%% 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 09:47:42} %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}{Matematick\U{e1} anal\U{fd}za III online - Komplexn\U{e9} \U{10d}\U{ed}sla a funkcie komplexnej premennej - Komplexn\U{e9} \U{10d}\U{ed}sla a algebraick\U{e9} oper\U{e1}cie s nimi\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{Determinanty} \begin{center} \begin{tabular}{|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{aindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{aindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{A4.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{A4.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{OA4.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{OA4.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{CA4.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{CA4.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{INDA.tex}}% %BeginExpansion \msihyperref{Index}{}{}{INDA.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \subsection{Determinanty} Nech $\mathbf{A}$ je \v{s}tvorcov\'{a} matica stup\v{n}a $n$, potom budeme pou\v{z}\'{\i}va\v{t} nasleduj\'{u}ce ozna\v{c}enia: $\mathbf{A}_{jk}$ \ \ \thinspace -- matica, ktor\'{a} vznikne z matice $% \mathbf{A}$ vy\v{s}krtnut\'{\i}m $j$-teho riadku a $k$-teho st\'{l}pca $% \left( n\geq 2\right) ,$ $\mathbf{A}_{jk;\,rs}$ -- matica, ktor\'{a} vznikne z matice $\mathbf{A}$ vy% \v{s}krtnut\'{\i}m $j$-teho a $k$-teho riadku a $r$-teho a $s$-teho st\'{l}% pca $\left( n\geq 3\right) $. \begin{example} Nech \begin{equation*} \mathbf{A}=\left( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9% \end{array}% \right) , \end{equation*}% $\,$potom napr. \begin{equation*} \mathbf{A}_{11}=\left( \begin{array}{cc} 5 & 6 \\ 8 & 9% \end{array}% \right) ,\,\mathbf{A}_{23}=\left( \begin{array}{cc} 1 & 2 \\ 7 & 8% \end{array}% \right) ,\,\mathbf{A}_{12;13}=(8).\square \end{equation*} \end{example} \begin{definition} \textsl{Determinantom~\v{s}tvorcovej~matice}~$\mathbf{A}=(a_{jk})_{n}^{n}$% ~naz\'{y}vame~\v{c}\'{\i}slo,~ktor\'{e}~ozna\v{c}ujeme~$\det \mathbf{A}$~a definujeme~takto: 1. ak~$n=1,\,\det \mathbf{A}=\det (a_{11})=a_{11},\,$ 2. ak $n\geq 2,$% \begin{equation*} \det \,\mathbf{A}=a_{11}\det \,\mathbf{A}_{11}-a_{12}\,\det \,\mathbf{A}% _{12}+a_{13}\,\det \,\mathbf{A}_{13}+\cdots +(-1)^{1+n}a_{1n}\,\det \,% \mathbf{A}_{1n}= \end{equation*}% \begin{equation*} =\dsum\limits_{k=1}^{n}(-1)^{1+k}a_{1k}\,\det \,\mathbf{A}_{1k}\text{ \ naz% \'{y}vame aj \ \emph{rozvoj determinantu pod\v{l}a prv\'{e}ho riadku}.} \end{equation*} \end{definition} \begin{quote} Determinant matice $\mathbf{A}=(a_{jk})_{n}^{n}$, ktor\'{a} m\'{a} riadky $% \mathbf{r}_{1},\mathbf{r}_{2},\ \ldots ,\mathbf{r}_{n}$ a st\'{l}pce $% \mathbf{s}_{1},\mathbf{s}_{2},\ \ldots ,\mathbf{s}_{n}$ budeme tie\v{z} ozna% \v{c}ova\v{t} symbolmi \begin{equation*} |\mathbf{A}|,\quad \left\vert \begin{array}{rrrr} a_{11} & a_{12} & \ldots , & a_{1n} \\ a_{21} & a_{22} & \ldots , & a_{2n} \\ \multicolumn{4}{c}{\dotfill} \\ a_{n1} & a_{n2} & \ldots , & a_{nn}% \end{array}% \right\vert ,\quad \left\vert \begin{array}{c} \mathbf{r}_{1} \\ \mathbf{r}_{2} \\ \vdots \\ \mathbf{r}_{n}% \end{array}% \right\vert ,\quad |\mathbf{s}_{1},\mathbf{s}_{2},\ \ldots ,\mathbf{s}_{n}| \end{equation*} \end{quote} \begin{example} \begin{equation*} \left\vert \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22}% \end{array}% \right\vert =a_{11}\,\det \,\mathbf{A}_{11}-a_{12}\,\det \,\mathbf{A}% _{12}=a_{11}a_{22}-a_{12}a_{21}, \end{equation*}% \begin{equation*} \left\vert \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}% \end{array}% \right\vert =a_{11}\left\vert \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33}% \end{array}% \right\vert -a_{12}\left\vert \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33}% \end{array}% \right\vert +a_{13}\left\vert \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32}% \end{array}% \right\vert = \end{equation*}% \begin{equation*} =a_{11}(a_{22}a_{33}-a_{32}a_{23})-a_{12}(a_{21}a_{33}-a_{31}a_{23})+a_{13}(a_{21}a_{32}-a_{31}a_{22})= \end{equation*}% \begin{equation*} =a_{11}a_{22}a_{33}+a_{12}a_{31}a_{23}+a_{13}a_{21}a_{32}-(a_{11}a_{32}a_{23}+a_{12}a_{21}a_{33}+a_{13}a_{31}a_{22}).\square \end{equation*} \end{example} \begin{quote} Na v\'{y}po\v{c}et determinantu matice \emph{tretieho stup\v{n}a} m\^{o}\v{z}% eme pou\v{z}i\v{t} \emph{Sarusovo pravidlo}. Je to sch\'{e}ma, v ktorej je nazna\v{c}en\'{e}. ktor\'{e} prvky matice je potrebn\'{e} vyn\'{a}sobi\v{t} a ako tieto s\'{u}\v{c}iny s\v{c}\'{\i}ta\v{t}. Jednou takou sch\'{e}mou je \FRAME{ftbpF}{2.7198in}{1.2211in}{0pt}{}{}{det.gif}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.7198in;height 1.2211in;depth 0pt;original-width 2.6775in;original-height 1.1874in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'det.gif';file-properties "XNPEU";}% } v ktorej prvky matice s\'{u} zn\'{a}zornen\'{e} kr\'{u}\v{z}kami. Prvky posp% \'{a}jan\'{e} \v{c}iarami sa vyn\'{a}sobia a od s\'{u}\v{c}tu s\'{u}\v{c}% inov prvej skupiny sa od\v{c}\'{\i}ta s\'{u}\v{c}et s\'{u}\v{c}inov druhej skupiny. \end{quote} \begin{example} Vypo\v{c}\'{\i}tajme determinant pou\v{z}it\'{\i}m Sarusovho pravidla \begin{equation*} \left\vert \begin{array}{ccc} 1 & 0 & 3 \\ 3 & 2 & 5 \\ 1 & 1 & 2% \end{array}% \right\vert . \end{equation*} \end{example} \begin{solution} M\'{a}me \begin{equation*} \left\vert \begin{array}{ccc} 1 & 0 & 3 \\ 3 & 2 & 5 \\ 1 & 1 & 2% \end{array}% \right\vert =1\cdot 2\cdot 2+3\cdot 1\cdot 3+5\cdot 0\cdot 1-(1\cdot 2\cdot 3+1\cdot 5\cdot 1+3\cdot 0\cdot 2)=2.\square \end{equation*} \end{solution} \begin{theorem} Nech $\mathbf{A}=(a_{jk})_{n}^{n}$, $n\geq 2$, potom% \begin{equation*} \det \mathbf{A}=a_{11}\det \,\mathbf{A}_{11}-a_{21}\,\det \,\mathbf{A}% _{21}+a_{31}\,\det \,\mathbf{A}_{31}+\cdots +(-1)^{n+1}a_{n1}\,\det \,% \mathbf{A}_{n1}= \end{equation*}% \begin{equation*} =\dsum\limits_{j=1}^{n}(-1)^{j+1}a_{j1}\,\det \,\mathbf{A}_{j1}-\text{\emph{% rozvoj determinantu pod\v{l}a prv\'{e}ho st\'{l}pca.}} \end{equation*} \end{theorem} \begin{tabular}{|l|} \hline \textbf{% %TCIMACRO{\hyperref{\hyperref{D\^{o}kaz}{}{}{DA411.tex}}{}{}{DA411.tex}}% %BeginExpansion \msihyperref{% \msihyperref{D\^{o}kaz}{}{}{DA411.tex}}{}{}{DA411.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{1} \begin{example} Vypo\v{c}\'{\i}tajme determinant\ \begin{equation*} \left\vert \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 3 & 2 & 1 \\ 0 & -1 & 2 & 1 \\ 2 & 5 & 2 & 2% \end{array}% \right\vert . \end{equation*} \end{example} \begin{solution} V prvom st\'{l}pci je ve\v{l}a n\'{u}l, preto na v\'{y}po\v{c}et pou\v{z}% ijeme rozvoj pod\v{l}a prv\'{e}ho st\'{l}pca. \begin{equation*} \left\vert \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 3 & 2 & 1 \\ 0 & -1 & 2 & 1 \\ 2 & 5 & 2 & 2% \end{array}% \right\vert =1\cdot \left\vert \begin{array}{rrr} 3 & 2 & 1 \\ -1 & 2 & 1 \\ 5 & 2 & 2% \end{array}% \right\vert -0\cdot \left\vert \begin{array}{rrr} 1 & 1 & 1 \\ -1 & 2 & 1 \\ 5 & 2 & 2% \end{array}% \right\vert +0\cdot \left\vert \begin{array}{rrr} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 5 & 2 & 2% \end{array}% \right\vert -2\cdot \left\vert \begin{array}{rrr} 1 & 1 & 1 \\ 3 & 2 & 1 \\ -1 & 2 & 1% \end{array}% \right\vert = \end{equation*}% \begin{equation*} =\left\vert \begin{array}{rrr} 3 & 2 & 1 \\ -1 & 2 & 1 \\ 5 & 2 & 2% \end{array}% \right\vert -2\cdot \left\vert \begin{array}{rrr} 1 & 1 & 1 \\ 3 & 2 & 1 \\ -1 & 2 & 1% \end{array}% \right\vert =12-2+10-(10+6-4)-2(2+6-1-(-2+2+3))= \end{equation*}% \begin{equation*} =20-12-2(7-3)=0.\square \end{equation*} \end{solution} \begin{theorem} Determinat hornej resp. dolnej trojuholn\'{\i}kovej matice sa rovn\'{a} s% \'{u}\v{c}inu prvkov na jej hlavnej diagon\'{a}le. \end{theorem} \begin{tabular}{|l|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA412.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA412.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{2} \begin{theorem} Pre ka\v{z}d\'{u} \v{s}tvorcov\'{u} maticu $\mathbf{A}$ \ \begin{equation*} \det \,\mathbf{A}^{T}=\det \,\mathbf{A.} \end{equation*} \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA413.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA413.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{3} \begin{quote} Transponovan\'{\i}m matice sa z jej riadkov stan\'{u} st\'{l}pce. Preto, na z% \'{a}klade predch\'{a}dzaj\'{u}cej vety, ak determinant matice z\'{a}visi ur% \v{c}it\'{y}m sp\^{o}sobom od jej riadkov, rovnako z\'{a}vis\'{\i} aj od jej st\'{l}pcov. Z tohto d\^{o}vodu budeme vlastnosti determinantov formulova% \v{t} aj dokazova\v{t} len pre riadky. Pre st\'{l}pce si ich dok\'{a}\v{z}% ete sformulova\v{t} sami. \end{quote} \begin{theorem} Nech $\mathbf{A}=(a_{jk})_{n}^{n}$, $n\geq 2,$ potom pre ka\v{z}d\'{e} $j\in \{1,\ldots ,n\}$ \begin{equation*} \det \,\mathbf{A}=(-1)^{j+1}a_{j1}\det \,\mathbf{A}_{j1}+(-1)^{j+2}a_{j2}\,% \det \,\mathbf{A}_{j2}+\cdots +(-1)^{j+n}a_{jn}\,\det \,\mathbf{A}_{jn}= \end{equation*}% \begin{equation*} =\dsum\limits_{k=1}^{n}(-1)^{j+k}a_{jk}\,\det \,\mathbf{A}_{jk}\quad \text{- \emph{rozvoj determinantu pod\v{l}a }}j\text{-\emph{teho riadku.}} \end{equation*} \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA414.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA414.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{4} \begin{example} Vypo\v{c}\'{\i}tajme \begin{equation*} \left\vert \begin{array}{cccc} 1 & a & 1 & 1 \\ 1 & b & 2 & 1 \\ 1 & c & 1 & 2 \\ 2 & d & 1 & 2% \end{array}% \right\vert . \end{equation*} \end{example} \begin{solution} Rozvi\v{n}me determinant pod\v{l}a druh\'{e}ho st\'{l}pca: \begin{equation*} \left\vert \begin{array}{cccc} 1 & a & 1 & 1 \\ 1 & b & 2 & 1 \\ 1 & c & 1 & 2 \\ 2 & d & 1 & 2% \end{array}% \right\vert =-a\left\vert \begin{array}{ccc} 1 & 2 & 1 \\ 1 & 1 & 2 \\ 2 & 1 & 2% \end{array}% \right\vert +b\left\vert \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 2 & 1 & 2% \end{array}% \right\vert -c\left\vert \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 2 & 1 & 2% \end{array}% \right\vert +d\left\vert \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2% \end{array}% \right\vert = \end{equation*}% \begin{equation*} =-3a+b+d.\,\square \end{equation*} \begin{theorem} Ak niektor\'{y} riadok \v{s}tvorcovej matice $\mathbf{A}$ je nulov\'{y}, tak $\det \,\mathbf{A}=0.$ \end{theorem} \end{solution} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA415.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA415.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{5} \begin{theorem} Nech matica $\mathbf{B}$ vznikne zo \v{s}tvorcovej matice $\mathbf{A}$ vz% \'{a}jomnou z\'{a}menou dvoch riadkov, potom $\ \det \,\mathbf{B}=-\det \,% \mathbf{A}.$ \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA416.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA416.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{6} \begin{theorem} Ak \v{s}tvorcov\'{a} matica $\mathbf{A}$ m\'{a} dva riadky rovnak\'{e}, tak $% \det \,\mathbf{A}=0$. \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA417.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA417.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{7} \begin{example} \begin{equation*} \left\vert \begin{array}{ccc} 1, & 2, & -1 \\ 0, & 3, & 1 \\ 4, & -1, & 0% \end{array}% \right\vert \overset{\mathbf{s}_{1}:=:\mathbf{s}_{3}}{=}-\left\vert \begin{array}{ccc} -1, & 2, & 1 \\ 1, & 3, & 0 \\ 0, & -1, & 4% \end{array}% \right\vert \overset{\mathbf{r}_{1}:=:\mathbf{r}_{2}}{=}\left\vert \begin{array}{ccc} 1, & 3, & 0 \\ -1, & 2, & 1 \\ 0, & -1, & 4% \end{array}% \right\vert .\square \end{equation*}% \newline \end{example} \begin{theorem} Nech matica $\mathbf{B}$ vznikne zo \v{s}tvorcovej matice $\mathbf{A}$ vyn% \'{a}soben\'{\i}m jej $j$-teho riadku \v{c}\'{\i}slom $\alpha $. Potom $\ \det \,\mathbf{B}=\alpha \det \,\mathbf{A},$ \v{c}i\v{z}e \begin{equation*} \left\vert \begin{array}{c} \mathbf{r}_{1} \\ \vdots \\ \mathbf{r}_{j-1} \\ \alpha \mathbf{r}_{j} \\ \mathbf{r}_{j+1} \\ \vdots \\ \mathbf{r}_{n}% \end{array}% \right\vert =\alpha \left\vert \begin{array}{c} \mathbf{r}_{1} \\ \vdots \\ \mathbf{r}_{j-1} \\ \mathbf{r}_{j} \\ \mathbf{r}_{j+1} \\ \vdots \\ \mathbf{r}_{n}% \end{array}% \right\vert \end{equation*} \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA418.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA418.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{8} \begin{example} \newline \begin{equation*} 6\left\vert \begin{array}{ccc} \frac{2}{3}, & 1, & \frac{3}{2} \\[3pt] -\frac{1}{3}, & -1, & 0 \\[3pt] 0, & 2, & -2% \end{array}% \right\vert \overset{\mathbf{r}_{1}:=2\mathbf{r}_{1}}{=}3\left\vert \begin{array}{ccc} \frac{4}{3}, & 2, & 3 \\[3pt] -\frac{1}{3}, & -1, & 0 \\[3pt] 0, & 2, & -2% \end{array}% \right\vert \overset{\mathbf{s}_{1}:=3\mathbf{s}_{1}}{=}\left\vert \begin{array}{ccc} 4, & 2, & 3 \\ -1, & -1, & 0 \\ 0, & 2, & -2% \end{array}% \right\vert .\square \end{equation*}% \newline \end{example} \begin{theorem} \newline \begin{equation*} \left\vert \begin{array}{c} \mathbf{r}_{1} \\ \vdots \\ \mathbf{r}_{j-1} \\ \mathbf{r}_{j}^{\prime }+\mathbf{r}_{j}^{\prime \prime } \\ \mathbf{r}_{j+1} \\ \vdots \\ \mathbf{r}_{n}% \end{array}% \right\vert =\left\vert \begin{array}{c} \mathbf{r}_{1} \\ \vdots \\ \mathbf{r}_{j-1} \\ \mathbf{r}_{j}^{\prime } \\ \mathbf{r}_{j+1} \\ \vdots \\ \mathbf{r}_{n}% \end{array}% \right\vert +\left\vert \begin{array}{c} \mathbf{r}_{1} \\ \vdots \\ \mathbf{r}_{j-1} \\ \mathbf{r}_{j}^{\prime \prime } \\ \mathbf{r}_{j+1} \\ \vdots \\ \mathbf{r}_{n}% \end{array}% \right\vert \end{equation*} \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA419.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA419.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{9} \begin{example} \newline \begin{equation*} \left\vert \begin{array}{ccc} 2 & 1 & 3 \\ -1 & -1 & 0 \\ 0 & 2 & -4% \end{array}% \right\vert +\left\vert \begin{array}{ccc} 2 & -1 & 3 \\ -1 & 1 & 0 \\ 0 & -2 & -4% \end{array}% \right\vert =\left\vert \begin{array}{ccc} 2 & 1-1 & 3 \\ -1 & -1+1 & 0 \\ 0 & 2-2 & -4% \end{array}% \right\vert =\left\vert \begin{array}{ccc} 2 & 0 & 3 \\ -1 & 0 & 0 \\ 0 & 0 & -4% \end{array}% \right\vert =0.\square \end{equation*}% \newline \end{example} \begin{theorem} Nech matica $\mathbf{B}$ vznikne zo \v{s}tvorcovej matice \begin{equation*} \mathbf{A}=\left( \begin{array}{c} \mathbf{r}_{1} \\ \vdots \\ \mathbf{r}_{n}% \end{array}% \right) \end{equation*}% nahraden\'{\i}m riadku $\mathbf{r}_{j}$ riadkom $\alpha \mathbf{r}_{j}+\beta \mathbf{r}_{k}$, kde $\alpha ,\ \beta $ s\'{u} \v{l}ubovo\v{l}n\'{e} \v{c}% \'{\i}sla, $j,k\in \{1,\ldots ,n\},\ j\neq k.$ Potom \begin{equation*} \det \,\mathbf{B}=\alpha \,\det \,\mathbf{A}. \end{equation*} \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA4110.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA4110.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{10} \begin{example} Vypo\v{c}\'{\i}tajme \begin{equation*} \left\vert \begin{array}{cccc} 2 & -1 & 3 & 1 \\ 4 & -3 & 4 & 2 \\ 3 & 5 & -1 & 2 \\ 6 & 10 & 1 & 3% \end{array}% \right\vert . \end{equation*} \end{example} \begin{solution} Pomocou ERO a ESO vytvor\'{\i}me v niektorom riadku alebo st\'{l}pci \v{c}o najviac n\'{u}l a potom determinant rozvinieme pod\v{l}a neho: \begin{equation*} \begin{array}{l} \left\vert \begin{array}{cccc} 2, & -1, & 3, & 1 \\ 4, & -3, & 4, & 2 \\ 3, & 5, & -1, & 2 \\ 6, & 10, & 1, & 3% \end{array}% \right\vert \overset{\mathbf{r}_{2}:=\mathbf{r}_{2}-2\mathbf{r}_{1}}{\overset% {\mathbf{r}_{4}:=\mathbf{r}_{4}-2\mathbf{r}_{3}}{\overset{\mathbf{r}_{3}:=2% \mathbf{r}_{3}-3\mathbf{r}_{1}}{=}}}\frac{1}{2}\left\vert \begin{array}{cccc} 2, & -1, & 3, & 1 \\ 0, & -1, & -2, & 0 \\ 0, & 13, & -11, & 1 \\ 0, & 0, & 3, & -1% \end{array}% \right\vert =\frac{1}{2}\,2\left\vert \begin{array}{ccc} -1, & -2, & 0 \\ 13, & -11, & 1 \\ 0, & 3, & -1% \end{array}% \right\vert = \\[30pt] \hspace*{15mm}\overset{\mathbf{s}_{2}:=\mathbf{s}_{2}+3\mathbf{s}_{3}}{=}% \left\vert \begin{array}{ccc} -1, & -2, & 0 \\ 13, & -8, & 1 \\ 0, & 0, & -1% \end{array}% \right\vert =-\left\vert \begin{array}{cc} -1, & -2 \\ 13, & -8% \end{array}% \right\vert =-(8+26)=-34.\square% \end{array}% \end{equation*}% \newline \end{solution} \begin{example} Vypo\v{c}\'{\i}tajme \begin{equation*} \left\vert \begin{array}{ccccc} a, & x, & x, & x, & x \\ x, & b, & x, & x, & x \\ x, & x, & c, & x, & x \\ x, & x, & x, & d, & x \\ x, & x, & x, & x, & e% \end{array}% \right\vert . \end{equation*} \end{example} \begin{solution} Pomocou ERO uprav\'{\i}me maticu na horn\'{u} trojuholn\'{\i}kov\'{u}. \begin{equation*} \left\vert \begin{array}{ccccc} a, & x, & x, & x, & x \\ x, & b, & x, & x, & x \\ x, & x, & c, & x, & x \\ x, & x, & x, & d, & x \\ x, & x, & x, & x, & e% \end{array}% \right\vert \overset{\mathbf{r}_{j}:=\mathbf{r}_{j}-\mathbf{r}_{5}}{\overset{% j=1,2,3,4}{=}}\left\vert \begin{array}{ccccc} a-x, & 0, & 0, & 0, & x-e \\ 0, & b-x, & 0, & 0, & x-e \\ 0, & 0, & c-x, & 0, & x-e \\ 0, & 0, & 0, & d-x, & x-e \\ x, & x, & x, & x. & e% \end{array}% \right\vert = \end{equation*}% Pokra\v{c}ujme \'{u}pravou $\mathbf{r}_{5}:=\mathbf{r}_{5}-\dfrac{x}{a-x}% \mathbf{r}_{1}-\dfrac{x}{b-x}\mathbf{r}_{2}-\dfrac{x}{c-x}\mathbf{r}_{3}-% \dfrac{x}{d-x}\mathbf{r}_{4}$ za predpokladu, \v{z}e~$x\neq a,b,c,d$. \begin{equation*} =\left\vert \begin{array}{ccccc} a-x, & 0, & 0, & 0, & x-e \\ 0, & b-x, & 0, & 0, & x-e \\ 0, & 0, & c-x, & 0, & x-e \\ 0, & 0, & 0, & d-x, & x-e \\ 0, & 0, & 0, & 0. & e-\dfrac{x(x-e)}{a-x}-\dfrac{x(x-e)}{b-x}-\dfrac{x(x-e)}{% c-x}-\dfrac{x(x-e)}{d-x}% \end{array}% \right\vert = \end{equation*}% \begin{equation*} =(a-x)(b-x)(c-x)(d-x)\left( e+\dfrac{x(e-x)}{a-x}+\dfrac{x(e-x)}{b-x}+\dfrac{% x(e-x)}{c-x}+\dfrac{x(e-x)}{d-x}\right) . \end{equation*}% Ost\'{a}va n\'{a}m vypo\v{c}\'{\i}ta\v{t} determinant v pr\'{\i}padoch, ke% \v{d} $x\in \{a,b,c,d\}$. Urob\'{\i}me to napr. pre $x=b$, v~ostatn\'{y}ch pr% \'{\i}padoch by sme postupovali rovnako:% \begin{equation*} \left\vert \begin{array}{ccccc} a, & b, & b, & b, & b \\ b, & b, & b, & b, & b \\ b, & b, & c, & b, & b \\ b, & b, & b, & d, & b \\ b, & b, & b, & b, & e% \end{array}% \right\vert \overset{\mathbf{r}_{j}:=\mathbf{r}_{j}-\mathbf{r}_{2}}{\overset{% j=1,3,4,5}{=}}\left\vert \begin{array}{ccccc} a-b, & 0, & 0, & 0, & 0 \\ b, & b, & b, & b, & b \\ 0, & 0, & c-b, & 0, & 0 \\ 0, & 0, & 0, & d-b, & 0 \\ 0, & 0, & 0, & 0, & e-b% \end{array}% \right\vert \overset{\mathbf{s}_{1}:=\mathbf{s}_{1}-\mathbf{s}_{2}}{=} \end{equation*}% \begin{equation*} =\left\vert \begin{array}{ccccc} a-b, & 0, & 0, & 0, & 0 \\ 0, & b, & b, & b, & b \\ 0, & 0, & c-b, & 0, & 0 \\ 0, & 0, & 0, & d-b, & 0 \\ 0, & 0, & 0, & 0, & e-b% \end{array}% \right\vert =(a-b)b(c-b)(d-b)(e-b).\square \newline \end{equation*} \end{solution} \begin{theorem} Ak s\'{u} riadky \v{s}tvorcovej matice $\mathbf{A}$ line\'{a}rne z\'{a}visl% \'{e}, tak $\det \,\mathbf{A}=0.$ \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA4111.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA4111.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{11} \begin{theorem} Matica $\mathbf{A}$ je regul\'{a}rna pr\'{a}ve vtedy, ke\v{d} $\ \det \,% \mathbf{A}\neq 0.$ \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA4112.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA4112.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{12} \begin{example} Zistime,~\v{c}i~s\'{u}~p\"{a}tice~$% (1,1,1,1,1),~(1,2,3,4,5),~(1,4,9,16,25),~(1,8,27,64,125)$,\newline $(1,16,81,256,625)$ line\'{a}rne z\'{a}visl\'{e} alebo nez\'{a}visl\'{e}. \end{example} \begin{solution} Zap\'{\i}\v{s}me p\"{a}tice ako riadky matice $\mathbf{A}$ a vypo\v{c}\'{\i}% tajme jej determinant.% \begin{equation*} \det \,\mathbf{A}=\left\vert \begin{array}{ccccc} 1, & 1, & 1, & 1, & 1 \\ 1, & 2, & 3, & 4, & 5 \\ 1, & 4, & 9, & 16, & 25 \\ 1, & 8, & 27, & 64, & 125 \\ 1, & 16, & 81, & 256, & 625% \end{array}% \right\vert \overset{\mathbf{s}_{k}:=\mathbf{s}_{k}-\mathbf{s}_{k-1}}{% \overset{k=5,4,3,2}{=}}\left\vert \begin{array}{ccccc} 1, & 0, & 0, & 0, & 0 \\ 1, & 1, & 1, & 1, & 1 \\ 1, & 3, & 5, & 7, & 9 \\ 1, & 7, & 19, & 37, & 61 \\ 1, & 15, & 65, & 175, & 369% \end{array}% \right\vert = \end{equation*}% \begin{equation*} =\left\vert \begin{array}{cccc} 1, & 1, & 1, & 1 \\ 3, & 5, & 7, & 9 \\ 7, & 19, & 37, & 61 \\ 15, & 65, & 175, & 369% \end{array}% \right\vert =\left\vert \begin{array}{cccc} 1, & 0, & 0, & 0 \\ 3, & 2, & 2, & 2 \\ 7, & 12, & 18, & 24 \\ 15, & 50, & 110, & 194% \end{array}% \right\vert =\left\vert \begin{array}{ccc} 2, & 2, & 2 \\ 12, & 18, & 24 \\ 50, & 110, & 194% \end{array}% \right\vert = \end{equation*}% \begin{equation*} =\left\vert \begin{array}{ccc} 2, & 0, & 0 \\ 12, & 6, & 6 \\ 50, & 60, & 84% \end{array}% \right\vert =2\left\vert \begin{array}{cc} 6, & 6 \\ 60, & 84% \end{array}% \right\vert =2\left\vert \begin{array}{cc} 6, & 0 \\ 60, & 24% \end{array}% \right\vert =288. \end{equation*}% Determinant matice $\mathbf{A}$ je r\^{o}zny od nuly, \v{c}i\v{z}e $\mathbf{A% }$ je regul\'{a}rna matica, preto dan\'{e} p\"{a}tice s\'{u} line\'{a}rne nez% \'{a}visl\'{e}.\ $\square $ \end{solution} \begin{theorem} Nech $\mathbf{A},\ \mathbf{B}$ s\'{u} \v{s}tvorcov\'{e} matice stup\v{n}a $n$ a nech $\mathbf{D}$ je regul\'{a}rna matica, potom \/1)~$\det \,\mathbf{A}\mathbf{B}=\det \,\mathbf{A}\,\det \,\mathbf{B}$ 2) $\det \,\mathbf{D}^{-1}=(\det \,\mathbf{D})^{-1}\newline $ \end{theorem} \begin{theorem} Nech $\mathbf{A}=(a_{jk})$ je \v{s}tvorcov\'{a} matica stup\v{n}a $n$ a nech $r,s\in \{1,\ 2,\ \ldots ,\ n\}$, potom \begin{equation*} \sum_{k=1}^{n}\left( -1\right) ^{r+k}a_{sk}\det \,\mathbf{A}_{rk}=\left\{ \begin{array}{l} \det \,\mathbf{A},\text{\ ak\ }r=s \\ 0,\text{\ ak\ }r\neq s% \end{array}% \right. \end{equation*} \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA4113.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA4113.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{13} \begin{theorem} Nech $\mathbf{A}=(a_{jk})_{n}^{n}$ je regul\'{a}rna matica, $\mathbf{A}% ^{-1}=(c_{jk})_{n}^{n}$ je k nej inverzn\'{a} matica, potom pre ka\v{z}d\'{e} $j,k\in \{1,\ \ldots ,\ n\}$ \begin{equation*} c_{jk}=(-1)^{j+k}\frac{\det \,\mathbf{A}_{kj}}{\det \,\mathbf{A}}. \end{equation*} \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA4114.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA4114.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{14} \begin{example} Vypo\v{c}\'{\i}tajme~$\mathbf{A}^{-1}$~(pokia\v{l}~existuje),~ak \begin{equation*} \mathbf{A}=\left( \begin{array}{ccc} -2 & -6 & -3 \\ -2 & -7 & -2 \\ 1 & 2 & 1% \end{array}% \right) . \end{equation*} \end{example} \begin{solution} \begin{equation*} \det \,\mathbf{A}=\left\vert \begin{array}{ccc} -2 & -6 & -3 \\ -2 & -7 & -2 \\ 1 & 2 & 1% \end{array}% \right\vert =\left\vert \begin{array}{ccc} -2 & -6 & -1 \\ -2 & -7 & 0 \\ 1 & 2 & 0% \end{array}% \right\vert =-\left\vert \begin{array}{cc} -2, & -7 \\ 1, & 2% \end{array}% \right\vert =-(-4+7)=-3\neq 0. \end{equation*}% Matica $\mathbf{A}$ je regul\'{a}rna, preto k nej inverzn\'{a} matica existuje. \begin{equation*} \det \,\mathbf{A}_{11}=\left\vert \begin{array}{cc} -7 & -2 \\ 2 & 1% \end{array}% \right\vert =-3,\,\det \,\mathbf{A}_{12}=\left\vert \begin{array}{cc} -2 & -2 \\ 1 & 1% \end{array}% \right\vert =0,\,\det \,\mathbf{A}_{13}=\left\vert \begin{array}{cc} -2 & -7 \\ 1 & 2% \end{array}% \right\vert =3, \end{equation*}% \begin{equation*} \det \,\mathbf{A}_{21}=\left\vert \begin{array}{cc} -6 & -3 \\ 2 & 1% \end{array}% \right\vert =0,\,\det \,\mathbf{A}_{22}=\left\vert \begin{array}{cc} -2 & -3 \\ 1 & 1% \end{array}% \right\vert =1,\,\det \,\mathbf{A}_{23}=\left\vert \begin{array}{cc} -2 & -6 \\ 1 & 2% \end{array}% \right\vert =2, \end{equation*}% \begin{equation*} \det \,\mathbf{A}_{31}=\left\vert \begin{array}{cc} -6 & -3 \\ -7 & -2% \end{array}% \right\vert =-9,\,\det \,\mathbf{A}_{32}=\left\vert \begin{array}{cc} -2 & -3 \\ -2 & -2% \end{array}% \right\vert =-2,\,\det \,\mathbf{A}_{33}=\left\vert \begin{array}{cc} -2 & -6 \\ -2 & -7% \end{array}% \right\vert =2.\vspace{6pt} \end{equation*}% \begin{equation*} \mathbf{A}^{-1}=\dfrac{1}{\det \mathbf{A}}\left( \begin{array}{ccc} \det \mathbf{A}_{11}, & -\det \mathbf{A}_{12}, & \det \mathbf{A}_{13} \\ -\det \mathbf{A}_{21}, & \det \mathbf{A}_{22}, & -\det \mathbf{A}_{23} \\ \det \mathbf{A}_{31}, & -\det \mathbf{A}_{32}, & \det \mathbf{A}_{33}% \end{array}% \right) ^{T}=-\dfrac{1}{3}\left( \begin{array}{ccc} -3, & 0, & 3 \\ 0, & 1, & -2 \\ -9, & 2, & 2% \end{array}% \right) ^{T}= \end{equation*}% \begin{equation*} =\dfrac{1}{3}\left( \begin{array}{ccc} 3, & 0, & 9 \\ 0, & -1, & -2 \\ -3, & 2, & -2% \end{array}% \right) =\left( \begin{array}{ccc} 1, & 0, & 3 \\[1pt] 0, & -\frac{1}{3}, & -\frac{2}{3} \\[3pt] -1, & \frac{2}{3}, & -\frac{2}{3}% \end{array}% \right) .\square \newline \end{equation*} \end{solution} \begin{theorem} (Cramerovo pravidlo) Nech $\mathbf{A}$ je regul\'{a}rna matica stup\v{n}a $% n,\,\mathbf{B}$ je matica typu $n\times 1$. Potom s\'{u}stava line\'{a}rnych rovn\'{\i}c $\mathbf{A}\mathbf{X}=\mathbf{B}$ m\'{a} jedin\'{e} rie\v{s}% enie: \begin{equation*} \frac{1}{\det \mathbf{A}}\left( \det \mathcal{A}_{1},\det \mathcal{A}_{2},\ \ldots ,\det \mathcal{A}_{n}\right) \end{equation*}% kde $\mathcal{A}_{j}$ je matica, ktor\'{a} vznikne z matice $\mathbf{A}$ nahraden\'{\i}m $j$-teho st\'{l}pca pravou stranou $\mathbf{B}$ s\'{u}stavy. \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA4115.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA4115.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{15} \begin{example} Rie\v{s}me s\'{u}stavu line\'{a}rnych rovn\'{\i}c s parametrom $\beta \in \mathbf{R}$ \begin{equation*} \begin{array}{rr@{\ +\ }rr@{\ +\ }rl@{\ =\ }l} \beta x & + & y & + & z & = & 1 \\ x & + & \beta y & + & z & = & \beta \\ x & + & y & + & \beta z & = & \beta ^{2}% \end{array}% \end{equation*} \end{example} \begin{solution} V pr\'{\i}padoch, kedy je matica s\'{u}stavy regul\'{a}rna, vyrie\v{s}ime s% \'{u}stavu Cramerov\'{y}m pravidlom, v ostatn\'{y}ch pr\'{\i}padoch Gaussovou elimina\v{c}nou met\'{o}dou. \begin{equation*} D=\left\vert \begin{array}{ccc} \beta & 1 & 1 \\ 1 & \beta & 1 \\ 1 & 1 & \beta% \end{array}% \right\vert =\beta ^{3}-3\beta +2. \end{equation*}% Pre racion\'{a}lne korene $\dfrac{p}{q}$ rovnice $\beta ^{3}-3\beta +2=0$ plat\'{\i} $\dfrac{p}{q}\in \{\pm 1,\pm 2\}$. Hornerovou sch\'{e}mou zist% \'{\i}me, ~\v{z}e~1~je~dvojn\'{a}sobn\'{y}m~a~-2~jednoduch\'{y}m~kore\v{n}% om.~Potom\ $D=\beta ^{3}-3\beta +2=(\beta -1)^{2}(\beta +2)$. 1. Nech $\beta \neq 1,-2$. \begin{equation*} D_{1}=\left\vert \begin{array}{ccc} 1 & \beta & 1 \\ \beta & \beta & 1 \\ \beta ^{2} & 1 & \beta% \end{array}% \right\vert =\beta ^{2}+\beta +\beta ^{3}-\beta ^{3}-1-\beta ^{3}=-\beta ^{3}+\beta ^{2}+\beta -1=-(\beta -1)^{2}(\beta +1) \end{equation*}% \begin{equation*} D_{2}=\left\vert \begin{array}{ccc} \beta & 1 & 1 \\ 1 & \beta & 1 \\ 1 & \beta ^{2} & \beta% \end{array}% \right\vert =\beta ^{2}-2\beta +1=(\beta -1)^{2} \end{equation*}% \begin{equation*} D_{3}=\left\vert \begin{array}{ccc} \beta & 1 & 1 \\ 1 & \beta & \beta \\ 1 & 1 & \beta ^{2}% \end{array}% \right\vert =\beta ^{4}-2\beta ^{2}+1=(\beta ^{2}-1)^{2}=(\beta -1)^{2}(\beta +1)^{2} \end{equation*}% Rie\v{s}en\'{\i}m s\'{u}stavy je trojica% \begin{equation*} \left( \frac{D_{1}}{D},\ \frac{D_{2}}{D},\ \frac{D_{3}}{D}\right) =\left( -% \frac{\beta +1}{\beta +2},\ \frac{1}{\beta +2},\ \frac{(\beta +1)^{2}}{\beta +2}\right) \end{equation*} 2. Nech $\beta =1$. \begin{equation*} \left( \begin{array}{ccc|c} 1, & 1, & 1 & 1 \\ 1, & 1, & 1 & 1 \\ 1, & 1, & 1 & 1% \end{array}% \right) \sim \left( \begin{array}{ccc|c} 1, & 1, & 1 & 1 \\ 0, & 0, & 0 & 0 \\ 0, & 0, & 0 & 0% \end{array}% \right) \end{equation*}% Rie\v{s}en\'{\i}m s\'{u} v\v{s}etky trojice $(1-s-t,\ s,\ t)$, kde $s,t\in \mathbf{R}$. 3. Nech $\beta =-2$. \begin{equation*} \left( \begin{array}{ccc|c} -2 & 1 & 1 & 1 \\ 1 & -2 & 1 & -2 \\ 1 & 1 & -2 & 4% \end{array}% \right) \sim \left( \begin{array}{ccc|c} 1 & 1 & -2 & 4 \\ 0 & 3 & -3 & 9 \\ 0 & -3 & 3 & -6% \end{array}% \right) \sim \left( \begin{array}{ccc|c} 1 & 1 & -2 & 4 \\ 0 & 3 & -3 & 9 \\ 0 & 0 & 0 & 3% \end{array}% \right) \end{equation*}% V~tomto~pr\'{\i}pade~s\'{u}stava~nem\'{a}~rie\v{s}enie. $\square $\newline \end{solution} \begin{center} \begin{tabular}{|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{aindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{aindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{A4.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{A4.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{OA4.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{OA4.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{CA4.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{CA4.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{Determinanty} \end{document}