Example Matrix usage
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Vago sin Voz
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Abstract
Spanish homework of precepts for Linear Algebra
\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}
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\usepackage{sectsty}
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%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}