
En s'inspirant de l'exercice précédent, on peut écrire que si:
\[
U_{n+1} = AU_n+B,
\]
alors:
\[
U_n = A^n(U_0-C)+C\quad,\quad C = (I_2-A)^{-1}B.
\]
Ainsi, dans cet exercice, on a:
\begin{align*}
\forall n\in\N,\quad X_n& =
\begin{pmatrix}-8 & -6\\9 &13\end{pmatrix}^n
\left[
\begin{pmatrix}1\\1\end{pmatrix}
-
\begin{pmatrix}9 & 6\\-9 & -12\end{pmatrix}^{-1}\begin{pmatrix}1\\-1\end{pmatrix}
\right]
+\begin{pmatrix}9 & 6\\-9 & -12\end{pmatrix}^{-1}\begin{pmatrix}1\\-1\end{pmatrix}
\end{align*}
\`A l'aide de XCAS (par exemple), on trouve
\[ \begin{pmatrix}-8 & -6\\9 & 13\end{pmatrix}=\begin{pmatrix}
2 & -1\\-1 & 3\end{pmatrix}\begin{pmatrix}-5 & 0\\0&10\end{pmatrix}
\begin{pmatrix}\frac{3}{5} & \frac{1}{5}\\\frac{1}{5} & \frac{2}{5}\end{pmatrix}\]
et
\[
\begin{pmatrix}9 & 6\\-9 & -12\end{pmatrix}^{-1} = \begin{pmatrix}\frac{2}{9} & \frac{1}{9}\\-\frac{1}{6}&-\frac{1}{6}\end{pmatrix}.
\]
Donc,
\begin{align*}
\begin{pmatrix}1\\1\end{pmatrix}
-
\begin{pmatrix}9 & 6\\-9 & -12\end{pmatrix}^{-1}\begin{pmatrix}1\\-1\end{pmatrix} &
= \begin{pmatrix}1-\frac{1}{9}\\1\end{pmatrix}=\begin{pmatrix}\frac{8}{9}\\1\end{pmatrix}.
\end{align*}
Ainsi,\renewcommand{\arraystretch}{1.2}
\begin{align*}
\forall n\in\N,\quad X_n& =
\begin{pmatrix}2 & -1\\-1 & 3\end{pmatrix}
\begin{pmatrix}(-5)^n & 0\\0&10^n\end{pmatrix}
\begin{pmatrix}\frac{3}{5} & \frac{1}{5}\\\frac{1}{5} & \frac{2}{5}\end{pmatrix}
\begin{pmatrix}\frac{8}{9}\\1\end{pmatrix}+
\begin{pmatrix}\frac{2}{9} & \frac{1}{9}\\-\frac{1}{6}&-\frac{1}{6}\end{pmatrix}
\begin{pmatrix}1\\-1\end{pmatrix}\\
& = \begin{pmatrix}2(-5)^n & -10^n\\-(-5)^n & 3\times10^n\end{pmatrix}
\begin{pmatrix}\frac{3}{5} & \frac{1}{5}\\\frac{1}{5} & \frac{2}{5}\end{pmatrix}
\begin{pmatrix}\frac{8}{9}\\1\end{pmatrix}+
\begin{pmatrix}\frac{1}{9} \\ 0\end{pmatrix}\\
& = \begin{pmatrix}\frac{6}{5}(-5)^n-\frac{1}{5}\times10^n & \frac{2}{5}(-5)^n-\frac{2}{5}\times10^n \\ -\frac{3}{5}(-5)^n+\frac{3}{5}\times10^n & -\frac{1}{5}(-5)^n + \frac{6}{5}\times10^n\end{pmatrix}
\begin{pmatrix}\frac{8}{9}\\1\end{pmatrix}+\begin{pmatrix}\frac{1}{9} \\ 0\end{pmatrix}\\
& = \begin{pmatrix}\frac{22}{5}(-5)^n-\frac{26}{45}\times10^n+\frac{1}{9}\\
-\frac{11}{15}(-5)^n+\frac{26}{15}\times10^n \end{pmatrix}
\end{align*}
On en conclut alors que :
\[
\forall n\in\N,\qquad \left\lbrace\begin{array}{l}
u_n=\frac{22}{5}(-5)^n-\frac{26}{45}\times10^n+\frac{1}{9}\\
v_n=-\frac{11}{15}(-5)^n+\frac{26}{15}\times10^n
\end{array}\right.
\]
