IB Maths AI HL
Transition Matrices & Markov Chains
Paper 1 & 2
~8 min read
Powers of Transition Matrices
One step is Ts0; n steps is Tns0. Raising the matrix to a power jumps the chain several time-steps ahead in one go — and the entries of Tn are themselves probabilities: row i, column j is the chance of being in state i after n steps given you started in state j. The GDC does the heavy lifting; diagonalisation handles an unknown n.
📘 What you need to know
- n steps: sn = Tns0 — raise T to the power, then multiply by the start vector.
- Entries of Tn: row i, col j = P(state i after n steps | state j now). Columns still sum to 1.
- Use the GDC for any numeric power; just enter T and compute Tn.
- Unknown power n: diagonalise T = PDP−1, then Tn = PDnP−1 (in the booklet).
- Every transition matrix has an eigenvalue of 1 — useful for diagonalisation and steady states.
- Expected numbers: multiply sn by the fixed population N.
Reading an entry of Tⁿ
Just like T itself, the power Tn is a transition matrix — but for n steps at once. The same “row = to, column = from” reading applies.
State after n steps
sn = Tns0
Given in the exam formula booklet ✓
Example reading: if t2,3 of T is “sunny tomorrow given raining today”, then t2,3 of T5 is “sunny in 5 days given raining today”. Same position, five steps ahead.
🤔 Why does multiplying matrices add the steps?
Multiplying T by T sums over every possible “middle state”: (going A→?→C) totals the probabilities of all one-step stops in between. That’s exactly the law of total probability over two steps — so T2 gives two-step probabilities, Tn gives n-step ones. The columns of Tn still sum to 1 because you always land somewhere.
Two routes to a power
How you raise T depends on whether the power is a number or an unknown n.
numeric power
use the GDC
Enter T, compute T4, T7, etc. directly. Fast and exact.
unknown power n
diagonalise
Write T = PDP−1, then Tn = PDnP−1 with Dn = each eigenvaluen.
Diagonalised power
Tn = PDnP−1
In the booklet — D = eigenvalues, P = eigenvectors ✓
🧭 Recipe — probability / expected number after n steps
- Set up T with a consistent state order; confirm each column sums to 1.
- Count the steps: how many time intervals from the start to the target day?
- Raise to that power on the GDC: Tn (or diagonalise for unknown n).
- For a single conditional probability, read the right entry of Tn (row = target state, column = start state).
- For a population, compute sn = Tns0 and multiply by N.
🧠 Count the gaps, not the days
Monday to Friday isn’t 5 — it’s 4 steps (Mon→Tue→Wed→Thu→Fri). Always count the arrows between days, not the days themselves, when choosing the power.
Worked examples
All five use the cat sanctuary (1000 cats). States are B (brushed) and B′ (not brushed), with T = (columns = current B, B′; rows = next B, B′).
WE 1How many steps to Friday?
Hippo is brushed on Monday. How many time steps is Friday, and which power of T do you need?
count the gaps Mon→Tue→Wed→Thu→Fri
4 steps → use T⁴
five days, but only four transitions between them.
WE 2Find P(brushed Friday | brushed Monday)
Compute T4 on the GDC and read off the probability Hippo is brushed on Friday.
T⁴ on the GDC, read row B / column B
current = B, future = B → top-left entry
P = 0.6424 WE 3Interpret a different entry
What does the entry in row B, column B′ of T4 (= 0.4023) represent?
row B = brushed next, column B′ = not brushed now
0.4023
P(brushed in 4 days | not brushed today) = 0.4023.
WE 4Set up the population step
On Monday 700 of the 1000 cats are brushed. Write s0 and the power needed for the following Monday.
700 of 1000 brushed → probabilities 0.7 and 0.3
Mon→Mon = 7 steps → use T⁷ WE 5Expected number brushed next Monday
Find the expected number of cats brushed on the following Monday.
expected numbers = N × T⁷s₀, with N = 1000
≈ 515 cats brushed
take the top (B) entry and round to a whole cat. 💡 Top tips
- Count gaps, not days — Monday→Friday is 4 steps, Monday→next Monday is 7.
- GDC for numeric powers; reach for diagonalisation only when the power is an unknown n.
- Reading Tn: row = future state, column = starting state — same rule as T.
- Columns of Tn still sum to 1 — a quick sanity check on your power.
- Eigenvalue 1 always exists — handy for the diagonalisation and the steady state coming next.
- Expected numbers = multiply the probability vector by N, then round at the end.
⚠ Common mistakes
- Off-by-one power — using T5 for Monday→Friday instead of T4.
- Reading the wrong entry — swapping row and column, so “to/from” are reversed.
- Rounding too early — round only the final probability or count, not intermediate matrix entries.
- Forgetting × N — leaving the answer as a probability when a number of members is asked for.
- Raising s0 to a power — it’s T that’s powered, then multiplied by s0.
Next up — Steady State & Long-term Probabilities, the last topic in this unit. As n grows, Tn settles to a matrix with identical columns, and sn tends to a fixed steady-state vector s with Ts = s. You’ll find it both by taking a large power on the GDC and exactly, as the eigenvector for eigenvalue 1 scaled to sum to 1 — using that “eigenvalue of 1 always exists” fact from this page.
Need help with Transition Matrices & Markov Chains?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
Book Free Session →