Example Matrix usage
Author:
Vago sin Voz
Last Updated:
11 anni fa
License:
Creative Commons CC BY 4.0
Abstract:
Spanish homework of precepts for Linear Algebra
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
\documentclass[12pt,letterpaper]{article}
\usepackage[utf8]{inputenc}
\usepackage[spanish]{babel}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage[left=1cm,right=1cm,top=2cm,bottom=1cm]{geometry}
\author{Sergio Salinas}
%Hipervinculos
\usepackage{hyperref}
%Color
\usepackage{color}
\definecolor{nred}{RGB}{174,49,54}
\definecolor{nblue}{RGB}{86,99,146}
\definecolor{nalgo}{RGB}{188,139,76}
\usepackage{sectsty}
\sectionfont{\color{nred}}
\subsectionfont{\color{nblue}}
\subsubsectionfont{\color{nalgo}}
%Cabeceras
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhead[L]{Sistemas de ecuaciones}
\fancyhead[C]{Licenciatura en Ciencia de la Computación}
\fancyhead[R]{USACH}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\U}{\mathbb{U}}
\newcommand{\Sol}{\mathbb{S}}
\newcommand{\M}{\mathbb{M}}
\newcommand{\X}{\mathbb{X}}
%Función, crea una matriz 3x4 con los corchetes ya puestos y la linea de separación entre A y b
%Se llama así \maa{1}{1}{1}{1}{2}{2}{2}{2}{3}{3}{3}{3}, esto va a crear una matriz donde la primera fila sean 1, la segunda solo 2 y la tercera 3, con los corchetes y la separación entre la matriz de las incognitas y el igual.
\newcommand{\maa}[8]{\left[ \begin{array}{ccc|c}
#1 & #2 & #3 & #4\\
#5 & #6 & #7 & #8\\
\maacont}
\newcommand{\maacont}[4]{#1 & #2 & #3 & #4\\
\end{array}\right]}
%Otra función, crea una matriz columna 3x1 , la idea se esta función es enviar como primer parametro el igual, y el segundo y el tercero mostrar las modificaciones de las filas ex: \inter{=}{f_1 - f_2} {f_3 - f_1}
\newcommand{\inter}[3]{\begin{array}{c}
#1\\
#2\\
#3\\
\end{array}}
\begin{document}
\begin{titlepage}
\begin{center}
{\Large {Licenciatura en Ciencia de la Computación} }
\includegraphics[scale=1]{usach.png}
\\[1cm]
{\Huge \textsc{Sistemas de ecuaciones}}\\[0.7cm]
{\Huge Álgebra I}\\[2cm]
\begin{minipage}[l]{0.4\textwidth}
\begin{flushleft}
\textbf{\textsf{Profesor:}}\\
\large Miguel Ángel Muñoz Jara\\
\linespread{4}
\large .\\
\end{flushleft}
\end{minipage}
\begin{minipage}[l]{0.4\textwidth}
\begin{flushright}
\textbf{\textsf{Integrantes:}}\\
\linespread{1}
\large Pablo Corrales\\Matías Fuentes\\Sergio Salinas\\Claudio Saji
\end{flushright}
\end{minipage}
\end{center}
\end{titlepage}
\section*{Introducción}
El siguiente trabajo es una solución a la guía de sistemas de ecuaciones de álgebra II, esta guía puede ser encontrada \href{http://palillo.usach.cl/coordinacion_algebra2/pdf/guias/2014/guia2_sistemas_20142.pdf}{aquí}.
\section*{Soluciones}
\begin{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 1 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Reduzca a su forma escalonada por filas de las siguientes matrices.
(1) $\begin{pmatrix}
2 & 1 & -3 & -2 \\
4 & 2 & 0 & 0 \\
1 & -2 & 3 & 4 \\
0 & 3 & 1 & 1 \\
\end{pmatrix} \inter{=}{f_2 - 2f_1}{2f_3-f_1} \begin{pmatrix}
2 & 1 & -3 & -2 \\
0 & 0 & 6 & 4 \\
0 & -5 & 9 & 10 \\
0 & 3 & 1 & 1 \\
\end{pmatrix} \inter{=}{f_2 \rightleftharpoons f_4}{} \begin{pmatrix}
2 & 1 & -3 & -2 \\
0 & 3 & 1 & 1 \\
0 & -5 & 9 & 10 \\
0 & 0 & 6 & 4 \\
\end{pmatrix}\\ \inter{=}{3f_3 + 5f_2}{} \begin{pmatrix}
2 & 1 & -3 & -2 \\
0 & 3 & 1 & 1 \\
0 & 0 & 32 & 35 \\
0 & 0 & 6 & 4 \\
\end{pmatrix} \inter{=}{16f_4 - 3f_3}{} \begin{pmatrix}
2 & 1 & -3 & -2 \\
0 & 3 & 1 & 1 \\
0 & 0 & 32 & 35 \\
0 & 0 & 0 & -41 \\
\end{pmatrix}
$
(2) $\begin{pmatrix}
2 & 1 & 0 & 3 \\
3 & 4 & 2 & 1 \\
4 & 2 & 0 & 1 \\
\end{pmatrix} \inter{=}{2f_2-3f_1}{f_3-2f_1} \begin{pmatrix}
2 & 1 & 0 & 3 \\
0 & 5 & 4 & -7 \\
0 & 0 & 0 & -5 \\
\end{pmatrix}
$
(3) $\begin{pmatrix}
3 & 2 & 0 & 4 \\
5 & 1 & 1 & 3 \\
\end{pmatrix} \inter{=}{3f_2-5f_1}{} \begin{pmatrix}
3 & 2 & 0 & 4 \\
0 & -7 & 3 & -11 \\
\end{pmatrix}
$
(4) $\begin{pmatrix}
1 & -2 & 3 & 3 \\
2 & 3 & 1 & 2 \\
0 & 1 & -2 & 2 \\
\end{pmatrix} \inter{=}{f_2-2f_1}{} \begin{pmatrix}
1 & -2 & 3 & 3 \\
0 & 7 & -5 & -4 \\
0 & 1 & -2 & 2 \\
\end{pmatrix} \inter{=}{7f_3-f_2}{} \begin{pmatrix}
1 & -2 & 3 & 3 \\
0 & 7 & -5 & -4 \\
0 & 0 & -9 & 18 \\
\end{pmatrix}
$
\item Usando el teorema del rango determine si los siguiente sistemas tienen o no solución, en caso afirmativo, determine la solución o las soluciones.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 2.5 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(5) $
\begin{array}{ccccccc|}
3x &+& 4y & & & =& 0\\
2x &-& y &+& 3z &=& 0\\
4x &+& 9y &-&3z&=&0\\ \hline
\end{array}
$\\
\[
\begin{array}{ccccc}
\left[ \begin{array}{ccc|c}
3 &4&0&0\\
2&-1&3&0\\
4&9&-3&0\\
\end{array} \right] & \begin{array}{c}
=\\
f_2 + f_3\\
\end{array} & \left[ \begin{array}{ccc|c}
3 &4&0&0\\
6&8&0&0\\
4&9&-3&0\\
\end{array} \right] & \begin{array}{c}
=\\
f_2 :2\\
\end{array} & \left[ \begin{array}{ccc|c}
3 &4&0&0\\
3&4&0&0\\
4&9&-3&0\\
\end{array} \right]\\
&\begin{array}{c}
=\\
f_3 \rightleftharpoons f_1\\
\end{array} & \left[ \begin{array}{ccc|c}
4&9&-3&0\\
3 &4&0&0\\
3&4&0&0\\
\end{array} \right] & \begin{array}{c}
=\\
f_1 - f_2\\
f_3 - f_2\\
\end{array} & \left[ \begin{array}{ccc|c}
1&5&-3&0\\
3 &4&0&0\\
0&0&0&0\\
\end{array} \right]\\
& \begin{array}{c}
=\\
f_3 - 3f_1\\
\end{array} & \left[ \begin{array}{ccc|c}
1&5&-3&0\\
0 &-1&3&0\\
0&0&0&0\\
\end{array} \right]\\
\end{array}
\]
$$\therefore \hbox{Como } \rho(A) = \rho(A/b) < 3 \hbox{, entonces el sistema tiene infinitas soluciones.}$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 2.6 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(6) $
\begin{array}{ccccccc|}
3x &+& 4y & -&7z & =& 6\\
2x &-& y &+& 8z &=& 2\\
6x &+& 4y &-&14z&=&5\\ \hline
\end{array}
$\\
\[
\begin{array}{ccccccc}
\maa{3}{4}{-7}{6}{2}{1}{8}{2}{6}{4}{-14}{5} & \inter{=}{f_1 - f_2}{f_3 - 2f_1} & \maa{1}{3}{-15}{4}{2}{1}{8}{2}{0}{-4}{0}{1} & \inter{=}{f_2-2f_1}{} & \maa{1}{3}{-15}{4}{0}{-5}{38}{-6}{0}{-4}{0}{1}\\
& \inter{=}{f_2-f_3}{} & \maa{1}{3}{-15}{4}{0}{1}{38}{-7}{0}{-4}{0}{1} & \inter{=}{f_3 + 4f_2}{} & \maa{1}{3}{-15}{4}{0}{1}{38}{-7}{0}{0}{152}{27}\\
\end{array}
\]
$$ \therefore \hbox{Como }\rho(A) = \rho(a/b) = n \hbox{ entonces el sistema tiene solución única.}$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 2.7 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(7) $
\begin{array}{lcccccc|}
x &+& 2y &+& 3z & =& 6\\
3x &+& 4y &+& 5z &=& 2\\
5x &+& 4y &+&3z&=&-18\\ \hline
\end{array}
$\\
\[
\begin{array}{ccccc}
\left[ \begin{array}{ccc|c}
1 &2&3&6\\
3&4&5&2\\
5&4&3&-18\\
\end{array} \right] & \begin{array}{c}
=\\
f_2 - 3\cdot f_1\\
f_3 - 5\cdot f_1\\
\end{array} & \left[ \begin{array}{ccc|c}
1 &2&3&6\\
0&-2&-4&-16\\
0&-6&-12&-48\\
\end{array} \right] & \begin{array}{c}
=\\
f_3 - 3\cdot f_2\\
\end{array} & \left[ \begin{array}{ccc|c}
1&2&3&6\\
0&-2&-4&-16\\
0&0&0&0\\
\end{array} \right]\\
\end{array}
\]
$$\therefore \hbox{Como } \rho(A) = \rho(A/b) < 3 \hbox{, entonces el sistema tiene infinitas soluciones.}$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 2.8 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(8) $
\begin{array}{lcccccc|}
x_1 &+& 2x_2 &+& 3x_3 & =& 2\\
x_1 &-& x_2 &+& x_3 &=& 0\\
x_1 &+& 3x_2 &-& x_3 &=& -2\\
3x_1 &+& 4x_2 &+&3x_3&=&0\\ \hline
\end{array}
$\\
\[
\begin{array}{ccccc}
\left[ \begin{array}{ccc|c}
1&2&3&2\\
1&-1&1&0\\
1&3&-1&-2\\
3&4&3&0\\
\end{array} \right] & \begin{array}{c}
=\\
f_2 - f_1\\
f_3 - f_1\\
f_4 - 3\cdot f_1\\
\end{array} & \left[ \begin{array}{ccc|c}
1 &2&3&2\\
0&-3&-2&-2\\
0&1&-4&-4\\
0&-2&-6&-6\\
\end{array} \right] & \begin{array}{c}
=\\
f_3 \rightleftharpoons f_2\\
f_4\cdot \dfrac{1}{2}
\end{array} & \left[ \begin{array}{ccc|c}
1&2&3&2\\
0&1&-4&-4\\
0&-3&-2&-2\\
0&-1&-3&-3\\
\end{array} \right]\\
&\begin{array}{c}
=\\
f_3 + 3f_2\\
f_4+f_2
\end{array} & \left[ \begin{array}{ccc|c}
1&2&3&2\\
0&1&-4&-4\\
0&0&-14&-14\\
0&0&-7&-7\\
\end{array} \right] & \begin{array}{c}
=\\
f_4 - \dfrac{1}{2}\cdot f_3\\
\end{array} & \left[ \begin{array}{ccc|c}
1&2&3&2\\
0&1&-4&-4\\
0&0&-14&-14\\
0&0&0&0\\
\end{array} \right]\\
\end{array}
\]
$$\therefore \hbox{Como } \rho(A) = \rho(A/b) = 3 \hbox{, entonces el sistema admite solución única}$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 2.9 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(9) $
\begin{array}{ccccccccc|}
1x_1 &+& 2x_2 &+& 3x_3 &-& 1x_4 &=& 0\\
1x_1 &-& 1x_2 &+& 1x_3 &+& 2x_4 &=& 4\\
1x_1 &+& 5x_2 &+& 5x_3 &-& 4x_4 &=& -4\\
1x_1 &+& 8x_2 &+& 7x_3 &-& 7x_4 &=& -8\\ \hline
\end{array}
$\\
\[
\begin{array}{ccccc}
\left[ \begin{array}{cccc|c}
1&2&3&-1&0\\
1&-1&1&2&4\\
1&5&5&-4&-4\\
1&8&7&-7&-8\\
\end{array} \right]
& \begin{array}{cc}
=\\
f_2 - f_1\\
f_3 - f_1\\
f_4 - f_1\\
\end{array}
& \left[ \begin{array}{cccc|c}
1&2&3&-1&0\\
0&-3&-2&3&4\\
0&3&2&-3&-4\\
0&6&4&-6&-8\\
\end{array} \right]
& \begin{array}{cc}
=\\
f_2 *-1\\
f_4 :-2\\
\end{array}
& \left[ \begin{array}{cccc|c}
1&2&3&-1&0\\
0&-3&-2&3&4\\
0&-3&-2&3&4\\
0&-3&-2&3&4\\
\end{array} \right]\\\\
& \begin{array}{cc}
=\\
f_3 - f_2\\
f_4 - f_2\\
\end{array}
& \left[ \begin{array}{cccc|c}
1&2&3&-1&0\\
0&-3&-2&3&4\\
0&0&0&0&0\\
0&0&0&0&0\\
\end{array} \right]\\
\end{array}
\]
Considerando, entonces, el teorema de Roché-Frobenius, comprobamos que el rango de la matriz incompleta es igual al rango de la matriz ampliada -completa-.
$$\therefore \hbox{Como, } \rho(A) = \rho(A/b) < 4 \hbox{; el sistema tiene infinitas soluciones.}$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 2.10 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(10)$
\begin{array}{ccccccc|}
(1-i)x &-& iy &+& 2z &=& 0\\
2x &+& (1+i)y &+& z &=& 0\\
x &+& y &+& z &=& -i\\ \hline
\end{array}
$\\
\[
\begin{array}{ccccc}
\left[ \begin{array}{ccc|c}
(1-i)&-i&2&0\\
2&(1+i)&1&0\\
1&1&1&-i\\
\end{array} \right]
& \begin{array}{c}
=\\
f_2 - (1+i(f_1))\\
f_3 - (\frac{1}{2}+\frac{i}{2}(f_1))\\
\end{array}
& \left[ \begin{array}{ccc|c}
(1-i)&-i&2&0\\
0&2i&-(1+2i)&0\\
0&\frac{1}{2}+\frac{i}{2}&-i&-i\\
\end{array} \right] \\\\
& \begin{array}{cc}
=\\
(1-i)*f_3 -\frac{-i}{2}*f_2\\
\end{array} & \left[\begin{array}{ccc|c}
(1-i)&-i&2&0\\
0&2i&-(1+2i)&0\\
0&0&\frac{-3i}{2}&-(1+i)\\
\end{array} \right] \\\\
& \begin{array}{cc}
=\\
\frac{2}{-3i}*f_3\\
\end{array} & \left[ \begin{array}{ccc|c}
(1-i)&-i&2&0\\
0&2i&-(1+2i)&0\\
0&0&1&\frac{2}{3}-\frac{2i}{3}\\
\end{array} \right]\\\\
& \begin{array}{cc}
=\\
\frac{1}{2}*(f_2+1+2i(f_3))\\
\end{array} & \left[ \begin{array}{ccc|c}
(1-i)&-i&2&0\\
0&1&0&\frac{1}{3}-\frac{i}{3}\\
0&0&1&\frac{2}{3}-\frac{2i}{3}\\
\end{array} \right]\\\\
& \begin{array}{cc}
=\\
f_1-(\frac{1}{i}*f_2 + 2*f_3)\\
\end{array} & \left[ \begin{array}{ccc|c}
(1-i)&0&0&\frac{-1}{3}+\frac{5i}{3}\\
0&1&0&\frac{1}{3}-\frac{i}{3}\\
0&0&1&\frac{2}{3}-\frac{2i}{3}\\
\end{array} \right]\\\\
& \begin{array}{cc}
=\\
f_1*\frac{1}{1-i}\\
\end{array} & \left[ \begin{array}{ccc|c}
1&0&0&\frac{-3}{3}+\frac{2i}{3}\\
0&1&0&\frac{1}{3}-\frac{i}{3}\\
0&0&1&\frac{2}{3}-\frac{2i}{3}\\
\end{array} \right]\\\\
\end{array}
\]
Considerando, entonces, el teorema de Roché-Frobenius, comprobamos que el rango de la matriz incompleta es igual al rango de la matriz ampliada -completa-.
$$\therefore \hbox{Como, } \rho(A) = \rho(A/b) = 3 \hbox{; el sistema tiene solución única:}$$
\makebox[\textwidth]{
x=-1+$\frac{2i}{3}$\\ y=$\frac{1}{3}-\frac{i}{3}$\\ z=$\frac{2}{3}-\frac{2i}{3}$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 3 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Dado el sistema lineal
$$
\begin{array}{ccccccc|}
kx &+& y & & & =& 1\\
x &+& ky & & & =& 1\\
\hline
\end{array} \; (\ast)
$$\\
Determine los conjuntos
$$\begin{array}{l}
S_1 = \lbrace k \in \R | (\ast) \hbox{ tiene soluciones} \rbrace\\
S_2 = \lbrace k \in \R | (\ast) \hbox{ tiene infinitas soluciones}\rbrace\\
S_3 = \lbrace k \in \R | (\ast) \hbox{ no tiene solución} \rbrace\\
\end{array}$$
La matriz será:
\[
\left [
\begin{array}{cc|c}
k&1&1\\
1&k&1\\
\end{array}\right ]
\inter{=}{f_1\rightleftharpoons f_2}{} \left [ \begin{array}{cc|c}
1&k&1\\
k&1&1\\
\end{array}\right ] \inter{=}{f_2 - kf_1}{}
\left [ \begin{array}{cc|c}
1&k&1\\
0&1-k^2&1-k\\
\end{array} \right ]
\]
Así, las soluciones respectivas serán:
\[
\begin{array}{l}
S_1 = \lbrace k \in \R |k \not= 1 \wedge k \not= -1 \rbrace\\
S_2 = \lbrace k \in \R |k = 1 \rbrace\\
S_3 = \lbrace k \in \R |k = -1 \rbrace\\
\end{array}
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 4 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Dado el sistema lineal
$$
\begin{array}{ccccccc|}
x &-& by & -&cz & =& 0\\
-ax &+& y &-& cz &=& 0\\
-ax &-& by &+&z&=&0\\ \hline
\end{array} \; (\ast)
$$\\
Demuestre que si ($\ast$) no tiene solución única entonces se verifica
$$\dfrac{a}{a+1} + \dfrac{b}{b+1} + \dfrac{c}{c+1} = 1$$
Para que el sistema ($\ast$) no tenga sol única entonces $\rho(A) < 3$, esto se puede comprobar fácilmente viendo cuando det(A) = 0.
$$ |A| = \left | \begin{array}{ccc}
1 & -b & -c\\
-a & 1 & -c\\
-a & -b & 1\\
\end{array} \right | = 1-2abc -ac -bc -ab\\
$$
\[
\begin{array}{rcl}
|A| &=& 0\\
1-2abc -ac -bc -ab &=& 0\\
1 &=& 2abc + ac + bc + ab \; /+a+b+c+bc+ac+ab+abc\\
a+b+c+bc+ac+ab+abbc+1 &=& 3abc+2ac+2abc+2ab+a+b+c\\
(1+b+a+ab)(1+c) &=& a(1+c+b+bc) + b(1+c+a+ac) + c(1+b+a+ab)\\
(1+a)(1+b)(1+c) &=& a(1+b)(1+c) + b(1+a)(1+c) + c(1+a)(1+b) \\
1 &=& \dfrac{a}{a+1} + \dfrac{b}{b+1} + \dfrac{c}{c+1}\\
\dfrac{a}{a+1} + \dfrac{b}{b+1} + \dfrac{c}{c+1} &=& 1\\
\end{array}
\]
$$ \therefore \hbox{Se ha demostrado que la afirmación es válida} $$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 5 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Dado el sistema lineal $$
\begin{array}{lcccccc|}
x_1 &+& x_2 &-& x_3 & =& 2\\
x_1 &+& 2x_2 &+& x_3 &=& 3\\
x_1 &+& x_2 &+&(a^2 - 4)x_3&=&a\\ \hline
\end{array} \; (\ast)
$$\\
$$\begin{array}{l}
S_1 = \lbrace a \in \R | (\ast) \hbox{ tiene solución única} \rbrace\\
S_2 = \lbrace a \in \R | (\ast) \hbox{ tiene infinitas soluciones} \rbrace\\
S_3 = \lbrace a \in \R| (\ast) \hbox{ no tiene solución} \rbrace\\
\end{array}$$
\[
\begin{array}{ccccc}
\left[ \begin{array}{ccc|c}
1&1&-1&2\\
1&2&1&3\\
1&1&a^2 - 4&a\\
\end{array} \right] & \begin{array}{c}
=\\
f_2 - f_1\\
f_3 - f_1\\
\end{array} & \left[ \begin{array}{ccc|c}
1&1&-1&2\\
0&1&2&1\\
0&0&a^2 - 3&a - 2\\
\end{array} \right]
\end{array}
\]
$\therefore$ Si $a^2 - 3 = 0$ , o sea, $a = \sqrt{3}$ $\vee$ $a = -\sqrt{3}$ $\rho(A) \neq \rho(A/b)$\\ $\hbox{ entonces el sistema no admite solución}$\\
Si $a^2 - 3 \neq 0$ , o sea, $a \neq \sqrt{3}$ $\wedge$ $a \neq -\sqrt{3}$ $\rho(A) = \rho(A/b) = 3$\\ $\hbox{ entonces el sistema admite solución única}$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 6 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Dado el sistema lineal $$
\begin{array}{ccccccc|}
(1-\lambda)x &+& 1y &+& 1z &=& 0\\
2x &+& (2-\lambda)y &+& 2z &=& 0\\
1x &+& 1y &+& (1-\lambda)z &=& 0\\ \hline
\end{array}
$$\\
$$\begin{array}{l}
S = \lbrace \lambda \in \R | (\ast) \hbox{ tiene infinitas soluciones} \rbrace\\
\end{array}$$
\[
\begin{array}{ccccc}
\left| \begin{array}{ccc}
(1-\lambda)&1&1\\
2&(2-\lambda)&2\\
1&1&(1-\lambda)\\
\end{array} \right|
& \begin{array}{cc}
=\\
\end{array}
& \begin{array}{c}
(1-\lambda)^2*(2-\lambda)+2+2\\-\\(2+\lambda)+2(1-\lambda)+2(2-\lambda)
\end{array}
& \begin{array}{cc}
=\\
\end{array}
& \begin{array}{c}
\lambda^3-4\lambda^2
\end{array}\\\\
\end{array}
\]
$$\lambda^3-4\lambda^2 = 0 $$$$
\lambda(\lambda^2-4\lambda) = 0 \setminus \lambda = 0$$$$
\lambda^2-4\lambda = 0 \setminus \lambda = 0$$$$
\lambda-4 = 0 \setminus \lambda = 4$$
Considerando, entonces, por el método de Cramer: Dado que la determinante en el numerador es cero, solo si la ecuación encontrada en la determinande de la matriz es igual a cero, se darán soluciones infinitas. En efecto:
\[
\begin{array}{ccccc}
\left| \begin{array}{ccc}
0&1&1\\
0&(2-\lambda)&2\\
0&1&(1-\lambda)\\
\end{array} \right|
& \begin{array}{cc}
=\\
\end{array}
& \left| \begin{array}{ccc}
(1-\lambda)&0&1\\
2&0&2\\
1&0&(1-\lambda)\\
\end{array} \right|
& \begin{array}{cc}
=\\
\end{array}
& \left| \begin{array}{ccc}
(1-\lambda)&1&0\\
2&(2-\lambda)&0\\
1&1&0\\
\end{array} \right|
\\\\
\begin{array}{ccc}
0+0+0-(0+0+0)
\end{array}
& \begin{array}{cc}
=\\
\end{array}
& \begin{array}{ccc}
0+0+0-(0+0+0)
\end{array}
& \begin{array}{cc}
=\\
\end{array}
& \begin{array}{ccc}
0+0+0-(0+0+0)
\end{array}
\\
\begin{array}{ccc}
0
\end{array}
& \begin{array}{cc}
=\\
\end{array}
& \begin{array}{ccc}
0
\end{array}
& \begin{array}{cc}
=\\
\end{array}
& \begin{array}{ccc}
0
\end{array}
\end{array}
\]
$$\therefore \hbox{Como, } \rho(A) = \rho(A/b) < 3 \hbox{; el sistema tiene infinitas soluciones si:}$$$$
\mathbb{S}= \{ \lambda \in \mathbb{R}, \quad \lambda = 0 \vee \lambda = 4\}$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 7 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Si A = $\left[ \begin{array}{ccccc}
1 & -1 & 0 & 0 &0\\
-1 & 1 & 0 & 0 & 0\\
0 & 0 & 5 & 0 & 0\\
0 & 0 & 0 & 3 & -1\\
0 & 0 & 0 & 1 &3\\
\end{array} \right ]$ Entonces determine el conjunto.
$$ \Sol = \lbrace \lambda \in \R | AX = \lambda X \rbrace$$
\[ A = \left[ \begin{array}{ccccc}
1 & -1 & 0 & 0 &0\\
-1 & 1 & 0 & 0 & 0\\
0 & 0 & 5 & 0 & 0\\
0 & 0 & 0 & 3 & -1\\
0 & 0 & 0 & 1 &3\\
\end{array} \right ] \inter{=}{f_2 + f_1}{f_4-3f_5} \left[ \begin{array}{ccccc}
1 & -1 & 0 & 0 &0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 5 & 0 & 0\\
0 & 0 & 0 & & -10\\
0 & 0 & 0 & 1 &3\\
\end{array} \right ] = \left[ \begin{array}{ccccc}
1 & -1 & 0 & 0 &0\\
0 & 0 & 5 & 0 & 0\\
0 & 0 & 0 & 1 &3\\
0 & 0 & 0 & 0 & -10\\
0 & 0 & 0 & 0 & 0\\
\end{array} \right ]
\]
Por recomendación del profesor X será una matriz columna con las incógnitas del sistema. Por lo que el sistema queda como.
\[
\left[ \begin{array}{ccccc}
1 & -1 & 0 & 0 &0\\
0 & 0 & 5 & 0 & 0\\
0 & 0 & 0 & 1 &3\\
0 & 0 & 0 & 0 & -10\\
0 & 0 & 0 & 0 & 0\\
\end{array} \right] \left [ \begin{array}{c}
x_1\\
x_2\\
x_3\\
x_4\\
x_5\\
\end{array} \right ] = \left [ \begin{array}{c}
\lambda x_1\\
\lambda x_2\\
\lambda x_3\\
\lambda x_4\\
\lambda x_5\\
\end{array} \right ]
\]
\[AX = \left[ \begin{array}{ccccc}
1 & -1 & 0 & 0 &0\\
0 & 0 & 5 & 0 & 0\\
0 & 0 & 0 & 1 &3\\
0 & 0 & 0 & 0 & -10\\
0 & 0 & 0 & 0 & 0\\
\end{array} \right] \left [ \begin{array}{c}
x_1\\
x_2\\
x_3\\
x_4\\
x_5\\
\end{array} \right ] = \left [ \begin{array}{c}
x_1 - x_2\\
5x_3\\
x_4 + 3x_5\\
-10x_5\\
0\\
\end{array} \right ] \Rightarrow \left [ \begin{array}{c}
x_1 - x_2\\
5x_3\\
x_4 + 3x_5\\
-10x_5\\
0\\
\end{array} \right ] = \left [ \begin{array}{c}
\lambda x_1\\
\lambda x_2\\
\lambda x_3\\
\lambda x_4\\
\lambda x_5\\
\end{array} \right ]
\]
\[ \begin{array}{ccc|}
x_1 - x_2 &=& \lambda x_1\\
5x_3 &=& \lambda x_2\\
x_4 + 3x_5 &=& \lambda x_3\\
-10x_5 &=& \lambda x_4\\
0 &=&\lambda x_5\\ \hline
\end{array} \Rightarrow \begin{array}{ccc|}
(1- \lambda) x_1 - x_2 &=& 0\\
- \lambda x_2 + 5x_3 &=& 0\\
x_4 - \lambda x_3+ 3x_5 &=& 0\\
-\lambda x_4 -10x_5 &=&0 \\
\lambda x_5 &=& 0\\ \hline
\end{array}
\]
Despejando las variables del sistema se tiene que: $ x_1 = x_2 =x_3 = x_4 =x_5 = 0$\\
$$\therefore \Sol = \lbrace \lambda \in \R \rbrace$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 8 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Dado el sistema lineal
$$
\begin{array}{lcccccc|}
dx &+& (2d-1)y &+& (d+2)z &=&1\\
0x &+& (d-1)y &+& (d-3)z &=& 1+d\\
dx &+& (3d-2)y &+&(3d+1)z&=&2-d\\ \hline
\end{array}
$$
\begin{enumerate}
\item Determine el conjunto
$$\Sol = \lbrace d \in \R | (\ast) \hbox{ tiene solución} \rbrace$$
\item Para d $\in \Sol$, (Si $\Sol \not= \oslash$), determine el conjunto.
$$\X = \left\lbrace X =\left [ \begin{array}{c}
x\\
y\\
z\\
\end{array}\right ] \in \M_\R (3x1) | \hbox{ X es solución de }(\ast) \right\rbrace $$
\end{enumerate}
\[
\begin{array}{ccccc}
\left[ \begin{array}{ccc|c}
d&2d-1&d+2&1\\
0&d-1&d-3&1+d\\
d&3d-2&3d+1&2-d\\
\end{array} \right] & \begin{array}{c}
=\\
f_3 - (f_1 + f_2)\\
\end{array} & \left[ \begin{array}{ccc|c}
d&2d-1&d+2&1\\
0&d-1&d-3&1+d\\
0&0&d+2&-2d\\
\end{array} \right]\\
&\begin{array}{c}
=\\
f_2 - \dfrac{2d-1}{d-1}\cdot f_1\\
\end{array} & \left[ \begin{array}{ccc|c}
d&0&d+2&1\\
0&d-1&-(d+2)&\dfrac{d(d-2)}{d-1}\\
0&0&d+2&-2d\\
\end{array} \right]\\
&\begin{array}{c}
=\\
f_1 - f_3\\
f_2 + f_3\\
\end{array} & \left[ \begin{array}{ccc|c}
d&0&0&1+2d\\
0&d-1&0&\dfrac{d^2}{1-d}\\
0&0&d+2&-2d\\
\end{array} \right]
\end{array}
\]
$$\therefore \mathbb{S}=\lbrace d \in \mathbb{R} \vert d \neq -2 \wedge d \neq 0 \wedge d \neq 1 \rbrace$$\\\\
Al dividir: $f_1/d \wedge f_2/(d-1) \wedge f_3/(d+2)$ Obtenemos:\\\\
$\left[ \begin{array}{ccc|c}
1&0&0&\dfrac{1+2d}{d}\\
0&1&0&\dfrac{-d^2}{(d-1)^2}\\
0&0&1&\dfrac{-2d}{d+2}\\
\end{array} \right]$
$\Rightarrow \mathbb{X}=\left( \begin{array}{c}
\dfrac{1+2d}{d}\\\\
\dfrac{-d^2}{(d-1)^2}\\\\
\dfrac{-2d}{d+2}\\\\
\end{array}
\right)$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 9 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Dado el sistema de ecuaciones lineales $$
\begin{array}{ccccccc|}
3x &-& 3y &+& 1z &=& 1\\
x &+& 0y &+& 3z &=& a+1\\
ax &+& 1y &+& 0z &=& -a\\ \hline
\end{array} \; (\ast)
$$\\
$$\begin{array}{l}
S_1 = \lbrace a \in \R | (\ast) \hbox{ tiene solución única} \rbrace\\
S_2 = \lbrace a \in \R | (\ast) \hbox{ tiene infinitas soluciones} \rbrace\\
S_3 = \lbrace a \in \R | (\ast) \hbox{ no tiene solución} \rbrace\\
\end{array}$$
\[
\begin{array}{ccccc}
\left| \begin{array}{ccc}
3&-3&1\\
1&0&3\\
a&1&0\\
\end{array} \right|
& \begin{array}{cc}
=\\
\end{array}
& \begin{array}{c}
0-9a+1-0+9-0
\end{array}
& \begin{array}{cc}
=\\
\end{array}
& \begin{array}{c}
-9a+8
\end{array}\\
\end{array}
\]
$$a \neq \frac{8}{9} \setminus \hbox{Descartamos indeterminancia.}$$
Considerando, entonces, por el método de Cramer: Dado que la determinante en el numerador es cero, solo si la ecuación encontrada en la determinande de la matriz es igual a cero, se darán soluciones infinitas. En efecto:
\[
\begin{array}{ccccc}
\left| \begin{array}{ccc}
1&-3&1\\
a+1&0&3\\
-a&1&0\\
\end{array} \right|
& \begin{array}{cc}
;\\
\end{array}
& \left| \begin{array}{ccc}
3&1&1\\
1&a+1&3\\
a&-a&0\\
\end{array} \right|
& \begin{array}{cc}
;\\
\end{array}
& \left| \begin{array}{ccc}
3&-3&1\\
1&0&a+1\\
a&1&-a\\
\end{array} \right|
\\\\
\begin{array}{ccc}
0+9a+a+1-3
\end{array}
& \begin{array}{cc}
;\\
\end{array}
& \begin{array}{ccc}
0+3a-a-a^2-a+9a
\end{array}
& \begin{array}{cc}
;\\
\end{array}
& \begin{array}{ccc}
-3a^2-3a+3-3a-3-3a
\end{array}
\\
\begin{array}{ccc}
10a-2
\end{array}
& \begin{array}{cc}
;\\
\end{array}
& \begin{array}{ccc}
8a-a^2+2
\end{array}
& \begin{array}{cc}
;\\
\end{array}
& \begin{array}{ccc}
2+9a-3a^2
\end{array}
\\
\begin{array}{ccc}
\to a= \frac{2}{10}
\end{array}
& \begin{array}{cc}
;\\
\end{array}
& \begin{array}{ccc}
\to a= 4\pm 3\sqrt{2}
\end{array}
& \begin{array}{cc}
;\\
\end{array}
& \begin{array}{ccc}
\to a= \frac{1}{6}*(9\pm \sqrt{105})
\end{array}
\end{array}
\]
Con este paso, además encontramos las soluciones únicas para este sistema:
\\
\makebox[\textwidth]{
x=$\frac{10a-2}{-9a+8}$\\ y=$\frac{-a^2+8a+2}{-9a+8}$\\ z=$\frac{-3a^2+9a+2}{-9a+8}$} \par
Finalmente, si evaluamos la determinante de la matriz ampliada, obtendremos:
\[
\begin{array}{ccccc}
\left| \begin{array}{ccc}
-3&1&1\\
0&3&a+1\\
1&0&-a\\
\end{array} \right|
& \begin{array}{cc}
=\\
\end{array}
&\begin{array}{cccc}
9a+a+1+0-3
\end{array}
& \begin{array}{cc}
=\\
\end{array}
&\begin{array}{cccc}
10a-2
\end{array}
\end{array}
\]
Luego, si 10a-2 $\to$ a $\neq$ 2/10, entonces el conjunto solución es vacío.
$$\therefore \hbox{Como, } \rho(A) = \rho(A/b) < 4 \hbox{; el sistema tiene:}$$$$
\hbox{Una única solucion para }\mathbb{S}_1= \{ a \in \mathbb{R}, \quad a = \frac{8}{9}\}$$$$
\hbox{Infinitas soluciones con }\mathbb{S}_2= \{ a \in \mathbb{R}, \quad a = \frac{1}{5} \wedge a= 4\pm 3\sqrt{2} \wedge a= \frac{1}{6}*(9\pm \sqrt{105} \}$$$$
\hbox{Solución vacía si }\mathbb{S}_3= \{ a \in \mathbb{R}, \quad a \neq \frac{1}{5} \}$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 10 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Dado el sistema $$
\begin{array}{lcccccc|}
x &+&my&+&z&=&1\\
mx&+&y&+&(m-1)z&=&m\\
x&+&y&-&z&=&m+1\\ \hline
\end{array} \; (\ast)
$$\\
$$\begin{array}{l}
S_1 = \lbrace m \in \R | (\ast) \hbox{ tiene solución única} \rbrace\\
S_2 = \lbrace m \in \R | (\ast) \hbox{ tiene infinitas soluciones} \rbrace\\
S_3 = \lbrace m \in \R | (\ast) \hbox{ no tiene solución} \rbrace\\
\end{array}$$
\[
\begin{array}{ccccc}
\left[ \begin{array}{ccc|c}
1&m&1&1\\
m&1&m-1&m\\
1&1&1&m+1\\
\end{array} \right] & \begin{array}{c}
=\\
f_1\leftrightharpoons f_3
\end{array} & \left[ \begin{array}{ccc|c}
1&1&1&m+1\\
m&1&m-1&m\\
1&m&1&1\\
\end{array} \right]\\\\
&\begin{array}{c}
=\\
f_2 - mf_1\\
f_3-f_1
\end{array} & \left[ \begin{array}{ccc|c}
1&1&1&m+1\\
0&1-m&-1&m-m(m+1)\\
0&m-1&0&-m\\
\end{array} \right]\\\\
&\begin{array}{c}
=\\
f_3 + f_2\\
\end{array} & \left[ \begin{array}{ccc|c}
1&1&1&m+1\\
0&1-m&-1&m-m(m+1)\\
0&0&-1&-m(m+1)\\
\end{array} \right]\\\\
&\begin{array}{c}
=\\
f_1 - \dfrac{1}{1-m}f_2\\
\end{array} & \left[ \begin{array}{ccc|c}
1&0&\dfrac{2-m}{1-m}&\dfrac{1}{1-m}\\
0&1-m&-1&m-m(m+1)\\
0&0&-1&-m(m+1)\\
\end{array} \right]\\\\
&\begin{array}{c}
=\\
f_1 + \dfrac{2-m}{1-m}f_3\\
\end{array} & \left[ \begin{array}{ccc|c}
1&0&0&\dfrac{m^3-m^2-2m+1}{1-m}\\
0&1-m&-1&m-m(m+1)\\
0&0&-1&-m(m+1)\\
\end{array} \right]\\\\
&\begin{array}{c}
=\\
f_2-f_3\\
\end{array} & \left[ \begin{array}{ccc|c}
1&0&0&\dfrac{m^3-m^2-2m+1}{1-m}\\
0&1-m&0&m\\
0&0&-1&-m(m+1)\\
\end{array} \right]\\\\
&\begin{array}{c}
=\\
\dfrac{1}{1-m}f_2\\\\
-1f_3\\
\end{array} & \left[ \begin{array}{ccc|c}
1&0&0&\dfrac{m^3-m^2-2m+1}{1-m}\\
0&1&0&\dfrac{m}{1-m}\\
0&0&1&m(m+1)\\
\end{array} \right]\\
\end{array}
\]
Solución única\\
$$\mathbb{S}_1=\lbrace m \in \mathbb{R} \vert m \neq 1 \rbrace$$\\\\
No tiene Solución\\
$$\mathbb{S}_2=\lbrace m \in \mathbb{R} \vert m=1 \rbrace$$\\\\
Infinitas soluciones\\
$$\mathbb{S}_3=\lbrace m \in \mathbb{R} \vert m\in \phi \rbrace$$\\\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 11 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Considere el siguiente sistema lineal
$$
\begin{array}{ccccccc|}
x_1 &+& x_2 &-& x_3 &=& 2\\
2x_1 &+& 2x_2 &-& x_3 &=& 3\\
x_1 &-& x_2 &+& (a^2+1)x_3 &=& a\\ \hline
\end{array} \; (\ast)
$$\\
Determine los siguientes conjuntos
$$\begin{array}{l}
S_1 = \lbrace a \in \R | (\ast) \hbox{ Tiene solución} \rbrace\\
S_2 = \lbrace a \in \R | (\ast) \hbox{ Tiene solución única}\rbrace\\
S_3 = \lbrace a \in \R | (\ast) \hbox{ No tiene solución} \rbrace\\
\end{array}$$
\[
\maa{1}{1}{-1}{2}{2}{2}{-1}{3}{1}{-1}{a^2+1}{a} \inter{=}{f_2-2f_1}{f_3-f_1} \maa{1}{1}{-1}{2}{0}{0}{1}{-1}{0}{-2}{a^2+2}{a-2} \inter{=}{f_2 \rightleftharpoons f_3}{} \maa{1}{1}{-1}{2}{0}{-2}{a^2+2}{a-2}{0}{0}{1}{-1}
\]
Como $x_3 = -1$ entonces se tiene lo siguiente
\begin{minipage}[l]{0.4\textwidth}
\begin{flushleft}
\[ \begin{array}{rcl}
-2x_2 - (a^2 + 2) &=& a-2\\
-2x_2 &=& a-2+a^2 + 2\\
x_2 &=& - \dfrac{a^2+a}{2}\\
\end{array}
\]
\end{flushleft}
\end{minipage}
\begin{minipage}[l]{0.4\textwidth}
\begin{flushright}
\[ \begin{array}{rcl}
x_1 + x_2 + 1 &=& 2\\
x_1 - \dfrac{a^2+a}{2} +1 &=& 2\\
x_1 &=& 1 + \dfrac{a^2+a}{2}\\
\end{array}
\]
\end{flushright}
\end{minipage}
$$ \therefore
\begin{array}{l}
S_1 = \lbrace a \in \R \rbrace\\
S_2 = \lbrace a \in \R \rbrace\\
S_3 \hbox{ no existe ya que para ningún valor dea se puede anular una fila de A}
\end{array}
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Ejercicio 12 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Considere el siguiente sistema de ecuaciones:
$$
\begin{array}{ccccccc|}
ax &+& by & +&2z & =& 1\\
ax &+& (2b-1)y &+& 3z &=& 1\\
ax &+& by &+&(b+3)z&=&2\\ \hline
\end{array} \; (\ast)
$$\\
$$\begin{array}{l}
S_1 = \lbrace (a,b) \in \R^2 | (\ast) \hbox{ tiene solución única} \rbrace\\
S_2 = \lbrace (a,b) \in \R^2 | (\ast) \hbox{ tiene infinitas soluciones} \rbrace\\
S_3 = \lbrace (a,b) \in \R^2 | (\ast) \hbox{ no tiene solución} \rbrace\\
\end{array}$$
\[ \maa{a}{b}{2}{1}{a}{2b-1}{3}{1}{a}{b}{b+3}{2b-1} \inter{=}{f_2-f_1}{f_3-f_2} \maa{a}{b}{2}{1}{0}{b-1}{1}{0}{0}{0}{b+1}{2(b-1)} \]
$\therefore$ Del sistema escalonado se pueden ver las siguientes soluciones.
$$\begin{array}{l}
S_1 = \lbrace (a,b) \in \R^2 |a \not= 0 \wedge b \not= -1 \rbrace\\
S_2 = \lbrace (a,b) \in \R^2 | a \not= 0 \wedge b = 1 \rbrace\\
S_3 = \lbrace (a,b) \in \R^2 | b = -1 \rbrace\\
\end{array}$$
\end{enumerate}
\end{document}