{N}^2$ .

D

S, P and R.

This option has the wrong order for the top two teams and incorrectly lists R as third. The correct order based on the scores is P ( $3.5$ ), S ( $3.0$ ), then Q ( $2.0$ ).

Q12

2024

QCAA

Paper 2

4 marks

Q12

A system of linear equations is given by

$x - 2y - 2z = -6$
$-3x - y + z = 2$
$2x + 3y - 5z = 10$

Q12a

1 mark

Express the system of equations as a matrix equation of the form $AX = B$ , where $A$ is a $3 \times 3$ matrix and both $X$ and $B$ are $3 \times 1$ column vectors.

Reveal Answer

$\begin{bmatrix} 1 & -2 & -2 \\ -3 & -1 & 1 \\ 2 & 3 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -6 \\ 2 \\ 10 \end{bmatrix}$

Marking Criteria

Descriptor	Marks
correctly expresses the equations of the form $AX = B$	1

Q12b

1 mark

Use matrix algebra to express $X$ in terms of $A$ and $B$ .

Reveal Answer

$X = A^{-1}B$

Marking Criteria

Descriptor	Marks
correctly expresses X in terms of A and B	1

Q12c

1 mark

Use your result from Question 12b) to determine the solution of the system of equations.

Reveal Answer

Using GDC

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2 \\ 3 \\ -1 \end{bmatrix}$

Marking Criteria

Descriptor	Marks
determines solution to equations	1

Q12d

1 mark

Verify your result from Question 12c) using one of the given linear equations.

Reveal Answer

Substituting $x=-2, y=3, z=-1$ into $x-2y-2z=-6$

$(-2)-2(3)-2(-1)=-6$

So result is verified.

Marking Criteria

Descriptor	Marks
verifies result by substituting result from 12c) into one of the given linear equations	1

Q6

2024

QCAA

Paper 1

1 mark

Q6

1 mark

Players P, Q, R and S played each other once in a competition where there were no draws.
Only the following results are known.

Player P defeated players Q and R.
Player Q defeated two players.
Players R and S each defeated one player.
Based on these results, a dominance matrix $N$ was partially constructed as shown.

$N = \begin{matrix} & P & Q & R & S \\ P & 0 & 1 & 1 & 0 \\ Q & \square & \square & \square & \square \\ R & 0 & 0 & 0 & 1 \\ S & 1 & 0 & 0 & 0 \end{matrix}$

The completed matrix $N$ is

A

$\begin{bmatrix} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

B

$\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

C

$\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

D

$\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

Reveal Answer

A

$\begin{bmatrix} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

This matrix indicates that Q defeated P (row 2, column 1 is 1). However, the problem states that P defeated Q, so the entry $N_{QP}$ must be 0.

B

$\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

Correct Answer

Since P defeated Q, Q lost to P ( $N_{QP}=0$ ). For Q to have defeated exactly two players, Q must have won against the remaining opponents R and S, making the second row $[0, 0, 1, 1]$ .

C

$\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

The sum of the second row is 1, which implies Q won only one game. This contradicts the problem statement that player Q defeated two players.

D

$\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$

The matrix shows a 1 on the main diagonal ( $N_{QQ}=1$ ), which implies Q defeated themselves. In a dominance matrix, the diagonal entries must always be 0.

Q16

2022

QCAA

Paper 1

7 marks

Q16

Consider this system of equations that corresponds to three planes.

$x + 5y = 1 + 2z$
$x + z = 3y + 3$
$8y - \lambda = 3z$

Q16a

4 marks

Use a Gaussian technique to determine the value of $\lambda$ for which this system of equations has infinitely many solutions.

Reveal Answer

Rearranging equations:
$x + 5y - 2z = 1$
$x - 3y + z = 3$
$8y - 3z = \lambda$
Expressing in matrix form:
$\begin{bmatrix} 1 & 5 & -2 & 1 \\ 1 & -3 & 1 & 3 \\ 0 & 8 & -3 & \lambda \end{bmatrix}$
$R_2' = R_2 - R_1$
$\begin{bmatrix} 1 & 5 & -2 & 1 \\ 0 & -8 & 3 & 2 \\ 0 & 8 & -3 & \lambda \end{bmatrix}$
$R_3' = R_3 + R_2$
$\begin{bmatrix} 1 & 5 & -2 & 1 \\ 0 & -8 & 3 & 2 \\ 0 & 0 & 0 & \lambda + 2 \end{bmatrix}$
For $\lambda = -2$ there are infinitely many solutions.

Marking Criteria

Descriptor	Marks
correctly rearranges the three equations	1
establishes an augmented matrix	1
establishes a row of zeros in the row containing $\lambda$	1
determines a value of $\lambda$	1

Q16b

3 marks

Use the result from Question 16a) to determine the infinitely many solutions. Express your answer in the form of a vector equation of a line.

Reveal Answer

Using $\begin{bmatrix} 1 & 5 & -2 & 1 \\ 0 & -8 & 3 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
Letting $z = k (k \in \mathbb{R})$
Row 2: $-8y + 3z = 2$
$8y - 3k = -2 \Rightarrow y = \frac{3k-2}{8}$
Row 1: $x + 5y - 2z = 1$
$x + 5(\frac{3k-2}{8}) - 2k = 1$
$x + \frac{15k-10}{8} - \frac{16k}{8} = 1 \Rightarrow x = \frac{k}{8} + \frac{9}{4}$
The solutions in vector form are:
$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \frac{1}{8} \\ \frac{3}{8} \\ 1 \end{bmatrix} k + \begin{bmatrix} \frac{9}{4} \\ -\frac{1}{4} \\ 0 \end{bmatrix}$

Marking Criteria

Descriptor	Marks
Expresses y in terms of a parameter	1
Expresses x in terms of a parameter	1
Determines the infinite solutions expressed in the form of a vector equation of a line	1

Q4

2023

QCAA

Paper 1

1 mark

Q4

1 mark

The age-specific population distribution of a particular species of animal is shown.

Age (years)	0–1	1–2	2–3	3–4
Female population	94	82	37	6
Breeding rate	0	1.3	0.9	0.2
Survival rate	0.6	0.8	0.4	0

The Leslie matrix based on this data is

A

$\begin{bmatrix} 94 & 82 & 37 & 6 \\ 0.6 & 0 & 0 & 0 \\ 0 & 0.8 & 0 & 0 \\ 0 & 0 & 0.4 & 0 \end{bmatrix}$

B

$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 1.3 & 0 & 0 & 0 \\ 0 & 0.9 & 0 & 0 \\ 0 & 0 & 0.2 & 0 \end{bmatrix}$

C

$\begin{bmatrix} 0.6 & 0.8 & 0.4 & 0 \\ 1.3 & 0 & 0 & 0 \\ 0 & 0.9 & 0 & 0 \\ 0 & 0 & 0.2 & 0 \end{bmatrix}$

D

$\begin{bmatrix} 0 & 1.3 & 0.9 & 0.2 \\ 0.6 & 0 & 0 & 0 \\ 0 & 0.8 & 0 & 0 \\ 0 & 0 & 0.4 & 0 \end{bmatrix}$

Reveal Answer

A

$\begin{bmatrix} 94 & 82 & 37 & 6 \\ 0.6 & 0 & 0 & 0 \\ 0 & 0.8 & 0 & 0 \\ 0 & 0 & 0.4 & 0 \end{bmatrix}$

This matrix incorrectly uses the female population counts ( $94, 82, 37, 6$ ) in the first row instead of the breeding rates.

B

$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 1.3 & 0 & 0 & 0 \\ 0 & 0.9 & 0 & 0 \\ 0 & 0 & 0.2 & 0 \end{bmatrix}$

This matrix contains incorrect values in the first row and places the breeding rates in the sub-diagonal where survival rates belong.

C

$\begin{bmatrix} 0.6 & 0.8 & 0.4 & 0 \\ 1.3 & 0 & 0 & 0 \\ 0 & 0.9 & 0 & 0 \\ 0 & 0 & 0.2 & 0 \end{bmatrix}$

This option swaps the correct positions; it places survival rates in the first row and breeding rates in the sub-diagonal, which is the reverse of the standard Leslie matrix structure.

D

$\begin{bmatrix} 0 & 1.3 & 0.9 & 0.2 \\ 0.6 & 0 & 0 & 0 \\ 0 & 0.8 & 0 & 0 \\ 0 & 0 & 0.4 & 0 \end{bmatrix}$

Correct Answer

This is the correct Leslie matrix. The first row contains the breeding rates ( $0, 1.3, 0.9, 0.2$ ), and the sub-diagonal contains the survival rates ( $0.6, 0.8, 0.4$ ) representing the probability of surviving to the next age class.

Q12

2023

QCAA

Paper 1

5 marks

Q12

5 marks

Given $A = \begin{pmatrix} 1 & -2 \\ 1 & 2 \end{pmatrix}$ , $B = \begin{pmatrix} 0 & 2 \\ 1 & 3 \end{pmatrix}$ and $C = \begin{pmatrix} -1 & -1 \\ 0 & 3 \end{pmatrix}$ , determine $X$ in the matrix equation $XA - XC = B$ .

Reveal Answer

$XA - XC = B$
$X(A-C) = B$
$X = B(A-C)^{-1}$

$X = \begin{pmatrix} 0 & 2 \\ 1 & 3 \end{pmatrix} \left[ \begin{pmatrix} 1 & -2 \\ 1 & 2 \end{pmatrix} - \begin{pmatrix} -1 & -1 \\ 0 & 3 \end{pmatrix} \right]^{-1}$
$= \begin{pmatrix} 0 & 2 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 1 & -1 \end{pmatrix}^{-1}$
$= \begin{pmatrix} 0 & 2 \\ 1 & 3 \end{pmatrix} \frac{1}{-1} \begin{pmatrix} -1 & 1 \\ -1 & 2 \end{pmatrix}$
$= \begin{pmatrix} 0 & 2 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & -2 \end{pmatrix}$
$= \begin{pmatrix} 2 & -4 \\ 4 & -7 \end{pmatrix}$

Marking Criteria

Descriptor	Marks
Correctly recognises the need to use X as a common factor	1
Expresses X as the subject of the equation	1
Represents X in terms of two matrices	1
Calculates the inverse of an appropriate matrix	1
Determines X	1

Q19

2024

QCAA

Paper 2

6 marks

Q19

6 marks

An experiment researching the population changes of a certain species of insect was conducted over a four-week period. The insect has two distinct stages in its two-week lifespan. Each stage is approximately one week in length.

A constant proportion of females survive from stage 1 into stage 2.

The ratio of the reproduction rate for females in stage 2 to females in stage 1 is 2:1. All offspring are born into stage 1.

The number of females in each stage was measured initially and then again after two weeks as shown.

Female population	Stage 1	Stage 2
Initially	48	32
After two weeks	25	21

Use a matrix approach to estimate the total number of females after four weeks.

Reveal Answer

Let the birth rate for Stage 1 females be $x$ and the survival rates for Stage 1 females to Stage 2 females be $y$ .
Let L be the Leslie matrix for this species.

$L = \begin{bmatrix} x & 2x \\ y & 0 \end{bmatrix}$

Let $P_n$ represent the population of the species after $n$ weeks.

$P_0 = \begin{bmatrix} 48 \\ 32 \end{bmatrix}$

$P_2 = L^2 P_0$

$\begin{bmatrix} 25 \\ 21 \end{bmatrix} = \begin{bmatrix} x & 2x \\ y & 0 \end{bmatrix}^2 \begin{bmatrix} 48 \\ 32 \end{bmatrix}$

$= \begin{bmatrix} x^2+2xy & 2x^2 \\ xy & 2xy \end{bmatrix} \begin{bmatrix} 48 \\ 32 \end{bmatrix}$

$= \begin{bmatrix} 48x^2+96xy+64x^2 \\ 48xy+64xy \end{bmatrix}$

Equating parts

$112x^2 + 96xy = 25 \dots (1)$

$112xy = 21 \dots (2)$

From (2)

$xy = \frac{21}{112} \dots (3)$

Substituting into (1)

$112x^2 + 96(\frac{21}{112}) = 25$

$x=0.25$ (reject -ive solution as $x>0$ )

Using 3, $y=0.75$

$P_4 = L^4 P_0$

$= \begin{bmatrix} 0.25 & 0.5 \\ 0.75 & 0 \end{bmatrix}^4 \begin{bmatrix} 48 \\ 32 \end{bmatrix}$

$= \begin{bmatrix} 13.6 \\ 12.6 \end{bmatrix}$

Approximate total number of females at the conclusion of the experiment is 26.

Marking Criteria

Descriptor	Marks
correctly determines an appropriate Leslie matrix	1
determines a matrix equation linking the initial population with the population after two weeks	1
determines two simultaneous equations in terms of the relevant birth and survival rates	1
determines appropriate values of x and y	1
determines the total number of females at the conclusion of the experiment	1
shows logical organisation of a fully attempted solution, communicating key steps	1

Q6

2021

VCAA

Paper 1

4 marks

Q6

4 marks

Consider the three vectors $\vec{a} = -\mathbf{i} + 6\mathbf{j} - 3\mathbf{k}$ , $\vec{b} = 2\mathbf{i} - 8\mathbf{j} + 5\mathbf{k}$ and $\vec{c} = 3\mathbf{i} + 2\mathbf{j} + |1 - p^2|\mathbf{k}$ , where $p$ is a real constant.

Find the values of $p$ for which the three vectors are linearly independent.

Reveal Answer

$p \in R \setminus \{-\sqrt{5}, \sqrt{5}\}$

Marking Criteria

Descriptor	Marks
Sets up a valid equation or system of equations to test for linear dependence or independence (e.g., setting the determinant of the matrix of vectors to zero or setting up $m\vec{a} + n\vec{b} = \vec{c}$ )	1
Simplifies the equation to isolate the term involving $p$ (e.g., obtaining $\|1-p^2\| = 4$ )	1
Finds the values of $p$ for which the vectors are linearly dependent ( $p = \pm\sqrt{5}$ )	1
Correctly states the values of $p$ for which the vectors are linearly independent ( $p \in R \setminus \{-\sqrt{5}, \sqrt{5}\}$ or equivalent)	1

Q12

2022

QCAA

Paper 2

5 marks

Q12

A scientist collects data for a species of tree frog in a protected area. Details for the female tree frog population are shown in the table.

Age (years)	0–1	1–2	2–3	3–4
Population in Year 1	150	101	84	62
Birth (breeding) rate	0.4	0.7	0.5	0.1
Survival rate	0.6	0.3	0.2	0

The scientist uses a Leslie matrix model to make predictions about the female tree frog population.

Q12a

1 mark

State the initial population matrix.

Reveal Answer

Initial population matrix $= N_1 = \begin{bmatrix} 150 \\ 101 \\ 84 \\ 62 \end{bmatrix}$

Marking Criteria

Descriptor	Marks
Correctly states the initial population matrix	1

Q12b

1 mark

Determine the Leslie matrix.

Reveal Answer

Leslie matrix $= L = \begin{bmatrix} 0.4 & 0.7 & 0.5 & 0.1 \\ 0.6 & 0 & 0 & 0 \\ 0 & 0.3 & 0 & 0 \\ 0 & 0 & 0.2 & 0 \end{bmatrix}$

Marking Criteria

Descriptor	Marks
Correctly determines the Leslie matrix	1

Q12c

3 marks

A species is considered to be endangered if the female population in a restricted area is predicted to fall to less than 125 in the next 20 years.
Determine whether this species of tree frog is considered to be endangered.

Reveal Answer

Consider the population in Year 20
Using matrix facility of GDC
$N_{20} = L^{19}N_1$
$\approx \begin{bmatrix} 62.7 \\ 39.7 \\ 12.6 \\ 2.7 \end{bmatrix}$

Female population in Year 20 $\approx 119$

The female population is less than 125 within the 20-year period so the species is considered to be endangered.

Marking Criteria

Descriptor	Marks
Calculates a matrix representing the female population within a 20-year period	1
Calculates female population for a year within a 20-year period	1
Makes a suitable decision whether the species is considered endangered	1

Q9

2022

QCAA

Paper 2

1 mark

Q9

1 mark

Consider the matrix equation.

$\mathbf{X} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$

Matrix $\mathbf{X}$ is

A

$\begin{bmatrix} 0 & 1 & 1 \\ 1 & -1 & 2 \\ -1 & 0 & 2 \end{bmatrix}$

B

$\begin{bmatrix} 0 & 1 & -1 \\ 1 & -1 & 0 \\ 1 & 2 & 2 \end{bmatrix}$

C

$\begin{bmatrix} 2 & 2 & 1 \\ 4 & 3 & 3 \\ 5 & 5 & 5 \end{bmatrix}$

D

$\begin{bmatrix} 2 & 4 & 5 \\ 2 & 3 & 5 \\ 1 & 3 & 5 \end{bmatrix}$

Reveal Answer

A

$\begin{bmatrix} 0 & 1 & 1 \\ 1 & -1 & 2 \\ -1 & 0 & 2 \end{bmatrix}$

Correct Answer

To solve for $\mathbf{X}$ in the equation $\mathbf{X}\mathbf{A} = \mathbf{B}$ , you must right-multiply both sides by $\mathbf{A}^{-1}$ , giving $\mathbf{X} = \mathbf{B}\mathbf{A}^{-1}$ . Calculating the inverse of the second matrix and performing the multiplication yields this result.

B

$\begin{bmatrix} 0 & 1 & -1 \\ 1 & -1 & 0 \\ 1 & 2 & 2 \end{bmatrix}$

This matrix is the result of $\mathbf{A}^{-1}\mathbf{B}$ . This would be the correct solution if the equation were $\mathbf{A}\mathbf{X} = \mathbf{B}$ (left-multiplication), but the unknown $\mathbf{X}$ is on the left side of the product.

C

$\begin{bmatrix} 2 & 2 & 1 \\ 4 & 3 & 3 \\ 5 & 5 & 5 \end{bmatrix}$

This matrix is the result of multiplying $\mathbf{A}\mathbf{B}$ . This is incorrect because solving for $\mathbf{X}$ requires using the inverse of $\mathbf{A}$ , and matrix multiplication is generally not commutative.

D

$\begin{bmatrix} 2 & 4 & 5 \\ 2 & 3 & 5 \\ 1 & 3 & 5 \end{bmatrix}$

This matrix is the result of multiplying $\mathbf{B}\mathbf{A}$ . This is incorrect because to isolate $\mathbf{X}$ , you must multiply by the inverse matrix $\mathbf{A}^{-1}$ , not the original matrix $\mathbf{A}$ .

Q2

2022

QCAA

Paper 2

1 mark

Q2

1 mark

The win/draw/loss results after a netball competition involving five teams is represented in matrix M.

Losing teams

$\begin{matrix} & P & Q & R & S & T \\ P & 0 & 1 & 2 & 0 & 2 \\ Q & 1 & 0 & 0 & 1 & 1 \\ R & 0 & 2 & 0 & 0 & 0 \\ S & 2 & 1 & 2 & 0 & 2 \\ T & 0 & 1 & 2 & 0 & 0 \end{matrix}$

Key: Team P drew with Team Q, defeated Team R and Team T, and lost to Team S

The model $\mathbf{M} + \mathbf{M}^2 + \mathbf{M}^3$ is used to rank the teams. The final positions from first to fifth are

A

S, Q, P, R, T

B

S, Q, P, T, R

C

S, P, Q, T, R

D

S, P, Q, R, T

Reveal Answer

A

S, Q, P, R, T

Correct Answer

Calculating the row sums of the matrix sum $\mathbf{M} + \mathbf{M}^2 + \mathbf{M}^3$ yields the scores: S=97, Q=61, P=59, R=38, and T=37. This results in the ranking order S, Q, P, R, T.

B

S, Q, P, T, R

This option incorrectly ranks Team T above Team R. The calculated total score for R is 38, which is slightly higher than T's score of 37.

C

S, P, Q, T, R

This option incorrectly ranks Team P above Team Q and Team T above Team R. The calculations show Q (61) > P (59) and R (38) > T (37).

D

S, P, Q, R, T

This option incorrectly ranks Team P above Team Q. The total scores derived from the model indicate that Team Q (61) is ranked higher than Team P (59).

Q11

2025

QCAA

Paper 1

5 marks

Q11

Consider the system of linear equations represented using the augmented matrix shown.

\left[ \begin{array}{ccc|c} 1 & -1 & -1 & -6 \\ -2 & 1 & 1 & 1 \\ 0 & 0 & 4 & 4 \end{array} \right] \begin{array}{l} R_1 \\ R_2 \\ R_3 \end{array}

Key: $R_1$ represents the row 1 values.

Q11a

1 mark

Modify the augmented matrix using the row operation shown.

\left[ \begin{array}{ccc|c} 1 & -1 & -1 & -6 \\ \square & \square & \square & \square \\ 0 & 0 & 4 & 4 \end{array} \right] \begin{array}{l} R_1 \\ R_2' = R_2 + 2R_1 \\ R_3 \end{array}

Key: $R_2' = R_2 + 2R_1$ indicates that the new row 2 values are equal to the sum of the existing row 2 and twice row 1 values.

Reveal Answer

Given

\begin{bmatrix} 1 & -1 & -1 & | & -6 \\ -2 & 1 & 1 & | & 1 \\ 0 & 0 & 4 & | & 4 \end{bmatrix} \begin{matrix} R_1 \\ R_2 \\ R_3 \end{matrix}

The augmented matrix is

\begin{bmatrix} 1 & -1 & -1 & | & -6 \\ 0 & -1 & -1 & | & -11 \\ 0 & 0 & 4 & | & 4 \end{bmatrix} \begin{matrix} R_1 \\ R_2' = R_2 + 2R_1 \\ R_3 \end{matrix}

Marking Criteria

Descriptor	Marks
correctly determines the new row 2 values in the augmented matrix	1

Q11b

3 marks

Given the row 1 values represent the equation $x - y - z = -6$ , use your result from Question 11a) to determine the solution of the system of linear equations.

Reveal Answer

From $R_3$
$4z = 4 \Rightarrow z = 1$

From $R_2'$

\begin{align*} -y - z &= -11\\ -y - 1 &= -11\\ y &= 10 \end{align*}

From $R_1$

\begin{align*} x - y - z &= -6\\ x - 10 - 1 &= -6\\ x &= 5 \end{align*}

Marking Criteria

Descriptor	Marks
correctly determines required value of $z$	1
determines required value of $x$	1
determines required value of $y$	1

Q11c

1 mark

The system of linear equations is geometrically represented by three planes.

Use your result from Question 11b) to describe a geometrical interpretation of your solution of the system of linear equations.

Reveal Answer

The planes intersect at a point.

Marking Criteria

Descriptor	Marks
describes a geometrical interpretation based on prior mathematical reasoning	1

Q14

2021

QCAA

Paper 2

5 marks

Q14

The Tasmanian thornbill is a species of bird that has an average life span of three years. Female thornbills do not reproduce in their first year, but produce an average of four female offspring in each of their second and third years. The survival rate of each age group is estimated as 25% in their first year and 30% in their second year. A Leslie matrix, L, modelling the population distribution of the Tasmanian thornbill, has been partially completed.

$\mathbf{L} = \begin{bmatrix} 0 & 4 & 4 \\ x & 0 & 0 \\ 0 & y & 0 \end{bmatrix}$

Q14a

1 mark

State the values of $x$ and $y$ .

Reveal Answer

$x = 0.25$
$y = 0.3$

Marking Criteria

Descriptor	Marks
Correctly states the values of x and y	1

Q14b

4 marks

At the start of 2021, a study began into the population of Tasmanian thornbills. The study:

estimated that the initial female population was 510 in their first year, 480 in their second year and 420 in their third year
found that the ratio of male to female was approximately 1:2.

Estimate the total population of Tasmanian thornbills at the start of 2025.

Reveal Answer

Number of females at the start of 2021
$N_0 = \begin{bmatrix} 510 \\ 480 \\ 420 \end{bmatrix}$

Number of females at the start of 2025
$N_4 = L^4 N_0$
$= \begin{bmatrix} 2166 \\ 938.25 \\ 81.45 \end{bmatrix}$

Total female population $\approx 2166 + 938 + 81$
$\approx 3185$

Total population $\approx \frac{3}{2} \times 3185 \approx 4778$

Marking Criteria

Descriptor	Marks
Correctly identifies the initial matrix	1
Determines female population in their first, second and third years at the start of 2025	1
Determines total female population at the start of 2025	1
Determines total population at the start of 2025, rounded to a whole number	1

Q6

2024

SCSA

Paper 1

6 marks

Q6b

The third equation in part (a) on page 8 is changed to $2x - ky + z = 11$ where $k$ is a real constant. The first two equations remain unchanged.

Ryan decided to solve this changed system of equations and obtained correctly the statement $(k + 1)y = -4$ .

Q6a

3 marks

Solve the system of equations:
$x + y + z = 4$
$x - y - z = 2$
$2x - 3y + z = 11$ .

Reveal Answer

Consider $(1)+(2): \quad \therefore 2x = 6 \quad \therefore x = 3$

Using $x = 3:$
$(1): \quad y+z = 1$
$(3): \quad -3y+z = 5$
$(1)-3: \quad 4y = -4$
$\therefore y = -1, z = 2$

Hence the solution is $x = 3$ , $y = -1$ , $z = 2$

Marking Criteria

Descriptor	Marks
eliminates a variable correctly using an appropriate technique	1
solves correctly for the first variable	1
solves correctly for the second and third variables	1

Q6b

1 mark

Determine the value of the constant $k$ so that the changed system of equations does not have a unique solution.

Reveal Answer

$(k+1)y = -4$

Hence, for there to be no unique solution we require $k+1 = 0$
i.e. $k = -1$

Marking Criteria

Descriptor	Marks
states that $k = -1$	1

Q6c

2 marks

For the value of $k$ determined from part (b), state the geometric interpretation of the solution of the three simultaneous equations.

Reveal Answer

For $k = -1$ there is NO solution to the simultaneous equations.

Since the planes represented by the equations are NOT parallel, then the geometric significance is that the non-parallel planes have no intersection.

i.e. the planes in pairs intersect in lines, but that these lines do not intersect.

Marking Criteria

Descriptor	Marks
states that there is no intersection	1
states that the planes are not parallel OR states the planes in pairs intersect in parallel lines	1

QCAA Specialist Mathematics Further matrices

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