\documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} \usepackage{amsmath} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Thursday, February 21, 2002 14:53:39} %TCIDATA{LastRevised=Monday, October 03, 2005 09:50:47} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{AllPages= %F=36,\PARA{038
Scientific Notebook: On Line Mathematics\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} \section{Determinanty} \subsection{D\^{o}kaz vety} (matematickou indukciou vzh\v{l}adom na stupe\v{n} matice $\mathbf{A}$) Nech $\mathbf{A}=(a_{jk})_{n}^{n}$. 1. $n=1$% \begin{equation*} \det \,\mathbf{A}^{T}=\det \,(a_{11})^{T}=\det \,(a_{11})=\det \,\mathbf{A.} \end{equation*} 2. Nech veta plat\'{\i} pre v\v{s}etky matice stup\v{n}a $n-1$, kde $n\geq 2$% . V\v{s}imnime si, \v{z}e transponovan\'{a} matica k $\mathbf{A}_{jk}$ je $(% \mathbf{A}_{jk})^{T}=\mathbf{A}_{kj}^{T}$. Rozvoj determinantu matice $% \mathbf{A}^{T}$ pod\v{l}a prv\'{e}ho st\'{l}pca je% \begin{equation*} \det \,\mathbf{A}^{T}=a_{11}\det \,\mathbf{A}_{11}^{T}-a_{12}\,\det \,% \mathbf{A}_{21}^{T}+\cdots +(-1)^{n+1}a_{1n}\,\det \,\mathbf{A}_{n1}^{T}= \end{equation*}% \begin{equation*} =a_{11}\det \,(\mathbf{A}_{11})^{T}-a_{12}\,\det \,(\mathbf{A}% _{12})^{T}+\cdots +(-1)^{n+1}a_{1n}\,\det \,(\mathbf{A}_{1n})^{T}. \end{equation*}% Matice $\mathbf{A}_{1k}$ s\'{u} stup\v{n}a $n-1$, preto pod\v{l}a induk\v{c}n% \'{e}ho predpokladu je $\det \,(\mathbf{A}_{1k})^{T}=\det \,\mathbf{A}_{1k}$ a pre determinant matice $\mathbf{A}^{T}$ dostaneme \begin{equation*} \det \,\mathbf{A}^{T}=a_{11}\det \,\mathbf{A}_{11}-a_{12}\,\det \,\mathbf{A}% _{12}+\cdots +(-1)^{n+1}a_{1n}\,\det \,\mathbf{A}_{1n}=\det \,\mathbf{A.\;}% \blacksquare \end{equation*} \begin{center} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{Sp\"{a}\v{t}}{}{}{A41.tex#3}}% %BeginExpansion \msihyperref{Sp\"{a}\v{t}}{}{}{A41.tex#3}% %EndExpansion } \\ \hline \end{tabular} \end{center} \end{document}