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\QTR{small}{Matematika 1 online - S\U{fa}stavy line\U{e1}rnych rovn\U{ed}c - Gaussova elimina\U{10d}n\U{e1} met\U{f3}da rie\U{161}enia s\U{fa}stav line\U{e1}rnych rovn\U{ed}c\dotfill \thepage }}
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\newtheorem{case}[theorem]{Case}
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\newtheorem{conclusion}[theorem]{Conclusion}
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\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}
\begin{document}
\author{A. U. Thor}
\title{Lab Report}
\date{The Date }
\maketitle
\begin{abstract}
A Laboratory report created with Scientific Notebook
\end{abstract}
\section{S\'{u}stavy line\'{a}rnych rovn\'{\i}c}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\textbf{%
%TCIMACRO{\hyperref{Obsah}{}{}{aindex.tex}}%
%BeginExpansion
\msihyperref{Obsah}{}{}{aindex.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Obsah kapitoly}{}{}{A2.tex}}%
%BeginExpansion
\msihyperref{Obsah kapitoly}{}{}{A2.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{A23.tex}}%
%BeginExpansion
\msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{A23.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{OA2.tex}}%
%BeginExpansion
\msihyperref{Ot\'{a}zky}{}{}{OA2.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{CA2.tex}}%
%BeginExpansion
\msihyperref{Cvi\v{c}enia}{}{}{CA2.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Index}{}{}{INDA.tex}}%
%BeginExpansion
\msihyperref{Index}{}{}{INDA.tex}%
%EndExpansion
} \\ \hline
\end{tabular}
\end{center}
\subsection{Gaussova elimina\v{c}n\'{a} met\'{o}da rie\v{s}enia s\'{u}stav
line\'{a}rnych rovn\'{\i}c}
T\'{a}to met\'{o}da spo\v{c}\'{\i}va v \'{u}prave danej s\'{u}stavy line\'{a}%
rnych rovn\'{\i}c pomocou element\'{a}rnych \'{u}prav na s\'{u}stavu s \v{n}%
ou ekvivaletn\'{u}, ktorej rie\v{s}enie vieme \v{l}ah\v{s}ie n\'{a}js\v{t}.
Takou je s\'{u}stava line\'{a}rnych rovn\'{\i}c, ktorej roz\v{s}\'{\i}ren%
\'{a} matica je stup\v{n}ovit\'{a} alebo e\v{s}te lep\v{s}ie redukovan\'{a}
stup\v{n}ovit\'{a}. Ke\v{d}\v{z}e s\'{u}stava line\'{a}rnych rovn\'{\i}c je
jednozna\v{c}ne ur\v{c}en\'{a} svojou roz\v{s}\'{\i}renou maticou, je v\'{y}%
hodn\'{e} zapisova\v{t} s\'{u}stavy pomocou nich. Uvedomme si e\v{s}te, \v{z}%
e vykona\v{t} na s\'{u}stave (2) element\'{a}rnu \'{u}pravu
\emph{vz\'{a}jomn\'{a} v\'{y}mena} $j$\emph{-tej a} $k$\emph{-tej rovnice,}
resp.
\emph{nahradenie} $j$\emph{-tej rovnice jej} $\alpha $ \emph{n\'{a}sobkom,}
resp.
\emph{nahradenie} $j$\emph{-tej rovnice s\'{u}\v{c}tom tejto rovnice a} $%
\beta $\emph{-n\'{a}sobku} $k$\emph{-tej rovnice, }resp.
\emph{vy\v{s}krtnutie} $j$\emph{-tej rovnice zo s\'{u}stavy }
a potom nap\'{\i}sa\v{t} roz\v{s}\'{\i}ren\'{u} maticu tejto novej s\'{u}%
stavy je to ist\'{e}, ako na roz\v{s}\'{\i}renej matici s\'{u}stavy (2)
vykona\v{t}
ERO $\mathbf{a}_{j}:=:\mathbf{a}_{k},$ resp.
ERO $\mathbf{a}_{j}:=\alpha \mathbf{a}_{j},$ resp.
ERO $\mathbf{a}_{j}:=\mathbf{a}_{j}+\beta \mathbf{a}_{k},$ resp.
vy\v{s}krtnutie $j$-t\'{e}ho riadku z roz\v{s}\'{\i}renej matice (\emph{%
pozor, to nie je ERO!}). Toto n\'{a}m umo\v{z}\v{n}uje z\'{\i}ska\v{t} stup%
\v{n}ovit\'{u} alebo redukovan\'{u} stup\v{n}ovit\'{u} roz\v{s}\'{\i}ren\'{u}
maticu s\'{u}stavy ekvivaletnej so s\'{u}stavou (1) z jej roz\v{s}\'{\i}%
renej matice pou\v{z}it\'{\i}m ERO. Pri t\'{y}chto \'{u}prav\'{a}ch budeme
pou\v{z}\'{\i}va\v{t} symboly $\sim ,\ \overset{r}{\sim }$ aj v pr\'{\i}%
pade, ke\v{d} pou\v{z}ijeme vy\v{s}krtnutie riadku.
Ak pri \'{u}prave roz\v{s}\'{\i}renej matice vznikne riadok
\begin{equation*}
(0,0,\ldots ,0\mid b),\text{ \ kde \ }b\neq 0
\end{equation*}%
znamen\'{a} to, \v{z}e v prisl\'{u}chaj\'{u}cej s\'{u}stave je rovnica
\begin{equation*}
0\,x_{1}+0\,x_{2}+\cdots +0\,x_{n}=b
\end{equation*}%
ktor\'{a} nem\'{a} rie\v{s}enie, a teda ani s\'{u}stava (1) nem\'{a} rie\v{s}%
enie.
\begin{example}
Uk\'{a}\v{z}eme, \v{z}e s\'{u}stava line\'{a}rnych rovn\'{\i}c
\begin{equation*}
\begin{array}{r@{\ }c@{\ }r@{\ }c@{\ }r@{\ }c@{\ }rr@{\ =\ }r}
x_{1} & - & x_{2} & & & + & x_{4} & = & 1 \\
3x_{1} & - & 3x_{2} & + & 2x_{3} & + & 6x_{4} & = & 5 \\
2x_{1} & - & 2x_{2} & + & 2x_{3} & + & 5x_{4} & = & 1%
\end{array}%
\end{equation*}%
nem\'{a} rie\v{s}enie.
\end{example}
\begin{solution}
Pomocou~ERO~upravujme~roz\v{s}\'{\i}ren\'{u}~maticu~s\'{u}stavy~na~stup\v{n}%
ovit\'{u}.~%
\begin{equation*}
\left(
\begin{array}{rrrr|r}
1 & -1 & 0 & 1 & 1 \\
3 & -3 & 2 & 6 & 5 \\
2 & -2 & 2 & 5 & 1%
\end{array}%
\right) \overset{\mathbf{r}_{2}:=\mathbf{r}_{2}-3\mathbf{r}_{1}}{\overset{%
\mathbf{r}_{3}:=\mathbf{r}_{3}-2\mathbf{r}_{1}}{\sim }}\left(
\begin{array}{rrrr|r}
1 & -1 & 0 & 1 & 1 \\
0 & 0 & 2 & 3 & 2 \\
0 & 0 & 2 & 3 & -1%
\end{array}%
\right) \overset{\mathbf{r}_{3}:=\mathbf{r}_{3}-\mathbf{r}_{2}}{\sim }\left(
\begin{array}{rrrr|r}
1 & -1 & 0 & 1 & 1 \\
0 & 0 & 2 & 3 & 2 \\
0 & 0 & 0 & 0 & -3%
\end{array}%
\right) .
\end{equation*}%
Z~tvaru posledn\'{e}ho~riadku~vypl\'{y}va,~\v{z}e~s\'{u}stava~nem\'{a}~rie%
\v{s}enie. $\square $
\end{solution}
Uva\v{z}ujme teraz situ\'{a}ciu, ke\v{d} m\'{a}me roz\v{s}\'{\i}ren\'{u}
maticu s\'{u}stavy upraven\'{u} na stup\v{n}ovit\'{u} maticu, ktor\'{a}
neobsahuje riadok $(0,0,\ldots ,0\mid b)$ s nenulov\'{y}m \v{c}\'{\i}slom $b$%
. V tom pr\'{\i}pade t\'{a}to matica, po vy\v{s}krtnut\'{\i} nulov\'{y}ch
riadkov (tie s\'{u} line\'{a}rnou kombin\'{a}ciou ostatn\'{y}ch), m\'{a}
tvar
\begin{equation*}
\mathbf{D}=\left(
\begin{array}{rrrrrrrrrr|r}
0, & \ldots , & 0, &
\begin{array}{|@{\;}c|}
\hline
d_{1k_{1}} \\ \hline
\end{array}%
\,, & \ldots , & d_{1k_{2}}, & \ldots , & d_{1k_{r}}, & \ldots , & d_{1k_{n}}
& u_{1} \\
0, & \ldots , & 0, & 0, & \ldots , &
\begin{array}{|@{\;}c|}
\hline
d_{2k_{2}} \\ \hline
\end{array}%
\,, & \ldots , & d_{2k_{r}}, & \ldots , & d_{2k_{n}} & u_{2} \\
\multicolumn{10}{c|}{\Large %
....................................................} & {\Large ..} \\
0, & \ldots , & 0, & 0, & \ldots , & 0, & \ldots , &
\begin{array}{|@{\;}c|}
\hline
d_{rk_{r}} \\ \hline
\end{array}%
\,, & \ldots , & d_{rk_{n}} & u_{r}%
\end{array}%
\right)
\end{equation*}%
V r\'{a}m\v{c}ekoch s\'{u} ved\'{u}ce prvky riadkov. V pr\'{\i}pade, \v{z}e
ved\'{u}ci prvok prv\'{e}ho riadku je hne\v{d} na jeho za\v{c}iatku, t.j. $%
k_{1}=1$, matica $\mathbf{D}$ nem\'{a} na za\v{c}iatku nulov\'{e} st\'{l}%
pce. Ke\v{d}\v{z}e ved\'{u}ce prvky riadkov sa pos\'{u}vaj\'{u} aspo\v{n} o
jedno miesto doprava a v matici nem\^{o}\v{z}e by\v{t} riadok $(0,0,\ldots
,0\mid b)$ s nenulov\'{y}m \v{c}\'{\i}slom $b$, mus\'{\i} plati\v{t}
\begin{equation*}
r\leq n,\ 1\leq k_{1}