Permutation and binary matrices: VCE General Mathematics Unit 4 Matrices

Syllabus dot point

“Recognise and use a permutation matrix as a binary matrix with exactly one 1 in each row and column, apply it to reorder the entries of a state matrix, and identify the matrix that reverses or repeats the reordering”

14 min readUpdated 23 Sep 2026

Quick answer

A binary matrix has only 0s and 1s. A permutation matrix is a square binary matrix with exactly one 1 in every row and every column. Pre-multiplying a column matrix by $P$ reorders it: the 1 in row $i$ , column $j$ puts element $j$ into position $i$ , and a 1 on the leading diagonal leaves that element where it is. $P^2$ applies the reordering twice, and $P^n = I$ means the original order returns after $n$ steps. The inverse of $P$ is its transpose, which undoes the reordering.

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What this dot point is asking
The answer
Binary matrices in other topics
Exam-style questions
Practice questions

What this dot point is asking

The Study Design lists "binary and permutation matrices, and their properties and applications" under Matrices and their applications, and names both in its list of matrix types. You are expected to:

recognise a binary matrix and a permutation matrix, and tell them apart from identity and diagonal matrices;
apply a permutation matrix to reorder the elements of a column (or row) matrix;
reverse a reordering with the inverse, which for a permutation matrix is its transpose;
use powers of a permutation matrix to repeat a reordering, and decide when the original order returns.

This is a small topic that turns up every year, usually as a one-mark multiple-choice item (2025 Examination 1 Question 25, $83\%$ correct; 2023 Examination 1 Question 26, $90\%$ ) and sometimes as a short-answer question in context (2023 Examination 2 Question 9c, $68\%$ ; 2025 Examination 2 Question 14, where parts b and c were answered correctly by only $27\%$ and $11\%$ ). The multiple-choice items are well done. The context questions are where marks go, because students multiply correctly but cannot say what $P^2$ or $P^4 = I$ means.

The answer

Binary matrices

A binary matrix is any matrix whose elements are all 0 or 1. Binary matrices record yes/no information: whether a student takes a subject, whether two people can communicate, whether one team beat another, whether two vertices of a graph are joined. Communication, dominance and adjacency matrices (for simple graphs) are all binary.

Because the elements are 0 or 1, row and column totals count things. Multiplying a binary matrix by a column of 1s gives its row totals; multiplying a row of 1s by it gives its column totals. That is the most common "application" of a binary matrix outside the permutation and dominance topics.

Permutation matrices

A permutation matrix is a square binary matrix with exactly one 1 in each row and exactly one 1 in each column. Every permutation matrix is binary; most binary matrices are not permutation matrices.

P = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} \text{ is a permutation matrix;} \qquad G = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \text{ is binary but not a permutation matrix.}

The identity matrix is a permutation matrix (it leaves everything in place). A permutation matrix is never a zero matrix, and it always has an inverse: its determinant is $1$ or $-1$ .

Definition

A permutation matrix is a square binary matrix with exactly one element equal to 1 in each row and each column. Multiplying a column matrix by it rearranges the elements without changing their values.

Reading a permutation matrix row by row

When a permutation matrix multiplies a column matrix, $PS$ , each row of $P$ picks exactly one element of $S$ :

\text{a 1 in row } i\text{, column } j \text{ of } P \text{ puts element } j \text{ of } S \text{ into position } i \text{ of } PS.

For $P$ above and $S = \begin{bmatrix} a & b & c \end{bmatrix}^T$ : row 1 has its 1 in column 2, so position 1 of $PS$ is $b$ ; row 2 picks column 3, giving $c$ ; row 3 picks column 1, giving $a$ . So $PS = \begin{bmatrix} b & c & a \end{bmatrix}^T$ .

A 1 on the leading diagonal (row $i$ , column $i$ ) keeps element $i$ in place. Counting the diagonal 1s tells you at a glance how many elements never move. In 2023 Examination 1 Question 26 the 1s in rows 1 and 5 were on the diagonal, so $t$ and $s$ stayed and $e, a, m$ moved.

Reading a movement matrix column by column

Some questions give a permutation matrix labelled like a transition matrix, with "this month" across the top and "next month" down the side (2023 Examination 2 Question 9c). Read it the transition-matrix way, by columns: the 1 in column $X$ sits in the row of the place $X$ goes to next. Both readings describe the same matrix; the labels tell you which question is being asked:

a column matrix of names in positions multiplied by $P$ : read rows ("position $i$ receives element $j$ ");
a from/to movement matrix: read columns ("from column, to row").

If you read a from/to matrix by rows you get the cycle backwards.

The inverse is the transpose

Undoing a reordering needs $P^{-1}$ . For a permutation matrix you never need the determinant formula:

P^{-1} = P^T.

Transposing turns each "row $i$ takes element $j$ " into "row $j$ takes element $i$ ", which sends every element back where it came from. You can check with CAS that $PP^T = I$ . For a swap of two elements (a matrix equal to its own transpose), $P^{-1} = P$ : doing the swap twice restores the order.

Powers repeat the reordering

If $S_{n+1} = PS_n$ , then $S_1 = PS_0$ , $S_2 = P^2S_0$ and in general $S_n = P^nS_0$ . So:

the order after two applications uses $P^2$ (day three of a program that starts on day one is $P^2X$ , not $P^3X$ );
if $P^n = I$ , then $S_n = S_0$ and the order repeats every $n$ steps.

The smallest such $n$ comes from the cycles of the permutation. Follow one element around until it returns: that is one cycle. The order returns when every cycle is complete, which is the lowest common multiple of the cycle lengths. A 3-cycle alone gives $n = 3$ ; a 2-cycle and a 3-cycle together give $\text{LCM}(2, 3) = 6$ .

Formula

$PS$ : a 1 in row $i$ , column $j$ puts element $j$ into position $i$ . Diagonal 1s mean "stays in place". $P^{-1} = P^T$ . With $S_{n+1} = PS_n$ : $S_n = P^nS_0$ . If $P^n = I$ the order repeats every $n$ steps, and the smallest such $n$ is the LCM of the cycle lengths.

The figure shows the permutation from 2025 Examination 2 Question 14, where seven activities are rearranged from one day to the next.

Worked examples: reading, reversing and repeating a permutation

Which three letters move? (2023 Examination 1 Question 26)

$P = \begin{bmatrix} 1&0&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&1&0&0&0 \\ 0&0&0&0&1 \end{bmatrix}$ and $Q = \begin{bmatrix} t & e & a & m & s \end{bmatrix}^T$ .

Step 1: read each row of $P$: Row 1 picks element 1 ( $t$ ). Row 2 picks element 3 ( $a$ ). Row 3 picks element 4 ( $m$ ). Row 4 picks element 2 ( $e$ ). Row 5 picks element 5 ( $s$ ).
Step 2: write the product: $PQ = \begin{bmatrix} t & a & m & e & s \end{bmatrix}^T$ .
Step 3: compare: $t$ and $s$ are unchanged (diagonal 1s in rows 1 and 5); $e$ , $a$ and $m$ have moved. Answer: B.

Following a movement matrix (2023 Examination 2 Question 9c)

Columns are "this month" $E, F, G, H, I$ ; rows are "next month". The matrix has 1s at (row $G$ , column $E$ ), (row $I$ , column $F$ ), (row $F$ , column $G$ ), (row $E$ , column $H$ ) and (row $H$ , column $I$ ).

Step 1: start at $I$ and read down column $I$ . The 1 is in row $H$ : the circus goes to $H$ .

Step 2: keep going. Column $H$ sends it to $E$ ; column $E$ to $G$ ; column $G$ to $F$ ; column $F$ back to $I$ .

Answer. I, H, E, G, F. All five towns appear before the circus returns to $I$ , so this is a single 5-cycle and the matrix to the power 5 is the identity.

A timetable that repeats (2025 Examination 2 Question 14)

Day one is $X = \begin{bmatrix} C & D & G & L & M & R & S \end{bmatrix}^T$ and each day's order is $P$ times the previous day's.

Step 1: the fixed activities: Rows 3, 4 and 5 of $P$ have their 1 on the leading diagonal, so gardening, lunch and music are always at 11 am, 12 pm and 1 pm.
Step 2: day two: Row 1 picks element 6 (R), row 2 picks element 7 (S), row 6 picks element 2 (D), row 7 picks element 1 (C): day two is R, S, G, L, M, D, C.
Step 3: day three: Apply the same rule to day two (or calculate $P^2X$ on CAS): D, C, G, L, M, S, R.
Step 4: the cycle: Follow cooking: 9 am on day one, 3 pm on day two, 10 am on day three, 2 pm on day four, 9 am again on day five. The four moving activities share one 4-cycle, so $P^4 = I$ . Day five is the same as day one, and the 40-day program is 10 full cycles of four days.

Marker's note: the report's answer for (c) was "Order of the activities rotates on a four-day cycle. There are 10 cycles completed in the 40-day program." Only $11\%$ gave an answer like this; most just restated that $P^4$ is the identity. Say what the identity does to the order.

Undoing a reordering

A list $A, B, C, D$ is reordered by $P = \begin{bmatrix} 0&0&1&0 \\ 1&0&0&0 \\ 0&0&0&1 \\ 0&1&0&0 \end{bmatrix}$ to give $C, A, D, B$ . Find the matrix that restores the list.

Step 1: transpose. $P^T = \begin{bmatrix} 0&1&0&0 \\ 0&0&0&1 \\ 1&0&0&0 \\ 0&0&1&0 \end{bmatrix}$ .

Step 2: check. $P^T$ applied to $C, A, D, B$ : row 1 picks element 2 ( $A$ ), row 2 picks element 4 ( $B$ ), row 3 picks element 1 ( $C$ ), row 4 picks element 3 ( $D$ ). The original list returns, so $P^{-1} = P^T$ .

When does the order come back?

$P = \begin{bmatrix} 0&1&0&0&0 \\ 1&0&0&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&1&0&0 \end{bmatrix}$ acts on $V, W, X, Y, Z$ each day.

Step 1: find the cycles: Rows 1 and 2 swap the first two people (length 2). Rows 3, 4 and 5 rotate the last three (length 3).
Step 2: combine: Both cycles are complete together after $\text{LCM}(2, 3) = 6$ days.
Step 3: check the tempting wrong answers: After 2 days the swap is undone but the rotation is not ( $V, W, Z, X, Y$ ); after 3 days the rotation is undone but the swap is not ( $W, V, X, Y, Z$ ). Only day 6 gives $V, W, X, Y, Z$ .

Common traps

Calling any 0-1 matrix a permutation matrix: Check every row and every column for exactly one 1. A matrix with a zero row or two 1s in a column is binary, not permutation.
Reading a movement matrix by rows: If the columns are "this month" and the rows "next month", the 1 in a column shows where that place goes. Reading rows gives the cycle in reverse.
Using the wrong power: Day three is two steps after day one: $P^2X$ . The state after $n$ steps is $P^nS_0$ .
Restating instead of interpreting: " $P^4$ is the identity matrix" is the given information. The mark is for "the order repeats every four days" (and, in context, how many cycles fit in the program).
Multiplying on the wrong side: $PS$ needs $S$ to be a column matrix. $SP$ with a column matrix $S$ is not defined; with a row matrix it reorders using the columns of $P$ , which is the same as applying $P^T$ .
Inverting the long way: You do not need the determinant: $P^{-1} = P^T$ .

Exam technique

Before multiplying, circle the 1 in each row (or column, for a from/to matrix) and write the element it selects beside the row. The reordered list can then be read straight off, and diagonal 1s show which elements are fixed. For "when does it repeat" questions, follow one element around its cycle, then check the others; the answer is the LCM of the cycle lengths. Confirm on CAS by raising $P$ to that power and checking you get the identity.

Binary matrices in other topics

Permutation matrices are only one kind of binary matrix. The same 0-1 idea appears in:

communication matrices, where a 1 means one person can send a message directly to another;
dominance matrices, where a 1 means one competitor beat another;
adjacency matrices of simple graphs in Networks, where a 1 means two vertices are joined (a graph with multiple edges or loops has entries larger than 1, so its adjacency matrix is not binary).

In each case the row and column totals count something, and powers of the matrix count multi-step links. The difference is that a permutation matrix has the strict one-per-row-and-column structure, so its powers just shuffle the order again, while the powers of a communication or dominance matrix grow.

Note

A permutation matrix is a grid of 0s and 1s with exactly one 1 in every row and every column. Multiplying a list by it shuffles the list: the 1 in each row points to the item that goes into that spot. A 1 on the diagonal means that item stays put. Doing the shuffle twice uses the matrix squared. If some power of the matrix is the identity, the list is back in its original order after that many shuffles. To undo a shuffle, swap the matrix's rows and columns (take the transpose).

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.

2025 VCAA Exam 1 Q25Past paper1 mark

Consider the matrix

G

where

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

. Which one of the following correctly describes matrix

G

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

Show worked answer →

Test each definition against $G$ .

Binary: every element is 0 or 1. True for $G$ .
Permutation: square, binary, and exactly one 1 in every row and every column. Row 2 has two 1s and row 3 has none, so $G$ is not a permutation matrix.
Identity: 1s on the leading diagonal and 0s elsewhere. The leading diagonal of $G$ is all 0s.
Diagonal: every element off the leading diagonal is 0. $G$ has non-zero elements off the diagonal.

The answer is A. $83\%$ answered correctly; $9\%$ chose B. Every permutation matrix is binary, but a binary matrix is a permutation matrix only if it passes the one-per-row-and-column test.

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

2023 VCAA Exam 1 Q26Past paper1 mark

Matrix

P

is a permutation matrix and matrix

Q

is a column matrix.

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}

Q = \begin{bmatrix} t \\ e \\ a \\ m \\ s \end{bmatrix}

. When

Q

is multiplied by

P

, which three letters change position? A.

t, e, a

e, a, m

a, m, s

m, s, t

e, a, s

Show worked answer →

" $Q$ is multiplied by $P$ " means $PQ$ ( $5 \times 5$ times $5 \times 1$ ). Read $P$ one row at a time: the column holding the 1 says which element of $Q$ goes into that position.

Row 1: 1 in column 1, so position 1 is $t$ .
Row 2: 1 in column 3, so position 2 is $a$ .
Row 3: 1 in column 4, so position 3 is $m$ .
Row 4: 1 in column 2, so position 4 is $e$ .
Row 5: 1 in column 5, so position 5 is $s$ .

$PQ = \begin{bmatrix} t & a & m & e & s \end{bmatrix}^T$ . The letters $t$ and $s$ stay (their rows have the 1 on the leading diagonal) and $e, a, m$ move: option B ( $90\%$ correct).

Source: VCAA 2023 General Mathematics Examination 1, Question 26, and the 2023 examination report.

2023 VCAA Exam 2 Q9cPast paper1 mark

The circus moves from one location to the next each month. It rotates through each of the five locations, before starting the cycle again. The following matrix displays the movement between the five locations (columns: this month

E, F, G, H, I

; rows: next month

E, F, G, H, I

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

The circus started in town

I

. What is the order in which the circus will visit the five towns?

Show worked answer →

This is a movement matrix, so read it by columns: the 1 in the column for "this month" sits in the row for "next month".

Column $I$ : the 1 is in row $H$ , so $I \to H$ .
Column $H$ : the 1 is in row $E$ , so $H \to E$ .
Column $E$ : the 1 is in row $G$ , so $E \to G$ .
Column $G$ : the 1 is in row $F$ , so $G \to F$ .
Column $F$ : the 1 is in row $I$ , back to the start.

Order: I, H, E, G, F (then back to I). $68\%$ scored the mark. Reading the rows instead gives the reverse cycle, I, F, G, E, H, which is the order in which the circus arrives from, not goes to.

(Part d, adding a sixth location, was invalidated by VCAA after an error was found in the question.)

Source: VCAA 2023 General Mathematics Examination 2, Question 9c, and the 2023 examination report.

2025 VCAA Exam 2 Q14Past paper3 marks

The early learning centre runs seven different activities during its 40-day holiday program: cooking (C), drama (D), gardening (G), lunch (L), music (M), reading (R) and sport (S). The timetabled order for day one (9 am to 3 pm) is C, D, G, L, M, R, S, shown in column matrix

X

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

is a permutation matrix: a column matrix of one day's order is multiplied by

P

to give the next day's order. a. State the activities that are always held at the same time on each day. b. Determine the order on day three. c.

P^4

is an identity matrix. Explain what this means for the order over the 40-day program.

Show worked answer →

(a) Rows 3, 4 and 5 of $P$ have their 1 on the leading diagonal, so those positions never change: gardening, lunch and music. (1 mark; $65\%$ .)

(b) Day two is $PX$ : row 1 takes element 6 (R), row 2 takes element 7 (S), rows 3 to 5 keep G, L, M, row 6 takes element 2 (D), row 7 takes element 1 (C). Day two: R, S, G, L, M, D, C. Day three is $P(PX) = P^2X$ : apply the same rule to day two to get D, C, G, L, M, S, R (drama, cooking, gardening, lunch, music, sport, reading). (1 mark; only $27\%$ . Day three needs $P^2$ , not $P^3$ .)

(c) $P^4 = I$ means $P^4X = X$ : the order on day five is the same as on day one, so the timetable repeats on a four-day cycle. The 40-day program is exactly 10 complete cycles. (1 mark; only $11\%$ .)

Source: VCAA 2025 General Mathematics Examination 2, Question 14, 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

Which of these matrices are permutation matrices? Give a reason for each.

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

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

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

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

Show worked solution →

Check that each row and each column contains exactly one 1.

$A$ : one 1 in every row and every column. Permutation matrix.
$B$ : every row has one 1, but column 2 has two 1s and column 3 has none. Not a permutation matrix (it would copy the second entry twice and lose the third).
$C$ : one 1 in every row and column. Permutation matrix (it swaps the first and third entries).
$D$ : row 1 has two 1s and row 3 has none. Not a permutation matrix.

All four are binary matrices. (2 marks: 1 for the correct pair $A$ and $C$ , 1 for the reasons.)

foundation3 marks

A coach lists four players as

S = \begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix}

and applies

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

. (a) Find

PS

. (b) Write down the matrix that reverses the reordering. (c) Show that it works.

Show worked solution →

(a) Row 1 picks element 3 ( $C$ ), row 2 picks element 1 ( $A$ ), row 3 picks element 4 ( $D$ ), row 4 picks element 2 ( $B$ ): $PS = \begin{bmatrix} C \\ A \\ D \\ B \end{bmatrix}$ . (1 mark)

(b) The inverse of a permutation matrix is its transpose: $P^{-1} = P^T = \begin{bmatrix} 0&1&0&0 \\ 0&0&0&1 \\ 1&0&0&0 \\ 0&0&1&0 \end{bmatrix}$ . (1 mark)

(c) $P^T$ applied to $C, A, D, B$ : row 1 picks element 2 ( $A$ ), row 2 picks element 4 ( $B$ ), row 3 picks element 1 ( $C$ ), row 4 picks element 3 ( $D$ ). The result is $A, B, C, D$ , the original order. (1 mark) A CAS check: $P P^T = I$ .

core2 marks

Let

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

. (a) Find

P \begin{bmatrix} 10 \\ 20 \\ 30 \end{bmatrix}

. (b) Find

\begin{bmatrix} 10 & 20 & 30 \end{bmatrix} P

. (c) Explain why the answers are in different orders.

Show worked solution →

(a) Row by row: $\begin{bmatrix} 20 \\ 30 \\ 10 \end{bmatrix}$ .

(b) Element $j$ of the product is the row matrix times column $j$ of $P$ . Column 1 has its 1 in row 3, so element 1 is $30$ ; column 2 has its 1 in row 1, giving $10$ ; column 3 has its 1 in row 2, giving $20$ : $\begin{bmatrix} 30 & 10 & 20 \end{bmatrix}$ .

(c) Pre-multiplying ( $PS$ ) reorders the rows of $S$ using the rows of $P$ ; post-multiplying ( $RP$ ) reorders the columns of $R$ using the columns of $P$ , which is the same as applying $P^T$ . (1 mark for (a) and (b), 1 mark for (c).)

core2 marks

Find the

4 \times 4

permutation matrix

Q

that changes the column matrix

\begin{bmatrix} W & X & Y & Z \end{bmatrix}^T

into

\begin{bmatrix} Z & W & X & Y \end{bmatrix}^T

, and state the smallest

n

for which

Q^n = I

Show worked solution →

Build $Q$ one row at a time: position 1 needs $Z$ (element 4), so row 1 has its 1 in column 4; position 2 needs $W$ (column 1); position 3 needs $X$ (column 2); position 4 needs $Y$ (column 3).

$Q = \begin{bmatrix} 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \end{bmatrix}$ . (1 mark)

Each application moves every letter one place down (and the last to the top), so after 4 applications every letter is home: $n = 4$ . (1 mark) Check: $Q^2$ gives $Y, Z, W, X$ , not yet the original.

exam3 marks

Five people V, W, X, Y and Z sit in a row,

S_0 = \begin{bmatrix} V & W & X & Y & Z \end{bmatrix}^T

. Each day they move according to

S_{n+1} = PS_n

, where

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

. (a) Find

S_1

. (b) Find the smallest

n

for which

S_n = S_0

. (c) Find

S_{19}

Show worked solution →

(a) $S_1 = \begin{bmatrix} W & V & Y & Z & X \end{bmatrix}^T$ . (1 mark)

(b) Split the movement into cycles. Positions 1 and 2 swap: a cycle of length 2. Positions 3, 4 and 5 rotate: a cycle of length 3. Everyone is home together only when both cycles are complete, after $\text{LCM}(2, 3) = 6$ days. (1 mark) Check: $S_2 = V, W, Z, X, Y$ and $S_3 = W, V, X, Y, Z$ ; neither is $S_0$ , but $S_6 = S_0$ .

(c) $19 = 3 \times 6 + 1$ , so $S_{19} = S_1 = \begin{bmatrix} W & V & Y & Z & X \end{bmatrix}^T$ . (1 mark)

exam3 marks

Four security guards rotate between stations N, E, S and W each week. The movement matrix (columns: this week N, E, S, W; rows: next week N, E, S, W) is

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

. (a) A guard starts at N. List the order of stations. (b) A fifth station X is added, visited after E and before W, with nothing else changed. Write the new

5 \times 5

movement matrix with columns and rows ordered N, E, S, W, X.

Show worked solution →

(a) Read columns: N $\to$ S (column N has its 1 in row S), S $\to$ E, E $\to$ W, W $\to$ N. Order: N, S, E, W then back to N. (1 mark)

(b) The new cycle is N $\to$ S $\to$ E $\to$ X $\to$ W $\to$ N. Put a 1 in row "to", column "from":

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

(2 marks: 1 for column E now sending to X, 1 for column X sending to W.) Check: every row and column still has exactly one 1, and the matrix now has order 5 (each guard returns after 5 weeks).

core2 marks

A binary matrix records which of four students (rows P, Q, R, T) study which of four subjects (columns: Maths, English, Art, Music), with 1 meaning "studies":

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

. (a) Calculate

B \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix}^T

and explain what it gives. (b) Calculate

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

and explain what it gives.

Show worked solution →

(a) Multiplying by a column of 1s adds each row: $\begin{bmatrix} 3 & 2 & 3 & 3 \end{bmatrix}^T$ . It is the number of subjects each student studies. (1 mark)

(b) A row of 1s in front adds each column: $\begin{bmatrix} 3 & 3 & 3 & 2 \end{bmatrix}$ . It is the number of students in each subject. (1 mark) Both totals add to $11$ , the number of 1s in $B$ .

Practise this

Sources & how we know this

VCE General Mathematics Examination 1 (2023) and examination report — VCAA (2023)
VCE General Mathematics Examination 2 (2023) and examination report — VCAA (2023)
VCE General Mathematics Examination 1 (2025) and examination report — VCAA (2025)
VCE General Mathematics Examination 2 (2025) and examination report — VCAA (2025)
VCE Mathematics Study Design: General Mathematics (from 2023) — VCAA
VCE General Mathematics: examination specifications, past examinations and reports — VCAA