VICGeneral MathematicsSyllabus dot point

How do transition matrices model step-by-step change in a system, and how are future states and steady states found?

Set up a transition matrix to model a Markov system, use it with an initial state matrix to find the state after n steps, identify the steady-state or long-run distribution, and apply a recurrence including the case with additions or removals

A focused answer to the VCE General Mathematics Unit 4 Matrices key-knowledge point on transition matrices. Building a column-based transition matrix, finding the state after n steps, the steady state, and the recurrence model with regular additions or removals.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Building the transition matrix
Stepping forward
The steady state
The model with additions or removals
Why this matters for the exams

What this dot point is asking

VCAA wants you to model a system that moves between a fixed set of states each step using a transition matrix. You set up the matrix from given percentages, multiply it by a state matrix to step forward, find the state after $n$ steps, and identify the long-run steady state. You also handle the extended model where a fixed amount is added or removed each step. This is the most distinctive part of the Matrices module and a reliable source of Exam 2 marks.

Building the transition matrix

Suppose a system has states $A$ and $B$ . If $80\%$ of $A$ stays in $A$ and $20\%$ moves to $B$ , while $30\%$ of $B$ moves to $A$ and $70\%$ stays in $B$ , the transition matrix is

T = \begin{bmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{bmatrix},

where the first column describes what happens to $A$ and the second column what happens to $B$ . Each column sums to $1$ because everything must go somewhere.

Stepping forward

Let $S_0$ be the initial state matrix (a column listing how many or what proportion start in each state). The state after one step is $S_1 = T S_0$ , and after $n$ steps

S_n = T^n S_0.

Worked example

State after two steps

A town has $1000$ people split between gym membership ( $A$ ) and no membership ( $B$ ), starting as $S_0 = \begin{bmatrix} 600 \\ 400 \end{bmatrix}$ , with $T = \begin{bmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{bmatrix}$ .

After one step: $S_1 = T S_0 = \begin{bmatrix} 0.8\times 600 + 0.3\times 400 \\ 0.2\times 600 + 0.7\times 400 \end{bmatrix} = \begin{bmatrix} 480 + 120 \\ 120 + 280 \end{bmatrix} = \begin{bmatrix} 600 \\ 400 \end{bmatrix}.$

After two steps: $S_2 = T S_1 = \begin{bmatrix} 0.8\times 600 + 0.3\times 400 \\ 0.2\times 600 + 0.7\times 400 \end{bmatrix} = \begin{bmatrix} 600 \\ 400 \end{bmatrix}.$

The totals stay at $1000$ because each column of $T$ sums to $1$ . Here $S_0$ already happens to be the steady state.

The steady state

For a regular transition matrix the state matrix settles toward a fixed steady-state (long-run) distribution that no longer changes from step to step. In practice you find it by computing $S_n = T^n S_0$ for a large $n$ (for example $n = 30$ or $n = 50$ ) on the calculator and reading the values, which stabilise. The steady state is independent of the starting distribution for a regular matrix.

The model with additions or removals

Many questions add or remove a fixed amount each step, giving the recurrence

S_{n+1} = T S_n + B,

where $B$ is a column matrix of the amounts added (positive) or removed (negative) from each state every step. This mirrors the mixed financial recurrence from Unit 3: a multiplicative transition plus a constant adjustment. You step it forward one period at a time, usually on the calculator, because $T^n$ alone no longer captures the constant term.

Worked example

One step with a fixed addition

With $T = \begin{bmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{bmatrix}$ , $S_0 = \begin{bmatrix} 500 \\ 500 \end{bmatrix}$ and $50$ new members added to state $A$ each step, $B = \begin{bmatrix} 50 \\ 0 \end{bmatrix}$ .

$S_1 = T S_0 + B = \begin{bmatrix} 0.9\times 500 + 0.2\times 500 \\ 0.1\times 500 + 0.8\times 500 \end{bmatrix} + \begin{bmatrix} 50 \\ 0 \end{bmatrix} = \begin{bmatrix} 450 + 100 \\ 50 + 400 \end{bmatrix} + \begin{bmatrix} 50 \\ 0 \end{bmatrix} = \begin{bmatrix} 600 \\ 450 \end{bmatrix}.$

The total has grown from $1000$ to $1050$ because $50$ were added.

Why this matters for the exams

Transition matrix questions are a signature of the Matrices module and almost always appear on Exam 2, where the calculator does the heavy multiplication and powers. Markers reward a correctly oriented matrix, the right recurrence, and an interpretation in context (numbers of people, proportions). The added-or-removed model links neatly back to the recurrence relations you met in Unit 3.

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2025 VCAA1 marksAn activity program uses a transition matrix K that gives the expected proportion of children who will change between the activities cooking (C), gardening (G) and music (M) from one week to the next. What do the values on the leading diagonal in matrix K indicate?

Show worked answer →

In a transition matrix, the entry in row X and column X (a leading-diagonal entry) gives the proportion of those currently doing activity X who are still doing activity X in the next step.

So the leading-diagonal values give the proportion (or percentage) of children who stay in the same activity, that is, who do not change activity, from one week to the next.

For example, the cooking-to-cooking entry is the fraction of this week's cooking children who choose cooking again next week. The mark is for stating that these values represent the children who remain in (do not change) their activity.