VCAA General Mathematics Matrices

15 sample questions with marking guides and sample answers

Q30
2025
VCAA
Paper 1
1 mark
Q30
1 mark

FF is a 4×44 \times 4 matrix.

The element in row ii and column jj of FF is fijf_{ij}.

The elements are determined by the rule fij=i2−jf_{ij} = i^2 - j.

How many of the elements in FF will be negative?

A

2

B

3

C

5

D

16

Reveal Answer
A

2

Incorrect. There are more than 2 negative elements; for i=1i=1, the elements f12f_{12}, f13f_{13}, and f14f_{14} are all negative.

B

3

Correct Answer

Correct. An element is negative when i2<ji^2 < j. Given 1≤i,j≤41 \le i, j \le 4, this is only true for i=1i=1 with j=2,3,4j=2, 3, 4, resulting in exactly 3 negative elements.

C

5

Incorrect. This overcounts the negative elements, perhaps by incorrectly evaluating i=2i=2 where i2−j=4−j≥0i^2 - j = 4 - j \ge 0 for all valid jj.

D

16

Incorrect. This represents the total number of elements in the 4×44 \times 4 matrix (4×4=164 \times 4 = 16), not just the negative ones.

Q31
2023
VCAA
Paper 1
1 mark
Q31
1 mark

A species of bird has a life span of three years.

The females in this species do not reproduce in their first year but produce an average of four female offspring in their second year, and three in their third year.

The Leslie matrix, LL, below is used to model the female population distribution of this species of bird.

L=[0430.20000.40]L = \begin{bmatrix} 0 & 4 & 3 \\ 0.2 & 0 & 0 \\ 0 & 0.4 & 0 \end{bmatrix}

The element in the second row, first column states that on average 20% of this population will

A

be female.

B

never reproduce.

C

survive into their second year.

D

produce offspring in their first year.

E

live for the entire lifespan of three years.

Reveal Answer
A

be female.

The Leslie matrix already models only the female population, so this element does not represent the proportion of females.

B

never reproduce.

This element represents a survival rate between age classes, not the proportion of birds that never reproduce.

C

survive into their second year.

Correct Answer

In a Leslie matrix, the subdiagonal elements represent survival rates. The element in the second row, first column (L2,1L_{2,1}) specifically represents the survival rate from the first age class to the second age class.

D

produce offspring in their first year.

The reproduction rate for the first year is represented by the element in the first row, first column (L1,1L_{1,1}), which is 0.

E

live for the entire lifespan of three years.

The probability of living the entire lifespan would be the product of the survival rates (0.2×0.4=0.080.2 \times 0.4 = 0.08), not just the survival rate from year one to year two.

Q30
2023
VCAA
Paper 1
1 mark
Q30
1 mark

How many of the following statements are true?

  • All square matrices have an inverse.
  • The inverse of a matrix could be the same as the transpose of that matrix.
  • If the determinant of a matrix is equal to zero, then the inverse does not exist.
  • It is possible to take the inverse of an identity matrix.
A

0

B

1

C

2

D

3

E

4

Reveal Answer
A

0

This is incorrect because there are exactly three true statements, not zero.

B

1

This is incorrect because there are exactly three true statements, not one.

C

2

This is incorrect because there are exactly three true statements, not two.

D

3

Correct Answer

This is correct because statements 2, 3, and 4 are true. Statement 1 is false because a square matrix must have a non-zero determinant to be invertible.

E

4

This is incorrect because the first statement is false. Square matrices with a determinant of zero (singular matrices) do not have an inverse.

Q13
2025
VCAA
Paper 2
4 marks
Q13

The three friends are staying at a resort classified as 'Luxury'.

At the start of every year, a classification board classifies resorts as either Luxury (LL), Mid-range (MM), Standard (SS) or Budget (BB).

The transition matrix, TT, shows the expected proportions of resorts that remain in the same classification or change classification from one year to the next.

this yearLMSBT=[0.90.30.100.10.6df00.1e0.10000.8]LMSB next year\begin{matrix} & \text{this year} \\ & \begin{matrix} L & M & S & B \end{matrix} \\ \end{matrix} \\ T = \begin{bmatrix} 0.9 & 0.3 & 0.1 & 0 \\ 0.1 & 0.6 & d & f \\ 0 & 0.1 & e & 0.1 \\ 0 & 0 & 0 & 0.8 \end{bmatrix} \begin{matrix} L \\ M \\ S \\ B \end{matrix} \text{ next year}
Q13a
1 mark

Interpret the meaning of element t14t_{14}.

Reveal Answer

No resorts are classified as Budget one year and then Luxury the next.

Marking Criteria
DescriptorMarks

Correctly interprets the meaning of element t14t_{14}, stating that no resorts are classified as Budget one year and then Luxury the next.

1
Q13b
1 mark

In the long term, what is predicted to happen to the number of resorts classified as Budget?

Reveal Answer

There will not be any resorts classified as Budget.

Marking Criteria
DescriptorMarks

Correctly predicts that there will not be any resorts classified as Budget in the long term.

1
Q13c
1 mark

Let S0S_0 be the state matrix that shows the number of resorts classified as either LL, MM, SS or BB at the start of 2024.

Let S1S_1 be the state matrix that shows the expected number of resorts classified as either LL, MM, SS or BB at the start of 2025.

S0=[401209070]S1=[811196456]S_0 = \begin{bmatrix} 40 \\ 120 \\ 90 \\ 70 \end{bmatrix} \quad S_1 = \begin{bmatrix} 81 \\ 119 \\ 64 \\ 56 \end{bmatrix}

Using this information, determine the value of dd in transition matrix TT.

Reveal Answer

0.4

Marking Criteria
DescriptorMarks

Determines the correct value of dd as 0.4.

1
Q13d
1 mark

What is the expected percentage growth in the number of Luxury resorts from the start of 2024 to the start of 2026?

Write your answer correct to one decimal place.

Reveal Answer

187.5%

Marking Criteria
DescriptorMarks

Calculates the correct expected percentage growth as 187.5%.

1
Q27
2024
VCAA
Paper 1
1 mark
Q27
1 mark

Consider the following matrix, where h≠0h \neq 0.

[4g8h]\begin{bmatrix} 4 & g \\ 8 & h \end{bmatrix}

The inverse of this matrix does not exist when gg is equal to

A

−2h-2h

B

h2\frac{h}{2}

C

hh

D

2h2h

Reveal Answer
A

−2h-2h

Incorrect. Setting g=−2hg = -2h results in a determinant of 20h20h, which is non-zero since h≠0h \neq 0, meaning the inverse would exist.

B

h2\frac{h}{2}

Correct Answer

Correct. A matrix does not have an inverse when its determinant is zero. Setting the determinant 4h−8g=04h - 8g = 0 and solving for gg yields g=h2g = \frac{h}{2}.

C

hh

Incorrect. Setting g=hg = h results in a determinant of −4h-4h, which is non-zero since h≠0h \neq 0, meaning the inverse would exist.

D

2h2h

Incorrect. Setting g=2hg = 2h results in a determinant of −12h-12h, which is non-zero since h≠0h \neq 0, meaning the inverse would exist.

Q31
2025
VCAA
Paper 1
1 mark
Q31
1 mark

A course requires students to pass two levels of study before they graduate.

Each level of study takes one year.

• Students must pass the first level (L1L1) before they move on to the second level (L2L2).
• Students must pass the second level before they become graduates (GG).

From one year to the next, it is known that

• 80% of students pass the first level of study and progress to the second level of study
• 60% of students in the second level of study become graduates the following year.

Students who do not pass a level of study repeat it in the following year. No student leaves the course before graduating.

What percentage of students are expected to graduate without repeating a level of study?

A

36%

B

48%

C

60%

D

64%

Reveal Answer
A

36%

This is incorrect. This would be the probability if both levels had a 60% pass rate (0.60×0.60=0.360.60 \times 0.60 = 0.36).

B

48%

Correct Answer

This is correct. To graduate without repeating, a student must pass both levels on their first attempt. The probability of this occurring is 0.80×0.60=0.480.80 \times 0.60 = 0.48, or 48%.

C

60%

This is incorrect. This is only the percentage of students who pass the second level, ignoring the requirement to also pass the first level without repeating.

D

64%

This is incorrect. This would be the probability if both levels had an 80% pass rate (0.80×0.80=0.640.80 \times 0.80 = 0.64).

Q26
2024
VCAA
Paper 1
1 mark
Q26
1 mark

A market stall sells three types of candles.

The cost of each type of candle is shown in matrix CC below.

C=[253243]C = \begin{bmatrix} 25 & 32 & 43 \end{bmatrix}

Towards the end of the day, the cost of each item is discounted by 15%.

Which one of the following expressions can be used to determine each discounted price?

A

0.15C0.15C

B

0.85C0.85C

C

8.5C8.5C

D

15C15C

Reveal Answer
A

0.15C0.15C

Multiplying by 0.15 calculates the amount of the discount itself, not the final discounted price.

B

0.85C0.85C

Correct Answer

A 15% discount means the final price is 85% of the original cost, which is calculated by multiplying the cost matrix CC by the scalar 0.85.

C

8.5C8.5C

Multiplying by 8.5 calculates 850% of the original price, which is a massive price increase rather than a discount.

D

15C15C

Multiplying by 15 calculates 1500% of the original price, rather than applying a 15% discount.

Q13
2024
VCAA
Paper 2
3 marks
Q13

A farmer has four fields for chickens to roam in during the day.
The fields are named North (NN), East (EE), South (SS) and West (WW).
The following transition matrix is used to determine the expected proportion of chickens that will either change fields from one day to the next, or stay in the same field on consecutive days.
There are initially 10000 chickens in each field.

this dayNESWT=NESW[0.300.120.300.200.140.280.250.440.240.400.120.170.320.200.330.19]next day\begin{matrix} & \text{this day} \\ & \begin{matrix} N & E & S & W \end{matrix} \\ T = \begin{matrix} N \\ E \\ S \\ W \end{matrix} & \begin{bmatrix} 0.30 & 0.12 & 0.30 & 0.20 \\ 0.14 & 0.28 & 0.25 & 0.44 \\ 0.24 & 0.40 & 0.12 & 0.17 \\ 0.32 & 0.20 & 0.33 & 0.19 \end{bmatrix} & \text{next day} \end{matrix}
Q13a
1 mark

After one day, how many chickens are expected to remain in the same field that they were in initially?

Reveal Answer

8900

Marking Criteria
DescriptorMarks

Correctly calculates the expected number of chickens to remain in the same field: 8900

1
Q13b
1 mark

After two days, 8962 chickens are expected to be in North field.
How many of these 8962 chickens are expected to have been in North field initially?

Reveal Answer

3000×0.3+1400×0.12+2400×0.3+3200×0.2=24283000 \times 0.3 + 1400 \times 0.12 + 2400 \times 0.3 + 3200 \times 0.2 = 2428

Marking Criteria
DescriptorMarks

Correctly calculates the expected number of chickens: 2428

1
Q13c
1 mark

Of the chickens expected to be in East field after three days, what is the percentage of chickens that were also expected to be in East field after both one day and two days?
Round your answer to the nearest whole number.

Reveal Answer

After 3 days 11292.8 in East field

After 1 day 11100 in East field

11100×0.28211292.8≈0.077\frac{11100 \times 0.28^2}{11292.8} \approx 0.077

therefore 8%.

Marking Criteria
DescriptorMarks

Correctly calculates the percentage, rounded to the nearest whole number: 8%

1
Q31
2025
VCAA
Paper 1
1 mark
Q31
1 mark

Consider the following four matrices.

A=[46−2−3]B=[428]C=[502531]D=[3−25]A = \begin{bmatrix} 4 & 6 \\ -2 & -3 \end{bmatrix} \quad B = \begin{bmatrix} 4 \\ 2 \\ 8 \end{bmatrix} \quad C = \begin{bmatrix} 5 & 0 \\ 2 & 5 \\ 3 & 1 \end{bmatrix} \quad D = \begin{bmatrix} 3 & -2 & 5 \end{bmatrix}

Which one of the following computations is defined?

A

D(A+B)D(A + B)

B

CTBDC^T B D

C

C+BDC + BD

D

(DA)T+D(DA)^T + D

Reveal Answer
A

D(A+B)D(A + B)

Matrix addition requires both matrices to have the exact same dimensions. Since AA is a 2×22 \times 2 matrix and BB is a 3×13 \times 1 matrix, the sum A+BA + B is undefined.

B

CTBDC^T B D

Correct Answer

The transpose CTC^T is 2×32 \times 3, BB is 3×13 \times 1, and DD is 1×31 \times 3. The product CTBC^T B results in a 2×12 \times 1 matrix, which can then be multiplied by the 1×31 \times 3 matrix DD to yield a valid 2×32 \times 3 matrix.

C

C+BDC + BD

The product BDBD results in a 3×33 \times 3 matrix, but CC is a 3×23 \times 2 matrix. Matrix addition is undefined for matrices of different dimensions.

D

(DA)T+D(DA)^T + D

Matrix multiplication requires the number of columns in the first matrix to equal the number of rows in the second. Since DD has 3 columns and AA has 2 rows, the product DADA is undefined.

Q32
2025
VCAA
Paper 1
1 mark
Q32
1 mark

A course requires students to pass two levels of study before they graduate.

Each level of study takes one year.

• Students must pass the first level (L1L1) before they move on to the second level (L2L2).
• Students must pass the second level before they become graduates (GG).

From one year to the next, it is known that

• 80% of students pass the first level of study and progress to the second level of study
• 60% of students in the second level of study become graduates the following year.

Students who do not pass a level of study repeat it in the following year. No student leaves the course before graduating.

At the start of 2025, the numbers of students enrolled in L1L1 and L2L2 of the course and who have graduated from the course are given by the recurrence relation

S0=[100000]L1L2,GSn+1=TSn+[k00]S_0 = \begin{bmatrix} 1000 \\ 0 \\ 0 \end{bmatrix} \begin{matrix} L1 \\ L2, \\ G \end{matrix} \quad S_{n+1} = TS_n + \begin{bmatrix} k \\ 0 \\ 0 \end{bmatrix}

where

• TT is the transition matrix showing the expected proportion of students moving between states L1L1, L2L2 and GG from one year to the next
• SnS_n is the state matrix nn years after the start of 2025.

To ensure that there are always students studying the course, new students are enrolled into the first level every year so that the number of students studying the first level remains at 1000.

kk is the number of new students enrolled at the first level each year.

At the start of which year will the expected number of students who have graduated from the course first exceed 5000?

A

2032

B

2033

C

2034

D

2035

Reveal Answer
A

2032

At the start of 2032 (n=7n=7), the expected number of graduates is approximately 4269, which has not yet exceeded 5000.

B

2033

Correct Answer

To maintain 1000 students in L1L1, k=800k = 800. By calculating the state matrices iteratively (Sn+1=TSn+KS_{n+1} = TS_n + K), the number of graduates at the start of 2033 (n=8n=8) is approximately 5068, making it the first year the total exceeds 5000.

C

2034

While the number of graduates does exceed 5000 at the start of 2034 (approximately 5867), it is not the first year this occurs.

D

2035

While the number of graduates does exceed 5000 at the start of 2035 (approximately 6667), it is not the first year this occurs.

Q26
2023
VCAA
Paper 1
1 mark
Q26
1 mark

Matrix PP is a permutation matrix and matrix QQ is a column matrix.

P=[1000000100000100100000001]Q=[teams]P = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \quad Q = \begin{bmatrix} t \\ e \\ a \\ m \\ s \end{bmatrix}

When QQ is multiplied by PP, which three letters change position?

A

t,e,at, e, a

B

e,a,me, a, m

C

a,m,sa, m, s

D

m,s,tm, s, t

E

e,a,se, a, s

Reveal Answer
A

t,e,at, e, a

Incorrect. The first row of PP has a 1 in the first column, meaning the first letter tt remains in its original position.

B

e,a,me, a, m

Correct Answer

Correct. Multiplying PP by QQ yields the column matrix [t,a,m,e,s]T[t, a, m, e, s]^T, showing that ee, aa, and mm are the letters that change positions.

C

a,m,sa, m, s

Incorrect. The fifth row of PP has a 1 in the fifth column, meaning the last letter ss remains in its original position.

D

m,s,tm, s, t

Incorrect. Both tt and ss remain in their original positions since the first and fifth rows of PP have 1s on the main diagonal.

E

e,a,se, a, s

Incorrect. The letter ss does not change position because the fifth row of PP is [0,0,0,0,1][0, 0, 0, 0, 1].

Q10
2025
VCAA
Paper 2
4 marks
Q10

Three friends from Melbourne, Australia, are planning an overseas holiday to Phuket, Thailand.

Matrix FF below shows the base cost, in dollars, of flying either one-way (OO) or return (RR) from Melbourne to Phuket on three separate airlines, Airline BB, Airline EE and Airline LL.

ORF=[4608905209506101100]BEL\begin{matrix} & O & R \\ \end{matrix} \\ F = \begin{bmatrix} 460 & 890 \\ 520 & 950 \\ 610 & 1100 \end{bmatrix} \begin{matrix} B \\ E \\ L \end{matrix}
Q10a
1 mark

Use the information in matrix FF to complete the table below.

AirlineOne-way base cost ($)Return base cost ($)
Airline BB  
Airline EE  
Airline LL  
Reveal Answer

  
460890
520950
6101100
Marking Criteria
DescriptorMarks

Completes the table correctly with the values from matrix FF:

AirlineOne-way base cost ($)Return base cost ($)
Airline BB460890
Airline EE520950
Airline LL6101100
1
Q10b (i)
1 mark

There is an additional fee for each airfare.

One-way flights have an additional fee of $30 and return flights have an additional fee of $60.

The cost with the additional fee can be calculated as follows

F+A=[49095055010106401160]F + A = \begin{bmatrix} 490 & 950 \\ 550 & 1010 \\ 640 & 1160 \end{bmatrix}

Write down matrix AA in the space below.

A=A =

Reveal Answer

[306030603060]\begin{bmatrix} 30 & 60 \\ 30 & 60 \\ 30 & 60 \end{bmatrix}

Marking Criteria
DescriptorMarks

Writes the correct matrix AA: [306030603060]\begin{bmatrix} 30 & 60 \\ 30 & 60 \\ 30 & 60 \end{bmatrix}

1
Q10b (ii)
1 mark

There will also be a 2% payment processing fee.

The total amount to be paid per flight, including this fee, is given by the following calculation

p×[49095055010106401160]p \times \begin{bmatrix} 490 & 950 \\ 550 & 1010 \\ 640 & 1160 \end{bmatrix}

Determine the value of pp.

Reveal Answer

1.02

Marking Criteria
DescriptorMarks

Determines the correct value of pp: 1.021.02

1
Q10c
1 mark

Airline BB has optional upgrades as shown in the table below. Prices are per person per one-way flight.

Seat selection ($)Extra baggage ($)Priority boarding ($)
303545

The three friends will select upgrades for their flights.

The total cost of the upgrades is $130.

The following calculation is performed to determine the total cost of all their upgrades.

[303545]×U=130\begin{bmatrix} 30 & 35 & 45 \end{bmatrix} \times U = 130

Where UU is a matrix representing the total number of each type of upgrade selected by the three friends.

Write down matrix UU in the space below.

U=U =

Reveal Answer

[220]\begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix}

Marking Criteria
DescriptorMarks

Writes the correct matrix UU: [220]\begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix}

1
Q28
2025
VCAA
Paper 1
1 mark
Q28
1 mark

A shop sells mattresses.

The mattresses come in four sizes, single (SS), double (DD), queen (QQ) and king (KK).

The mattresses come in two comfort levels, lush (LL) and firm (FF).

Matrix BB below shows the number of each type of mattress sold in February 2025.

B=[4515871233126918]LFB = \begin{bmatrix} 45 & 15 & 87 & 12 \\ 33 & 12 & 69 & 18 \end{bmatrix} \begin{matrix} L \\ F \end{matrix}
with columns S,D,Q,KS, D, Q, K.

Matrix CC gives the predicted number of each type of mattress that will be sold during February 2026.

C=[60201161644169224]LFC = \begin{bmatrix} 60 & 20 & 116 & 16 \\ 44 & 16 & 92 & 24 \end{bmatrix} \begin{matrix} L \\ F \end{matrix}
with columns S,D,Q,KS, D, Q, K.

Matrix CC could have been calculated using which one of the following equations?

A

C=310BC = \frac{3}{10} B

B

C=13BC = \frac{1}{3} B

C

C=1310BC = \frac{13}{10} B

D

C=43BC = \frac{4}{3} B

Reveal Answer
A

C=310BC = \frac{3}{10} B

Multiplying matrix BB by 310\frac{3}{10} would result in smaller values, such as 45×310=13.545 \times \frac{3}{10} = 13.5, which does not match the corresponding entry of 6060 in matrix CC.

B

C=13BC = \frac{1}{3} B

Multiplying matrix BB by 13\frac{1}{3} would result in smaller values, such as 45×13=1545 \times \frac{1}{3} = 15, which does not match the corresponding entry of 6060 in matrix CC.

C

C=1310BC = \frac{13}{10} B

Multiplying matrix BB by 1310\frac{13}{10} would result in 45×1310=58.545 \times \frac{13}{10} = 58.5, which does not match the corresponding entry of 6060 in matrix CC.

D

C=43BC = \frac{4}{3} B

Correct Answer

Multiplying each element in matrix BB by the scalar 43\frac{4}{3} yields the corresponding elements in matrix CC (e.g., 45×43=6045 \times \frac{4}{3} = 60 and 15×43=2015 \times \frac{4}{3} = 20).

Q26
2025
VCAA
Paper 1
1 mark
Q26
1 mark

Matrix KK is a square matrix.

Multiplying KK by its inverse will result in

A

a row matrix.

B

a column matrix.

C

the determinant.

D

an identity matrix.

Reveal Answer
A

a row matrix.

Multiplying two square matrices of the same dimensions results in another square matrix, not a single row matrix.

B

a column matrix.

Multiplying two square matrices of the same dimensions results in another square matrix, not a single column matrix.

C

the determinant.

The determinant is a scalar (a single number) associated with a square matrix, not the matrix product of a matrix and its inverse.

D

an identity matrix.

Correct Answer

By definition, multiplying a square matrix by its inverse always results in the identity matrix, denoted as II (i.e., KK−1=IK K^{-1} = I).

Q27
2025
VCAA
Paper 1
1 mark
Q27
1 mark

Consider the matrix EE where

E=[m−94n]E = \begin{bmatrix} m & -9 \\ 4 & n \end{bmatrix}

For the inverse of EE to exist, the values of mm and nn, respectively, cannot be

A

3 and 12

B

12 and 3

C

3 and -12

D

-3 and -12

Reveal Answer
A

3 and 12

Incorrect. The determinant of EE is mn−(−36)=mn+36mn - (-36) = mn + 36. For m=3m=3 and n=12n=12, the determinant is 7272, which is non-zero, meaning the inverse exists.

B

12 and 3

Incorrect. The determinant of EE is mn+36mn + 36. For m=12m=12 and n=3n=3, the determinant is 7272, which is non-zero, meaning the inverse exists.

C

3 and -12

Correct Answer

Correct. A matrix does not have an inverse if its determinant is zero. The determinant of EE is mn+36mn + 36, and for m=3m=3 and n=−12n=-12, the determinant is 3(−12)+36=03(-12) + 36 = 0.

D

-3 and -12

Incorrect. The determinant of EE is mn+36mn + 36. For m=−3m=-3 and n=−12n=-12, the determinant is (−3)(−12)+36=72(-3)(-12) + 36 = 72, which is non-zero, meaning the inverse exists.

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