Ajout de la semaine des cours du 7 au 13 février

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+%%        Version:  1.0
+%%        Created:  03/06/2024
+%%       Revision:  none
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+%%         Author:  YOUR NAME (), 
+%%   Organization:  
+%%      Copyright:  Copyright (c) 2024, YOUR NAME
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+%%          Notes:  
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+
+\title{TD du 13 février}
+\author{William Hergès\thanks{Sorbonne Université - Faculté des Sciences, Faculté des Lettres}}
+
+\begin{document}
+	\maketitle
+	\newpage
+	\section*{Exercice 1}
+	Trivial
+	\section*{Exercice 2}
+	Trivial
+	\section*{Exercice 3}
+	Trivial
+	\section*{Exercice 4}
+	On a :
+	\begin{align*}
+		2 &= a-b+c \\
+		1 &= a+b-c \\
+		3 &= -a+b+c
+	\end{align*}
+	où $(a,b,c)$ sont les coordonnées du vecteur dans la base.
+
+	D'où :
+	$$\begin{pmatrix} 1&-1&1&|&2\\1&1&-1&|&1\\-1&1&1&|&3 \end{pmatrix}\iff\begin{pmatrix} 1&-1&1&|&2\\0&2&-2&|&-1\\0&0&2&|&5\end{pmatrix}$$
+	Ainsi,
+	\begin{align*}
+		c &= 5/2 \\
+		b &= 2 \\
+		a &= 3/2
+	\end{align*}
+	\section*{Exercice 5}
+	Si $x=0$, alors $v = u + w$, donc ce n'est pas libre.\\
+	Si $x\neq 0$, alors $(uvx)$ est échelonné, donc elle est libre.
+
+	Il s'agit  d'une base si et seulement si $x\neq 0$.
+	\section*{Exercice 6}
+	\begin{lititle}
+		Matrice de changement de base.
+	\end{lititle}
+	On a $v$ un vecteur dans la base canonique. $\lambda$ est dans la base $A$ et $\mu$ dans la base $B$.
+
+	Ainsi, on a :
+	\begin{align*}
+		\mathrm{I}_n v &= A\lambda \\
+		\mathrm{I}_n v &= B\mu
+	\end{align*}
+	d'où :
+	$$ A\lambda = B\mu \iff \mu = B^{-1}A\lambda $$
+	Ainsi, $B^{-1}A$ est la matrice de passage de $A$ à $B$.
+	\section*{Exercice 7}
+	$\lambda v_1+v_2 = (\lambda x_1+x_2,\lambda y_1 + y_2)$\\
+	$\lambda y_1+y_2=\lambda x_1+x_2+\lambda+1 \neq \lambda x_1+x_2+1$\\
+	donc $A$ n'est pas un sev.
+
+	$\lambda v_1+v_2 = (\lambda x_1 + x_2, \lambda x_1 + x_2)$\\
+	donc $A_1$ est un sev.
+
+	$\lambda v_1+v_2 = (\lambda x_1+x_2,\lambda y_1 + y_2)$\\
+	$\lambda y_1 = \lambda^2 x^2 \neq x^2$\\
+	donc $A_2$ n'est pas un sev.
+\end{document}