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LinAlg2
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Change colors again and fix typo

This commit is contained in:
Anton Mosich 2022-06-09 23:44:55 +02:00
parent b158a0bc33
commit acc127d2e8
Signed by: Flugschwein
GPG Key ID: 9303E1C32E3A14A0
1 changed files with 7 additions and 9 deletions
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LinAlg2.tex
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@ -39,6 +39,10 @@
\newcommand\K{\ensuremath{\mathbb{K}}}
\newcommand\mapsfrom{\rotatebox{180}{$\mapsto$}}
\definecolor{pastellblau}{HTML}{5BCFFA}
\definecolor{pastellrosa}{HTML}{F5ABB9}
\definecolor{weiss}{HTML}{FFFFFF}
\theoremstyle{break}
\theoremseparator{:\smallskip}
\theoremindent=1em
@ -46,7 +50,7 @@
\theorembodyfont{\normalfont}
\theoreminframepreskip{0em}
\theoreminframepostskip{0em}
\newtcbox{theoremBox}{colback=Plum!17,colframe=Plum!87,boxsep=0pt,left=7pt,right=7pt,top=7pt,bottom=7pt}
\newtcbox{theoremBox}{colback=pastellrosa!17,colframe=pastellrosa!87,boxsep=0pt,left=7pt,right=7pt,top=7pt,bottom=7pt}
\def\theoremframecommand{\theoremBox}
\newshadedtheorem{theo}{Theorem}[section]
@ -55,7 +59,7 @@
\newshadedtheorem{lemma}[theo]{Lemma}
\newshadedtheorem{korollar}[theo]{Korollar}
\newshadedtheorem{folgerung}[theo]{Folgerung}
\newtcbox{definBox}{colback=Cerulean!17,colframe=Cerulean!94,boxsep=0pt,left=7pt,right=7pt,top=7pt,bottom=7pt}
\newtcbox{definBox}{colback=pastellblau!17,colframe=pastellblau!94,boxsep=0pt,left=7pt,right=7pt,top=7pt,bottom=7pt}
\def\theoremframecommand{\definBox}
\newshadedtheorem{defin}[theo]{Definition}
@ -81,10 +85,6 @@
\begin{document}
\definecolor{pastellblau}{HTML}{5BCFFA}
\definecolor{pastellrosa}{HTML}{F5ABB9}
\definecolor{weiss}{HTML}{FFFFFF}
\tikzset{%
-||-/.style={decoration={markings,
mark=at position 0.5 with {\draw[thick, -] (-.2,-.2) -- (0, .2);\draw[thick, -] (0, -.2) -- (.2, .2);}},
@ -273,7 +273,6 @@ $\pi = (2 3 1), f(X_1, X_2, X_3) = X_1-X_2+X_1X_3 \implies \pi f(X_1, X_2, X_3)
\end{proof}
\begin{folgerung}
Es gibt genau $\frac12n!$ gerade und $\frac12n!$ ungerade Permutationen in $S_n$
\end{folgerung}
\begin{proof}
@ -748,8 +747,7 @@ da obige Matrix aus $M_{ij}$ durch Spaltenadditionen hervorgeht.
\end{proof}
\begin{folgerung}
Sei $A\in\K^{n\times n}$ invertierbar. Sei $x\in\K^n$ die eindeutige Lösung des linearen Gleichunssystems
Sei $A\in\K^{n\times n}$ invertierbar. Sei $x\in\K^n$ die eindeutige Lösung des linearen Gleichungssystems
$Ax=b$. Dann gilt
\[
x_i= \det(A)^{-1} \det(a_{\_1}, \dots, \underbrace{b}_{i}, \dots, a_{\_n})