If a transition matrix column reads bmatrix 0.7 \\ 0.3 bmatrix, what does it say? [1 point]

United StatesPrecalculusSyllabus dot point

How do matrices model real situations such as transitions between states over time?

Topic 4.14 Matrices Modeling Contexts: use matrices to model transitions between states, and apply repeated multiplication to project the state forward in time.

A focused answer to AP Precalculus Topic 4.14, covering how a transition matrix models movement between states, how multiplying a state vector by the matrix advances one step, and how repeated multiplication projects the system forward.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

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

Have a quick question? Jump to the Q&A page

Quick answer

Many situations involve a population split among categories (states) that shifts over time according to fixed rates. A state vector lists the amounts in each category, and a transition matrix holds the fractions moving between them: its entry in row $i$ , column $j$ is the fraction of category $j$ that becomes category $i$ next step. Multiplying the state vector by the transition matrix, $\mathbf{s}_{n+1} = T\mathbf{s}_n$ , advances the system one time step. To project $k$ steps ahead, multiply by the matrix raised to the $k$ -th power: $\mathbf{s}_{n+k} = T^k\mathbf{s}_n$ , because each multiplication adds one step. Each column of the transition matrix should sum to $1$ when nothing leaves the system, so the total population is conserved. This makes matrices a compact tool for modelling step-by-step change between states.

Jump to a section

What this topic is asking
State vectors and transition matrices
Advancing the system
Long-run behavior
Try this

What this topic is asking

The College Board (Topic 4.14) wants you to use matrices to model contexts, especially transitions between states. A state vector records how a population is distributed among categories, a transition matrix encodes the rates of movement between them, and multiplying advances the system one step. Repeated multiplication projects the state forward in time.

State vectors and transition matrices

Reading the matrix correctly matters: the columns are the sources and the rows are the destinations, so each column describes where one category's members go.

Advancing the system

This repeated-multiplication structure is why matrix powers, rather than scalar multiples, drive the long-run behavior of the model.

Projecting a population two steps ahead

A transition matrix is $T = \begin{bmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{bmatrix}$ with initial state $\mathbf{s}_0 = \begin{bmatrix} 100 \\ 100 \end{bmatrix}$ . Find the state after two steps.

step 1 First step

$\mathbf{s}_1 = T\mathbf{s}_0 = \begin{bmatrix} 0.9(100) + 0.2(100) \\ 0.1(100) + 0.8(100) \end{bmatrix} = \begin{bmatrix} 90 + 20 \\ 10 + 80 \end{bmatrix} = \begin{bmatrix} 110 \\ 90 \end{bmatrix}$ .

step 2 Second step

$\mathbf{s}_2 = T\mathbf{s}_1 = \begin{bmatrix} 0.9(110) + 0.2(90) \\ 0.1(110) + 0.8(90) \end{bmatrix} = \begin{bmatrix} 99 + 18 \\ 11 + 72 \end{bmatrix} = \begin{bmatrix} 117 \\ 83 \end{bmatrix}$ .

step 3 Check the total is conserved

Each step keeps the total at $200$ : $110 + 90 = 200$ and $117 + 83 = 200$ . The columns of $T$ each sum to $1$ , so no population is lost.

step 4 Note the equivalent power

The same result comes from $T^2\mathbf{s}_0$ : squaring $T$ and multiplying once gives $\begin{bmatrix} 117 \\ 83 \end{bmatrix}$ , confirming that two steps equal multiplication by $T^2$ .

Long-run behavior

Over many steps, many transition models approach a stable distribution, a state vector that no longer changes when multiplied by $T$ . The proportions settle even though individuals keep moving, much as an exponential model approaches its limit. AP Precalculus introduces the step-by-step mechanism; the qualitative idea that the system tends toward a steady state is the modelling insight to carry forward.

A point worth stating once is the conservation check built into a transition matrix: when each column sums to $1$ , the total population is preserved at every step, because every member of each category goes somewhere. If a column sums to less than $1$ , some population leaves the system; if more, it grows. Verifying the column sums is the quickest way to confirm a transition matrix is set up correctly, and it connects this matrix model back to the proportion-and-rate reasoning that runs through the modelling topics of Units 2 and 3.

Try this

Q1. To advance a state three steps, what do you multiply by? [1 point]

Cue. $T^3$ , the transition matrix cubed; each multiplication is one step.

Q2. If a transition matrix column reads $\begin{bmatrix} 0.7 \\ 0.3 \end{bmatrix}$ , what does it say? [1 point]

Cue. Of that category, $70\%$ stay (go to row $1$ ) and $30\%$ move to the other category (row $2$ ); the column sums to $1$ .

Exam-style practice questions

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

AP 2025 (style)1 marksSection I, Part B (multiple choice, calculator allowed). A transition matrix

T

advances a state vector by one time step via

\mathbf{s}_{n+1} = T\mathbf{s}_n

. How do you find the state two steps ahead,

\mathbf{s}_{n+2}

? (A)

2T\mathbf{s}_n

(B)

T^2\mathbf{s}_n

(C)

T + T

(D)

\mathbf{s}_n + 2

Show worked answer →

The correct answer is (B), $T^2\mathbf{s}_n$ .

Each multiplication by $T$ advances one step: $\mathbf{s}_{n+1} = T\mathbf{s}_n$ and $\mathbf{s}_{n+2} = T\mathbf{s}_{n+1} = T(T\mathbf{s}_n) = T^2\mathbf{s}_n$ . Advancing $k$ steps multiplies by $T^k$ , not by $kT$ .

AP 2025 (style)4 marksSection II (free response, calculator allowed). In a town, each year

80\%

of city residents stay and

20\%

move to the suburbs, while

10\%

of suburb residents move to the city and

90\%

stay. The state vector is

\begin{bmatrix} \text{city} \\ \text{suburb} \end{bmatrix}

, starting at

\begin{bmatrix} 600 \\ 400 \end{bmatrix}

. (a) Write the transition matrix. (b) Find the populations after one year.

Show worked answer →

A 4-point question on a transition-matrix model.

(a) Transition matrix (2 points): the new city total is $0.8(\text{city}) + 0.1(\text{suburb})$ and the new suburb total is $0.2(\text{city}) + 0.9(\text{suburb})$ , so $T = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{bmatrix}$ .
(b) One year (2 points): $T\begin{bmatrix} 600 \\ 400 \end{bmatrix} = \begin{bmatrix} 0.8(600) + 0.1(400) \\ 0.2(600) + 0.9(400) \end{bmatrix} = \begin{bmatrix} 480 + 40 \\ 120 + 360 \end{bmatrix} = \begin{bmatrix} 520 \\ 480 \end{bmatrix}$ . After one year: $520$ in the city, $480$ in the suburbs.

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)