\textbf{Exercise 5 (5 points) — Candidates who have followed the speciality course}

In a given territory, we are interested in the coupled evolution of two species: the buzzards (the predators) and the voles (the prey).\\
Scientists model, for all natural integer $n$, this evolution by:

$$\left\{ \begin{aligned}
b _ { 0 } & = 1000 \\
c _ { 0 } & = 1500 \\
b _ { n + 1 } & = 0,3 b _ { n } + 0,5 c _ { n } \\
c _ { n + 1 } & = - 0,5 b _ { n } + 1,3 c _ { n }
\end{aligned} \right.$$

where $b _ { n }$ represents approximately the number of buzzards and $c _ { n }$ the approximate number of voles on June 1st of the year $2000 + n$ (where $n$ denotes a natural integer).

\begin{enumerate}
  \item We denote $A$ the matrix $\left( \begin{array} { c c } 0,3 & 0,5 \\ - 0,5 & 1,3 \end{array} \right)$ and, for all natural integer $n , U _ { n }$ the column matrix $\binom { b _ { n } } { c _ { n } }$.\\
a. Verify that $U _ { 1 } = \binom { 1050 } { 1450 }$ and calculate $U _ { 2 }$.\\
b. Verify that, for all natural integer $n , U _ { n + 1 } = A U _ { n }$.
  \item We are given the matrices $P = \left( \begin{array} { l l } 1 & 0 \\ 1 & 1 \end{array} \right) , T = \left( \begin{array} { c c } 0,8 & 0,5 \\ 0 & 0,8 \end{array} \right)$ and $I = \left( \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right)$.\\
We admit that $P$ has as its inverse a matrix $Q$ of the form $\left( \begin{array} { l l } 1 & 0 \\ a & 1 \end{array} \right)$ where $a$ is a real number.\\
a. Determine the value of $a$ by justifying.\\
b. We admit that $A = P T Q$. Prove that, for all non-zero integer $n$, we have
$$A ^ { n } = P T ^ { n } Q .$$
c. Prove using a proof by induction that, for all non-zero integer $n$,
$$T ^ { n } = \left( \begin{array} { c c } 
0,8 ^ { n } & 0,5 n \times 0,8 ^ { n - 1 } \\
0 & 0,8 ^ { n }
\end{array} \right)$$
  \item Lucie executes the algorithm below and obtains as output $N = 40$. What conclusion can Lucie state for the buzzards and the voles?
\begin{verbatim}
Initialization: N takes the value 0
        B takes the value 1000
        C takes the value 1500
    Processing: While B > 2 or C > 2
            N takes the value N + 1
            R takes the value B
            B takes the value  0,3 R + 0,5 C
            C takes the value -0,5 R + 1,3 C
        End While
        Output: Display N
\end{verbatim}
  \item We admit that, for all non-zero natural integer $n$, we have
$$U _ { n } = \binom { 1000 \times 0,8^n + 500 n \times 0,8^{n-1} }{ \text{(expression for } c_n\text{)}}$$
\end{enumerate}