Long-run Markov analysis requires computing $T^n$ for large $n$. Even $T^{10}$ by hand is impractical. CAS calculators handle this instantly.
Define matrices as named variables:
T := [[0.8, 0.3], [0.2, 0.7]]
S0 := [[600], [400]]
Then:
- $S_1 = T \cdot S_0$: T * S0
- $S_5 = T^5 \cdot S_0$: T^5 * S0
- $S_{100} = T^{100} \cdot S_0$: T^100 * S0
Method 1: Raise $T$ to a large power and read the columns (they converge).
Method 2: Solve $(T - I)S^ = \mathbf{0}$ subject to $s_1^ + s_2^* = 1$.
On CAS:
solve({0.8*x + 0.3*y = x, 0.2*x + 0.7*y = y, x + y = 1}, {x, y})
Returns $x = 0.6$, $y = 0.4$: the steady-state has 60% in state 1 and 40% in state 2.
Using the two-brand example ($T$ as above), $S_0 = (400, 600)^T$ (A=400, B=600).
CAS: T^50 * [[400],[600]] → approximately $(600, 400)^T$.
The system converges to the steady state regardless of initial conditions.
Markov models can use either:
- Proportions: state vector entries sum to 1
- Counts: state vector entries sum to total population
Both are valid; ensure column sums of $T$ = 1 and that you interpret the output correctly.
A valid (column-stochastic) transition matrix has:
- All entries between 0 and 1
- Each column summing exactly to 1
Use CAS to verify: colNorm(T) or manually sum each column.
States: Employed (E), Unemployed (U), Inactive (I).
$$T = \begin{pmatrix} 0.92 & 0.25 & 0.10 \ 0.05 & 0.60 & 0.15 \ 0.03 & 0.15 & 0.75 \end{pmatrix}$$
$S_0 = (8000, 1500, 500)^T$ (initial labour force distribution).
CAS: T^12 * S0 gives the distribution after 12 months.
STUDY HINT: Store the transition matrix and initial state vector as named variables in CAS at the start of each problem. This makes computing multiple powers quick and reduces the chance of re-entry errors.
EXAM TIP: For three-state systems, always compute $S_n$ using CAS. Attempting \$3 \times 3$ matrix powers by hand is error-prone and time-consuming.