Unit 4 · Topic 4.14

Matrices Modeling Contexts

A transition matrix holds the chances of moving between states in one step, with "from" columns that add up to 1.

Practice 1: Procedural and Symbolic FluencyPractice 3: Communication and Reasoning

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

Part 1 · Hook

Why this matters

A city's bike-share system moves bikes between stations every day: most stay put, some are ridden across town. Knowing only the chances of each move, a matrix can predict how many bikes each station will have tomorrow, next week and in the long run, and even how many it had yesterday.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\left[ \begin{matrix} 0.5 & 0.2 \\ 0.5 & 0.8 \end{matrix} \right]\left[ \begin{matrix} 100 \\ 200 \end{matrix} \right]$ ?

$\left[ \begin{matrix} 90 \\ 210 \end{matrix} \right]$
$\left[ \begin{matrix} 50 \\ 160 \end{matrix} \right]$
$\left[ \begin{matrix} 120 \\ 180 \end{matrix} \right]$
$\left[ \begin{matrix} 150 \\ 150 \end{matrix} \right]$

Show the answer

Row 1: $0.5(100) + 0.2(200) = 90$ . Row 2: $0.5(100) + 0.8(200)$ $= 210$ .

Correct: $\left[ \begin{matrix} 90 \\ 210 \end{matrix} \right]$ :
$\left[ \begin{matrix} 50 \\ 160 \end{matrix} \right]$ :
$\left[ \begin{matrix} 120 \\ 180 \end{matrix} \right]$ :
$\left[ \begin{matrix} 150 \\ 150 \end{matrix} \right]$ :

2. If $A\mathbf{v} = \mathbf{w}$ and $\det A \ne 0$ , how do you find $\mathbf{v}$ from $\mathbf{w}$ ?

$\mathbf{v} = A^{-1}\mathbf{w}$
$\mathbf{v} = A\mathbf{w}$
$\mathbf{v} = \mathbf{w}A^{-1}$
$\mathbf{v} = \mathbf{w} - A$

Show the answer

Multiply both sides on the left by $A^{-1}$ : $A^{-1}A\mathbf{v} = \mathbf{v}$ .

Correct: $\mathbf{v} = A^{-1}\mathbf{w}$ :
$\mathbf{v} = A\mathbf{w}$ :
$\mathbf{v} = \mathbf{w}A^{-1}$ :
$\mathbf{v} = \mathbf{w} - A$ :

Part 4 · See it

See it first

Each arrow is a daily chance. As a transition matrix with "from" columns and "to" rows:

\left[ \begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix} \right]

. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

List the chance of moving from each state (column) to each state (row)the transition matrix $T$ ; each column adds up to $1$
Multiply the current state vector by $T$ the expected state one step later: $\mathbf{x}_{n+1} = T\mathbf{x}_n$
Multiply by $T$ again and again $\mathbf{x}_n = T^n\mathbf{x}_0$ , the prediction $n$ steps ahead
Multiply by $T^{-1}$ insteadthe state one step earlier, as long as $T$ has an inverse

Part 6 · Key ideas

Key ideas

A transition matrix $T$ : the entry in row $i$ , column $j$ is the chance of moving from state $j$ to state $i$ in one step. Each column adds up to $1$ .
A state vector lists how many (or what share) are in each state. One step later: $\mathbf{x}_{n+1} = T\mathbf{x}_n$ ; after $n$ steps: $\mathbf{x}_n = T^n\mathbf{x}_0$ .
One step earlier: $\mathbf{x}_{n-1} = T^{-1}\mathbf{x}_n$ .
A steady state satisfies $T\mathbf{x} = \mathbf{x}$ . Over many steps the state vector usually settles toward it.

Part 7 · Misconception

A common mistake

The wrong idea: The first row of the bike matrix, $0.8$ and $0.3$ , lists what happens to the bikes at station A.

What actually happens: With this setup the columns are "from" and the rows are "to". Column 1, $0.8$ and $0.2$ , is what happens to bikes at A; row 1 collects every bike that ends at A. Always check which way your matrix is set up: each column adds up to 1 here.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

A two-station bike share

A bike-share system has $500$ bikes. Today $400$ are at station A and $100$ at station B. The diagram shows the chance that a bike moves or stays each day.

LevlPrep original diagram.

1. Which transition matrix, with "from" columns (A, B) and "to" rows (A, B), models one day?

$\left[ \begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix} \right]$
$\left[ \begin{matrix} 0.8 & 0.2 \\ 0.3 & 0.7 \end{matrix} \right]$
$\left[ \begin{matrix} 0.8 & 0.2 \\ 0.7 & 0.3 \end{matrix} \right]$
$\left[ \begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix} \right]$

Show the answer

Column 1 is a bike at A: stays $0.8$ , goes to B $0.2$ . Column 2 is a bike at B: goes to A $0.3$ , stays $0.7$ . Each column adds up to $1$ .

Correct: $\left[ \begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix} \right]$ : Columns $(0.8, 0.2)$ and $(0.3, 0.7)$ .
$\left[ \begin{matrix} 0.8 & 0.2 \\ 0.3 & 0.7 \end{matrix} \right]$ : This puts "from" in the rows; its columns add up to $1.1$ and $0.9$ .
$\left[ \begin{matrix} 0.8 & 0.2 \\ 0.7 & 0.3 \end{matrix} \right]$ : The chances for B are reversed: a bike at B stays with chance $0.7$ .
$\left[ \begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix} \right]$ : The stay and move chances are swapped.

2. How many bikes are expected at station A after one day?

No calculator. Type the exact value (a fraction, π, e or a root is fine), in bikes.

Show the answer

Row 1 of $T$ times $\left[ \begin{matrix} 400 \\ 100 \end{matrix} \right]$ : $0.8(400)$ $+ 0.3(100)$ $= 320$ $+ 30$ $= 350$ .

Answer: 350 bikes

3. How many bikes are expected at station A after two days?

No calculator. Type the exact value (a fraction, π, e or a root is fine), in bikes.

Show the answer

$\mathbf{x}_1 = \left[ \begin{matrix} 350 \\ 150 \end{matrix} \right]$ , so after two days A has $0.8(350)$ $+ 0.3(150)$ $= 280$ $+ 45$ $= 325$ bikes.

Answer: 325 bikes

4. In the long run, how many bikes are expected at station A?

No calculator. Type the exact value (a fraction, π, e or a root is fine), in bikes.

Show the answer

At a steady state $0.2a = 0.3b$ , so $a = 1.5b$ ; with $a + b = 500$ , $b = 200$ and $a = 300$ . Check: $0.8(300) + 0.3(200)$ $= 300$ .

Answer: 300 bikes

5. Why does each column of a transition matrix (with "from" columns) add up to 1?

Everything in a state ends the step in some state
Each row lists the states something can come from
The matrix has an inverse
The state vector has two entries

Show the answer

A column lists the chances for one starting state across every possible ending state, and something must end somewhere.

Correct: Everything in a state ends the step in some state: The chances for one starting state cover every outcome.
Each row lists the states something can come from: Rows collect arrivals; their sums can be anything.
The matrix has an inverse: Columns adding up to 1 does not make a matrix invertible.
The state vector has two entries: The number of entries does not set the sum.

6. Which could be a transition matrix with "from" columns?

$\left[ \begin{matrix} 0.6 & 0.1 \\ 0.4 & 0.9 \end{matrix} \right]$
$\left[ \begin{matrix} 0.6 & 0.4 \\ 0.6 & 0.4 \end{matrix} \right]$
$\left[ \begin{matrix} 0.5 & 1.2 \\ 0.5 & -0.2 \end{matrix} \right]$
$\left[ \begin{matrix} 0.3 & 0.3 \\ 0.3 & 0.3 \end{matrix} \right]$

Show the answer

Entries between $0$ and $1$ , and each column adds up to $1$ : $0.6 + 0.4$ and $0.1 + 0.9$ .

Correct: $\left[ \begin{matrix} 0.6 & 0.1 \\ 0.4 & 0.9 \end{matrix} \right]$ : Valid chances, columns add up to $1$ .
$\left[ \begin{matrix} 0.6 & 0.4 \\ 0.6 & 0.4 \end{matrix} \right]$ : Its columns add up to $1.2$ and $0.8$ .
$\left[ \begin{matrix} 0.5 & 1.2 \\ 0.5 & -0.2 \end{matrix} \right]$ : A chance cannot be $1.2$ or $-0.2$ .
$\left[ \begin{matrix} 0.3 & 0.3 \\ 0.3 & 0.3 \end{matrix} \right]$ : Its columns add up to $0.6$ .

Part 9 · Summary

Summary

A transition matrix holds the chances of moving between states in one step, with "from" columns that add up to 1. Multiplying a state vector by

T

moves it one step forward,

T^n

moves it

n

steps,

T^{-1}

moves it one step back, and a steady state is a vector that

T

leaves unchanged.

Part 10 · Up next

What comes next

This is the last topic in the course.

Part 11 · Connections

Connections

Back: Matrices as Functions (4.13)

Question set for this topic