Answering Questions That Require Multiple Applications of a Transition Matrix Revision Notes for VCE SSCE General Mathematics

Answering Questions That Require Multiple Applications of a Transition Matrix

Introduction to matrix recurrence relations

When working with transition matrices, we often need to predict what happens over multiple time periods. For example, if we're tracking whether people go to the movies or stay home each week, we might want to know:

How many people will be at the movies after $1$ week?
How many after $2$ weeks?
What about after $10$ weeks or more?

This is similar to financial problems where we track how an investment grows over time. Just as we use recurrence relations to model the growth of investments, we can use recurrence relations with matrices to model changes in populations or states over time.

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Understanding how to apply transition matrices multiple times is essential for making long-term predictions in real-world scenarios, from population movements to market behavior.

The initial state matrix

Before we can track changes, we need a starting point. This is called the initial state matrix, denoted $S_0$ . It tells us the situation at the beginning, before any transitions have occurred.

For example, if $200$ people initially go to the movies and $150$ stay at home, the initial state matrix would be:

$S_0 = \begin{bmatrix} 200 \\ 150 \end{bmatrix}$

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The initial state matrix $S_0$ is always your starting point. Every subsequent calculation builds from this foundation, so it's crucial to identify it correctly from the problem statement.

Matrix recurrence relations

A matrix recurrence relation is a formula that allows us to generate a sequence of state matrices, showing how the situation changes at each time step.

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The General Form of a Matrix Recurrence Relation:

$S_0 = \text{initial state matrix}$

$S_{n+1} = T \times S_n$

where:

$T$ is the transition matrix
$S_n$ is the state matrix after $n$ applications of the transition matrix (or $n$ time intervals)
$S_{n+1}$ is the state matrix after $n+1$ applications

This formula is the foundation for tracking changes over time.

How it works

To find what happens after each time period, we multiply the transition matrix by the current state matrix:

After $1$ time period: $S_1 = T \times S_0$
After $2$ time periods: $S_2 = T \times S_1$
After $3$ time periods: $S_3 = T \times S_2$

This pattern continues for as many time periods as we need to calculate. Each new state depends only on the previous state and the transition matrix.

Worked example: using recursion to generate successive state matrices

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Worked Example: Children at a Fair

Children at a fair can either go on the Ferris wheel or the Merry-go-round. At the start, $140$ children went on the Ferris wheel and $80$ children went on the Merry-go-round.

The initial state matrix is:

$S_0 = \begin{bmatrix} 140 \\ 80 \end{bmatrix}$

The transition matrix is:

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

Part 1: Finding the state after one turn

To predict the number of children on each ride after one turn, we calculate $S_1 = T \times S_0$ :

\begin{aligned} S_1 &= \begin{bmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{bmatrix} \begin{bmatrix} 140 \\ 80 \end{bmatrix} \\ &= \begin{bmatrix} 0.8 \times 140 + 0.3 \times 80 \\ 0.2 \times 140 + 0.7 \times 80 \end{bmatrix} \\ &= \begin{bmatrix} 112 + 24 \\ 28 + 56 \end{bmatrix} \\ &= \begin{bmatrix} 136 \\ 84 \end{bmatrix} \end{aligned}

After one turn, we predict that 136 children will go on the Ferris wheel and 84 will go on the Merry-go-round.

Part 2: Finding the state after three turns

To find the state after three turns, we need to work through each step:

After two turns, $S_2 = T \times S_1$ :

\begin{aligned} S_2 &= \begin{bmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{bmatrix} \begin{bmatrix} 136 \\ 84 \end{bmatrix} \\ &= \begin{bmatrix} 0.8 \times 136 + 0.3 \times 84 \\ 0.2 \times 136 + 0.7 \times 84 \end{bmatrix} \\ &= \begin{bmatrix} 134 \\ 86 \end{bmatrix} \end{aligned}

After three turns, $S_3 = T \times S_2$ :

\begin{aligned} S_3 &= \begin{bmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{bmatrix} \begin{bmatrix} 134 \\ 86 \end{bmatrix} \\ &= \begin{bmatrix} 0.8 \times 134 + 0.3 \times 86 \\ 0.2 \times 134 + 0.7 \times 86 \end{bmatrix} \\ &= \begin{bmatrix} 133 \\ 87 \end{bmatrix} \end{aligned}

After three turns, we predict 133 children on the Ferris wheel and 87 on the Merry-go-round.

The explicit rule for large values of n

While the recurrence relation works well for a few time steps, it becomes inefficient when we need to find the state after many transitions (for example, after $20$ or $40$ years). There is a more efficient method using an explicit rule.

Understanding the pattern

Notice the following relationship between successive states:

$S_1 = T \times S_0$
$S_2 = T \times S_1 = T \times (T \times S_0) = T^2 \times S_0$
$S_3 = T \times S_2 = T \times (T^2 \times S_0) = T^3 \times S_0$

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Key Pattern Recognition:

Each time we apply the transition matrix, we're effectively multiplying by $T$ one more time. This pattern reveals that $S_n = T^n \times S_0$ .

The explicit rule

If the recurrence relation is:

$S_0 = \text{initial state matrix}$

$S_{n+1} = T \times S_n$

then an explicit rule for finding $S_n$ is:

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The Explicit Rule:

$S_n = T^n \times S_0$

where:

$T$ is the transition matrix
$S_0$ is the initial state matrix
$n$ is the number of time intervals or applications of the transition matrix

This rule allows us to jump directly to any time period without calculating all the intermediate steps.

Worked example: modelling population growth and decay

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Worked Example: Population Changes Between Cities

A study was conducted to investigate the change in the populations of Geelong ( $G$ ) and Ballarat ( $B$ ) due to movement of people between the two cities.

At the start of the study, the populations were:

Geelong: $250\,000$ people
Ballarat: $110\,000$ people

The initial state matrix is:

$S_0 = \begin{bmatrix} 250\,000 \\ 110\,000 \end{bmatrix}$

The transition matrix describing population movements from year to year is:

$T = \begin{bmatrix} 0.9 & 0.15 \\ 0.1 & 0.85 \end{bmatrix}$

Part a: Predicting the population after one year

Using the recursion rule $S_1 = T \times S_0$ :

\begin{aligned} S_1 &= \begin{bmatrix} 0.9 & 0.15 \\ 0.1 & 0.85 \end{bmatrix} \begin{bmatrix} 250\,000 \\ 110\,000 \end{bmatrix} \\ &= \begin{bmatrix} 0.9 \times 250\,000 + 0.15 \times 110\,000 \\ 0.1 \times 250\,000 + 0.85 \times 110\,000 \end{bmatrix} \\ &= \begin{bmatrix} 225\,000 + 16\,500 \\ 25\,000 + 93\,500 \end{bmatrix} \\ &= \begin{bmatrix} 241\,500 \\ 118\,500 \end{bmatrix} \end{aligned}

The predicted population of Geelong after one year is 241,500.

Part b: Predicting populations after two, three and four years

Using the recursion rule $S_{n+1} = T \times S_n$ repeatedly:

After two years: $235\,125$
After three years: $230\,344$
After four years: $226\,758$

Part c: Predicting populations after 20, 30 and 40 years

Using the explicit rule $S_n = T^n \times S_0$ (calculations done with a calculator):

After twenty years: $216\,108$
After thirty years: $216\,006$
After forty years: $216\,000$

Part d: Long-term behaviour

Over time, the predicted population in Geelong is declining. Initially, the population falls rapidly from $250\,000$ to $235\,125$ in just one year. However, the rate of decline then slows, and the population appears to stabilise at around $216\,000$ people.

Understanding steady state and equilibrium

When we calculate population changes over many time periods, we often observe that the populations eventually become stable. This doesn't mean that nobody is moving between locations; rather, it means that the number moving in each direction is equal.

Calculating the steady state

In the Geelong and Ballarat example, when $n = 40$ :

Geelong's population: $216\,000$
Ballarat's population: $144\,000$

At $n = 41$ , we can check if the populations are in equilibrium:

Number of people moving from Geelong to Ballarat:

$0.1 \times 216\,000 = 21\,600$

Number of people moving from Ballarat to Geelong:

$0.15 \times 144\,000 = 21\,600$

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Understanding Equilibrium:

Since these two numbers are equal, the populations are at steady state or equilibrium state. The same number of people are moving in both directions, so the overall populations remain constant.

This is a key characteristic of steady state: movement continues, but the net effect is zero change in each population.

Exam tips

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Strategies for Success:

Always start with the initial state matrix $S_0$ - this is your foundation.
For a few time steps (like $1$ , $2$ , or $3$ periods), use the recurrence relation $S_{n+1} = T \times S_n$ and work step-by-step.
For many time steps (like $20$ or $40$ periods), use the explicit rule $S_n = T^n \times S_0$ with your calculator.
Check your dimensions: When multiplying matrices, make sure the dimensions are compatible. A $2 \times 2$ transition matrix times a $2 \times 1$ state matrix gives a $2 \times 1$ state matrix.
Interpret your results: Remember what each number in your state matrix represents. The question will usually ask about a specific state or population.
Look for patterns: If asked about long-term behaviour, calculate several values and observe whether populations are increasing, decreasing, or stabilising.

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Key Points to Remember:

The recurrence relation $S_{n+1} = T \times S_n$ allows you to generate state matrices step-by-step, where $S_0$ is the initial state matrix.
The explicit rule $S_n = T^n \times S_0$ provides a shortcut for finding the state matrix after $n$ transitions without calculating all intermediate steps.
Steady state or equilibrium state occurs when the number of items (people, objects, etc.) moving between states is equal in both directions, resulting in stable populations over time.
To solve these problems effectively, identify whether you need to use the recurrence relation (for a few steps) or the explicit rule (for many steps).
Always check that your final answer makes sense in the context of the problem - do the numbers represent realistic populations or proportions?

Answering Questions That Require Multiple Applications of a Transition Matrix (VCE SSCE General Mathematics): Revision Notes