%% This document created by Scientific Notebook (R) Version 3.5 %% Starting shell: article \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amsmath} \usepackage{amssymb} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Wednesday, February 10, 1999 13:29:48} %TCIDATA{LastRevised=Monday, October 03, 2005 09:42:38} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{Counters=arabic,1} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematika 1 online - S\U{fa}stavy line\U{e1}rnych rovn\U{ed}c - Matice - z\U{e1}kladn\U{e9} vlastnosti\dotfill \thepage }} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \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|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}{}{}{A21.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{A21.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{A23.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{A23.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{\hyperref{Ot\'{a}zky}{}{}{OA2.tex}}{}{}{OA2.tex}}% %BeginExpansion \msihyperref{% \msihyperref{Ot\'{a}zky}{}{}{OA2.tex}}{}{}{OA2.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{\hyperref{Cvi\v{c}enia}{}{}{CA2.tex}}{}{}{CA2.tex}}% %BeginExpansion \msihyperref{% \msihyperref{Cvi\v{c}enia}{}{}{CA2.tex}}{}{}{CA2.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{INDA.tex}}% %BeginExpansion \msihyperref{Index}{}{}{INDA.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \subsection{Matice - z\'{a}kladn\'{e} vlastnosti} \begin{definition} \ \textit{Maticou} typu $m\times n,\ m,n\in \mathbf{N},$ naz\'{y}vame tabu% \v{l}ku komplexn\'{y}ch (re\'{a}lnych) \v{c}\'{\i}sel \begin{equation*} \mathbf{A}=\left( \begin{tabular}{cccc} $a_{11}$ & $a_{12}$ & $\dots $ & $a_{1n}$ \\ $a_{21}$ & $a_{22}$ & $\dots $ & $a_{2n}$ \\ $\vdots $ & $\vdots $ & $\ddots $ & $\vdots $ \\ $a_{m1}$ & $a_{m2}$ & $\dots $ & $a_{mn}$% \end{tabular}% \right) \end{equation*}% \v{C}\'{\i}slo $a_{jk}\in \mathbf{C\,}\left( \mathbf{R}\right) $ sa naz\'{y}% va \emph{prvok\ matice}$\ \mathbf{A}\ $\emph{na}$\ (j,k)$\emph{-tom mieste}. Prvky $a_{11},a_{22},\ldots ,a_{pp}$, kde $p=\min \{m,\,n\}$, tvoria \emph{% hlavn\'{u} diagon\'{a}lu} a prvky $a_{1n},a_{2\;n-1},\ldots ,a_{p\;n-p+1}$ \emph{ved\v{l}aj\v{s}iu diagon\'{a}lu}. \end{definition} Budeme pou\v{z}\'{\i}va\v{t} \v{d}al\v{s}ie ozna\v{c}enia a n\'{a}zvy: $\ $ $\mathbf{C}^{m\times n}$ -- mno\v{z}ina v\v{s}etk\'{y}ch matic typu $m\times n$ s komplexn\'{y}mi prvkami, $\mathbf{R}^{m\times n}$ -- mno\v{z}ina v\v{s}etk\'{y}ch matic typu $m\times n$ s re\'{a}lnymi prvkami. Prvky $(a)\in \mathbf{C}^{1\times 1},\ a\in \mathbf{C}$ budeme pova\v{z}ova% \v{t} za rovnak\'{e}, preto aj $\mathbf{C}^{1\times 1}=\mathbf{C}$. Matice typu $n\times 1$ naz\'{y}vame \emph{st\'{l}pcov\'{e} }$n$\emph{-tice} alebo \emph{st\'{l}pcov\'{e} vektory}. Matice typu $1\times n$ s\'{u} n\'{a}m u% \v{z} dobre zn\'{a}me $n$-tice. Maticu $\mathbf{A,}$ ktorej prvky s\'{u} $% a_{jk}$ pre $j\in \{1.\ldots ,m\},\ k\in \{1,\ldots ,n\}$, budeme stru\v{c}% nej\v{s}ie zapisova\v{t} $(a_{jk})_{n}^{m}$ alebo, ke\v{d} bude jasn\'{e}, ak% \'{y} je po\v{c}et riadkov a st\'{l}pcov matice alebo tieto hodnoty nebud% \'{u} d\^{o}le\v{z}it\'{e}, len $(a_{jk})$. Matica, ktorej v\v{s}etky prvky s\'{u} nuly, sa naz\'{y}va \emph{nulov\'{a}} a ozna\v{c}ujeme ju $\mathbf{0}$ alebo $(0)_{n}^{m}$, ak chceme vyzna\v{c}i% \v{t} aj jej typ. $n$-tica \begin{equation*} (a_{j1},a_{j2},\ldots ,a_{jn}) \end{equation*}% sa naz\'{y}va $j$\emph{-t\'{y} riadok} a st\'{l}pcov\'{a} $m$-tica \begin{equation*} \left( \begin{array}{c} a_{1k} \\ a_{2k} \\ \vdots \\ a_{mk}% \end{array}% \right) \end{equation*}% $k$\emph{-t\'{y} st\'{l}pec} matice $\mathbf{A}=(a_{jk})_{n}^{m}$. \ Ak ozna\v{c}\'{\i}me $\mathbf{r}_{1},\mathbf{r}_{2},\ldots ,\mathbf{r}_{m}$ riadky a $\mathbf{s}_{1},\mathbf{s}_{2},\ldots ,\mathbf{s}_{n}$ st\'{l}pce matice $\mathbf{A}$ typu $m\times n$, tak budeme tie\v{z} p\'{\i}sa\v{t} \begin{equation*} \mathbf{A}=\left( \begin{array}{c} \mathbf{r}_{1} \\ \mathbf{r}_{2} \\ \vdots \\ \mathbf{r}_{m}% \end{array}% \right) ,\qquad \mathbf{A}=(\mathbf{s}_{1},\mathbf{s}_{2},\ldots ,\mathbf{s}% _{n}). \end{equation*} \begin{example} Pre maticu \begin{equation*} \mathbf{A}=\left( \begin{array}{rrrr} 1 & -5 & 6 & 7 \\ 3 & 2 & 1 & 0 \\ -4 & 5 & 1 & 3% \end{array}% \right) \in \mathbf{R}^{3\times 4}. \end{equation*}% je tret\'{\i} st\'{l}pec $\mathbf{s}_{3}=\left( \begin{array}{c} 6 \\ 1 \\ 1% \end{array}% \right) $ a druh\'{y} riadok $\mathbf{r}_{2}=(3,2,1,0).\square $\newline \end{example} \begin{definition} Dve matice $\mathbf{A}=(a_{jk}),\ \mathbf{B}=(b_{jk})$ sa rovnaj\'{u} $% \left( \mathbf{A}=\mathbf{B}\right) $\ , ak \textsl{s\'{u} rovnak\'{e}ho typu% } a \textsl{pre v\v{s}etky} $j,k$ \textsl{plat\'{\i} }$a_{jk}=b_{jk}$. \end{definition} \begin{definition} \textsl{Ved\'{u}cim prvkom} $n$-tice naz\'{y}vame jej \textsl{prv\'{u} nenulov\'{u} zlo\v{z}ku z\v{l}ava.} \end{definition} \begin{example} Ved\'{u}cim prvkom \v{s}estice \begin{equation*} (0,0,-2,0,3,1) \end{equation*}% je jej tretia zlo\v{z}ka $-2$. $\square $\newline \end{example} \begin{quote} Nulov\'{a} $n$-tica nem\'{a} ved\'{u}ci prvok. \end{quote} \begin{definition} Matica~% \begin{equation*} \mathbf{A}=\left( \begin{array}{c} \mathbf{r}_{1} \\ \mathbf{r}_{2} \\ \vdots \\ \mathbf{r}_{m}% \end{array}% \right) \end{equation*} sa~naz\'{y}va~\emph{stup\v{n}ovit\'{a}},~ak~pre~$j\in \{1,2,\ldots ,m-1\}$% ~plat\'{\i}: 1. Ved\'{u}ci prvok riadku $\mathbf{r}_{j+1}$ je posunut\'{y} aspo\v{n} o jedno miesto doprava vzh\v{l}adom k ved\'{u}cemu prvku~riadku~$\mathbf{r}% _{j} $. 2. Ak $\mathbf{r}_{j}=\mathbf{0},\,$ tak $\mathbf{r}_{j+1}=\mathbf{0}.$ \end{definition} \begin{example} Matica \begin{equation*} \left( \begin{array}{rrrrrr} 0 & 1 & -2 & 1 & -1 & 0 \\ 0 & 0 & 0 & 2 & 3 & -1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0% \end{array}% \right) \end{equation*}% je stup\v{n}ovit\'{a}, matica\ \begin{equation*} \left( \begin{array}{rrrrrr} 0 & 1 & 0 & 1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2 & -1 \\ 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 & 2 & 0% \end{array}% \right) \end{equation*} ~nie~je~stup\v{n}ovit\'{a}.~ \ $\square $\newline \end{example} \begin{definition} Matica $\mathbf{A}$ sa naz\'{y}va \emph{redukovan\'{a} stup\v{n}ovit\'{a}}, ak je stup\v{n}ovit\'{a}, ved\'{u}ci prvok ka\v{z}d\'{e}ho nenulov\'{e}ho riadku je 1 a v ka\v{z}dom st\'{l}pci, v ktorom sa nach\'{a}dza ved\'{u}ci prvok niektor\'{e}ho riadku, s\'{u} v\v{s}etky ostatn\'{e} prvky rovnaj\'{e} 0. \end{definition} \begin{example} Matica\medskip \begin{equation*} \left( \begin{array}{rrrrrr} 0 & 1 & 3 & 0 & 0 & 5 \\ 0 & 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0% \end{array}% \right) \medskip \end{equation*}% je~redukovan\'{a}~stup\v{n}ovit\'{a}. $\square $\newline \end{example} \begin{definition} \emph{Element\'{a}rnou riadkovou oper\'{a}ciou} (ERO) na matici \begin{equation*} \mathbf{A}=\left( \begin{array}{c} \mathbf{r}_{1} \\ \mathbf{r}_{2} \\ \vdots \\ \mathbf{r}_{m}% \end{array}% \right) \end{equation*}% naz\'{y}vame ka\v{z}d\'{u} z~nasleduj\'{u}cich~\'{u}prav~matice~$\mathbf{A}$: 1. \emph{Vz\'{a}jomn\'{a}~v\'{y}mena~dvoch~riadkov~matice}~$\mathbf{A}$.% \textbf{\ } 2. \emph{Nahradenie~jedn\'{e}ho~riadku~matice}~$\mathbf{A}$~\emph{% jeho~nenulov\'{y}m~n\'{a}sobkom.} 3. \emph{Nahradenie jedn\'{e}ho riadku s\'{u}\v{c}tom tohto riadku a \v{l}% ubovo\v{l}n\'{e}ho n\'{a}sobku in\'{e}ho riadku matice} $\mathbf{A}$. \end{definition} Jednotliv\'{e} ERO budeme ozna\v{c}ova\v{t} takto: $\mathbf{r}_{j}:=:\mathbf{r}_{k}$ -- vz\'{a}jomn\'{a} v\'{y}mena $j$-teho a $% k$-teho riadku $\mathbf{r}_{j}:=\alpha \mathbf{r}_{j}$ -- nahradenie $j$-teho riadku jeho $% \alpha $-n\'{a}sobkom $\mathbf{r}_{j}:=\mathbf{r}_{j}+\beta \mathbf{r}_{k}$ -- nahradenie $j$-teho riadku s\'{u}\v{c}tom tohto riadku a $\beta $-n\'{a}sobku $k$-teho riadku \begin{example} Upravte~maticu~$\left( \begin{array}{cccc} 1 & -3 & 2 & 1+i \\ -i & 1-2i & 3 & 5 \\ 0 & 3i & -2 & 4% \end{array}% \right) $~pomocou~ERO: 1. $\mathbf{r}_{2}:=:\mathbf{r}_{3}$ 2. $\mathbf{r}_{1}:=i\mathbf{r}_{1}$ 3. $\mathbf{r}_{3}:=\mathbf{r}_{3}+2\mathbf{r}_{1}$ \end{example} \begin{solution} \begin{equation*} \text{1. }\left( \begin{array}{cccc} 1 & -3 & 2 & 1+i \\ -i & 1-2i & 3 & 5 \\ 0 & 3i & -2 & 4% \end{array}% \right) \overset{\mathbf{r}_{2}:=:\mathbf{r}_{3}}{\longrightarrow }\left( \begin{array}{cccc} 1 & -3 & 2 & 1+i \\ 0 & 3i & -2 & 4 \\ -i & 1-2i & 3 & 5% \end{array}% \right) . \end{equation*}% \vspace{0pt}. \begin{equation*} \text{2. }\left( \begin{array}{cccc} 1 & -3 & 2 & 1+i \\ -i & 1-2i & 3 & 5 \\ 0 & 3i & -2 & 4% \end{array}% \right) \overset{\mathbf{r}_{1}:=\,i\mathbf{r}_{1}}{\longrightarrow }\left( \begin{array}{cccc} i & -3i & 2i & -1+i \\ -i & 1-2i & 3 & 5 \\ 0 & 3i & -2 & 4% \end{array}% \right) . \end{equation*}% \vspace{0pt}. \begin{equation*} \text{3. }\left( \begin{array}{cccc} 1 & -3 & 2 & 1+i \\ -i & 1-2i & 3 & 5 \\ 0 & 3i & -2 & 4% \end{array}% \right) \overset{\mathbf{r}_{3}:=\mathbf{r}_{3}+2\mathbf{r}_{1}}{% \longrightarrow }\left( \begin{array}{cccc} 1 & -3 & 2 & 1+i \\ -i & 1-2i & 3 & 5 \\ 2 & -6+3i & 2 & 6+2i% \end{array}% \right) .\,\square \newline \end{equation*} \end{solution} \begin{quote} Analogicky sa definuj\'{u} \emph{element\'{a}rne st\'{l}pcov\'{e} oper\'{a}% cie} (ESO) na matici $\mathbf{A}=(\mathbf{s}_{1},\mathbf{s}_{2},\ldots ,% \mathbf{s}_{n})$. Sta\v{c}\'{\i} v~defin\'{\i}cii ERO nahradi\v{t} riadky st% \'{l}pcami, pri\v{c}om \emph{s\'{u}\v{c}et st\'{l}pcov\'{y}ch vektorov} a \emph{n\'{a}sobok st\'{l}pcov\'{e}ho vektora \v{c}\'{\i}slom} s\'{u} definovan\'{e} takto: \begin{equation*} \left( \begin{array}{c} u_{1} \\ u_{2} \\ \vdots \\ u_{m}% \end{array}% \right) +\left( \begin{array}{c} v_{1} \\ v_{2} \\ \vdots \\ v_{m}% \end{array}% \right) =\left( \begin{array}{c} u_{1}+v_{1} \\ u_{2}+v_{2} \\ \vdots \\ u_{m}+v_{m}% \end{array}% \right) ,\qquad \alpha \left( \begin{array}{c} u_{1} \\ u_{2} \\ \vdots \\ u_{m}% \end{array}% \right) =\left( \begin{array}{c} \alpha u_{1} \\ \alpha u_{2} \\ \vdots \\ \alpha u_{m}% \end{array}% \right) \end{equation*}% ERO a ESO naz\'{y}vame spolo\v{c}n\'{y}m n\'{a}zvom \emph{element\'{a}rne oper\'{a}cie} (EO). \end{quote} \begin{definition} Hovor\'{\i}me, \v{z}e matica $\mathbf{B}$ vznikla z matice $\mathbf{A}$ kone% \v{c}n\'{y}m po\v{c}tom ERO (resp. ESO \v{c}i EO), ak existuj\'{u} tak\'{e} matice $\mathbf{A}_{1},\mathbf{A}_{2},\ldots ,\mathbf{A}_{p}$, \v{z}e $% \mathbf{A}=\mathbf{A}_{1},\mathbf{B}=\mathbf{A}_{p}$ a pre ka\v{z}d\'{e} $% j\in \{1,2,\ldots ,p-1\}$ matica $\mathbf{A}_{j+1}$ vznikla z matice $% \mathbf{A}_{j}$ jedinou ERO (resp. ESO \v{c}i EO). \end{definition} \begin{theorem} Ak matica $\mathbf{B}$ vznikne z matice $\mathbf{A}$ kone\v{c}n\'{y}m po\v{c}% tom ERO (resp. ESO \v{c}i EO), tak aj $\mathbf{A}$ vznikne z~matice $\mathbf{% B}$ kone\v{c}n\'{y}m po\v{c}tom ERO (resp. ESO \v{c}i EO). \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{\textbf{D\^{o}kaz}}{}{}{DA221.tex}}% %BeginExpansion \msihyperref{\textbf{D\^{o}kaz}}{}{}{DA221.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{1} \begin{definition} Dve matice $\mathbf{A},\mathbf{B}$ sa naz\'{y}vaj\'{u} \begin{equation*} \left. \begin{array}{l} \text{\emph{riadkovoekvivalentn\'{e}}} \\ \text{\emph{st\'{l}pcovoekvivalentn\'{e}}} \\ \text{\emph{ekvivalentn\'{e}}}% \end{array}% \right\} \text{ ak }\mathbf{B}\text{ vznikne z }\mathbf{A}\text{ kone\v{c}n% \'{y}m po\v{c}tom }\left\{ \begin{array}{l} \text{ERO} \\ \text{ESO} \\ \text{EO}% \end{array}% \right. \end{equation*}% a ozna\v{c}ujeme to v porad\'{\i} $\mathbf{A}\overset{r}{\sim }\mathbf{B},\ \mathbf{A}\overset{s}{\sim }\mathbf{B},\ \mathbf{A}\sim \mathbf{B}$. \end{definition} \begin{example} Matice \begin{equation*} \mathbf{A}=\left( \begin{array}{rrr} 0 & 2 & 4 \\ 1 & 0 & 3 \\ 3 & 1 & 12% \end{array}% \right) ,\mathbf{B}=\left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 1% \end{array}% \right) \end{equation*}% s\'{u} riadkovo ekvivalentn\'{e}, lebo \begin{equation*} \mathbf{A}=\left( \begin{array}{rrr} 0 & 2 & 4 \\ 1 & 0 & 3 \\ 3 & 1 & 12% \end{array}% \right) \underset{\mathbf{r}_{1}:=:\mathbf{r}_{2}}{\overset{r}{\sim }}\left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 2 & 4 \\ 3 & 1 & 12% \end{array}% \right) \underset{\mathbf{r}_{3}:=\,\mathbf{r}_{3}-3\mathbf{r}_{1}}{\overset{% r}{\sim }}\left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 2 & 4 \\ 0 & 1 & 3% \end{array}% \right) \underset{\mathbf{r}_{3}:=2\mathbf{r}_{3}}{\overset{r}{\sim }} \end{equation*}% \begin{equation*} \underset{\mathbf{r}_{3}:=2\mathbf{r}_{3}}{\overset{r}{\sim }}\left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 2 & 4 \\ 0 & 2 & 6% \end{array}% \right) \underset{\mathbf{r}_{3}:=\mathbf{r}_{3}-\mathbf{r}_{2}}{\overset{r}{% \sim }}\left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 2% \end{array}% \right) \underset{\mathbf{r}_{3}:=\frac{1}{2}\mathbf{r}_{3}}{\overset{r}{% \sim }}\left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 1% \end{array}% \right) =\mathbf{B.\;\square } \end{equation*}% $\newline $ \end{example} Z d\^{o}vodu \'{u}spory \v{c}asu a miesta je vhodn\'{e} pri \'{u}prave matice pomocou EO zap\'{\i}sa\v{t} nov\'{u} maticu nie po ka\v{z}dej vykonanej EO, ale a\v{z} po nieko\v{l}k\'{y}ch. \'{U}pravu matice $\mathbf{A} $ m\^{o}\v{z}eme krat\v{s}ie zap\'{\i}sa\v{t} napr. takto: \begin{equation*} \mathbf{A}=\left( \begin{array}{rrr} 0 & 2 & 4 \\ 1 & 0 & 3 \\ 3 & 1 & 12% \end{array}% \right) \underset{\mathbf{r}_{3}:=\mathbf{r}_{3}-3\mathbf{r}_{1}}{\underset{% \mathbf{r}_{1}:=:\mathbf{r}_{2}}{\overset{r}{\sim }}}\left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 2 & 4 \\ 0 & 1 & 3% \end{array}% \right) \underset{\mathbf{r}_{3}:=\frac{1}{2}\mathbf{r}_{3}}{\underset{% \mathbf{r}_{3}:=2\mathbf{r}_{3}-\mathbf{r}_{2}}{\overset{r}{\sim }}}\left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 1% \end{array}% \right) =\mathbf{B} \end{equation*}% Namiesto posledn\'{y}ch dvoch \'{u}prav $\mathbf{r}_{3}:=2\mathbf{r}_{3}-% \mathbf{r}_{2},\ \mathbf{r}_{3}:=\frac{1}{2}\mathbf{r}_{3}$ sme mohli pou% \v{z}i\v{t} len jednu $\mathbf{r}_{3}:=\mathbf{r}_{3}-\frac{1}{2}\mathbf{r}% _{2}$. \begin{theorem} Ka\v{z}d\'{a} matica je riadkovo ekvivaletn\'{a} s niektorou stup\v{n}ovitou a tie\v{z} s niektorou redukovanou stup\v{n}ovitou maticou. \end{theorem} D\^{o}kaz pre \v{l}ubovo\v{l}n\'{u} maticu nebudeme robi\v{t}. Uk\'{a}\v{z}% eme si v\v{s}ak postup pri \'{u}prave konkr\'{e}tnej matice $\mathbf{A}$ na stup\v{n}ovit\'{u} resp. redukovan\'{u} stup\v{n}ovit\'{u} maticu. \begin{example} N\'{a}jdime stup\v{n}ovit\'{u} a redukovan\'{u} stup\v{n}ovit\'{u} maticu riadkovo ekvivaletn\'{u} s maticou% \begin{equation*} \mathbf{A}=\left( \begin{array}{rrrr} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & -1 & -1 & 0 \\ 1 & 1 & -1 & -1 \\ 2 & 2 & 0 & -1% \end{array}% \right) . \end{equation*} \end{example} \begin{solution} Najprv v matici $\mathbf{A}$ vyh\v{l}ad\'{a}me riadok, ktor\'{e}ho ved\'{u}% ci prvok sa nach\'{a}dza najviac v\v{l}avo, a vymen\'{\i}me tento riadok s prv\'{y}m. Potom pripo\v{c}\'{\i}tame vhodn\'{e} n\'{a}sobky prv\'{e}ho riadku k ostatn\'{y}m riadkom tak, aby v st\'{l}pci pod ved\'{u}cim prvkom prv\'{e}ho riadku boli len nuly. \begin{equation*} \mathbf{A}=\left( \begin{array}{rrrr} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & -1 & -1 & 0 \\ 1 & 1 & -1 & -1 \\ 2 & 2 & 0 & -1% \end{array}% \right) \underset{\mathbf{r}_{2}:=:\mathbf{r}_{5}}{\underset{\mathbf{r}% _{1}:=:\mathbf{r}_{3}}{\overset{r}{\sim }}}\left( \begin{array}{rrrr} 1 & -1 & -1 & 0 \\ 2 & 2 & 0 & -1 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0% \end{array}% \right) \underset{\mathbf{r}_{4}:=\mathbf{r}_{4}-\mathbf{r}_{1}}{\underset{% \mathbf{r}_{2}:=\mathbf{r}_{2}-2\mathbf{r}_{1}}{\overset{r}{\sim }}}\left( \begin{array}{rrrr} 1 & -1 & -1 & 0 \\ 0 & 4 & 2 & -1 \\ 0 & 1 & 1 & 0 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & 0 & 0% \end{array}% \right) =\mathbf{B} \end{equation*}% Maticu $\mathbf{B}$ upravujeme tak ako maticu $\mathbf{A}$, s t\'{y}m rozdielom. \v{z}e \'{u}lohu prv\'{e}ho riadku tu preber\'{a} riadok druh\'{y}% , \v{d}alej potom riadok tret\'{\i} at\v{d} a\v{z} nakoniec dostaneme stup% \v{n}ovit\'{u} maticu. \begin{equation*} \mathbf{B}\underset{\mathbf{r}_{4}:=\mathbf{r}_{4}-2\mathbf{r}_{2}}{\underset% {\mathbf{r}_{3}:=\mathbf{r}_{3}-4\mathbf{r}_{2}}{\underset{\mathbf{r}_{2}:=:% \mathbf{r}_{3}}{\overset{r}{\sim }}}}\left( \begin{array}{rrrr} 1 & -1 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 0 & 0% \end{array}% \right) \underset{\mathbf{r}_{4}:=\mathbf{r}_{4}-\mathbf{r}_{3}}{\overset{r}{% \sim }}\left( \begin{array}{rrrr} 1 & -1 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0% \end{array}% \right) =\mathbf{D} \end{equation*}% Matica $\mathbf{D}$ je stup\v{n}ovit\'{a} a riadkovo ekvivalentn\'{a} s maticou $\mathbf{A}$. Redukovan\'{u} stup\v{n}ovit\'{u} maticu dostaneme z matice $\mathbf{D}$ takto: Vhodn\'{e} n\'{a}sobky posledn\'{e}ho nenulov\'{e}% ho riadku pripo\v{c}\'{\i}tame k riadkom nad n\'{\i}m tak, aby sa vynulovali prvky nad ved\'{u}cim prvkom posledn\'{e}ho nenulov\'{e}ho riadku. Takto pokra\v{c}ujeme s predposledn\'{y}m nenulov\'{y}m riadkom at\v{d}, a\v{z} v% \v{s}etky prvky nad ved\'{u}cimi prvkami bud\'{u} nuly. Nakoniec ka\v{z}d% \'{y} nenulov\'{y} riadok vyn\'{a}sob\'{\i}me prevr\'{a}tenou hodnotou jeho ved\'{u}ceho prvku.% \begin{equation*} \mathbf{D}\underset{\mathbf{r}_{2}:=2\mathbf{r}_{2}+\mathbf{r}_{3}}{\underset% {\mathbf{r}_{1}:=2\mathbf{r}_{1}-\mathbf{r}_{3}}{\overset{r}{\sim }}}\left( \begin{array}{rrrr} 2 & -2 & 0 & 1 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0% \end{array}% \right) \underset{\mathbf{r}_{1}:=\mathbf{r}_{1}+\mathbf{r}_{2}}{\overset{r}{% \sim }}\left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0% \end{array}% \right) \underset{\mathbf{r}_{3}:=-\frac{1}{2}\mathbf{r}_{3}}{\underset{% \mathbf{r}_{2}:=\frac{1}{2}\mathbf{r}_{2}}{\underset{\mathbf{r}_{1}:=\frac{1% }{2}\mathbf{r}_{1}}{\overset{r}{\sim }}}}\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -\frac{1}{2} \\[2pt] 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0% \end{array}% \right) =\mathbf{E.} \end{equation*}% $\mathbf{E}$~je~redukovan\'{a}~stup\v{n}ovit\'{a}~matica~riadkovo~ekvivaletn% \'{a}~s~maticou~$\mathbf{A}$. $\square $ \end{solution} \begin{definition} \emph{Trasponovanou maticou} k matici $\mathbf{A}=(a_{jk})_{n}^{m}$ naz\'{y}% vame maticu $\mathbf{A}^{T}=(a_{kj}^{\prime })_{m}^{n}$, kde $% a_{kj}^{\prime}=a_{jk}$ pre v\v{s}etky $j,k$. \end{definition} \begin{example} Transponovanou~maticou~k~matici \begin{equation*} \mathbf{A}=\left( \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 0 & -1 & -2% \end{array}% \right) \end{equation*}% ~\ je \begin{equation*} \mathbf{A}^{T}=\left( \begin{array}{rrr} 1 & 5 & 9 \\ 2 & 6 & 0 \\ 3 & 7 & -1 \\ 4 & 8 & -2% \end{array}% \right) .\square \end{equation*}% \newline \end{example} \begin{center} \begin{tabular}{|c|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}{}{}{A21.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{A21.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{A23.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{A23.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{\hyperref{Ot\'{a}zky}{}{}{OA2.tex}}{}{}{OA2.tex}}% %BeginExpansion \msihyperref{% \msihyperref{Ot\'{a}zky}{}{}{OA2.tex}}{}{}{OA2.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{\hyperref{Cvi\v{c}enia}{}{}{CA2.tex}}{}{}{CA2.tex}}% %BeginExpansion \msihyperref{% \msihyperref{Cvi\v{c}enia}{}{}{CA2.tex}}{}{}{CA2.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{INDA.tex}}% %BeginExpansion \msihyperref{Index}{}{}{INDA.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematika 1} \section{S\'{u}stavy line\'{a}rnych rovn\'{\i}c} \end{document}