Matrix types, order, the transpose and constructing matrices for VCE General Mathematics Unit 4 Matrices

Syllabus dot point

“Use the order of a matrix, element notation and the types of matrices (row, column, square, diagonal, symmetric, triangular, zero, binary, permutation and identity) and the transpose, represent tabular information as a matrix, and construct a matrix from a rule for its ij element”

VCEGeneral MathematicsUnit 4 Matrices15 min readUpdated 23 Sep 2026

Quick answer

A matrix has order rows $\times$ columns and $a_{ij}$ is the element in row $i$ , column $j$ . Know the ten types: row, column, square, zero, identity, diagonal, symmetric ( $A^T = A$ ), triangular, binary and permutation; many matrices are several types at once. The transpose turns rows into columns. To build a matrix from a rule, substitute each row and column number, row by row.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

Before you can multiply matrices, invert them or use them to model populations, you need the language. VCAA wants you to state the order of a matrix, read and write elements using $a_{ij}$ notation, recognise every type of matrix named in the study design (row, column, square, diagonal, symmetric, triangular, zero, binary, permutation and identity), find the transpose, store tabular information in a matrix and interpret any element in context, and construct a matrix from a rule for its $ij$ element.

These skills are examined directly in Examination 1 almost every year (2023 and 2025 each had several of them) and are assumed in every Examination 2 matrices question.

The answer

Order and elements

A matrix is a rectangular array of numbers. Its order is written rows $\times$ columns. A matrix with 3 rows and 4 columns has order $3 \times 4$ and contains $3 \times 4 = 12$ elements.

Each element is named by its position: $a_{ij}$ is the element of matrix $A$ in row $i$ and column $j$ . The row always comes first.

In this café sales matrix, $s_{23}$ sits in row 2 (tea) and column 3 (Wednesday), so $s_{23} = 18$ means 18 teas were sold on Wednesday. Swapping the subscripts gives a different element: $s_{32}$ is row 3 (juice), column 2 (Tuesday), which is 11.

Storing tabular information

Any table of numbers can be written as a matrix, as long as you keep track of what the rows and columns mean. The labels are not part of the matrix, so a question will tell you (or you should state) what each row and column represents.

Sales, stock or prices: rows might be products and columns might be days, shops or sizes.
Networks: rows and columns might both be towns, and the element records a road, a distance or a communication link (this leads to adjacency, communication and dominance matrices).
Populations: a column matrix can hold the numbers in each group, which is the state matrix used later with transition and Leslie matrices.

Once a table is a matrix, matrix operations do the bookkeeping: multiplying a sales matrix by a price matrix gives revenue, and so on (see the matrix arithmetic page).

The types of matrix

The study design names ten types. Several matrices belong to more than one type, and exam questions test exactly that.

Type	Definition	Example
Row matrix	Exactly one row, order $1 \times n$	$\begin{bmatrix} 2 & 5 & 7 \end{bmatrix}$
Column matrix	Exactly one column, order $m \times 1$	$\begin{bmatrix} 4 \\ 1 \end{bmatrix}$
Square matrix	Same number of rows and columns, $n \times n$	$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
Zero matrix $O$	Every element is 0 (any order)	$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
Identity matrix $I$	Square; 1s on the leading diagonal, 0s elsewhere	$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Diagonal matrix	Square; every element off the leading diagonal is 0	$\begin{bmatrix} 4 & 0 \\ 0 & -1 \end{bmatrix}$
Symmetric matrix	Square; $a_{ij} = a_{ji}$ , so $A^T = A$	$\begin{bmatrix} 1 & 3 \\ 3 & 2 \end{bmatrix}$
Triangular matrix	Square; all 0s below (upper) or above (lower) the leading diagonal	$\begin{bmatrix} 2 & 5 \\ 0 & 3 \end{bmatrix}$
Binary matrix	Every element is 0 or 1 (any order)	$\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$
Permutation matrix	Square binary matrix with exactly one 1 in each row and each column	$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

The leading diagonal (main diagonal) runs from the top-left element to the bottom-right: $a_{11}, a_{22}, a_{33}, \ldots$ . It only exists for square matrices, which is why diagonal, symmetric, triangular, identity and permutation matrices must all be square.

Key fact

Order is rows $\times$ columns and $a_{ij}$ is row $i$ , column $j$ . Diagonal, symmetric, triangular, identity and permutation matrices must be square. Every diagonal matrix is also symmetric and triangular, and the identity matrix is diagonal, symmetric, triangular, binary and a permutation matrix all at once.

How the types overlap:

Identity $\subset$ diagonal $\subset$ symmetric, and diagonal matrices are also both upper and lower triangular.
Permutation $\subset$ binary, and permutation matrices are square. The identity is the permutation matrix that leaves everything in place.
A zero matrix that is square is diagonal, symmetric, triangular and binary as well. (It is not a permutation matrix, because it has no 1s.)

The transpose

The transpose of $A$ , written $A^T$ , is formed by turning the rows of $A$ into columns: row 1 of $A$ becomes column 1 of $A^T$ , row 2 becomes column 2, and so on. Equivalently $\left(A^T\right)_{ij} = a_{ji}$ .

A = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \qquad\Rightarrow\qquad A^T = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}

If $A$ has order $m \times n$ , then $A^T$ has order $n \times m$ . A matrix is symmetric exactly when it equals its own transpose, $A^T = A$ . A distance table between towns is symmetric (the distance from P to Q equals the distance from Q to P), and so is the adjacency matrix of an undirected graph. On a CAS, the transpose is a menu command, which is handy for checking symmetry in a large matrix.

Constructing a matrix from a rule

Sometimes a matrix is defined by a formula for its elements, such as $a_{ij} = 2i - j$ or $f_{ij} = i^2 - j$ . To build it:

Use the order to decide the ranges: for a $3 \times 2$ matrix, $i = 1, 2, 3$ and $j = 1, 2$ .
Work row by row, substituting each $j$ in turn.
For a conditional rule (" $a_{ij} = 1$ if $i < j$ , otherwise 0"), check the condition for each position. $i < j$ means above the leading diagonal, $i = j$ is on it, and $i > j$ is below it.

For example, with $b_{ij} = 2i - j$ and order $3 \times 3$ :

B = \begin{bmatrix} 1 & 0 & -1 \\ 3 & 2 & 1 \\ 5 & 4 & 3 \end{bmatrix}.

Useful patterns: any rule that stays the same when $i$ and $j$ are swapped (like $a_{ij} = i + j$ or $a_{ij} = |i - j|$ ) produces a symmetric matrix; " $= 0$ when $i \ne j$ " produces a diagonal matrix; " $= 0$ when $i > j$ " produces an upper triangular matrix.

Worked examples: storing, reading, classifying and constructing

Storing a table and interpreting elements

A café sells coffee, tea and juice. Sales from Monday to Thursday are: coffee 42, 38, 45, 51; tea 15, 12, 18, 14; juice 9, 11, 7 and 13.

Build the matrix. One row per drink, one column per day:

S = \begin{bmatrix} 42 & 38 & 45 & 51 \\ 15 & 12 & 18 & 14 \\ 9 & 11 & 7 & 13 \end{bmatrix}

Order. $3 \times 4$ .

Interpret. $s_{14} = 51$ is the number of coffees sold on Thursday. $s_{32} = 11$ is the number of juices sold on Tuesday. The total tea sales are the sum of row 2: $15 + 12 + 18 + 14 = 59$ .

Marker's note: when a question asks what an element "indicates", answer in context, with both the row meaning and the column meaning.

Classifying matrices

Classify $P = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$ and $Q = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ .

$P$ . Square; zeros off the leading diagonal, so diagonal; therefore also symmetric and triangular. Not binary (it contains 5s), so not identity or permutation. (It is $5I$ , a scalar multiple of the identity.)

$Q$ . Square and binary, with exactly one 1 in every row and every column: a permutation matrix. Not symmetric: $q_{13} = 1$ but $q_{31} = 0$ . Not triangular or diagonal.

Marker's note: in multiple-choice questions every option can sound plausible; test each definition against the actual elements.

Constructing from a conditional rule

A $4 \times 4$ matrix has $a_{ij} = |i - j|$ . Write it down and describe it.

Build. Row 1: $0, 1, 2, 3$ . Row 2: $1, 0, 1, 2$ . Row 3: $2, 1, 0, 1$ . Row 4: $3, 2, 1, 0$ .

A = \begin{bmatrix} 0 & 1 & 2 & 3 \\ 1 & 0 & 1 & 2 \\ 2 & 1 & 0 & 1 \\ 3 & 2 & 1 & 0 \end{bmatrix}

Describe. It is square and symmetric (swapping $i$ and $j$ does not change $|i - j|$ ), with zeros on the leading diagonal. It is exactly the distance matrix for four stations equally spaced 1 km apart along a straight line.

Marker's note: before building a large matrix, look for a shortcut like symmetry. It halves the work and gives a built-in check.

Counting elements that satisfy a condition

A $4 \times 4$ matrix has $f_{ij} = i^2 - j$ . How many elements are negative?

Reason by rows. Row $i$ is $i^2 - 1, i^2 - 2, i^2 - 3, i^2 - 4$ . For $i = 1$ these are $0, -1, -2, -3$ : three negatives. For $i = 2$ the smallest is $4 - 4 = 0$ , not negative. Rows 3 and 4 are all positive.

Answer. 3 negative elements.

Marker's note: this is 2025 Examination 1 Question 30. Writing out the whole matrix takes 30 seconds and removes any doubt about the boundary case (0 is not negative).

Common traps

Swapping row and column: $a_{23}$ is row 2, column 3. Reading it as column 2, row 3 is the most common element error.
Writing the order as columns by rows: A matrix with 2 rows and 5 columns is $2 \times 5$ , not $5 \times 2$ .
Calling a non-square matrix diagonal, symmetric or triangular: These types need a leading diagonal, which only square matrices have.
Calling any binary matrix a permutation matrix: A permutation matrix needs exactly one 1 in every row and every column.
Thinking a matrix can only be one type: The identity is diagonal, symmetric, triangular, binary and a permutation matrix. Multiple-choice questions often reward spotting every type that applies.
Getting the ranges wrong in a rule: For a $3 \times 2$ matrix, $i$ goes to 3 and $j$ goes to 2.

Exam technique

For any element question, write the row meaning and the column meaning next to the matrix before answering. For classification questions, check in this order: is it square? binary? what is on and off the leading diagonal? is it equal to its transpose? For a rule question, write out the full matrix even if only one element or a count is asked for; in Examination 1 your CAS can also build it (define the rule and use the matrix constructor), which is a fast check.

Note

A matrix is just a table of numbers without the labels. Its size is given as rows first, then columns, and each number has an address: $a_{23}$ means "go to row 2, then along to column 3". Different shapes and patterns get names: a single row, a single column, a square, all zeros, only numbers down the diagonal, or a mirror-image pattern. Some tables have a recipe for their numbers, like "row number times 2, minus column number", and you build them by filling in the recipe one spot at a time.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA-style1 mark

The daily maximum temperatures at a town for two weeks are stored in a

2 \times 7

matrix

M

. Row 1 is week 1 and row 2 is week 2; columns 1 to 7 are Monday to Sunday. Week 1: 20, 17, 23, 20, 18, 19, 30. Week 2: 29, 27, 28, 21, 20, 20, 22 (degrees Celsius). Element

m_{21}

indicates that: A. the temperature was 29 °C on Monday in week 2 B. the temperature was 17 °C on Tuesday in week 1 C. the lowest temperature for these two weeks was 17 °C D. the highest temperature for these two weeks was 29 °C E. week 2 had a higher average maximum temperature than week 1

Show worked answer →

The subscripts are row first, then column. $m_{21}$ is in row 2 (week 2) and column 1 (Monday), and its value is 29. So it tells us the temperature was 29 °C on Monday of week 2: option A.

Option B is $m_{12}$ (row 1, column 2), the classic row and column swap. Options C, D and E describe the whole matrix, which no single element can do. $89\%$ of students answered correctly.

Source: VCAA 2023 General Mathematics Examination 1, Question 25, and the 2023 examination report.

2025 VCAA-style1 mark

Consider the matrix

G = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

. Which one of the following correctly describes

G

? A. a binary matrix B. a permutation matrix C. an identity matrix D. a diagonal matrix

Show worked answer →

Every element is 0 or 1, so $G$ is a binary matrix: option A.

It is not a permutation matrix, because row 2 contains two 1s and row 3 contains none (a permutation matrix has exactly one 1 in every row and every column). It is not the identity (the identity has 1s on the leading diagonal and 0s elsewhere) and it is not diagonal (it has non-zero elements off the leading diagonal). $83\%$ of students answered correctly.

Source: VCAA 2025 General Mathematics Examination 1, Question 25, and the 2025 examination report.

2025 VCAA-style1 mark

F

is a

4 \times 4

matrix. The element in row

i

and column

j

F

f_{ij}

, and the elements are determined by the rule

f_{ij} = i^2 - j

. How many of the elements in

F

will be negative? A. 2 B. 3 C. 5 D. 16

Show worked answer →

Build $F$ row by row. In row $i$ every element starts from $i^2$ and subtracts the column number $j = 1, 2, 3, 4$ :

F = \begin{bmatrix} 0 & -1 & -2 & -3 \\ 3 & 2 & 1 & 0 \\ 8 & 7 & 6 & 5 \\ 15 & 14 & 13 & 12 \end{bmatrix}

Only row 1 has negative elements ( $-1$ , $-2$ and $-3$ ). Row 2 reaches 0 in column 4, but 0 is not negative. There are 3 negative elements: option B. $70\%$ of students answered correctly.

Source: VCAA 2025 General Mathematics Examination 1, Question 30, and the 2025 examination report.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A café records the number of coffees, teas and juices sold on Monday, Tuesday, Wednesday and Thursday in matrix

S

, with a row for each drink (in that order) and a column for each day (in that order): coffee 42, 38, 45, 51; tea 15, 12, 18, 14; juice 9, 11, 7, 13. (a) State the order of

S

. (b) What does

s_{23}

represent, and what is its value?

Show worked solution →

(a) There are 3 rows (drinks) and 4 columns (days), so $S$ is a $3 \times 4$ matrix. (1 mark) The order is always rows by columns.

(b) $s_{23}$ is row 2 (tea), column 3 (Wednesday): the number of teas sold on Wednesday, which is 18. (1 mark)

foundation3 marks

Classify each matrix using as many of these words as apply: row, column, square, zero, identity, diagonal, symmetric, triangular, binary. (a)

\begin{bmatrix} 4 & 0 \\ 0 & -1 \end{bmatrix}

(b)

\begin{bmatrix} 2 & 5 & 7 \end{bmatrix}

(c)

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

Show worked solution →

(a) Square, diagonal (all non-zero elements are on the leading diagonal), and therefore also symmetric and triangular (both upper and lower). Not binary, because of the 4 and $-1$ . (1 mark)

(b) A row matrix, order $1 \times 3$ . Not square, so none of the square-matrix types apply. (1 mark)

(c) The $3 \times 3$ identity matrix $I$ . It is also square, diagonal, symmetric, triangular, binary and a permutation matrix. (1 mark)

foundation2 marks

Write down the transpose of

A = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}

and state its order.

Show worked solution →

Rows become columns: row 1 of $A$ becomes column 1 of $A^T$ , and so on.

A^T = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}

(1 mark) $A$ is $3 \times 2$ , so $A^T$ is $2 \times 3$ . (1 mark)

core2 marks

Construct the

3 \times 3

matrix

B

whose elements are given by

b_{ij} = 2i - j

Show worked solution →

Substitute each row number $i$ and column number $j$ :

Row 1: $2(1) - 1 = 1$ , $2(1) - 2 = 0$ , $2(1) - 3 = -1$
Row 2: $4 - 1 = 3$ , $4 - 2 = 2$ , $4 - 3 = 1$
Row 3: $6 - 1 = 5$ , $6 - 2 = 4$ , $6 - 3 = 3$

B = \begin{bmatrix} 1 & 0 & -1 \\ 3 & 2 & 1 \\ 5 & 4 & 3 \end{bmatrix}

(2 marks: 1 for correct method, 1 for all nine elements.)

core2 marks

3 \times 3

matrix

C

is defined by

c_{ij} = 1

i < j

and

c_{ij} = 0

otherwise. (a) Write down

C

. (b) Which two types of matrix is

C

Show worked solution →

(a) A 1 appears only where the row number is less than the column number, which is above the leading diagonal:

C = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

(1 mark)

(b) It is binary (only 0s and 1s) and upper triangular (every element below the leading diagonal is 0). (1 mark)

core2 marks

The road distances, in km, between towns P, Q and R are P to Q 12, P to R 20 and Q to R 9. (a) Write a

3 \times 3

matrix

D

of these distances, with rows and columns in the order P, Q, R, and 0 for a town to itself. (b) Explain why

D

is symmetric.

Show worked solution →

(a)

D = \begin{bmatrix} 0 & 12 & 20 \\ 12 & 0 & 9 \\ 20 & 9 & 0 \end{bmatrix}

(1 mark)

(b) The distance from P to Q is the same as from Q to P, so $d_{ij} = d_{ji}$ for every pair: the matrix is a mirror image across its leading diagonal, which means $D^T = D$ . (1 mark)

exam1 mark

Matrix

K

is a

3 \times 2

matrix whose elements are given by

k_{ij} = (i - j)^2

. The sum of all the elements of

K

is: A. 4 B. 5 C. 6 D. 7

Show worked solution →

K = \begin{bmatrix} (1-1)^2 & (1-2)^2 \\ (2-1)^2 & (2-2)^2 \\ (3-1)^2 & (3-2)^2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 4 & 1 \end{bmatrix}

The sum is $0 + 1 + 1 + 0 + 4 + 1 = 7$ : option D.

The common error is building a $2 \times 3$ matrix by mistake (3 columns instead of 3 rows). The order $3 \times 2$ means $i$ runs from 1 to 3 and $j$ from 1 to 2.

exam1 mark

How many of the following statements are true? (1) Every diagonal matrix is symmetric. (2) Every symmetric matrix is diagonal. (3) The transpose of a

2 \times 5

matrix is a

5 \times 2

matrix. (4) The identity matrix is a permutation matrix. A. 1 B. 2 C. 3 D. 4

Show worked solution →

(1) True. A diagonal matrix has zeros everywhere off the diagonal, so it matches its mirror image.
(2) False. $\begin{bmatrix} 1 & 3 \\ 3 & 2 \end{bmatrix}$ is symmetric but not diagonal.
(3) True. Transposing swaps the numbers of rows and columns.
(4) True. The identity has exactly one 1 in each row and each column.

Three statements are true: option C.

Practise this

Sources & how we know this

VCE General Mathematics Examination 1 (2023) — VCAA (2023)
VCE General Mathematics Examination 1 (2025) — VCAA (2025)
VCE Mathematics Study Design: General Mathematics (from 2023) — VCAA
VCE General Mathematics: examination specifications, past examinations and reports — VCAA