[A2/H1/E1] Solved first exercise

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diff --git a/Year 2/Algebra/Homework1/Exercise1.pdf b/Year 2/Algebra/Homework1/Exercise1.pdf
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diff --git a/Year 2/Algebra/Homework1/Exercise1.tex b/Year 2/Algebra/Homework1/Exercise1.tex
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+% Document MUST be built with XeLaTeX!
+\documentclass[11pt,a4paper]{article}
+
+\usepackage[margin=1.5cm]{geometry}
+
+% Поддръжка на български език
+% Thanks to https://tex.stackexchange.com/a/136926
+\usepackage{fontspec}
+\defaultfontfeatures{Ligatures=TeX}
+\setmainfont{CMU Serif}
+\setsansfont{CMU Sans Serif}
+
+\usepackage[bulgarian,english]{babel}
+
+\usepackage{amsmath}            % Математически символи
+\usepackage{amssymb}
+\usepackage{mathtools}
+\usepackage{amsfonts}           % Още математически символи
+\usepackage{cancel}             % Зачертаване
+\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
+\usepackage{indentfirst}        % Индентация на параграф след секция
+
+\usepackage{enumitem}
+\setlist[enumerate]{label=\arabic*.}
+
+\renewcommand{\thesection}{Решение}
+\renewcommand{\thesubsection}{(\roman{subsection})}
+
+\newcommand{\im}{\operatorname{im}}
+
+\let\oldforall\forall
+\renewcommand{\forall}{\; \oldforall}
+
+\let\oldexist\exists
+\renewcommand{\exists}{\; \oldexist}
+
+\setlist[enumerate]{label=\Roman*.}
+
+\usepackage[pdfauthor={Камен Димитров Младенов, КН Летен семестър, 2023},
+		    pdftitle={Домашна работа 1 по ДАА},
+		    pdfsubject={},
+		    pdfkeywords={},
+		    colorlinks=true,
+		    urlcolor=blue]{hyperref}
+
+\title{\vspace{-1.5cm}Домашно №1, Задача 1 по Алгебра 2 на Компютърни Науки}
+\author{Камен Димитров Младенов, Група 6}
+\date{7 Април 2023}
+
+\begin{document}
+
+\maketitle
+
+\subsection{Подточка}
+
+Ще докажем, че $G$ е подгрупа на $GL_2(\mathbb{R})$, съгласно
+\begin{align*}
+	\forall A = \left( \begin{array}{cc}
+		a & c \\
+		0 & b
+	\end{array} \right) \in G, \text{ знаем че } a, b \in \mathbb{R}^* \implies \det(A) = a.b - c.0 = ab \neq 0
+\end{align*}
+
+Тоест ще докажем, че $\forall A,B \in G \Rightarrow A.B \in G, B^{-1} \in G$. Нека
+\begin{align*}
+	A = \left( \begin{array}{cc}
+		a_1 & c_1 \\
+		0 & b_1
+	\end{array} \right) \qquad
+	B = \left( \begin{array}{cc}
+		a_2 & c_2 \\
+		0 & b_2
+	\end{array} \right)
+\end{align*}
+
+\begin{align*}
+	A.B =
+	\left( \begin{array}{cc}
+		a_1 & c_1 \\
+		0 & b_1
+	\end{array} \right)
+	\left( \begin{array}{cc}
+		a_2 & c_2 \\
+		0 & b_2
+	\end{array} \right) = 
+	\left( \begin{array}{cc}
+		a_1 a_2 & a_1 c_2 + c_1 b_2 \\
+		0 & b_1 b_2
+	\end{array} \right) = C
+\end{align*}
+\begin{align*}
+	a_1 a_2, b_1 b_2 \in \mathbb{R}^* \implies C \in G \text{ и } \det(C) \neq 0
+\end{align*}
+\begin{align*}
+	B^{-1} = \left( \begin{array}{cc}
+		a_2 & c_2 \\
+		0 & b_2
+	\end{array} \right)^{-1} = \frac{1}{\det(B)}
+	\left( \begin{array}{cc}
+		B_{11} & B_{12} \\
+		B_{21} & B_{22}
+	\end{array} \right)^\top = \frac{1}{\det(B)}
+	\left( \begin{array}{cc}
+		b_2 & 0 \\
+		c_2 & a_2
+	\end{array} \right)^\top = \frac{1}{\det(B)}
+	\left( \begin{array}{cc}
+		b_2 & c_2 \\
+		0 & a_2
+	\end{array} \right)
+\end{align*}
+\begin{align*}
+	a_2, b_2 \in \mathbb{R}^* \text{ и } \det(B) \neq 0 \implies B^{-1} \in G \text{ и } \det(B^{-1}) \neq 0
+\end{align*}
+
+\subsection{Подточка}
+
+Ще докажем, че изображението $\varphi$ е хомоморфизъм на групи, като
+\begin{align*}
+	\varphi : G \rightarrow (\mathbb{R}^*, .), \quad \forall
+	\left( \begin{array}{cc}
+		a & c \\
+		0 & b
+	\end{array} \right) \in G: \;
+	\varphi
+	\left( \begin{array}{cc}
+		a & c \\
+		0 & b
+	\end{array} \right) = a^2b
+\end{align*}
+\begin{align*}
+	\varphi \left(
+	\left( \begin{array}{cc}
+		a_1 & c_1 \\
+		0 & b_1
+	\end{array} \right)
+	\left( \begin{array}{cc}
+		a_2 & c_2 \\
+		0 & b_2
+	\end{array} \right)
+	\right) & = 
+	\varphi \left(
+	\left( \begin{array}{cc}
+		a_1 a_2 & a_1 c_2 + c_1 b_2 \\
+		0 & b_1 b_2
+	\end{array} \right) 
+	\right) = (a_1a_2)^2(b_1b_2) = (a_1^2b_1)(a_2^2b_2) \\
+	& = \varphi \left(
+	\left( \begin{array}{cc}
+		a_1 & c_1 \\
+		0 & b_1
+	\end{array} \right) \right)
+	\varphi \left(
+	\left( \begin{array}{cc}
+		a_2 & c_2 \\
+		0 & b_2
+	\end{array} \right) \right) \implies \varphi \text{ е хомоморфизъм}
+\end{align*}
+\begin{align*}
+	\forall a \in \mathbb{R}^*, \forall c \in \mathbb{R}: \; \varphi \left(
+	\left( \begin{array}{cc}
+		a & c \\
+		0 & a^{-1}
+	\end{array} \right) \right) = a^2a^{-1} = a \implies \varphi \text{ е върху}
+\end{align*}
+
+Забелязваме, че
+\begin{align*}
+	\forall a \in \mathbb{R}^* \forall c \in \mathbb{R}: \; \varphi \left(
+	\left( \begin{array}{cc}
+		a & c \\
+		0 & a^{-2}
+	\end{array} \right) \right) = a^2a^{-2} = 1
+	\implies \ker(\varphi) = \left\{ \left( \left. \begin{array}{cc}
+		a & c \\
+		0 & a^{-2}
+	\end{array} \right) \right| a \in \mathbb{R}^*, c \in \mathbb{R} \right\} = H
+\end{align*}
+\begin{align*}
+	\forall a, b \in \mathbb{R}^*, \forall c \in \mathbb{R}: \; \varphi
+	\left( \begin{array}{cc}
+		a & c \\
+		0 & b
+	\end{array} \right) = a^2b \in (\mathbb{R}^*, .) \implies \im(\varphi) = (\mathbb{R}^*, .)
+\end{align*}
+
+От теоремата за хомоморфизмите, следва че $G / H = G / \ker(\varphi) \simeq \im (\varphi) = (\mathbb{R}^*, .)$
+	
+\end{document}
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