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\newcommand{\Det}[1]{\det\left(#1\right)}

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

\newtheorem{problem}{Problem}
\theoremstyle{definition}
\newtheorem*{solution}{\c C\"oz\"um}
\theoremstyle{remark}
\newtheorem*{remark}{Not}


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%\let\endsolution=\endcomment



\pagestyle{empty}
\begin{document}
\subtitle{S\i nav \itshape \c C\"oz\"umleri}
\title{Lineer Cebir (MAT \newstylenums{114})}
\author{David Pierce, MSGS\"U}
\date{31 Mart 2017}
%\publishers{Matematik B\"ol\"um\"u, MSGS\"U}
%\mbox{}\\
%\url{dpierce@msgsu.edu.tr}\\
%\url{http://mat.msgsu.edu.tr/~dpierce/}

\maketitle\thispagestyle{empty}

S\i nav, a\c sa\u g\i daki s\"ozlerle ba\c slad\i:
\begin{quote}\relscale{0.9}
\c C\"oz\"um y\"ontemlerinizi d\"u\c s\"unerek se\c cin.
%Dikkatle \c cal\i\c s\i n
\c C\"oz\"umlerinizi net bir \c sekilde yaz\i n.
M\"umk\"unse cevaplar\i n\i z\i\ kontrol edin.
\.Iyi \c cal\i\c smalar dilerim!
\end{quote}
\"Ozel olarak Problem 2 ve Problem 4'\"un cevaplar\i\
kontrol edilebilir, dolay\i s\i yla yanl\i\c s cevaplar
s\i f\i r puan alabilir.
Her problem, 8 puand\i r.

\begin{problem}
  $A=
    \begin{pmatrix}
      1&1&1&1&1\\
      0&2&2&2&2\\
      0&0&3&3&3\\
      0&0&0&4&4\\
      0&0&0&0&5
    \end{pmatrix}$ ve
    $B=\begin{pmatrix}
      1&2&3&4&5\\
      0&2&3&4&5\\
      0&0&3&4&5\\
      0&0&0&4&5\\
      0&0&0&0&5
    \end{pmatrix}$ ise $\det(AB)$ hesaplay\i n.
\end{problem}

\begin{solution}
  $\det(AB)=\det A\det B=(5!)^2=(120)^2=14400$.
\end{solution}

\begin{remark}
  Determinant g\"ondermesi \c carp\i msal oldu\u gundan
  $AB$ \c carp\i m\i n\i\ bulmak gerekmez.
\end{remark}
\newpage


\begin{problem}
%\setlength{\arraycolsep}{0.1em}
  $\left\{\begin{array}{rrrrrrrrrrr}
     x&+& 6y&+& z&-&  t&-& u&=& 0\\
    -x&-& 6y& &  &+& 5t&+&2u&=& 1\\
      & &   & &-z&-& 4t& &  &=& 1\\
   -2x&-&12y& &  &+&10t&+&2u&=&-2
  \end{array}\right\}$
  sistemi veriliyor.
  \begin{compactenum}[(a)]
  \item
    Sistemi \c c\"oz\"un.
  \item
    Kar\c s\i l\i k gelen homojen sistemin temel \c c\"oz\"umlerini verin.
  \end{compactenum}
\end{problem}

\begin{solution}
  \begin{compactenum}[(a)]
    \item
  \begin{gather*}
    \begin{pmatrix}
    1&  6& 1&-1&-1& 0\\
   -1& -6& 0& 5& 2& 1\\
    0&  0&-1&-4& 0& 1\\
   -2&-12& 0&10& 2&-2
    \end{pmatrix}\xrightarrow[2R_1+R_4]{R_1+R_2}
        \begin{pmatrix}
    1&6& 1&-1&-1& 0\\
    0&0& 1& 4& 1& 1\\
    0&0&-1&-4& 0& 1\\
    0&0& 2& 8& 0&-2
        \end{pmatrix}\xrightarrow[-2R_2+R_4]{R_2+R_3}\\
  \begin{pmatrix}
    1&6&1&-1&-1& 0\\
    0&0&1& 4& 1& 1\\
    0&0&0& 0& 1& 2\\
    0&0&0& 0&-2&-4
  \end{pmatrix}\xrightarrow{2R_3+R_4}
  \begin{pmatrix}
    1&6&1&-1&-1& 0\\
    0&0&1& 4& 1& 1\\
    0&0&0& 0& 1& 2\\
    0&0&0& 0& 0& 0
  \end{pmatrix}\xrightarrow[R_3+R_1]{-R_3+R_2}\\
  \begin{pmatrix}
    1&6&1&-1&0& 2\\
    0&0&1& 4&0&-1\\
    0&0&0& 0&1& 2\\
    0&0&0& 0&0& 0
  \end{pmatrix}\xrightarrow{-R_2+R_1}
    \begin{pmatrix}
    1&6&0&-5&0&3 \\
    0&0&1& 4&0&-1\\
    0&0&0& 0&1& 2\\
    0&0&0& 0&0& 0
    \end{pmatrix},
  \end{gather*}
  dolay\i s\i yla sistemin \c c\"oz\"umleri,
  \begin{equation*}
    \begin{pmatrix}
      x\\y\\z\\t\\u
    \end{pmatrix}
    =
    \begin{pmatrix}
      -6y+5t+3\\y\\-4t-1\\t\\2
    \end{pmatrix}
    =y
    \begin{pmatrix}
      -6\\1\\0\\0\\0
    \end{pmatrix}
    +t
    \begin{pmatrix}
      5\\0\\-4\\1\\0
    \end{pmatrix}
    +
    \begin{pmatrix}
      3\\0\\-1\\0\\2
    \end{pmatrix}.
  \end{equation*}
\item
$    \begin{pmatrix}
      -6\\1\\0\\0\\0
    \end{pmatrix}$
    ve
$    \begin{pmatrix}
      5\\0\\-4\\1\\0
    \end{pmatrix}$.
  \end{compactenum}
\end{solution}

\begin{remark}
  Cevaplar kontrol edilebilir: $A=    \begin{pmatrix}
    1&  6& 1&-1&-1\\
   -1& -6& 0& 5& 2\\
    0&  0&-1&-4& 0\\
   -2&-12& 0&10& 2
  \end{pmatrix}$ ise
  \begin{align*}
    A\begin{pmatrix}
      -6\\1\\0\\0\\0
    \end{pmatrix}&=
    \begin{pmatrix}
      0\\0\\0\\0
    \end{pmatrix},&
A\begin{pmatrix}
      5\\0\\-4\\1\\0
\end{pmatrix}&=\begin{pmatrix}
      0\\0\\0\\0
    \end{pmatrix},&
A\begin{pmatrix}
      3\\0\\-1\\0\\2
\end{pmatrix}&=
\begin{pmatrix}
  0\\1\\1\\-2
\end{pmatrix}
  \end{align*}
\end{remark}

\newpage
\begin{problem}
  $A=\begin{pmatrix}
    0&5&-1\\
    18&2&9\\
    26&3&13
  \end{pmatrix}$ ise $A\inv$, $\det A$, ve $\Adj A$ bulun.
\end{problem}

\begin{solution}
  \begin{gather*}
    \Adj A
	=\begin{pmatrix}
	\Det{\begin{matrix} 2&9\\3&13\end{matrix}}
        &-\Det{\begin{matrix}5&-1\\3&13\end{matrix}}
        &\Det{\begin{matrix}5&-1\\2&9\end{matrix}}\\
	-\Det{\begin{matrix}18&9\\26&13\end{matrix}}
        &\Det{\begin{matrix}0&-1\\26&13\end{matrix}}
        &-\Det{\begin{matrix}0&-1\\18&9\end{matrix}}\\
	\Det{\begin{matrix}18&2\\26&3\end{matrix}}
        &-\Det{\begin{matrix}0&5\\26&3\end{matrix}}
        &\Det{\begin{matrix}0&5\\18&2\end{matrix}}
	\end{pmatrix}
	=\begin{pmatrix}-1&-68&47\\0&26&-18\\2&130&-90\end{pmatrix}.
  \end{gather*}
  Kontrol ederiz:
  \begin{equation*}
    \Adj A\cdot A
    =\begin{pmatrix}-1&-68&47\\0&26&-18\\2&130&-90\end{pmatrix}
\begin{pmatrix}
    0&5&-1\\
    18&2&9\\
    26&3&13
\end{pmatrix}
=
\begin{pmatrix}
  -2&0&0\\0&-2&0\\0&0&-2
\end{pmatrix},
  \end{equation*}
  dolay\i s\i yla
  \begin{align*}
    \det A&=-2,&
    A\inv&=\frac1{\det A}\Adj A
    =\begin{pmatrix}\nicefrac12&34&\nicefrac{-47}2\\
    0&-13&9\\-1&-65&45\end{pmatrix}.
  \end{align*}
\end{solution}

\begin{remark}
  Di\u ger y\"ontem daha uzundur:
\begin{gather*}
    \begin{pmatrix}
     0&5&-1&1&0&0\\
    18&2& 9&0&1&0\\
    26&3&13&0&0&1
    \end{pmatrix}\xrightarrow{R_1\leftrightarrow R_2}
    \begin{pmatrix}
    18&2& 9&0&1&0\\
     0&5&-1&1&0&0\\
    26&3&13&0&0&1
    \end{pmatrix}\xrightarrow{\frac1{18}R_1}\\
    \begin{pmatrix}
    1&\nicefrac19&\nicefrac12&0&\nicefrac1{18}&0\\
     0&5&-1&1&0&0\\
    26&3&13&0&0&1
    \end{pmatrix}\xrightarrow{-26R_1+R_3}
    \begin{pmatrix}
    1&\nicefrac19&\nicefrac12&0&\nicefrac1{18}&0\\
     0&5&-1&1&0&0\\
     0&\nicefrac19&0&0&\nicefrac{-13}9&1
    \end{pmatrix}\xrightarrow{9R_3}\\
    \begin{pmatrix}
    1&\nicefrac19&\nicefrac12&0&\nicefrac1{18}&0\\
     0&5&-1&1&0&0\\
     0&1&0&0&-13&9
    \end{pmatrix}\xrightarrow{-5R_3+R_2}
    \begin{pmatrix}
    1&\nicefrac19&\nicefrac12&0&\nicefrac1{18}&0\\
     0&0&-1&1&65&-45\\
     0&1&0&0&-13&9
    \end{pmatrix}\xrightarrow{R_2\leftrightarrow R_3}\\
    \begin{pmatrix}
    1&\nicefrac19&\nicefrac12&0&\nicefrac1{18}&0\\
    0&1&0&0&-13&9\\
         0&0&-1&1&65&-45
    \end{pmatrix}\xrightarrow{-R_3}
    \begin{pmatrix}
    1&\nicefrac19&\nicefrac12&0&\nicefrac1{18}&0\\
    0&1&0&0&-13&9\\
         0&0&1&-1&-65&45
    \end{pmatrix}\xrightarrow{-\frac12R_3+R_1}\\
    \begin{pmatrix}
    1&\nicefrac19&0&\nicefrac12&\nicefrac{293}9&\nicefrac{-45}2\\
    0&1&0&0&-13&9\\
         0&0&1&-1&-65&45
    \end{pmatrix}\xrightarrow{-\frac19R_2+R_1}
    \begin{pmatrix}
    1&0&0&\nicefrac12&34&\nicefrac{-47}2\\
    0&1&0&0&-13&9\\
         0&0&1&-1&-65&45
    \end{pmatrix},\\
    A\inv
    =    \begin{pmatrix}
    \nicefrac12&34&\nicefrac{-47}2\\
    0&-13&9\\
    -1&-65&45
  \end{pmatrix},\qquad
  \det A=(-1)^3\cdot\frac19\cdot18=-2,\\
  \Adj A
  =\det A\cdot A\inv
  =    \begin{pmatrix}
    -1&-68&47\\
    0&26&-18\\
    2&130&90
  \end{pmatrix}.
  \end{gather*}
\end{remark}

\newpage

\begin{problem}
%\setlength{\arraycolsep}{0.1em}
  $a$'n\i n ve $b$'nin hangi de\u gerleri i\c cin
  \begin{math}\left\{\begin{array}{*7r}
        x&+&2y&-&     z&=&1\\
      -2x&+&ay&+&   2 z&=&b\\
         & & y&+&(a-1)z&=&0
    \end{array}\right\}
  \end{math}
  sisteminin
  \begin{enumerate}[(a)]
  \item
    tek bir \c c\"oz\"um\"u vard\i r?
  \item
    birden fazla \c c\"oz\"um\"u vard\i r?
  \item
    hi\c c \c c\"oz\"um\"u yoktur?
  \end{enumerate}
\end{problem}

\begin{solution}
  \begin{multline*}
    \begin{pmatrix}
      1&2&-1&1\\-2&a&2&b\\0&1&a-1&0
    \end{pmatrix}\xrightarrow{2R_1+R_2}
    \begin{pmatrix}
      1&2&-1&1\\0&a+4&0&b+2\\0&1&a-1&0
    \end{pmatrix}\xrightarrow{R_2\leftrightarrow R_3}\\
    \begin{pmatrix}
      1&2&-1&1\\0&1&a-1&0\\0&a+4&0&b+2
    \end{pmatrix}\xrightarrow{-(a+4)R_2+R_3}
    \begin{pmatrix}
      1&2&-1&1\\0&1&a-1&0\\0&0&-(a+4)(a-1)&b+2
    \end{pmatrix}
  \end{multline*}
  dolay\i s\i yla
  \begin{comment}
    

  \begin{compactitem}
  \item
    $a=-4$ ve $b\neq-2$ ise \c c\"oz\"um yoktur.
  \item
    $a=-4$ ve $b=-2$ ise birden fazla \c c\"oz\"um vard\i r.
  \item
    \c Simdi $a\neq-4$ olsun.
    \begin{equation*}
    \begin{pmatrix}
      1&2&-1&1\\0&1&a-1&0\\0&a+4&0&b+2
    \end{pmatrix}\xrightarrow{-(a+4)R_2+R_3}
        \begin{pmatrix}
      1&2&-1&1\\0&1&a-1&0\\0&0&-(a-1)(a+4)&b+2
    \end{pmatrix}.
    \end{equation*}
    \begin{compactitem}[$*$]
    \item
      $a=1$ ve $b\neq-2$ ise \c c\"oz\"um yoktur.
    \item
      $a=1$ ve $b=-2$ ise birden fazla \c c\"oz\"um vard\i r.
    \item
      $a\neq 1$ ise tek bir \c c\"oz\"um vard\i r.
    \end{compactitem}
  \end{compactitem}

  \end{comment}
  \begin{compactenum}[(a)]
  \item
    $a\notin\{-4,1\}$ ise
    tek bir \c c\"oz\"um\"u vard\i r.
  \item
    $a\in\{-4,-1\}$ ve $b=-2$ ise
    birden fazla \c c\"oz\"um\"u vard\i r.
  \item
    $a\in\{-4,-1\}$ ve $b\neq-2$ ise
    hi\c c \c c\"oz\"um\"u yoktur.
  \end{compactenum}  
\end{solution}
\end{document}
