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\newcommand{\inv}{^{-1}}
\DeclareMathOperator{\Adj}{Ek}
\newcommand{\Det}[1]{\det\left(#1\right)}

\usepackage{nicefrac}
\newcommand{\matfrac}[2]{\nicefrac{#1}{#2}}

\theoremstyle{definition}
\newtheorem*{problem}{Problem}
\theoremstyle{remark}
\newtheorem*{solution}{\c C\"oz\"um}

\begin{document}
\title{Ekmatris ve ters}
\subtitle{Lineer Cebir}
\author{David Pierce}
\date{24 Mart 2017}
\publishers{Matematik B\"ol\"um\"u, MSGS\"U\\
%\mbox{}\\
%\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}

\maketitle

\begin{problem}
  $A=\begin{pmatrix}1&3&-3\\4&-1&-1\\-2&1&-2\end{pmatrix}$ ise
  $\Adj A$ ve $A\inv$ 
  %matrisinin ekmatrisini ve tersini
  bulun.
\end{problem}

\begin{solution}
\.Iki y\"ontemden biri kullan\i labilir.
\begin{asparaenum}
\item
\"Once tan\i m\i ndan ekmatris hesaplanabilir:
\begin{multline*}
\Adj A
	=\Adj\begin{pmatrix}1&3&-3\\4&-1&-1\\-2&1&-2\end{pmatrix}\\
	=\begin{pmatrix}
	 \Det{\begin{matrix}-1&-1\\ 1&-2\end{matrix}}&-\Det{\begin{matrix}3&-3\\ 1&-2\end{matrix}}&\Det{\begin{matrix}3&-3\\-1&-1\end{matrix}}\\
	-\Det{\begin{matrix} 4&-1\\-2&-2\end{matrix}}& \Det{\begin{matrix}1&-3\\-2&-2\end{matrix}}&-\Det{\begin{matrix}1&-3\\ 4&-1\end{matrix}}\\
	 \Det{\begin{matrix} 4&-1\\-2& 1\end{matrix}}&-\Det{\begin{matrix}1& 3\\-2& 1\end{matrix}}&\Det{\begin{matrix}1& 3\\ 4&-1\end{matrix}}
	\end{pmatrix}
	=\begin{pmatrix}3&3&-6\\10&-8&-11\\2&-7&-13\end{pmatrix}.
\end{multline*}
Kontrol ederiz:
\begin{multline*}
\begin{pmatrix}3&3&-6\\10&-8&-11\\ 2&-7&-13\end{pmatrix}
\cdot A
=\begin{pmatrix}3&3&-6\\10&-8&-11\\ 2&-7&-13\end{pmatrix}
\begin{pmatrix}1&3&-3\\ 4&-1& -1\\-2& 1& -2\end{pmatrix}\\
=\begin{pmatrix} 3+12+12& 9-3- 6&- 9-3+12\\
                10-32+22&30+8-11&-30+8+22\\
								 2-28+26& 6+7-13&- 6+7+26\end{pmatrix}
=\begin{pmatrix}27&0&0\\0&27&0\\0&0&27\end{pmatrix}.
\end{multline*}
Sonu\c c olarak $\det A=27$, dolay\i s\i yla
\begin{equation*}
A\inv=\frac1{27}\Adj A
=\begin{pmatrix}\matfrac  3 {27}&\matfrac  3 {27}&\matfrac{- 6}{27}\\
                \matfrac{10}{27}&\matfrac{-8}{27}&\matfrac{-11}{27}\\
                \matfrac  2 {27}&\matfrac{-7}{27}&\matfrac{-13}{27}\end{pmatrix}.
\end{equation*}
\item
Gauss--Jordan indirgemesiyle
\begin{multline*}
\left(\begin{array}{c|c}A&I\end{array}\right)
=	\begin{pmatrix} 1& 3&-3&1&0&0\\
                  4&-1&-1&0&1&0\\
								 -2& 1&-2&0&0&1\end{pmatrix}
\xrightarrow[2R_1+R_3]{-4R_1+R_2}
\begin{pmatrix}1&  3&-3& 1&0&0\\
               0&-13&11&-4&1&0\\
   						 0&  7&-8& 2&0&1\end{pmatrix}
\xrightarrow{-\frac1{13}R_2}\\
\begin{pmatrix}1&3&          -3     &        1    &          0     &0\\
               0&1&\matfrac{-11}{13}&\matfrac4{13}&\matfrac{-1}{13}&0\\
							 0&7&          -8     &        2    &          0     &1\end{pmatrix}
\xrightarrow{-7R_2+R_3}
\begin{pmatrix}1&3&         -3      &          1    &           0     &0\\
               0&1&\matfrac{-11}{13}&\matfrac  4 {13}&\matfrac{-1}{13}&0\\
							 0&0&\matfrac{-27}{13}&\matfrac{-2}{13}&\matfrac  7 {13}&1\end{pmatrix}
\xrightarrow{-\frac{13}{27}R_3}\\
\begin{pmatrix}1&3&          -3     &        1    &          0     &           0     \\
               0&1&\matfrac{-11}{13}&\matfrac4{13}&\matfrac{-1}{13}&           0     \\
							 0&0&           1     &\matfrac2{27}&\matfrac{-7}{27}&\matfrac{-13}{27}\end{pmatrix}
\xrightarrow[3R_3+R_1]{\frac{11}{13}R_3+R_2}
\begin{pmatrix}1&3&0&\matfrac{33}{27}&\matfrac{-21}{27}&\matfrac{-39}{27}\\
               0&1&0&\matfrac{10}{27}&\matfrac {-8}{27}&\matfrac{-11}{27}\\
               0&0&1&\matfrac  2 {27}&\matfrac {-7}{27}&\matfrac{-13}{27}\end{pmatrix}
\xrightarrow{-3R_2+R_1}\\
\begin{pmatrix}1&0&0&\matfrac  3 {27}&\matfrac   3 {27}&\matfrac{ -6}{27}\\
               0&1&0&\matfrac{10}{27}&\matfrac {-8}{27}&\matfrac{-11}{27}\\
               0&0&1&\matfrac  2 {27}&\matfrac {-7}{27}&\matfrac{-13}{27}\end{pmatrix}.
\end{multline*}
Tekrar kontrol ederiz:
\begin{equation*}
\begin{pmatrix}\matfrac  3 {27}&\matfrac   3 {27}&\matfrac{ -6}{27}\\
               \matfrac{10}{27}&\matfrac {-8}{27}&\matfrac{-11}{27}\\
               \matfrac  2 {27}&\matfrac {-7}{27}&\matfrac{-13}{27}\end{pmatrix}
\begin{pmatrix}1&3&-3\\4&-1&-1\\-2&1&-2\end{pmatrix}
=I,							
\end{equation*}
dolay\i s\i yla
\begin{equation*}
A\inv
=\begin{pmatrix}\matfrac  3 {27}&\matfrac   3 {27}&\matfrac{ -6}{27}\\
                \matfrac{10}{27}&\matfrac {-8}{27}&\matfrac{-11}{27}\\
                \matfrac  2 {27}&\matfrac {-7}{27}&\matfrac{-13}{27}\end{pmatrix}.
\end{equation*}
Kulland\i\u g\i m\i z sat\i r i\c slemlerden
$\det A=\left(-\frac{27}{13}\right)\cdot(-13)=27$,
dolay\i s\i yla
\begin{equation*}
\Adj A=\det A\cdot A\inv
=\begin{pmatrix}  3 &   3 &{ -6}\\
                {10}& {-8}&{-11}\\
                  2 & {-7}&{-13}\end{pmatrix}.
\end{equation*}
\end{asparaenum}
\end{solution}

Yukar\i daki problemin kayna\u g\i:
Ko\c c \&\ Esin, \emph{Do\u grusal Cebir} (Ankara: 2014), sayfa 127.

\begin{problem}
  Herhangi kare matris yaz\i p ekmatrisini ve (varsa) tersini bulun!
\end{problem}

\end{document}