Repeated transitions | Year 12

01 / LEARN

Understand the idea.

The exponent counts transitions

The exponent counts transitions from the initial state $x_{0}$ : $x_{1} = T x_{0}$ , $x_{2} = T^{2} x_{0}$ and generally $x_{n} = T^{n} x_{0}$ . This requires the same matrix to apply to every equal time period.

A matrix power alone is still a transformation. Multiply it by the initial state to predict amounts. Retain full precision internally and check non-negative counts and the preserved total after forecasting; early rounding can accumulate error.

A changing rule changes the product

If the first period uses $T$ and the second uses $U$ , the state becomes $U (T x_{0}) = U T x_{0}$ . Read the timeline from the state vector outward: the earliest transformation acts first and is written closest to $x_{0}$ .

Changing the factor order usually changes the forecast. Constant powers are inappropriate when seasonal behaviour or a policy changes the transition rule. Specify which matrix applies in each period and distinguish a finite forecast from a long-term claim.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Three periods

Use rows (0.8,0.3),(0.2,0.7) and initial state (100,0). Find $x_{3}$ .

Worked solution

1
Apply the first transition.
$x_{1} = (\begin{array}{cccccccccc} 0.8 (100) + 0.3 (0) \\ 0.2 (100) + 0.7 (0) \end{array}) = (\begin{array}{cccccccccc} 80 \\ 20 \end{array})$
The initial state is $x_{0} = {(100, 0)}^{𝖳}$ ; the first multiplication gives the state after one period.
2
Apply the same rule a second time.
$x_{2} = (\begin{array}{cccccccccc} 0.8 (80) + 0.3 (20) \\ 0.2 (80) + 0.7 (20) \end{array}) = (\begin{array}{cccccccccc} 70 \\ 30 \end{array})$
The second period starts from the first forecast, not from the original state.
3
Apply the third transition.
$x_{3} = (\begin{array}{cccccccccc} 0.8 (70) + 0.3 (30) \\ 0.2 (70) + 0.7 (30) \end{array}) = (\begin{array}{cccccccccc} 65 \\ 35 \end{array})$
Three applications of the unchanged matrix are equivalent to $T^{3} x_{0}$ .
4
Check conservation.
$80 + 20 = 70 + 30 = 65 + 35 = 100$
Every predicted state preserves the initial total of 100.

Answer

(65,35).

EXAMPLE 02

A policy changes

First use rows $T$ : (0.8,0.3),(0.2,0.7), then rows $U$ : (0.6,0.1),(0.4,0.9), from (100,0).

Worked solution

1
Apply the first-period matrix T.
$T x_{0} = (\begin{array}{cccccccccc} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}) (\begin{array}{cccccccccc} 100 \\ 0 \end{array}) = (\begin{array}{cccccccccc} 80 \\ 20 \end{array})$
The T policy acts on the initial state before the policy changes.
2
Apply the second-period matrix U.
$U (T x_{0}) = (\begin{array}{cccccccccc} 0.6 (80) + 0.1 (20) \\ 0.4 (80) + 0.9 (20) \end{array}) = (\begin{array}{cccccccccc} 50 \\ 50 \end{array})$
The U policy acts on the distribution produced by T.
3
State the ordered product and total.
$x_{2} = U T x_{0} = (\begin{array}{cccccccccc} 50 \\ 50 \end{array}), 50 + 50 = 100$
The earliest matrix is closest to the column state vector, and the closed model preserves the total.

Answer

Final state (50,50), using $U T x_{0}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A state is measured Monday morning. What predicts Thursday morning under daily $T$ ?

Show answer & reasoning

$T^{3} x_{0}$ .

Monday to Tuesday, Tuesday to Wednesday and Wednesday to Thursday are three daily transitions. The Thursday prediction is therefore $T^{3} x_{0}$ , with Monday morning as $x_{0}$ .

Keep practising

6 worksheet questions · 21 marks

Open worksheet Worked solutions

Read the questions online

Use rows $(\begin{array}{cccccccccc} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array})$ and initial $(100, 0)$ . Find $x_{2}$ .
3 marks
For the same matrix, begin at $(0, 100)$ . Find $x_{1}, x_{2}$ .
3 marks
For that matrix, verify that $(60, 40)$ is unchanged after two periods.
3 marks
A swap matrix has rows $(\begin{array}{cccccccccc} 0 & 1 \\ 1 & 0 \end{array})$ . Start at $(40, 10)$ . Find the states after 5 and 6 periods.
3 marks
Use $T$ : $(\begin{array}{cccccccccc} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array})$ , and $U$ : $(\begin{array}{cccccccccc} 0.6 & 0.1 \\ 0.4 & 0.9 \end{array})$ , with $x_{0} = (100, 0)$ .
6 marks
1. (a)
  Calculate $U T x_{0}$ .
  3 marks
2. (b)
  Calculate $T U x_{0}$ and explain the difference.
  3 marks
Name two checks after computing $T^{20} x_{0}$ for a closed 500-person model.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Use rows $(\begin{array}{cccccccccc} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array})$ and initial $(100, 0)$ . Find $x_{2}$ .
3 marks
First obtain $(80, 20)$ .
Multiply once more: $(64 + 6, 16 + 14)$ .
$(70, 30)$ .
For the same matrix, begin at $(0, 100)$ . Find $x_{1}, x_{2}$ .
3 marks
First multiply the initial state.
Then A is $0.8 (30) + 0.3 (70) = 45$ and B is 55.
$(30, 70)$ , then $(45, 55)$ .
For that matrix, verify that $(60, 40)$ is unchanged after two periods.
3 marks
A: $0.8 (60) + 0.3 (40) = 60$ .
B: $0.2 (60) + 0.7 (40) = 40$ ; repeat the same calculation.
Both predictions are $(60, 40)$ .
A swap matrix has rows $(\begin{array}{cccccccccc} 0 & 1 \\ 1 & 0 \end{array})$ . Start at $(40, 10)$ . Find the states after 5 and 6 periods.
3 marks
Odd powers swap entries; even powers restore them.
After 5: $(10, 40)$ ; after 6: $(40, 10)$ .
Use $T$ : $(\begin{array}{cccccccccc} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array})$ , and $U$ : $(\begin{array}{cccccccccc} 0.6 & 0.1 \\ 0.4 & 0.9 \end{array})$ , with $x_{0} = (100, 0)$ .
6 marks
a) $T x_{0} = (80, 20)$ , then apply $U$ .
b) $U x_{0} = (60, 40)$ , which $T$ leaves unchanged.
1. (a)
  Calculate $U T x_{0}$ .
  3 marks
  $T x_{0} = (80, 20)$ , then apply $U$ .
  $(50, 50)$ .
2. (b)
  Calculate $T U x_{0}$ and explain the difference.
  3 marks
  $U x_{0} = (60, 40)$ , which $T$ leaves unchanged.
  $(60, 40)$ ; it reverses the policies.
a) $(50, 50)$ .; b) $(60, 40)$ ; it reverses the policies.
Name two checks after computing $T^{20} x_{0}$ for a closed 500-person model.
3 marks
Also check state order and whether the same rule is justified for 20 periods.
Nonnegative entries and total 500, apart from small rounding error.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics matrix-models option. This is the offered fifth-topic option, alongside the four compulsory topics; a school-designed Open Topic is an alternative, not prescribed content in this chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

The exponent counts transitions

A changing rule changes the product

See the method in action.

Three periods

Apply the first transition.

Apply the same rule a second time.

Apply the third transition.

Check conservation.

A policy changes

Apply the first-period matrix T.

Apply the second-period matrix U.

State the ordered product and total.

Try it yourself.

Check your understanding

Keep practising