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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}
+
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\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}
+}
+
+\pdfsuppresswarningpagegroup=1
+
+\colorlet{defn-color}{DarkBlue}
+\colorlet{props-color}{Blue}
+\colorlet{warn-color}{Red}
+\colorlet{exemple-color}{Green}
+\colorlet{corol-color}{Orange}
+\newenvironment{defn-leftbar}{%
+ \def\FrameCommand{{\color{defn-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{warn-leftbar}{%
+ \def\FrameCommand{{\color{warn-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{exemple-leftbar}{%
+ \def\FrameCommand{{\color{exemple-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{props-leftbar}{%
+ \def\FrameCommand{{\color{props-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{corol-leftbar}{%
+ \def\FrameCommand{{\color{corol-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+
+\def \freespace {1em}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{defn-color}Définition~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{defn-leftbar},%
+ postfoothook=\end{defn-leftbar},%
+]{better-defn}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{warn-color}Attention\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{warn-leftbar},%
+ postfoothook=\end{warn-leftbar},%
+]{better-warn}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+notebraces={}{},%
+headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{exemple-color}Exemple~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{exemple-leftbar},%
+ postfoothook=\end{exemple-leftbar},%
+]{better-exemple}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{props-color}Proposition~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{props-leftbar},%
+ postfoothook=\end{props-leftbar},%
+]{better-props}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{props-color}Théorème~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{props-leftbar},%
+ postfoothook=\end{props-leftbar},%
+]{better-thm}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{corol-color}Corollaire~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{corol-leftbar},%
+ postfoothook=\end{corol-leftbar},%
+]{better-corol}
+
+\declaretheorem[style=better-defn]{defn}
+\declaretheorem[style=better-warn]{warn}
+\declaretheorem[style=better-exemple]{exemple}
+\declaretheorem[style=better-corol]{corol}
+\declaretheorem[style=better-props, numberwithin=defn]{props}
+\declaretheorem[style=better-thm, sibling=props]{thm}
+\newtheorem*{lemme}{Lemme}%[subsection]
+%\newtheorem{props}{Propriétés}[defn]
+
+\newenvironment{system}%
+{\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}{}
+
+\newenvironment{lititle}%
+{\vspace{7mm}\LobsterTwo \large}%
+{\\}
+
+\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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+%%=====================================================================================
+%%
+%% 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}
+
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\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}
+}
+
+\pdfsuppresswarningpagegroup=1
+
+\colorlet{defn-color}{DarkBlue}
+\colorlet{props-color}{Blue}
+\colorlet{warn-color}{Red}
+\colorlet{exemple-color}{Green}
+\colorlet{corol-color}{Orange}
+\newenvironment{defn-leftbar}{%
+ \def\FrameCommand{{\color{defn-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{warn-leftbar}{%
+ \def\FrameCommand{{\color{warn-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{exemple-leftbar}{%
+ \def\FrameCommand{{\color{exemple-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{props-leftbar}{%
+ \def\FrameCommand{{\color{props-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{corol-leftbar}{%
+ \def\FrameCommand{{\color{corol-color}\vrule width 3pt} \hspace{10pt}}%
+ \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+
+\def \freespace {1em}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{defn-color}Définition~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{defn-leftbar},%
+ postfoothook=\end{defn-leftbar},%
+]{better-defn}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{warn-color}Attention\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{warn-leftbar},%
+ postfoothook=\end{warn-leftbar},%
+]{better-warn}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+notebraces={}{},%
+headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{exemple-color}Exemple~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{exemple-leftbar},%
+ postfoothook=\end{exemple-leftbar},%
+]{better-exemple}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{props-color}Proposition~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{props-leftbar},%
+ postfoothook=\end{props-leftbar},%
+]{better-props}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{props-color}Théorème~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{props-leftbar},%
+ postfoothook=\end{props-leftbar},%
+]{better-thm}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
+ headpunct=,%
+ bodyfont=\sffamily,%
+ headformat=\color{corol-color}Corollaire~\NUMBER\hfill\NOTE\smallskip\linebreak,%
+ preheadhook=\vspace{\freespace}\begin{corol-leftbar},%
+ postfoothook=\end{corol-leftbar},%
+]{better-corol}
+
+\declaretheorem[style=better-defn]{defn}
+\declaretheorem[style=better-warn]{warn}
+\declaretheorem[style=better-exemple]{exemple}
+\declaretheorem[style=better-corol]{corol}
+\declaretheorem[style=better-props, numberwithin=defn]{props}
+\declaretheorem[style=better-thm, sibling=props]{thm}
+\newtheorem*{lemme}{Lemme}%[subsection]
+%\newtheorem{props}{Propriétés}[defn]
+
+\newenvironment{system}%
+{\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}{}
+
+\newenvironment{lititle}%
+{\vspace{7mm}\LobsterTwo \large}%
+{\\}
+
+\renewenvironment{proof}{$\square$ \footnotesize\textit{Démonstration.}}{\begin{flushright}$\blacksquare$\end{flushright}}
+
+\title{TD du 30 janvier}
+\author{William Hergès\thanks{Sorbonne Université - Faculté des Sciences, Faculté des Lettres}}
+
+\begin{document}
+ \maketitle
+ \tableofcontents
+ \newpage
+ \section*{Exercice 1}
+ \begin{enumerate}
+ \item On a :
+ $$ \begin{pmatrix} 1&1\\2&-3 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 0\\0 \end{pmatrix} $$
+ $$ \begin{pmatrix} 0&-2&-3\\ -2&3&0\\3&0&2 \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} 1\\2\\3 \end{pmatrix} $$
+ $$ \begin{pmatrix} 1&1&1&0\\1&1&0&1\\0&0&1&t \end{pmatrix} \begin{pmatrix} x\\y\\z\\t \end{pmatrix} = \begin{pmatrix} 0\\-1\\1 \end{pmatrix} $$
+ \item On fait l'opération inverse (flemme de le faire).
+ \end{enumerate}
+ \section*{Exercice 2}
+ \begin{enumerate}
+ \item $y=\frac{3}{2}$ et $x=\frac{1}{5}$
+ \item $\begin{pmatrix} -1&1&|~1\\ 1&1&|~2 \end{pmatrix}$
+ \item En faisant $l_1+l_2$, on obtient $2y=3$, ce qui nous donne les mêmes $x$ et $y$
+ \item Son rang est $2$.
+ \end{enumerate}
+ \section*{Exercice 3}
+ Refaire ces transformations chez soi. $S_1$ n'a pas de solution (car on a une ligne $0\neq 0$). $S_2$ possède une infinité de solutions. $S_3$ possède une infinité de solutions.
+
+ On a que la quatrième ligne est la somme des deux premières, on peut donc la supprimer en écrivant $0=0$.
+ $$ \begin{pmatrix} 1&-1&1&1&|~3\\ 5&2&-1&-3&|~5\\ -3&-4&3&2&|~1\\ 0&0&0&0&|~0 \end{pmatrix} \iff \begin{pmatrix} 1&-1&1&1&|~3\\ 0&7&-6&-8&|~-10\\ 0&-7&6&5&|~10\\ 0&0&0&0&|~0 \end{pmatrix} $$
+ $$ \iff\begin{pmatrix} 1&-1&1&1&|~3\\ 0&7&-6&-8&|~-10\\ 0&0&0&-3&|~0\\ 0&0&0&0&|~0 \end{pmatrix}$$
+
+ L'ensemble solution de $S_3$ est donc $\left\{ \left( \frac{11-z}{7},\frac{6z-10}{7}, z, 0 \right), z\in\mathbb{R} \right\} $
+
+ Technique de la matrice échelonnée réduite pour trouver les inconnues :
+ \begin{enumerate}
+ \item on fait en sorte que tous les pivots soient égaux à $1$ ;
+ \item on refait des combinaisons linéaires pour que tous les coefficiants (à part les pivots) soient nuls (c'est plus simple de commencer par le dernier pivot).
+ \end{enumerate}
+ Il n'est pas possible de faire en sorte que tous les coefficiants soient nuls si le rang de la matrice n'est pas assez élevée. Dans ce cas, on a que les coefficiants nuls doivent forcément être ceux dans les colonnes des pivots.
+ \section*{Exercice 5}
+ \begin{enumerate}
+ \item $\begin{pmatrix} -1&-2&3&1&|~b_1\\1&1&2&2&|~b_2\\3&4&1&3&|~b_3 \end{pmatrix} $
+ \item On a que $L_1 = 2L_2-L_3$, donc $b_1=2b_2-b_3$ si le système admet au moins une solution.
+ \item On a donc que $\begin{pmatrix} 1&1&2&2&|~b_2\\3&4&1&3&|~b_3\\0&0&0&0&|~~0 \end{pmatrix}\iff \begin{pmatrix} 1&1&2&2&|~b_2\\0&1&-5&-3&|~b_3-3b_2\\0&0&0&0&|~~0 \end{pmatrix} $. Donc, $x_3$ et $x_4$ sont libres.
+ $$x_2 = b_3-3b_2+5x_3+3x_4$$
+ et
+ $$ x_1 = b_3+4b_2-7x_3-6x_4 $$
+ $$ S = \left\{ (b_4+4b_2-7x_3-6x_4,b_3-3b_2+5x_3+3x_4,x_3,x_4), (x_3,x_4)\in\mathbb{R}^2 \right\} $$
+ \item $\mathrm{rg}(S) = 2$
+ \end{enumerate}
+ \section*{Exercice 8}
+ Si $A^{-1}$ existe, alors il existe $v_1,v_2$ et $v_3$ tel que $Av_1 = \begin{pmatrix} 1\\0\\0 \end{pmatrix}$, $Av_2 = \begin{pmatrix} 0\\1\\0 \end{pmatrix}$ et $Av_3 = \begin{pmatrix} 0\\0\\1 \end{pmatrix}$.
+
+ Résolvons $\begin{pmatrix} 1&1&1&|~b_1\\2&1&1&|~b_2\\1&2&1&|~b_3 \end{pmatrix}$. On a donc :
+ $$ \begin{pmatrix} 1&1&1&|~b_1\\2&1&1&|~b_2\\1&2&1&|~b_3 \end{pmatrix} = \begin{pmatrix} 1&1&1&|~b_1\\0&-1&-1&|~b_2-2b_1\\0&1&0&|~b_3-b_1 \end{pmatrix} = \begin{pmatrix} 1&1&1&|~b_1\\0&0&-1&|~b_2-3b_1+b_3\\0&1&0&|~b_3-b_1 \end{pmatrix} $$
+ Donc, $y=b_3-b_1$, $z=-b_2+3b_1-b_3$ et $x=-b_2-b_1$.
+ Ainsi,
+ $$ v_1= \begin{pmatrix} -1\\-1\\3 \end{pmatrix}\quad v_2= \begin{pmatrix} -1\\0\\-1 \end{pmatrix} \quad v_3 = \begin{pmatrix} 0\\1\\-1 \end{pmatrix} $$
+ D'où $$ A^{-1} = \begin{pmatrix} -1&-1&0\\-1&0&1\\3&-1&-1 \end{pmatrix} $$
+ \section*{Exercice 11}
+ Le système associé à ce problème est :
+ $$ \begin{pmatrix} 3&4&5&|~66\\7&4&3&|~74\\8&8&9&|136 \end{pmatrix} $$
+\end{document}
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