Repeated transitions | Year 12

01 / LEARN

The idea, step by step

The exponent counts transitions

The initial state is $x_{0}$ . After one period it is $T x_{0}$ ; after two it is $T (T x_{0}) = T^{2} x_{0}$ . The formula $x_{n} = T^{n} x_{0}$ requires the same transition rule for each equal time period. Monday morning to Thursday morning is three daily transitions.

For large powers, use technology while retaining full precision internally. Check nonnegative entries and the preserved total. The matrix $T^{n}$ is still a transformation; it becomes a predicted population only after multiplication by the initial state.

A changing rule changes the product

If the first period uses $T$ and the second uses $U$ , the final state is $U T x_{0}$ . The earliest transition acts closest to the column state vector. Reversing the factors usually describes a different timeline.

Seasonal behaviour or a policy change can invalidate a constant-rate forecast. Use a new matrix only with a stated explanation of its entries. Distinguish a finite forecast from a long-run claim, and avoid rounding the intermediate state repeatedly.

A common trap

Using $T^{3}$ as if it were a state vector.

Instead: Multiply the power by the initial state, with compatible labels and dimensions.

02 / WORKED EXAMPLES

See the thinking

Example 1

Three periods

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

$x_{1} = (80, 20)$ .
$x_{2} = (70, 30)$ .
$x_{3} = (65, 35)$ ; each total is 100.

(65,35).

Example 2

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).

First $T x_{0} = (80, 20)$ .
Then $U (80, 20) = (50, 50)$ .

Final state (50,50), using

U T x_{0}

Check your understanding

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

Show answer

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

Three daily transitions occur.

03 / WORKSHEET

Your turn

6 questions · 21 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Use rows (0.8,0.3),(0.2,0.7) 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 (0,1),(1,0). Start at (40,10). Find the states after 5 and 6 periods.
3 marks
Use $T$ : (0.8,0.3),(0.2,0.7), and $U$ : (0.6,0.1),(0.4,0.9), 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 (0.8,0.3),(0.2,0.7) 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 (0,1),(1,0). 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 $T$ : (0.8,0.3),(0.2,0.7), and $U$ : (0.6,0.1),(0.4,0.9), 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.

Checking saved progress…

Continue learning

Long-term behaviour and steady states

Next lesson

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 ↗

Back to Matrix models