\section*{Exercise 3 -- Candidates who have followed the specialization course}
The manager of a website, composed of three web pages numbered 1 to 3 and linked together by hypertext links, wishes to predict the frequency of connection to each of his web pages.

Statistical studies have allowed him to notice that:
\begin{itemize}
  \item If an internet user is on page no. 1, then he will go either to page no. 2 with probability $\frac{1}{4}$, or to page no. 3 with probability $\frac{3}{4}$.
  \item If an internet user is on page no. 2, then either he will go to page no. 1 with probability $\frac{1}{2}$ or he will stay on page no. 2 with probability $\frac{1}{4}$, or he will go to page no. 3 with probability $\frac{1}{4}$.
  \item If an internet user is on page no. 3, then either he will go to page no. 1 with probability $\frac{1}{2}$, or he will go to page no. 2 with probability $\frac{1}{4}$, or he will stay on page no. 3 with probability $\frac{1}{4}$.
\end{itemize}

For every natural number $n$, we define the following events and probabilities:\\
$A_n$: ``After the $n$-th navigation, the internet user is on page no. 1'' and we denote $a_n = P(A_n)$.\\
$B_n$: ``After the $n$-th navigation, the internet user is on page no. 2'' and we denote $b_n = P(B_n)$.\\
$C_n$: ``After the $n$-th navigation, the internet user is on page no. 3'' and we denote $c_n = P(C_n)$.

\begin{enumerate}
  \item Show that, for every natural number $n$, we have $a_{n+1} = \frac{1}{2}b_n + \frac{1}{2}c_n$.
\end{enumerate}

We admit that, similarly, $b_{n+1} = \frac{1}{4}a_n + \frac{1}{4}b_n + \frac{1}{4}c_n$ and $c_{n+1} = \frac{3}{4}a_n + \frac{1}{4}b_n + \frac{1}{4}c_n$.\\
Thus:
$$\left\{\begin{aligned}
a_{n+1} &= \frac{1}{2}b_n + \frac{1}{2}c_n \\
b_{n+1} &= \frac{1}{4}a_n + \frac{1}{4}b_n + \frac{1}{4}c_n \\
c_{n+1} &= \frac{3}{4}a_n + \frac{1}{4}b_n + \frac{1}{4}c_n
\end{aligned}\right.$$

\begin{enumerate}
  \setcounter{enumi}{1}
  \item For every natural number $n$, we set $U_n = \begin{pmatrix}a_n\\b_n\\c_n\end{pmatrix}$. $U_0 = \begin{pmatrix}a_0\\b_0\\c_0\end{pmatrix}$ represents the initial situation, with $a_0 + b_0 + c_0 = 1$.\\
Show that, for every natural number $n$, $U_{n+1} = MU_n$ where $M$ is a $3\times 3$ matrix that you will specify.\\
Deduce that, for every natural number $n$, $U_n = M^n U_0$.
  \item Show that there exists a unique column matrix $U = \begin{pmatrix}x\\y\\z\end{pmatrix}$ such that: $x + y + z = 1$ and $MU = U$.
  \item A computer algebra system has made it possible to obtain the expression of $M^n$, $n$ being a non-zero natural number:
$$M^n = \begin{pmatrix}
\frac{1}{3} + \frac{\left(\frac{-1}{2}\right)^n \times 2}{3} & \frac{1}{3} + \frac{\left(\frac{-1}{2}\right)^n}{-3} & \frac{1}{3} + \frac{\left(\frac{-1}{2}\right)^n}{-3} \\
\frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\
\frac{5}{12} + \frac{\left(-\left(\frac{-1}{2}\right)^n\right)\times 2}{3} & \frac{5}{12} + \frac{-\left(\frac{-1}{2}\right)^n}{-3} & \frac{5}{12} + \cdots
\end{pmatrix}$$
\end{enumerate}