A rabbit moves in a burrow composed of three galleries, denoted $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{C}$, in each of which it is confronted with a particular stimulus. Each time it is subjected to a stimulus, the rabbit either stays in the gallery where it is or changes gallery. This constitutes a step.
Let $n$ be a natural number. We denote by $a_n$ the probability of the event: ``the rabbit is in gallery A at step $n$''. We denote by $b_n$ the probability of the event: ``the rabbit is in gallery B at step $n$''. We denote by $c_n$ the probability of the event: ``the rabbit is in gallery C at step $n$''.
At step $n = 0$, the rabbit is in gallery A.
A previous study of the rabbit's reactions to different stimuli allows us to model its movements by the following system: $$\left\{ \begin{aligned}
a_{n+1} &= \frac{1}{3}a_n + \frac{1}{4}b_n \\
b_{n+1} &= \frac{2}{3}a_n + \frac{1}{2}b_n + \frac{2}{3}c_n \\
c_{n+1} &= \frac{1}{4}b_n + \frac{1}{3}c_n
\end{aligned} \right.$$
The objective of this exercise is to estimate in which gallery the rabbit has the greatest probability of being found in the long term.
Part A
Using a spreadsheet, we obtain the following table of values:
| A | B | C | D |
| 1 | $n$ | $a_n$ | $b_n$ | $c_n$ |
| 2 | 0 | 1 | 0 | 0 |
| 3 | 1 | 0,333 | 0,667 | 0 |
| 4 | 2 | 0,278 | 0,556 | 0,167 |
| 5 | 3 | 0,231 | 0,574 | 0,194 |
| 6 | 4 | 0,221 | 0,571 | 0,208 |
It is admitted that $P$ is invertible and that $$P^{-1} = \frac{1}{121}\begin{pmatrix} 120 & 1 \\ -1 & 1 \end{pmatrix}$$
Determine the matrix $D$ defined by $D = P^{-1}AP$.
Prove that, for every natural number $n$, $A^n = PD^nP^{-1}$.
It is admitted henceforth that, for every natural number $n$, $$A^n = \frac{1}{121}\begin{pmatrix} 120 + 0{,}395^n & 1 - 0{,}395^n \\ 120(1 - 0{,}395^n) & 1 + 120 \times 0{,}395^n \end{pmatrix}$$
Deduce an expression for $a_n$ as a function of $n$.
Determine the limit of the sequence $(a_n)$. Conclude.
Part B - Study of a second medium
In this part, we consider a second medium (medium 2), in which we do not know the probability that an atom passes from the excited state to the stable state. We denote by $a$ this probability assumed to be constant. We know, on the other hand, that at each nanosecond, the probability that an atom passes from the stable state to the excited state is 0.01.
- Give, as a function of $a$, the transition matrix $M$ in medium 2.
- After a very long time, in medium 2, the proportion of excited atoms stabilizes around $2\%$. It is admitted that there exists a unique vector $X$, called the stationary state, such that $XM = X$, and that $X = (0.98 \quad 0.02)$. Determine the value of $a$.