
\documentclass{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%TCIDATA{OutputFilter=LATEX.DLL}
%TCIDATA{Version=5.50.0.2890}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{Created=Saturday, December 30, 2006 12:55:45}
%TCIDATA{LastRevised=Saturday, December 30, 2006 13:08:01}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
%TCIDATA{CSTFile=40 LaTeX article.cst}

\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]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


Aufgabe 125

Gegeben sei das eindeutig l\"{o}sbare lineare Gleichungssystem $\ A\cdot 
\overrightarrow{x}=\overrightarrow{b}$ mit 

$A=\left( 
\begin{array}{cccccc}
4 & -1 & 0 & -1 & 0 & 0 \\ 
-1 & 4 & -1 & 0 & -1 & 0 \\ 
0 & -1 & 4 & 0 & 0 & -1 \\ 
-1 & 0 & 0 & 4 & -1 & 0 \\ 
0 & -1 & 0 & -1 & 4 & -1 \\ 
0 & 0 & -1 & 0 & -1 & 4%
\end{array}%
\right) $, $\overrightarrow{b}=\left( 
\begin{array}{c}
2 \\ 
1 \\ 
2 \\ 
2 \\ 
1 \\ 
2%
\end{array}%
\right) $

a.\qquad Sei $\overrightarrow{x}^{\left( 0\right) }=\overrightarrow{0}$.
Berechnen Sie die N\"{a}herungsl\"{o}sung $\overrightarrow{x}^{\left(
3\right) }$\ des Systems, die man nach 3 Schritten des
Gesamtschrittverfahrens erh\"{a}lt.

b.

\end{document}
