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\newcommand{\Z}{\mathbb Z}
\newcommand{\lspan}[1]{\langle#1\rangle}
\newcommand{\inv}{^{-1}}
\DeclareMathOperator{\Adj}{Ek}
\newcommand{\Det}[1]{\det\left(#1\right)}
\DeclareMathOperator{\sutun}{\text{\normalfont s\"ut}}
\newcommand{\sut}[1]{\sutun(#1)}
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\newtheorem{problem}{Problem}
\theoremstyle{definition}
\newtheorem*{solution}{\c C\"oz\"um}
%\theoremstyle{remark}
\newtheorem*{remark}{Not}

\usepackage{verbatim}
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%\let\endsolution=\endcomment



\pagestyle{empty}
\begin{document}
\subtitle{Final S\i nav\i}
\title{Lineer Cebir (MAT \newstylenums{114})}
\author{David Pierce, MSGS\"U}
\date{7 Haziran 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}

L\"utfen:
\begin{compactenum}
\item
\c C\"oz\"um y\"ontemlerinizi d\"u\c s\"unerek se\c cin.
\item
\c C\"oz\"umlerinizi net bir \c sekilde yaz\i n.
\item
M\"umk\"unse cevaplar\i n\i z\i\ kontrol edin.
\end{compactenum}
\.Iyi \c cal\i\c smalar dilerim!

\begin{problem}[9 puan]
$P_n=\{f\in\R[x]\colon\deg(f)<n\}$ olmak \"uzere $\Phi$,
$P_3$ uzay\i ndan $P_4/P_1$ b\"ol\"um uzay\i na giden
\begin{equation*}
\Phi(f)=\int f
\end{equation*}
ko\c sulunu sa\u glayan lineer d\"on\"u\c s\"um\"u olsun.
\"Orne\u gin $\Phi(x^2)=\frac13x^3+P_1$.
\begin{compactitem}
\item
$\{1,x,x^2\}$, $P_3$ uzay\i n\i n bir $B$ taban\i d\i r,
ve 
\item
$\{x+P_1,x^2+P_1,x^3+P_1\}$, 
$P_4/P_1$ b\"ol\"um uzay\i n\i n bir $C$ taban\i d\i r.
\end{compactitem}
\begin{enumerate}[(a)]
\item
$\left[\Phi(f)\right]_C=A[f]_B$
ko\c sulunu sa\u glayan $A$ matrisini bulun.
\item
$P_3$ i\c cin
$\left[\Phi(f)\right]_C=[f]_D$
ko\c sulunu sa\u glayan $D$ taban\i n\i\ bulun.
\end{enumerate}
		\end{problem}

\begin{solution}
\begin{asparaenum}[(a)]
\item
$A=
\left[\begin{array}{c|c|c}
\left[\int 1\right]_C&
\left[\int x\right]_C&
\left[\int x^2\right]_C
\end{array}\right]$
\begin{equation*}
=
\left[\begin{array}{c|c|c}
\left[x+P_1\right]_C&
\left[\frac12x^2+P_1\right]_C&
\left[\frac13x^3+P_1\right]_C
\end{array}\right]
=
\begin{bmatrix}
1&0&0\\
0&\nicefrac12&0\\
0&0&\nicefrac13
\end{bmatrix}.
\end{equation*}
\item
Baz\i\ $g_i$ i\c cin $D=\{g_1,g_2,g_3\}$, ve
$I=
\left[\begin{array}{c|c|c}
\left[\int g_1\right]_C&
\left[\int g_2\right]_C&
\left[\int g_3\right]_C
\end{array}\right]$, dolay\i s\i yla
\begin{align*}
	\int g_1&=x+P_1,&
	\int g_2&=x^2+P_1,&
	\int g_3&=x^3+P_1,
\end{align*}
ve sonu\c c olarak $D=\{1,2x,3x^2\}$.
\end{asparaenum}
\end{solution}

\begin{remark}
$\mathcal B$, sonlu-boyutlu bir $V$ vekt\"or uzay\i n\i n bir
%$\{v_1,\dots,v_n\}$ 
taban\i\ olsun;
$\mathcal C$, sonlu-boyutlu bir $W$ vekt\"or uzay\i n\i n bir 
%$\{w_1,\dots,w_m\}$ 
taban\i\ olsun;
ve $L$, $V$'den $W$'ye giden do\u grusal bir d\"on\"u\c s\"um\"u olsun.
Ko\c c--Esin kitab\i ndaki Teorem 7.2.1'e g\"ore
ve 5 May\i s 2017 tarihli ``Vektor Uzaylar\i'' notlar\i mdaki 31 Teorem'e g\"ore,
bir $A$ matrisi i\c cin,
$V$'nin her $\bm v$ eleman\i\ i\c cin
\begin{equation*}
[L(\bm v)]_{\mathcal C}=A[\bm v]_{\mathcal B}.
\end{equation*}
Ayr\i ca $\mathcal B=\{\bm v_1,\dots,\bm v_n\}$ ise
$A=\left[\begin{array}{c|c|c}[L(\bm v_1)]_{\mathcal C}&\cdots&[L(\bm v_n)]_{\mathcal C}\end{array}\right]$.
\end{remark}

%\newpage

\begin{problem}[12 puan]
  $\R$ \"uzerinde
	\begin{align*}
		A&=
  \begin{bmatrix}
      1&-2& 1& 0&-1&0\\
      2&-4& 2&-1&-3&0\\
     -2& 4&-2& 1& 3&0\\
     -3& 6&-3& 1& 4&1
  \end{bmatrix},&
    \left[
      \begin{array}{c|c|c|c}
        \bm e_1&\bm e_2&\bm e_3&\bm e_4
      \end{array}
      \right]
			&=I	
	\end{align*}
  olsun.
  \begin{enumerate}[(a)]
  \item
    Hangi $k$
    \begin{equation*}
     \R^4=\lspan{\sut A\cup\{\bm e_k\}}
    \end{equation*}
    ko\c sulunu sa\u glar?
  \item
    $A$'n\i n s\"utun uzay\i---yani $\sut A$---i\c cin bir taban bulun.
  \item
    $\{\bm x\in\R^6\colon A\bm x=\bm0\}$ uzay\i n\i n bir taban\i\ bulun.
  \item
    $A\bm x=\bm e_1$ sistemini \c c\"oz\"un.
  \end{enumerate}
\end{problem}

\begin{remark}
Tan\i ma g\"ore 
$\bm e_1=\left[\begin{smallmatrix}1\\0\\0\\0\end{smallmatrix}\right]$,
$\bm e_2=\left[\begin{smallmatrix}0\\1\\0\\0\end{smallmatrix}\right]$,
$\bm e_3=\left[\begin{smallmatrix}0\\0\\1\\0\end{smallmatrix}\right]$,
$\bm e_4=\left[\begin{smallmatrix}0\\0\\0\\1\end{smallmatrix}\right]$.
Problemin (a) \c s\i kk\i nda hangi $k$ i\c cin 
$\left[\begin{array}{c|c}A&\bm e_k\end{array}\right]$ matrisinin s\"utun uzay\i n\i n
$\R^4$ oldu\u gunu \"o\u grenmek isteriz.
Bunun i\c cin $\left[\begin{array}{c|c}A&I\end{array}\right]$ matrisini sat\i rca indirgeyebiliriz.
Asl\i nda $A$'n\i n son s\"utununun $\bm e_4$ oldu\u gundan
$\left[\begin{array}{c|c|c|c}A&\bm e_1&\bm e_2&\bm e_3\end{array}\right]$ matrisini sat\i rca indirgemek yeter.
Bu indirgemeyi di\u ger \c s\i klar i\c cin kullanabiliriz.
\end{remark}

\begin{solution}
$\left[\begin{array}{c|c|c|c}A&\bm e_1&\bm e_2&\bm e_3\end{array}\right]=$
  \begin{multline*}
    \begin{bmatrix}
      1&-2& 1& 0&-1&0& 1&0& 0\\
      2&-4& 2&-1&-3&0& 0&1& 0\\
     -2& 4&-2& 1& 3&0& 0&0& 1\\
     -3& 6&-3& 1& 4&1& 0&0& 0
    \end{bmatrix}
		\xrightarrow[\substack{2R_1+R_3\\3R_1+R_4}]{-2R_1+R_2}
    \begin{bmatrix}
      1&-2& 1& 0&-1&0& 1&0& 0\\
      0& 0& 0&-1&-1&0&-2&1& 0\\
      0& 0& 0& 1& 1&0& 2&0& 1\\
      0& 0& 0& 1& 1&1& 3&0& 0
    \end{bmatrix}		\\
		\xrightarrow[R_2+R_4]{R_2+R_3}
    \begin{bmatrix}
      1&-2& 1& 0&-1&0& 1&0& 0\\
      0& 0& 0&-1&-1&0&-2&1& 0\\
      0& 0& 0& 0& 0&0& 0&1& 1\\
      0& 0& 0& 0& 0&1& 1&1& 0
    \end{bmatrix}
		\xrightarrow{R_3\leftrightarrow R_4}
    \begin{bmatrix}
      1&-2& 1& 0&-1&0& 1&0& 0\\
      0& 0& 0&-1&-1&0&-2&1& 0\\
      0& 0& 0& 0& 0&1& 1&1& 0\\
      0& 0& 0& 0& 0&0& 0&1& 1
    \end{bmatrix}		
		\end{multline*}
		\begin{enumerate}[(a)]
		\item
		Yukar\i daki indirgemeye g\"ore
		$\left[\begin{array}{c|c}A&\bm e_1\end{array}\right]\rightarrow
    \left[\begin{smallmatrix}
      1&-2& 1& 0&-1&0& 1\\
      0& 0& 0&-1&-1&0&-2\\
      0& 0& 0& 0& 0&1& 1\\
      0& 0& 0& 0& 0&0& 0
    \end{smallmatrix}\right]$ dolay\i s\i yla $k\neq1$,
		ama
		$\left[\begin{array}{c|c}A&\bm e_2\end{array}\right]\rightarrow
    \left[\begin{smallmatrix}
      1&-2& 1& 0&-1&0& 0\\
      0& 0& 0&-1&-1&0& 1\\
      0& 0& 0& 0& 0&1& 1\\
      0& 0& 0& 0& 0&0& 1
    \end{smallmatrix}\right]$ dolay\i s\i yla $k=2$ olabilir.
		Benzer \c sekilde $k=3$ olabilir, ama $k\neq4$.
		K\i saca \fbox{$k\in\{2,3\}$.}
		\item
		$\{(1,2,-2,-3),(0,-1,1,1),(0,0,0,1)\}$,
		yani $\left\{
		\left[\begin{smallmatrix}1\\2\\-2\\-3\end{smallmatrix}\right],
		\left[\begin{smallmatrix}0\\-1\\1\\1\end{smallmatrix}\right],
		\left[\begin{smallmatrix}0\\0\\0\\1\end{smallmatrix}\right]
		\right\}$.
		\item
		Homojen sistemin serbest de\u gi\c skenleri $x_2$, $x_3$, ve $x_5$ oldu\u gundan
		\c c\"oz\"um uzay\i n\i n taban\i\
		$\{(*,1,0,*,0,*),(*,0,1,*,0,*),(*,0,0,*,1,*)\}$ bi\c ciminde olur ve
		\begin{equation*}
		\{(2,1,0,0,0,0),(-1,0,1,0,0,0),(1,0,0,-1,1,0)\}
		\end{equation*}
		olur.
                Bu cevap kontrol edilebilir:
$A
                  \left[\begin{smallmatrix}
                    2&-1&1\\1&0&0\\0&1&0\\0&0&-1\\0&0&1\\0&0&0
                  \end{smallmatrix}\right]
                  =
                  \left[\begin{smallmatrix}
                    0&0&0\\0&0&0\\0&0&0\\0&0&0\\
                  \end{smallmatrix}\right]$.
		\item
		\c C\"oz\"um k\"umesi
		$(1,0,0,2,0,1)+\lspan{(2,1,0,0,0,0),(-1,0,1,0,0,0),(1,0,0,-1,1,0)}$ k\"umesi,
yani \c c\"oz\"umler
$\left[\begin{smallmatrix}1\\0\\0\\2\\0\\1\end{smallmatrix}\right]
+x_2
\left[\begin{smallmatrix}2\\1\\0\\0\\0\\0\end{smallmatrix}\right]
+x_3
\left[\begin{smallmatrix}-1\\0\\1\\0\\0\\0\end{smallmatrix}\right]
+x_5
\left[\begin{smallmatrix}1\\0\\0\\-1\\1\\0\end{smallmatrix}\right]$
vekt\"orleridir.
                Bu cevap da kontrol edilebilir:
$A
                  \left[\begin{smallmatrix}
1\\0\\0\\2\\0\\1
                  \end{smallmatrix}\right]
                  =
                  \left[\begin{smallmatrix}
                    1\\0\\0\\0
                  \end{smallmatrix}\right]$.
\end{enumerate}
\end{solution}

%\newpage

\begin{problem}[6 puan]
Tekrar
\begin{equation*}
		A=
  \begin{bmatrix}
      1&-2& 1& 0&-1&0\\
      2&-4& 2&-1&-3&0\\
     -2& 4&-2& 1& 3&0\\
     -3& 6&-3& 1& 4&1
  \end{bmatrix}
\end{equation*}
olmak \"uzere
    $\sut A=\{\bm x\in\R^4\colon C\bm x=\bm0\}$
    ko\c sulunu sa\u glayan bir $C$ matrisi bulun.
\end{problem}

\begin{solution}
\begin{equation*}
  \begin{bmatrix}
      1& 0&0&a\\
      2&-1&0&b\\
     -2& 1&0&c\\
     -3& 1&1&d
  \end{bmatrix}
\xrightarrow{R_2+R_3}
  \begin{bmatrix}
      1& 0&0&a\\
      2&-1&0&b\\
      0& 0&0&b+c\\
     -3& 1&1&d
  \end{bmatrix}
\end{equation*}
dolay\i s\i yla
\begin{equation*}
\sut A=\{(a,b,c,d)\in\R^4\colon b+c=0\}
=\{\bm x\in\R^4\colon\begin{bmatrix}0&1&1&0\end{bmatrix}\bm x=\bm0\}
\end{equation*}
\end{solution}

\begin{remark}
  Bunun gibi problemler,
Ko\c c--Esin kitab\i ndaki \"Ornek 4.3.1, 4.3.2, ve 4.3.3'te \c c\"oz\"ul\"ur.
\end{remark}

\newpage

\begin{problem}[9 puan]
Ya iki-elemanl\i\ $\Z_2$ cismi \"uzerinde,
ya da $\R$ \"uzerinde (siz birini se\c cin),
\begin{equation*}
A=
	\begin{bmatrix}
    1&0&1&0\\
		0&1&0&0\\
		1&1&0&1\\
		1&1&1&1
  \end{bmatrix}
\end{equation*}
	olsun.
  $A$'n\i n tersini ve ek matrisini bulun.
		\end{problem}
		
\begin{solution}
$\R$ \"uzerinde:
  \begin{multline*}
    \left[
      \begin{array}{c|c}
        A&I
      \end{array}
      \right]=
	\begin{bmatrix}
    1&0&1&0&1&0&0&0\\
		0&1&0&0&0&1&0&0\\
		1&1&0&1&0&0&1&0\\
		1&1&1&1&0&0&0&1
  \end{bmatrix} 
	\xrightarrow[-R_2+R_4]{-R_2+R_3}
	\begin{bmatrix}
    1&0&1&0&1& 0&0&0\\
		0&1&0&0&0& 1&0&0\\
		1&0&0&1&0&-1&1&0\\
		1&0&1&1&0&-1&0&1
  \end{bmatrix} 
		\xrightarrow{-R_1+R_4}\\
	\begin{bmatrix}
    1&0&1&0& 1& 0&0&0\\
		0&1&0&0& 0& 1&0&0\\
		1&0&0&1& 0&-1&1&0\\
		0&0&0&1&-1&-1&0&1
  \end{bmatrix} 
	\xrightarrow{-R_4+R_3}
	\begin{bmatrix}
    1&0&1&0& 1& 0&0& 0\\
		0&1&0&0& 0& 1&0& 0\\
		1&0&0&0& 1& 0&1&-1\\
		0&0&0&1&-1&-1&0& 1
  \end{bmatrix} 
	\xrightarrow{-R_3+R_1}\\
	\begin{bmatrix}
    0&0&1&0& 0& 0&-1& 1\\
		0&1&0&0& 0& 1& 0& 0\\
		1&0&0&0& 1& 0& 1&-1\\
		0&0&0&1&-1&-1& 0& 1
  \end{bmatrix} 	
	\xrightarrow{R_1\leftrightarrow R_3}
	\begin{bmatrix}
		1&0&0&0& 1& 0& 1&-1\\
		0&1&0&0& 0& 1& 0& 0\\
    0&0&1&0& 0& 0&-1& 1\\
		0&0&0&1&-1&-1& 0& 1
  \end{bmatrix} 		
  \end{multline*}
	Kontrol edelim:
	\begin{equation*}
	\begin{bmatrix}
		 1& 0& 1&-1\\
		 0& 1& 0& 0\\
     0& 0&-1& 1\\
		-1&-1& 0& 1
  \end{bmatrix} 			
A=	
\begin{bmatrix}
		 1& 0& 1&-1\\
		 0& 1& 0& 0\\
     0& 0&-1& 1\\
		-1&-1& 0& 1
  \end{bmatrix} 			
	\begin{bmatrix}
    1&0&1&0\\
		0&1&0&0\\
		1&1&0&1\\
		1&1&1&1
  \end{bmatrix}
		=
			\begin{bmatrix}
			1&0&0&0\\
			0&1&0&0\\
			0&0&1&0\\
			0&0&0&1
  \end{bmatrix}
	\end{equation*}
	Ayr\i ca ($A$'n\i n indirgenmesinden) $\det A=-1$, dolay\i s\i yla
	\begin{align*}
A\inv&=		
\begin{bmatrix}
		 1& 0& 1&-1\\
		 0& 1& 0& 0\\
     0& 0&-1& 1\\
		-1&-1& 0& 1
  \end{bmatrix},&
	\Adj A&=\det A\cdot A\inv=
\begin{bmatrix}
		 -1& 0&-1& 1\\
		  0&-1& 0& 0\\
      0& 0& 1&-1\\
		  1& 1& 0&-1
  \end{bmatrix}.	
	\end{align*}
	$\Z_2$ \"uzerinde
	\begin{equation*}
	A\inv=
	\begin{bmatrix}
		 1& 0& 1& 1\\
		 0& 1& 0& 0\\
     0& 0& 1& 1\\
		 1& 1& 0& 1
  \end{bmatrix}=\Adj A.
	\end{equation*}	
\end{solution}

\begin{remark}
  $\Z_2$ \"uzerinde bu problem biraz daha kolayd\i r
  \c c\"unk\"u bu durumda $-1=1$.
  24 Mart 2017 tarihli ``Ekmatris ve ters'' notumdaki gibi
  $A$'n\i n tersinin yerine $A$'n\i n ek matrisi \"once hesaplanabilir.
  Herhangi durumda hata yapmadan hesaplama zor olabilir,
  ama cevaplar kolayl\i kla kontrol edilebilir.
\end{remark}

\end{document}