How many QCAA Specialist Mathematics Alternative Sequence questions cover Further matrices?

AusGrader has 16 QCAA Specialist Mathematics Alternative Sequence questions on Further matrices, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics Alternative Sequence — Further matrices Questions & Answers

Q14

2020

QCAA

Paper 2

4 marks

Q14

Consider the matrix equation $\mathbf{M}^3\mathbf{X} + \mathbf{N} = \mathbf{M}\mathbf{X} - \mathbf{N}$ .

Q14a

2 marks

Use matrix algebra to solve the equation for $\mathbf{X}$ .

Reveal Answer

\text{Rearrange the equation:}

\mathbf{M}^3\mathbf{X} + \mathbf{N} = \mathbf{M}\mathbf{X} - \mathbf{N}

\mathbf{M}^3\mathbf{X} - \mathbf{M}\mathbf{X} = -2\mathbf{N}

(\mathbf{M}^3 - \mathbf{M})\mathbf{X} = -2\mathbf{N}

\mathbf{X} = (\mathbf{M}^3 - \mathbf{M})^{-1}(-2\mathbf{N}) \ (\forall \det(\mathbf{M}^3 - \mathbf{M})^{-1} \neq 0)

Marking Criteria

Descriptor	Marks
correctly rearranges the equation and uses the common factor of X	1
correctly solves the equation for X	1

Q14b

1 mark

Given the matrices below, use the result from 14a) to calculate $\mathbf{X}$ .

$\mathbf{M} = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & 0 \\ 0 & -1 & 2 \end{bmatrix}$ and $\mathbf{N} = \begin{bmatrix} 60 \\ -5 \\ 12 \end{bmatrix}$

Reveal Answer

\mathbf{X} = \begin{bmatrix} 1 \\ 12 \\ 17 \end{bmatrix}

Marking Criteria

Descriptor	Marks
calculates X	1

Q14c

1 mark

Calculate $\det(\mathbf{M})$ .

Reveal Answer

\det(\mathbf{M}) = 10

Marking Criteria

Descriptor	Marks
correctly calculates the determinant of M	1

Q2

2020

QCAA

Paper 2

1 mark

Q2

1 mark

The Leslie matrix for a certain endangered species is given.

$\mathbf{L} = \begin{bmatrix} 0.8 & 2.4 & 0.3 \\ 0.4 & 0 & 0 \\ 0 & 0.55 & 0 \end{bmatrix}$

A group of the species was moved into a secure property at the start of 2018. The initial female population is given.

$\mathbf{N}_0 = \begin{bmatrix} 150 \\ 80 \\ 40 \end{bmatrix}$

The best estimate of the total female population at the start of 2025 is

A

3000

B

4000

C

5000

D

6000

Reveal Answer

A

3000

This underestimates the population. It is closer to the total population after 6 years ( $\mathbf{N}_6 \approx 2850$ ), rather than the required 7 years for the start of 2025.

B

4000

Correct Answer

The population at the start of 2025 is found by calculating $\mathbf{N}_7 = \mathbf{L}^7 \mathbf{N}_0$ . The sum of the resulting matrix elements is approximately 4221, which is best estimated as 4000.

C

5000

This overestimates the population. It likely results from a calculation error or using an incorrect power for the Leslie matrix.

D

6000

This overestimates the population. It is closer to the total population after 8 years ( $\mathbf{N}_8 \approx 6238$ ), rather than the required 7 years.

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 $\mathbf{X}\mathbf{A} = \mathbf{B}$ , we must multiply both sides on the right by the inverse of $\mathbf{A}$ . Computing $\mathbf{X} = \mathbf{B}\mathbf{A}^{-1}$ yields this correct matrix.

B

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

This is the result of computing $\mathbf{A}^{-1}\mathbf{B}$ . Because matrix multiplication is not commutative, multiplying on the left by $\mathbf{A}^{-1}$ does not correctly isolate $\mathbf{X}$ .

C

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

This matrix is the result of computing $\mathbf{A}\mathbf{B}$ . To solve for $\mathbf{X}$ , you must multiply $\mathbf{B}$ by the inverse of $\mathbf{A}$ , not by $\mathbf{A}$ itself.

D

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

This matrix is the result of computing $\mathbf{B}\mathbf{A}$ . To isolate $\mathbf{X}$ , you need to multiply $\mathbf{B}$ by the inverse of $\mathbf{A}$ ( $\mathbf{A}^{-1}$ ), rather than $\mathbf{A}$ .

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

Q7

2022

QCAA

Paper 2

1 mark

Q7

1 mark

The determinant of $\begin{bmatrix} 0 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ is

A

2

B

1

C

-1

D

-2

Reveal Answer

A

2

Incorrect. This result comes from incorrectly calculating the $2 \times 2$ minor determinant as $1$ instead of $-1$ .

B

1

Incorrect. This is not the correct determinant; it might result from an arithmetic error or incorrect cofactor expansion.

C

-1

Incorrect. This is the value of the $2 \times 2$ minor, but it must be multiplied by the coefficient $2$ from the first row.

D

-2

Correct Answer

Correct. Expanding along the first row gives $2 \cdot \det\left(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\right) = 2 \cdot (0 \cdot 1 - 1 \cdot 1) = -2$ .

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 $\mathbf{M}$ .

\begin{matrix} & & \text{Losing teams} \\ & & \begin{matrix} \text{P} & \text{Q} & \text{R} & \text{S} & \text{T} \end{matrix} \\ \mathbf{M} = \text{Winning teams} & \begin{matrix} \text{P} \\ \text{Q} \\ \text{R} \\ \text{S} \\ \text{T} \end{matrix} & \begin{bmatrix} 0 & 1 & 2 & 0 & 2 \\ 1 & 0 & 0 & 1 & 1 \\ 0 & 2 & 0 & 0 & 0 \\ 2 & 1 & 2 & 0 & 2 \\ 0 & 1 & 2 & 0 & 0 \end{bmatrix} \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

By calculating the row sums of $\mathbf{M} + \mathbf{M}^2 + \mathbf{M}^3$ , the total scores are S (97), Q (61), P (59), R (38), and T (37). Ordering these from highest to lowest gives the correct ranking of S, Q, P, R, T.

B

S, Q, P, T, R

This ranking incorrectly places T ahead of R. Calculating the row sums of $\mathbf{M} + \mathbf{M}^2 + \mathbf{M}^3$ shows that R has a higher total score (38) than T (37).

C

S, P, Q, T, R

This ranking incorrectly places P ahead of Q and T ahead of R. The row sums of $\mathbf{M} + \mathbf{M}^2 + \mathbf{M}^3$ reveal that Q (61) scored higher than P (59), and R (38) scored higher than T (37).

D

S, P, Q, R, T

This ranking incorrectly places P ahead of Q. Evaluating the model $\mathbf{M} + \mathbf{M}^2 + \mathbf{M}^3$ shows that Q has a higher total score (61) compared to P (59).

Q16

2021

QCAA

Paper 1

6 marks

Q16

Three planes intersect in the line $l$ .

$\begin{aligned}-x+2y+3z&=2\\x+3y+2z&=-7\\2x-y+mz&=n\end{aligned}$

Q16a

3 marks

Use a Gaussian technique of elimination to determine the values of $m$ and $n$ .

Reveal Answer

Expressing the equations as an augmented matrix:
$\begin{bmatrix} -1 & 2 & 3 & | & 2 \\ 1 & 3 & 2 & | & -7 \\ 2 & -1 & m & | & n \end{bmatrix} \quad \begin{matrix} R_1 \\ R_2 \\ R_3 \end{matrix}$
$\begin{bmatrix} -1 & 2 & 3 & | & 2 \\ 0 & 5 & 5 & | & -5 \\ 0 & 3 & m + 6 & | & n + 4 \end{bmatrix} \quad \begin{matrix} R_1 \\ R_2 \rightarrow R_2 + R_1 \\ R_3 \rightarrow R_3 + 2R_1 \end{matrix}$
$\begin{bmatrix} -1 & 2 & 3 & | & 2 \\ 0 & 1 & 1 & | & -1 \\ 0 & 3 & m + 6 & | & n + 4 \end{bmatrix} \quad \begin{matrix} R_1 \\ R_2 \rightarrow \frac{1}{5}R_2 \\ R_3 \end{matrix}$
$\begin{bmatrix} -1 & 2 & 3 & | & 2 \\ 0 & 1 & 1 & | & -1 \\ 0 & 0 & m + 3 & | & n + 7 \end{bmatrix} \quad \begin{matrix} R_1 \\ R_2 \\ R_3 \rightarrow R_3 - 3R_2 \end{matrix}$
As the planes intersect in a line, there are infinitely many solutions so the values in the last row must all be 0
$\therefore m = -3, n = -7$

Marking Criteria

Descriptor	Marks
correctly expresses the equations as an augmented matrix	1
establishes augmented matrix with two 0s in the third row	1
determines values of $m$ and $n$	1

Q16b

3 marks

Determine the equation of the line $l$ in Cartesian form.

Reveal Answer

Determining the equation of the line, $l$
Let $z = t$
From $R_2$ : $y + z = -1 \Rightarrow y = -1 - z$
$\therefore y = -1 - t \Rightarrow t = -y - 1$
From $R_1$ : $-x + 2y + 3z = 2 \Rightarrow x = 2y + 3z - 2$
$\therefore x = 2(-1 - t) + 3t - 2$
$\therefore x = t - 4 \Rightarrow t = x + 4$
Cartesian equation of the line is
$x + 4 = -y - 1 = z$

Marking Criteria

Descriptor	Marks
expresses $y$ in terms of a parameter	1
expresses $x$ in terms of a parameter	1
determines a Cartesian equation of the line	1

QCAA Specialist Mathematics Alternative Sequence Further matrices

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