Transition matrices (SACE Stage 2 General Mathematics)

Syllabus dot point

“Use transition matrices and an initial state vector to predict future states and to find the long-term steady state of a system.”

8 min readUpdated 2 Jun 2026

Quick answer

A transition matrix $T$ stores the proportions moving between categories each step. Multiply it by the current state $S_n$ to get the next state: $S_{n+1} = T S_n$ . After many steps the state usually settles to a steady state that stops changing.

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What this dot point is asking
Setting up a transition matrix
Predicting future states
The long-term steady state
Why a steady state exists
Interpreting the result in context
Exam-style questions

What this dot point is asking

You must build a transition matrix from movement data, predict the state after one or more steps, and find or interpret the long-term steady state.

Setting up a transition matrix

A system moves objects (customers, animals, voters) between a fixed set of categories at constant rates each period. The transition matrix $T$ records these rates. Using the convention $S_{n+1} = T S_n$ , the entry in row $i$ , column $j$ is the proportion moving from category $j$ into category $i$ . Each column therefore sums to 1, since everything currently in a category must go somewhere.

State vector and transition matrix

The state vector $S_n$ is a column listing the amount in each category after $n$ steps. The transition matrix $T$ holds the step-to-step proportions, with columns summing to 1. The recurrence is $S_{n+1} = T S_n$ , and after $n$ steps $S_n = T^n S_0$ .

Worked example

Two competing supermarkets

Each week 80% of shoppers at store A return to A and 20% switch to B; 90% at B return to B and 10% switch to A. Set up the transition matrix with order A then B.

Column A describes shoppers starting at A: 0.8 stay at A, 0.2 go to B. Column B describes shoppers starting at B: 0.1 go to A, 0.9 stay at B.

T = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{bmatrix}

Each column sums to 1, as it must.

Predicting future states

To advance one week, multiply the current state by $T$ . Repeat, or use $S_n = T^n S_0$ to jump ahead.

Worked example

One week ahead

Suppose 1000 shoppers start at A and 1000 at B, so $S_0 = \begin{bmatrix} 1000 \\ 1000 \end{bmatrix}$ . Find next week's distribution.

S_1 = T S_0 = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{bmatrix}\begin{bmatrix} 1000 \\ 1000 \end{bmatrix} = \begin{bmatrix} 0.8(1000) + 0.1(1000) \\ 0.2(1000) + 0.9(1000) \end{bmatrix} = \begin{bmatrix} 900 \\ 1100 \end{bmatrix}

After one week 900 shop at A and 1100 at B. The column totals still add to 2000, confirming no shoppers are lost.

The long-term steady state

As $n$ grows, the state vector usually settles to a fixed steady state $S$ that satisfies $S = TS$ and stops changing from week to week. You can find it by computing high powers $T^n S_0$ until the numbers stabilise, or by solving $S = TS$ with the constraint that the categories add to the total.

Steady state

The steady state $S$ satisfies $TS = S$ : applying the transition matrix leaves it unchanged. The total population stays constant, so for a population of $N$ split into two categories you solve $TS = S$ together with the two parts adding to $N$ .

Worked example

Long-run split of shoppers

Find the steady-state split of the 2000 shoppers. Let the steady state be $\begin{bmatrix} a \\ b \end{bmatrix}$ with $a + b = 2000$ .

From $TS = S$ , the first row gives $0.8a + 0.1b = a$ , so $0.1b = 0.2a$ , that is $b = 2a$ . Substituting into $a + b = 2000$ :

a + 2a = 2000 \implies 3a = 2000 \implies a \approx 667

So in the long run about 667 shop at A and about 1333 at B, and this split no longer changes week to week.

Why a steady state exists

The reason the system settles is that the transition matrix gently pulls the state toward a balance point. Whenever more leave a category than join it, that category shrinks; whenever more join than leave, it grows. The steady state is precisely where these flows balance for every category at once, so $TS = S$ . For a regular transition matrix - one where some power has all positive entries - this balance point is unique and is reached from any starting distribution, which is why the long-run market share does not depend on where the shoppers began. This is the key insight examiners want you to articulate when interpreting a steady state.

Interpreting the result in context

The steady state predicts the eventual market share regardless of where shoppers started, provided the transition rates stay constant. Examiners reward a sentence interpreting the numbers, for example "store B eventually holds about two-thirds of the market."

Use the column-sum check

After writing $T$ , add each column. If any column does not sum to 1 you have made an error, because everything in a category must move somewhere. This quick check catches most setup mistakes before you compute.

Common mistake

Do not mix up rows and columns when building $T$ under the $S_{n+1} = T S_n$ convention. With this convention the columns (not the rows) must sum to 1, and the entry in row $i$ , column $j$ is the proportion moving from $j$ into $i$ . Putting the "stay" and "leave" percentages in the wrong cells gives the wrong predictions.

Exam-style questions

Questions in the style of SACE Board 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.

Practice5 marks

Calculator-assumed. Each week 80% of shoppers at store A return to A and 20% switch to B; 90% at B return to B and 10% switch to A. Initially 1000 shop at A and 1000 at B. (a) Write the transition matrix

T

(order A then B). (b) Find the distribution after one week. (c) Find the long-term steady-state distribution of the 2000 shoppers.

Show worked answer →

(a) Columns describe where shoppers start: $T = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{bmatrix}$ , each column summing to 1. (1 mark)

(b) $S_1 = T S_0 = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{bmatrix}\begin{bmatrix} 1000 \\ 1000 \end{bmatrix} = \begin{bmatrix} 900 \\ 1100 \end{bmatrix}$ : 900 at A, 1100 at B. (2 marks)

(c) Steady state $\begin{bmatrix} a \\ b \end{bmatrix}$ with $a + b = 2000$ and $TS = S$ . The first row gives $0.8a + 0.1b = a$ , so $0.1b = 0.2a$ , i.e. $b = 2a$ . Then $3a = 2000$ , so $a \approx 667$ and $b \approx 1333$ . (2 marks)

Marks: one for $T$ with valid columns, two for the one-week product, two for solving the steady state.

Practice3 marks

Calculator-assumed. A transition matrix is

T = \begin{bmatrix} 0.7 & 0.4 \\ 0.3 & 0.6 \end{bmatrix}

with initial state

S_0 = \begin{bmatrix} 500 \\ 500 \end{bmatrix}

. Find the state after two steps,

S_2

Show worked answer →

First step: $S_1 = T S_0 = \begin{bmatrix} 0.7(500) + 0.4(500) \\ 0.3(500) + 0.6(500) \end{bmatrix} = \begin{bmatrix} 550 \\ 450 \end{bmatrix}$ . [1 mark]

Second step: $S_2 = T S_1 = \begin{bmatrix} 0.7(550) + 0.4(450) \\ 0.3(550) + 0.6(450) \end{bmatrix} = \begin{bmatrix} 565 \\ 435 \end{bmatrix}$ . [2 marks]

The total stays 1000 at each step, confirming no loss.

Practise this

Sources & how we know this

Stage 2 General Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 General Mathematics: external assessment — SACE Board