From d7c69934248f3fab4988c0d37c11feba25d653b8 Mon Sep 17 00:00:00 2001
From: Anhgelus Morhtuuzh <anhgelus@anhgelus.world>
Date: Fri, 31 Jan 2025 15:39:30 +0100
Subject: =?UTF-8?q?Ajout=20du=20d=C3=A9but=20du=20deuxi=C3=A8me=20semestre?=
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+%%=====================================================================================
+%%
+%%       Filename:  cours.tex
+%%
+%%    Description:  
+%%
+%%        Version:  1.0
+%%        Created:  03/06/2024
+%%       Revision:  none
+%%
+%%         Author:  YOUR NAME (), 
+%%   Organization:  
+%%      Copyright:  Copyright (c) 2024, YOUR NAME
+%%
+%%          Notes:  
+%%
+%%=====================================================================================
+\documentclass[a4paper, titlepage]{article}
+
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+\usepackage{textcomp}
+\usepackage[french]{babel}
+\usepackage{amsmath, amssymb}
+\usepackage{amsthm}
+\usepackage[svgnames]{xcolor}
+\usepackage{thmtools}
+\usepackage{lipsum}
+\usepackage{framed}
+\usepackage{parskip}
+\usepackage{titlesec}
+
+\renewcommand{\familydefault}{\sfdefault}
+
+% figure support
+\usepackage{import}
+\usepackage{xifthen}
+\pdfminorversion=7
+\usepackage{pdfpages}
+\usepackage{transparent}
+\newcommand{\incfig}[1]{%
+	\def\svgwidth{\columnwidth}
+	\import{./figures/}{#1.pdf_tex}
+}
+
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+]{better-corol}
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+\declaretheorem[style=better-props, numberwithin=defn]{props}
+\declaretheorem[style=better-thm, sibling=props]{thm}
+\newtheorem*{lemme}{Lemme}%[subsection]
+%\newtheorem{props}{Propriétés}[defn]
+
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+{\left\lbrace\begin{align}}%
+{\end{align}\right.}
+
+\newenvironment{AQT}{{\fontfamily{qbk}\selectfont AQT}}
+
+\usepackage{LobsterTwo}
+\titleformat{\section}{\newpage\LobsterTwo \huge\bfseries}{\thesection.}{1em}{}
+\titleformat{\subsection}{\vspace{2em}\LobsterTwo \Large\bfseries}{\thesubsection.}{1em}{}
+\titleformat{\subsubsection}{\vspace{1em}\LobsterTwo \large\bfseries}{\thesubsubsection.}{1em}{}
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+\newenvironment{lititle}%
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+\renewenvironment{proof}{$\square$ \footnotesize\textit{Démonstration.}}{\begin{flushright}$\blacksquare$\end{flushright}}
+
+\title{TD du 23 janvier 2025}
+\author{William Hergès\thanks{Sorbonne Université - Faculté des Sciences, Faculté des Lettres}}
+
+\begin{document}
+	\maketitle
+	\newpage
+	\section*{Exercice 1}
+	Les produits possibles sont $AX$, $BX$, $BA$, $AB$ et $DZ$, i.e.
+	$$ AX = \begin{pmatrix} 1&2\\ 2&1 \end{pmatrix}\begin{pmatrix} 1\\-1 \end{pmatrix} = \begin{pmatrix} -1\\1 \end{pmatrix} $$
+	$$ BX = \begin{pmatrix} -1&1\\0&1 \end{pmatrix}\begin{pmatrix} 1\\-1 \end{pmatrix} = \begin{pmatrix} -2\\-1 \end{pmatrix} $$
+	$$ BA = \begin{pmatrix} -1&1\\0&1 \end{pmatrix}\begin{pmatrix} 1&2\\2&1 \end{pmatrix}= \begin{pmatrix} 1&-1\\2&1 \end{pmatrix} $$
+	$$ AB =  \begin{pmatrix} 1&2\\2&1 \end{pmatrix}\begin{pmatrix} -1&1\\0&1 \end{pmatrix} = \begin{pmatrix} -1&3\\-2&3  \end{pmatrix} $$ 
+	$$ DZ = \begin{pmatrix} 1&2&3\\3&2&1\\2&1&3 \end{pmatrix} \begin{pmatrix} 0\\2\\3 \end{pmatrix} = \begin{pmatrix} 13\\7\\11  \end{pmatrix}  $$
+	\section*{Exercice 2}
+	\begin{enumerate}
+		\item On a :
+			$$ AB = \begin{pmatrix} 1&1\\0&1 \end{pmatrix} \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = \begin{pmatrix} -1&1\\-1&0 \end{pmatrix}  $$
+			$$ BA =  \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} \begin{pmatrix} 1&1\\0&1 \end{pmatrix}= \begin{pmatrix} 0&1\\-1&-1 \end{pmatrix}  $$
+			Donc $AB \neq BA$
+		\item On a :
+			$$ (A + B)^2 = \begin{pmatrix} 1&2\\-1&1 \end{pmatrix} \begin{pmatrix} 1&2\\-1&1 \end{pmatrix} = \begin{pmatrix} -1&3\\-2&-1 \end{pmatrix}  $$
+		\item On a :
+			$$ A^2 = \begin{pmatrix} 1&1\\0&1 \end{pmatrix} \begin{pmatrix} 1&1\\0&1 \end{pmatrix} = \begin{pmatrix} 1&2\\0&1 \end{pmatrix} $$
+			$$ B^2 = \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = \begin{pmatrix} -1&0\\0&-1 \end{pmatrix}  $$
+			Après calcul, on obtient que $(A+B)^2 = A^2+AB+BA+B^2 \neq A^2+2AB+B^2$.
+		\item Cherchons l'inverse de $A$ :
+			$$ ad-bc = 1-0 = 1 $$
+			Donc l'inverse existe, i.e.
+			$$ A^{-1} = \begin{pmatrix} 1&-1\\0&1 \end{pmatrix}  $$
+			Cherchons maintenant l'inverse de $B$ :
+			$$ ad-bc = 0+1 = 1 $$
+			Donc l'inverse existe, i.e.
+			$$ B^{-1} = \begin{pmatrix} 0&-1\\1&0 \end{pmatrix}  $$
+	\end{enumerate}
+	\section*{Exercice 3}
+	$$ A^t = \begin{pmatrix} -5&-1&3\\-4&0&4\\-3&1&5\\-2&2&6 \end{pmatrix}  $$
+	puis
+	$$ BA^t = \begin{pmatrix} -1&0&1&2\\ 2&1&0&-1\\ 1&0&-1&2 \end{pmatrix}A^t = \begin{pmatrix} -2&6&14\\-12&-4&4\\-8&2&10 \end{pmatrix}   $$
+	\section*{Exercice 4}
+	\begin{enumerate}
+		\item Ce vecteur $v$ peut être représentée par un complexe $z=a+ib$. Or il existe $r,\theta\in\mathbb{R}$ tel que $z=re^{i\theta}$, d'où $r\cos\theta+ri\sin\theta$.
+		\item On a 
+			$$ R_{\theta}\cdot v = r(\cos\phi\cos\theta-\sin\phi\sin\theta)+ir(\sin\phi\cos\theta+\cos\theta\sin\theta) $$
+			i.e.
+			$$ R_{\theta}\cdot v = r\cos(\phi+\theta)+ir\sin(\phi+\theta) $$
+	\end{enumerate}
+	\section*{Exercice 5}
+	On a :
+	$$ AT = \begin{pmatrix} 2&4&8\\ 1&2&5 \end{pmatrix} $$
+	$$ BT = \begin{pmatrix} 1&2&3\\-1&-2&-1 \end{pmatrix}  $$
+	$$ CT = \begin{pmatrix} 0&0&2\\1&2&1 \end{pmatrix}  $$
+	$$ DT = \begin{pmatrix} \frac{\sqrt{3}-1}{2}&\sqrt{3}-1&\frac{3\sqrt{3}-1}{2}\\\frac{1+\sqrt{3}}{2}&1+\sqrt{3}&\frac{3+\sqrt{3}}{2} \end{pmatrix}  $$
+	\section*{Exercice 6}
+	\begin{enumerate}
+		\item Soit $P_n: M^n\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} f_n\\f_{n+1} \end{pmatrix}$ avec $n\in\mathbb{N}^*$. Montrons que $P_n$ est vraie pour tout $n$.
+
+			\fbox{Initialisation} On a $M^1\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} 1\\1 \end{pmatrix} = \begin{pmatrix} f_{1}\\f_{2} \end{pmatrix}$. Donc $P_1$ est vraie.
+
+			\fbox{Hérédité} Posons $n\in\mathbb{N}^*$ tel que $P_n$ soit vraie. On a :
+			$$ M^n\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} f_{n}\\f_{n+1} \end{pmatrix}  $$
+			Donc,
+			$$ M^{n+1}\begin{pmatrix} 0\\1 \end{pmatrix} = M\begin{pmatrix} f_n\\f_{n+1} \end{pmatrix} $$
+			Or,
+			$$ M\begin{pmatrix} f_n\\f_{n+1} \end{pmatrix} = \begin{pmatrix} f_{n+1}\\f_n+f_{n+1} \end{pmatrix} $$
+			i.e.,
+			$$ M^{n+1}\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} f_{n+1}\\f_n+f_{n+1} \end{pmatrix} $$
+			Alors,
+			$$ M^{n+1}\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} f_{n+1}\\f_{n+2} \end{pmatrix} $$
+			Ainsi, $P_{n+1}$ est vraie 
+
+			\fbox{Conclusion} $P_n$ est vraie pour tout $n\in\mathbb{N}^*$.
+
+		\item Montrons que pour tout $n\in\mathbb{N}^*$, on a que
+			$$ P_n:\quad M^n = \begin{pmatrix} f_{n-1}&f_n\\f_n&f_{n+1} \end{pmatrix}  $$
+			est vraie.
+
+			\fbox{Initialisation} On a $M^1 = \begin{pmatrix} 0&1\\1&1 \end{pmatrix} = \begin{pmatrix} f_0&f_1\\f_1&f_2 \end{pmatrix}  $. Donc $P_1$ est vraie.
+
+			\fbox{Hérédité} Fixons $n$ dans $\mathbb{N}^*$ tel que $P_n$ soit vraie. On a :
+			$$ M^n = \begin{pmatrix} f_{n-1}&f_n\\f_n&f_{n+1} \end{pmatrix} $$
+			Donc,
+			$$ M^nM = \begin{pmatrix} f_{n-1}&f_n\\f_n&f_{n+1} \end{pmatrix} M $$
+			i.e.,
+			$$ M^{n+1} = \begin{pmatrix} f_{n}&f_{n+1}\\f_{n+1}&f_n+f_{n+1} \end{pmatrix} $$
+			Or,
+			$$ M^{n+1} = \begin{pmatrix} f_{n}&f_{n+1}\\f_{n+1}&f_{n+2} \end{pmatrix} $$
+			Ainsi, $P_{n+1}$ est vraie.
+
+			\fbox{Conclusion} $P_n$ est vraie pour tout $n\in\mathbb{N}^*$.
+	\end{enumerate}
+\end{document}
-- 
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