problem with old hebrew latex files in iso 8859-8 coding

[Uwe Brauer <oub@mat.ucm.es>]

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Hello 

A friend of mine posses a lot of hebrew latex files which still use
8859-8 coding. Understandable he wants to switch to UTF8 and xelatex.

The problem is that while his files are displayed perfectly in texstudio
(having iso 8859-8 enabled), which means the system has the requires
fonts installed, emacs fails to do so. It simply does not display the
hebrew chars with the correct fonts.

What also works is to run 

iconv -f ISO8859-8 -t UTF-8 file1.txt file2.txt

The resulting file file2.txt is displayed perfectly in GNU emacs.



What does work is to open the file in GNU emacs, save it as UTF-8, the
resulting file is still not correctly displayed, I even tried to save
the file as 8859-8 but then I received the following error



,----
| These default coding systems were tried to encode text
| in the buffer ‘doc2.txt’:
|   (hebrew-iso-8bit-unix (1 . 233) (2 . 228) (3 . 233) (7 . 238) (8 .
|   248) (9 . 231) (10 . 225) (12 . 229) (14 . 232) ...)
| However, each of them encountered characters it couldn’t encode:
|   hebrew-iso-8bit-unix cannot encode these: é ä é î ø ç á å è å ...
| 
| Click on a character (or switch to this window by ‘C-x o’
| and select the characters by RET) to jump to the place it appears,
| where ‘C-u C-x =’ will give information about it.
| 
| Select one of the safe coding systems listed below,
| or cancel the writing with C-g and edit the buffer
|    to remove or modify the problematic characters,
| or specify any other coding system (and risk losing
|    the problematic characters).
| 
|   utf-8 iso-8859-1 euc-jis-2004 euc-jp iso-2022-jp-2004 next cp858
|   cp850 windows-1252 iso-8859-15 gb18030 utf-7 utf-16
|   utf-16be-with-signature utf-16le-with-signature utf-16be utf-16le
|   iso-2022-7bit utf-8-auto utf-8-with-signature eucjp-ms utf-8-hfs
|   japanese-shift-jis-2004 japanese-iso-7bit-1978-irv ibm1047
|   utf-7-imap utf-8-emacs prefer-utf-8
`----

I attach the file, once as a original file, once with the latex commands
removed.

Any help is strongly appreciated.

Regards

Uwe Brauer 


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éäé 

îøçá å÷èåøé ëàùø 

å 

áñéñ ì  
äåëéçå àå äôøéëå àú äèòðåú äáàåú. 


ìëì å÷èåø 

ä÷áåöä 

áñéñ ì 

[-- Attachment #3: junk.tex --]
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\documentclass[11pt]{article}
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\usepackage[english,hebrew]{babel}
\usepackage{color}
%\usepackage{hebcal}
\usepackage{culmus}
\usepackage{courier}
\usepackage{ulem}
%\usepackage[HE8,OT1]{fontenc}
%\usepackage{ccfonts}
%\usepackage{float}
% \usepackage{color}
\usepackage{theorem}
\usepackage{fancyhdr}
\usepackage{esvect}
\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetkzobj{all}
\usetikzlibrary{calc}
%\usepackage{url}
%\usepackage{listings}
%\usepackage{a4wide}
%\renewcommand{\baselinestretch}{1.5}
%\usepackage{graphicx,epsf}

%\renewcommand{\labelenumi}{\alph{enumi}.}
\usepackage{amssymb,amsfonts,amsmath}
\usepackage{pgffor, ifthen}
\newcommand{\notes}[3][\empty]{%
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\newcommand{\setF}{{\mathord{\mathbb F}}}
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\newcommand{\setZ}{{\mathord{\mathbb Z}}}
\newcommand{\nk}[1] {\textcolor{red}{)#1 ð÷'(}}
\newcommand{\red}[1] {\textcolor{red}{#1}}
%
\newcommand{\br}[1]{\left[{#1}\right]} %square bracets
\newcommand{\brf}[1]{\left\{{#1}\right\}}%figure bracets


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%
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				x \\ x \\ x  \\
			\end{array}\right]}}}
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\def \ese {\end{array}\right)}

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\def \bd {\left|\begin{matrix}}
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	\def \ebm {\end{bmatrix}}
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\def \cis {\mathrm{cis}}
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	\def \em {\end{matrix}\right)}
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\def \pageoflines {\notes[24pt]{18}{\textwidth}}
\def \answer {\textbf{ôúøåï:}}

\def \S {\mathcal{S}}

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		\fontfamily{david}\selectfont \textbf{ôúøåï}}}
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%------------------------------------------------------------------------





\begin{document}
%\fontfamily{david}\selectfont



\begin{center}
{\textbf{ \Large{
 îèìä îñ' 5 àìâáøä 1îç'  \L{11102} \\
\quad \\
ìà ìäâùä
}}
\\}

\end{center}
%
%\textbf{äòøä:}
%áîèìä æå îåúø ìäùúîù áëì èòðä ùäåëéçå áîèìåú ÷åãîåú.



\begin{enumerate}

\item
éäé 
$(V, \setF)$
îøçá å÷èåøé ëàùø 
$\dim V=n$
å 
$B=\{\v_1, \ldots, \v_n\}$ 
áñéñ ì $V$.
äåëéçå àå äôøéëå àú äèòðåú äáàåú. 
\begin{enumerate}
\item 
ìëì å÷èåø 
$\y\in V$ 
ä÷áåöä 
$\{\y+\v_1, \ldots, \y+\v_n\}$ 
áñéñ ì $V$.

%\answer 
%
%äèòðä àéðä ðëåðä. ð÷ç 
%$\y=-\v_1$ 
%åàæ ä÷áåöä 
%\[\{\y+\v_1, \ldots, \y+\v_n\}=\{-\v_1+\v_1, \ldots, -\v_1+\v_n\}=\{\0, \ldots, -\v_1+\v_n\}\]
%îëéìä àú å÷èåø äàôñ ìëï äéà àéðä áú"ì )ëîåáï äéà âí ìà áñéñ(.

\item 
ìëì ñ÷ìøéí 
$\alpha_1, \ldots, \alpha_n\neq 0$
ä÷áåöä
$\{\alpha_1 \v_1, \ldots, \alpha_n \v_n\}$	
áñéñ ì 
$V$.

%\answer
%
%ä÷áåöä
%$\{\alpha_1 \v_1, \ldots, \alpha_n \v_n\}$	
%îú÷áìú îäáñéñ $B$ òì éãé ñãøä ùì ôòåìåú àìîðèøéåú. ìôé èòðä, ôòåìåú àìîðèøéåú ìà îùðåú úìåú ìéðàøéú àå ôøéùä ùì ÷áåöä. ëìï
%$\{\alpha_1 \v_1, \ldots, \alpha_n \v_n\}$
% âí áñéñ ì $V$.

\item 
áñòéó äæä áìáã
$V=\setR^n$
îøçá å÷èåøé îòì äùãä 
$\setR$ 
\usualop\ 
å 
${B=\{\v_1, \ldots, \v_n\}}$
áñéñ ì 
$\setR^n$.
òáåø
$A\in \setR^{n\times n}$ 
îèøéöä îñãø 
$n\times n$
ä÷áåöä 
$\{A\v_1, \ldots, A\v_n\}$ 
áñéñ ì $V$ àí åø÷ àí 
$A$ 
äôéëä.


%\answer
%
%ðæëéø ëé 
%$A$ 
%äôéëä àí åø÷ àí 
%$\Nu(A)=\{\0\}$.
%
%\textbf{ëéååï 1:}
%ððéç ëé $A$ äôéëä.
%éäéå 
%$\alpha_1, \ldots, \alpha_n\in \setR $
%ëê ù
%\begin{align*}
%\alpha_1 A\v_1+\cdots +\alpha_n A\v_n=\0.
%\end{align*}
%ðåáò ëé 
%\[A(\alpha_1 \v_1+\cdots +\alpha_n \v_n)=\0.\]
%îëéååï ù 
%$A$ 
%äôéëä ðåëì ìäëôéì áùðé äàâôéí áîèøéöä äåôëéú 
%$A^{-1}$
%åð÷áì
%\[A^{-1}A(\alpha_1 \v_1+\cdots +\alpha_n \v_n)=I(\alpha_1 \v_1+\cdots +\alpha_n \v_n)=\alpha_1 \v_1+\cdots +\alpha_n \v_n=\0\]
%àáì $B$ áú"ì )ëé äåà áñéñ( ìëï
%$\alpha_1=\cdots=\alpha_n=0$.
%ðåáò ëé 
%$A\v_1,\ldots, A\v_n$
%áú"ì àáì 
%$\dim V=n$ 
%ìëï 
%$A\v_1,\ldots, A\v_n$
%îäååéí áñéñ ì 
%$V$.
%
%\textbf{ëéååï 2:}
%
%ððéç ù 
%$A\v_1,\ldots, A\v_n$
%îäååéí áñéñ ì 
%$\setR^n$.
%òì îðú ìäøàåú ù $A$ äôéëä, ðøàä ù
%$\Nu(A)=\{\0\}$.
%ìùí ëê, éäé 
%$\x\in \Nu(A)$.
%îëéååï 
%$B$
%áñéñ ì 
%$\setR^n$, 
%äéà ôåøùú àú 
%$\setR^n$.
%ìëï ÷ééîéí 
%$\beta_1, \ldots, \beta_n\in \setR$ 
%ëê ù 
%$\beta_1\v_1+\cdots+\beta_n \v_n=\x$.
%ðëôéì á 
%$A$ 
%åð÷áì 
%\begin{align*}
%&A(\beta_1\v_1+\cdots+\beta_n \v_n)=A\x=\0\\
%\implies & \beta_1A\v_1+\cdots+\beta_n A\v_n=\0.
%\end{align*}
%îëéååï ù 
%$\{A\v_1,\ldots, A\v_n\}$
%áñéñ ì 
%$\setR^n$ 
%äéà áú"ì. ðåáò ëé 
%$\beta_1=\cdots=\beta_n=0$. 
%îú÷áì ù 
%$\x=\beta_1\v_1+\cdots+\beta_n\v_n=\0$.
%ðåáò ëé 
%$\Nu(A)=\{\0\}$ 
%ìëï 
%$A$ 
%äôéëä.

\item 
éäéå 
$\a_1=\bm a_{11}\\ 0\\0\\0 \em, \a_2=\bm a_{12}\\ a_{22}\\0\\0 \em, \a_3=\bm a_{13}\\ a_{23}\\a_{33}\\0 \em, \a_4=\bm a_{14}\\ a_{24}\\a_{34}\\a_{44} \em$
å÷èåøéí á 
$\setR^4$ 
ëàùø 
$a_{11}, a_{22}, a_{33}, a_{44}\neq	 0$. 
äåëéçå ùäåå÷èåøéí îäååéí áñéñ ì 
$\setR^4$.

%\answer
%
%ðñãø àú äåå÷èåøéí äðúåðéí áîèøéöä:
%\[A=(\a_1,\a_2,\a_3,\a_4)=
%\bm a_{11}&a_{12}&a_{13}&a_{14}\\
% 0&a_{22}&a_{23}&a_{24}\\
% 0&0&a_{33}&a_{34}\\
% 0&0&0&a_{44} \em.
%\]
%ëòú, 
%$\r(A)=4$
%ëé äî÷ãîéí 
%$a_{11}, a_{22}, a_{33}, a_{44}$ 
%ùåðéí îàôñ ìëï ëì àçã îåáéì åäîèøéöä îãåøâú. ìôé èòðä, ðåáò ù 
%$A\in \setR^{4\times 4}$
% äôéëä.
\end{enumerate}


	
\item 
éäé 
$(\setR^3, \setR)$ 
îøçá å÷èåøé òí äôòåìåú äøâéìåú åéäé 
$V=\{f\in \setR[x]\mid \deg f\leq 2\}$ 
îøçá äôåìéðåîéí îòì 
$\setR$ 
îîòìä ìëì äéåúø 
$2$.
ðâãéø ôåð÷öéä 
$T:\setR^3\to V$ 
òì éãé 	
$T\bm a\\b\\c \em=(a+c)x^2+(a-b)x+(b+c)$
ìëì 
$\bm a\\b\\c\em\in \setR^3$.

\begin{enumerate}
\item	
äåëéçå ù $T$ äòú÷ä ìéðàøéú.

%\answer
%
%éäéå
%$\u_1=\bm a_1\\b_1\\c_1\em, \u_2=\bm a_2\\b_2\\c_2\em\in \setR^3$
%åéäé 
%$\alpha\in \setR$. 
%àæé îú÷ééîéí 
%\begin{align*}T(\u_1+\u_2)&=T\bm a_1+a_2\\b_1+b_2\\c_1+c_2\em\\
%	&=(a_1+a_2+c_1+c_2)x^2+(a_1+a_2-b_1-b_2)x+(b_1+b_2+c_1+c_2)\\
%	&=(a_1+    c_1    )x^2+(a_1    -b_1    )x+b_1    +c_1+(a_2+    c_2    )x^2+(a_2    -b_2    )x+b_2    +c_2\\
%	&=T(\u_1)+T(\u_2)
%	\end{align*}
%	å 
%\begin{align*}T(\alpha \u_1)&=(\alpha a_1+   \alpha c_1    )x^2+(\alpha a_1    -\alpha b_1    )x+(\alpha b_1    +\alpha c_1    )\\
%	&=\alpha \Big(a_1+    c_1    )x^2+(a_1    -b_1    )x+b_1    +c_1\Big)\\
%	&=\alpha T(\u_1).
%\end{align*}
%ðåáò ëé 
%$T$ 
%äòú÷ä ìéðàøéú.

\item 
äøàå ëé 
$x^2+1\in \ip(T)$
åâí 
$-x+1\in \ip(T)$.

%\answer 
%
%ð÷ç 
%$a=b=1, c=0$ 
%åð÷áì 
%$T\bm 1\\1\\0 \em=(1+0)x^2+(1-1)x+1+0=x^2+1$ 
%ìëï 
%$x^2+1\in \ip(T)$.
%
%ð÷ç 
%$a=c=0, b=1$ 
%åð÷áì 
%$T\bm 0\\0\\1 \em=(0+0)x^2+(0-1)x+1+0=-x+1$ 
%ìëï 
%$-x+1\in \ip(T)$.


\item 
äåëéçå ëé 
$\dim (\ip(T))\geq 2$.

%\answer 
%
%áñòéó ä÷åãí øàéðå ëé 
%$x^2+1, -x+1\in \ip(T)$. 
%áøåø ùäí áú"ì ëé äîòìåú ùåðåú.
%îöã ùðé, 
%$\sp\{x^2+1, -x+1\}\subseteq \ip(T)$.
% ðåáò ëé 
%\[\dim (\ip(T))\geq \dim (\sp\{ x^2+1, -x+1 \})=2.\]

\item 
äøàå ëé 
$\bm 1\\1\\-1 \em\in \ker(T)$.


%\answer
%
%ðçùá 
%\[T\bm 1\\1\\-1 \em=(1-1)x^2+(1-1)x+1-1=0\]
%ìëï 
%$\bm 1\\1\\-1\em\in \ker T$.

\item 
äåëéçå ëé 
$\dim (\ker(T))\geq 1$.


%\answer
%
%ðåáò îäñòéó ä÷åãí ëé 
%$\sp\left\{ \bm 1\\1\\-1 \em \right\}\subseteq \ker T$
%ìëï 
%\[\dim(\ker T)\geq \dim \left(\sp\left\{ \bm 1\\1\\-1 \em \right\}\right)=1.\]

\item 
çùáå àú 
$\dim (\ker (T))$ 
åàú 
$\dim (\ip(T))$.


%\answer
%
%ìôé îùôè îú÷ééí 
%$\dim (\ker (T))+\dim (\ip(T))=\dim(\setR^3)=3$. 
%ìôé äñòéôéí ÷åãîéí 
%$\dim (\ker (T))\geq 1$ 
%å 
%$\dim (\ip(T))\geq 2$
%ìëï áäëøç 
%$\dim (\ker (T))= 1$ 
%å 
%$\dim (\ip(T))= 2$.

%\item 
%éäé 
%$B=\{x^2, x, 1\}$
%äáñéñ äñèðãøèé ùì 
%$V$.
%úäé  
%$A\in \setR^{3\times 3}$
%îèøéöä äî÷ééîú 
%$[T(\u)]_B=A\u$ 
%ìëì 
%$\u\in \setR^3$.
%\begin{enumerate}
%	\item 
%	îöàå àú 
%	$A$ 
%	áîôåøù.
% 
%%\answer
%%
%%ðëúåá 
%%$\u=\bm a\\b\\c \em$
%%åðçùá 
%%\[[T\u]_B=[(a+c)x^2+(a-b)x+(b+c)]_B=\bm a+c\\ a-b\\ b+c \em.\]
%%ðãøù ìîöåà îèøéöä 
%%$A$ 
%%ëê ù 
%%$A\u=A\bm a \\ b \\ c \em=\bm a+c\\ a-b\\ b+c \em$.
%%ðéúï ì÷çú 
%%\[A=\bm 1&0&1\\ 1&-1&0\\ 0&1&1 \em\]
%%ëé àæ 
%%\[A\bm a \\ b \\ c \em=\bm 1&0&1\\ 1&-1&0\\ 0&1&1 \em\bm a \\ b \\ c \em=\bm a+c\\ a-b\\ b+c \em.\]
%
%\item
%äøàå ëé 
%$\Nu(A)=\sp\left\{ \bm 1\\1\\-1 \em\right\}$.
%
%%\answer
%%
%%
%%ëòú, 
%%$\Nu(A)$ 
%%äåà àåñó äôúøåðåú ìîî"ì 
%%$A\x=\0$. 
%%ðôúåø: 
%%\begin{align*}
%%(A|\0)=\bse{ccc} 
%%1&0&1&0\\ 1&-1&0&0\\ 0&1&1&0 \ese &
%%\widesim{R_2\to R_2-R_1} 
%%\bse{ccc} 
%%1&0&1&0\\ 0&-1&-1&0\\ 0&1&1&0 \ese\\
%%&\widesim{R_3\to R_3+R_2} 
%%\bse{ccc} 
%%1&0&1&0\\ 0&-1&-1&0\\ 0&0&0&0 \ese
%%\widesim{R_2\to -R_2} 
%%\bse{ccc} 
%%1&0&1&0\\ 0&1&1&0\\ 0&0&0&0 \ese.
%%\end{align*}
%%
%%ðëúåá 
%%$c=t$ 
%%åàæ ð÷áì 
%%$a=b=-t$ 
%%ìëï éù ôúøåï ëììé 
%%$\bm a\\ b\\ c\em=\left\{ t\bm -1\\-1\\1 \em\mid t\in \setR \right\}$
%%ìëï 
%%\[\Nu(A)=\sp \left\{\bm -1\\-1\\1 \em\right\}=\sp \left\{\bm 1\\1\\-1 \em\right\}.\]
%
%\item 
%îöàå áñéñ ì 
%$\C(A)$.
%
%%\answer
%%
%%ìôé ääâãøä ùì îøçá òîåãåú
%%$\C(A)=\sp\left\{
%%\bm 1\\1\\0\em, \bm 0\\ -1\\ 1 \em, \bm 1\\0\\1 \em
%%\right\}.$
%%ëáø øàéðå ëé 
%%$\bm 1\\ 1\\ -1\em\in \Nu(A)$ 
%%ìëï 
%%\[\bm 0\\0\\0 \em=\bm 1&0&1\\ 1&-1&0\\ 0&1&1 \em\bm 1\\ 1\\ -1\em=1\bm 1\\1\\0\em+1\bm 0\\ -1\\ 1 \em+(-1)\bm 1\\0\\1 \em.\]
%%ìëï ðåëì ìëúåá àú äòîåãä äàçøåðä ëöéøåó ìéðàøé ùì äòîåãåú äàçøåú: 
%%\[\bm 1\\0\\1 \em=\bm 1\\1\\0\em+\bm 0\\ -1\\ 1 \em.\]
%%ðåáò ëé 
%%$\C(A)=\sp\left\{
%%\bm 1\\1\\0\em, \bm 0\\ -1\\ 1 \em, \bm 1\\0\\1 \em
%%\right\}=\sp\left\{
%%\bm 1\\1\\0\em, \bm 0\\ -1\\ 1 \em
%%\right\}.$
%%äåå÷èåøéí 
%%$\bm 1\\1\\0\em, \bm 0\\ -1\\ 1 \em$ 
%%äí áú"ì ëé äí àéðí î÷áéìéí ìëï äí îäååéí áñéñ ì 
%%$\C(A)$. 

%\end{enumerate}


\end{enumerate}


	\item 
	éäé 
	$\setF$ 
	ùãä. 
		\begin{enumerate}
		\item 
			äåëéçå ùàí 
			$A, B\in \setF^{n\times n}$ 
îèøéöåú ñéîèøéåú àæé
$AB$ 
ñéîèøéú àí åø÷ àí 
$AB=BA$.
		 

%\answer 
%
%ðúåï ù 
%$A, B$ 
%ñéîèøéåú ìëï 
%$A^T=A, B^T=B$. 
%ëòú, 
%$(AB)^T=B^TA^T=BA$. 
%ðéúï ìäñé÷ ëé 
%$AB$
%ñéîèøéú àí åø÷ àí 
%$(AB)^T=AB$ 
%àí åø÷ àí 
%$BA=AB$.		
		
		\item 
		äåëéçå ùàí 
		$A, B\in \setR^{3\times 3}$
		îèøéöåú ñéîèøéåú 
		ëê ù 
		$AB=\bm  2&0&0\\ 0&5&0\\ 0&0&-1 \em$
		àæé âí 
		$BA=\bm  2&0&0\\ 0&5&0\\ 0&0&-1 \em$.

%\answer
%
%äîèøéöä
%$AB=\bm  2&0&0\\ 0&5&0\\ 0&0&-1 \em$
%ñéîèøéú ëé äéà àìëñåðéú. ìôé äñòéó ä÷åãí ðåáò ëé 
%$BA=AB=\bm  2&0&0\\ 0&5&0\\ 0&0&-1 \em$.

\item 
úðå ãåâîä ìîèøéöåú 
$A, B\in \setR^{2\times 2}$
)ìà áäëøç ñéîèøéåú( ëê ù 
$AB$
ñéîèøéú å
${AB\neq BA}$.		

%\answer 
%
%ð÷ç 
%$A=\bm 1&1\\0&1 \em, B=\bm 1&0\\ 1&1 \em$.
%àæ
%$AB=\bm 1&1\\0&1 \em \bm 1&0\\ 1&1 \em=\bm 2&1\\1&1 \em$
%ñéîèøéú å 
%$BA=\bm 1&0\\ 1&1 \em \bm 1&1\\0&1 \em = \bm 1&1\\1&2 \em \neq AB$.

	\end{enumerate}

	\item \label{inverse-of-products}
	éäé 
	$\setF$
	ùãä åúäééðä 
	$A, B\in \setF^{n\times n}$
	îèøéöåú äôéëåú. 
	\begin{enumerate}
		\item 
		äåëéçå ù 
		$AB$ 
		äôéëä.

%\answer
%
%ìôé îùôè 
%$\det (AB)=\det A\det B$. 
%îëéååï ù 
%$A, B$ 
%äôéëåú îú÷ééí 
%$\det A, \det B\neq 0$. 
%ðåáò ëé 
%$\det (AB)\neq 0$ 
%ìëï 
%$AB$ 
%äôéëä.
		\item 
		äåëéçå ëé 
		$(AB)^{-1}=B^{-1}A^{-1}$.
		
%\answer		
%		
%òì îðú ìäøàåú ëé 
%$B^{-1}A^{-1}$ 
%ääåôëéú ùì $AB$ àðå ðçùá: 
%\[ (AB)(B^{-1}A^{-1})= A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I. \]		
%ìôé îùôè, îú÷ééí âí 
%$(B^{-1}A^{-1})(AB)=I$
%ìëï 
%$B^{-1}A^{-1}$
%àëï ääåôëéú ùì 
%$AB$.

\item 

àí 
$C\in \setF^{n\times n}$
âí äôéëä, îöàå ðåñçä
òáåø 
$(ABC)^{-1}$
ùáä îåôéòéí äîèøéöåú 
$A^{-1}, B^{-1}, C^{-1}$.

%\answer 
%
%ìôé äñòéó ä÷åãí îú÷ééí 
%$(ABC)^{-1}=((AB)C)^{-1}=C^{-1}(AB)^{-1}=C^{-1}B^{-1}A^{-1}$.
%		
		\item 		
		äåëéçå ù 
		$A^2$ 
		äôéëä.
		
%\answer 
%
%ðåáò éùéøåú îñòéó à' ëàùø 
%$B=A$
%ëé àæ
%$AB=A^2$.		
		
		\item
		äåëéçå ù
		$(A^2)^{-1}=(A^{-1})^2$.
		
%\answer	
%		
%		
%		ð÷ç 
%		$B=A$ 
%		åàæ ìôé ñòéó ùì äùàìä äðåëçéú á' ð÷áì
%		\[(A^2)^{-1}=(AA)^{-1}=(A^{-1})(A^{-1})=(A^{-1})^2.\]

		\item 
		àí 
		\[A=\bm -11&3&-4\\ -18&5&-7\\ 8&-2&3 \em, B=\bm 1&-1&1\\1&2&5\\ -2&4&1 \em\]
		çùáå àú 
		$AB$ 
		åàú 
		$B^{-1}A^{-1}$.
	îåîìõ ìäéòæø áñòéó á' áî÷åí ìçùá àú 
		$A^{-1}$ 
	åàú 
		$B^{-1}$ 
		áðôøã.
		
%\answer		
%		
%\[AB=\bm 0&1&0\\1&0&0\\0&0&1 \em.\]		
%ìôé ñòéó á' îú÷ééí 
%$B^{-1}A^{-1}=(AB)^{-1}=\bm 0&1&0\\1&0&0\\0&0&1 \em^{-1}$.	
%ðçùá àú ääåôëéú:
%\begin{align*}
%	\left(
%	\begin{array}{ccc|ccc}
%		0&1&0&1&0&0\\
%		1&0&0&0&1&0\\
%		0&0&1&0&0&1
%	\end{array} 	
%	\right) 	
%	&{\widesim{R_1\leftrightarrow R_2}}	
%	\left(
%	\begin{array}{ccc|ccc}
%1&0&0&0&1&0\\
%0&1&0&1&0&0\\
%0&0&1&0&0&1
%	\end{array} 	
%	\right).
%\end{align*}
%	ðåáò ëé 
%	$B^{-1}A^{-1}=\bm 0&1&0\\1&0&0\\0&0&1 \em^{-1}=\bm 0&1&0\\1&0&0\\0&0&1 \em$.
	
	\end{enumerate}	
	
	
	\item \label{eigen-of-inverse}
	éäé 
	$\setF$
	ùãä åúäé 
	$A\in \setF^{n\times n}$
	îèøéöä äôéëä. 
\begin{enumerate}
\item 
äåëéçå ùëì òøê òöîé ùì 
$A$ 
ùåðä îàôñ.

%\answer
%
%ðúåï ù 
%$A$ 
%äôéëä. ìëï, ìôé èòðä, 
%$\Nu(A)=\{\0\}$. 
%ðåáò ù 
%$A\x\neq \0$ 
%ìëì 
%$\x\neq \0$.
%ìëï ìà éúëï ù 
%$A\x=0\x$ 
%òáåø å÷èåø 
%$\x\neq \0$.
%ðåáò ù 
%$0$ 
%àéðå òøê òöîé ùì 
%$A$.



\item 
äåëéçå ùàí 
$\x$ 
å÷èåø òöîé ùì 
$A$ 
òí òøê òöîé 
$\lambda$ 
àæé 
$\x$ 
å÷èåø òöîé ùì 
$A^{-1}$
òí òøê òöîé 
$\frac{1}{\lambda}$.
ëàï 
$A^{-1}$ 
îñîï àú äîèøéöä ääåôëéú ùì $A$.

%\answer
%
%ìôé äðúåï îú÷ééí 
%$A\x=\lambda \x$. 
%ðëôéì àú ùðé äàâôéí á 
%$A^{-1}$ 
%åð÷áì 
%\[A^{-1}A\x=A^{-1}(\lambda \x)=\lambda A^{-1}\x.\]
%ëòú, ìôé ääâãøä ùì îèøéöä äåôëéú îú÷ééí 
%$A^{-1}A=I$ 
%ëàùø 
%$I$ 
%îèøéöú äéçéãä îñãø 
%$n\times n$.
%ðåáò ù 
%$A^{-1}A\x=I\x=\x$ 
%ìëï 
%$\x=\lambda A^{-1}\x$.
%ðéúï ìäñé÷ ëé 
%$A^{-1}\x=\frac{1}{\lambda}\x$
%ìëï 
%$\x$ 
%å÷èåø òöîé ùì 
%$A^{-1}$ 
%òí òøê òöîé 
%$\frac{1}{\lambda}$.

\item 
äåëéçå ùàí 
$\x$ 
å÷èåø òöîé ùì 
$A$ 
òí òøê òöîé 
$\lambda$ 
àæé 
$\x$ 
å÷èåø òöîé ùì 
$A^2$
òí òøê òöîé 
                    $\lambda^2$.

%\answer
%
%ìôé äðúåï 
%$A\x=\lambda\x$.
%ëòú,
%$A^2\x=AA\x=A\lambda \x=\lambda A\x=\lambda (\lambda \x)=\lambda^2\x$
%ìëï 
%$\x$ 
%å÷èåø òöîé ùì 
%$A^2$ 
%òí òøê òöîé 
%$\lambda^2$.


\end{enumerate}

\item 

	éäé 
	$\setF$
	ùãä åúäé 
	$A\in \setF^{n\times n}$
	îèøéöä äôéëä. 

\begin{enumerate}
	\item 
äåëéçå ëé 
$(A^T)^{-1}=(A^{-1})^T$.	


%\answer
%
%ëãé ìáãå÷ äàí 
%$(A^{-1})^T$
%ääåôëéú ùì 
%$A^T$,
%ðçùá àú 
%$A^T(A^{-1})^T$. 
%ëòú, 
%\begin{align*}
%A^T(A^{-1})^T&=(A^{-1}A)^T\\
%&=I^T=I.
%\end{align*}
%ìôé îùôè ðåáò ùâí 
%$(A^{-1})^TA^T=I$.
%ìëï àëï 
%$(A^{-1})^T$
%ääåôëéú ùì 
%$A^T$.

	\item 
äåëéçå ùàí 
$A$ 
ñéîèøéú àæé âí 
$A^{-1}$
ñéîèøéú.

%\answer
%
%àí $A$ ñéîèøéú àæ 
%$A^T=A$. 
%ìôé ñòéó ä÷åãí ð÷áì 
%\[(A^{-1})^T=(A^T)^{-1}=A^{-1}\]
%ìëï 
%$A^{-1}$
%ñéîèøéú.
\item 
àí 
\[A=\bm 3&0&5\\0&2&1\\5&1&9 \em\]
\begin{enumerate}
	\item
çùáå àú 
$A^{-1}$.

%\answer
%
%òì îðú ìçùá àú ääåëôéú ùì $A$ ðãøâ:
%
%\begin{align*}
%	\left(
%	\begin{array}{ccc|ccc}
%		3&0&5&1&0&0\\
%		0&2&1&0&1&0\\
%		5&1&9&0&0&1
%	\end{array} 	
%	\right) 	
%	&{\widesim{R_3\to 3R_3-5R_1}}	
%	\left(
%	\begin{array}{ccc|ccc}
%		3&0&5&1&0&0\\
%		0&2&1&0&1&0\\
%		0&3&2&-5&0&3
%	\end{array} 	
%	\right)\\[0.5em]
%	&{\widesim{R_3\to 2R_3-3R_2}}	
%	\left(
%	\begin{array}{ccc|ccc}
%		3&0&5&1&0&0\\
%		0&2&1&0&1&0\\
%		0&0&1&-10&-3&6
%	\end{array} 	
%	\right)\\[0.5em]
%		&{\widesim{R_1\to R_1-5R_2}}	
%		\left(
%		\begin{array}{ccc|ccc}
%			3&0&0&51&15&-30\\
%			0&2&1&0&1&0\\
%			0&0&1&-10&-3&6
%		\end{array} 	
%		\right)\\[0.5em]
%		&{\widesim{R_2\to R_2-R_3}}	
%		\left(
%		\begin{array}{ccc|ccc}
%			3&0&0&51&15&-30\\
%			0&2&0&10&4&-6\\
%			0&0&1&-10&-3&6
%		\end{array} 	
%		\right)\\[0.5em]
%		&{\widesim{R_1\to \frac{1}{3}R_1}}	
%		\left(
%		\begin{array}{ccc|ccc}
%			1&0&0&17&5&-10\\
%			0&2&0&10&4&-6\\
%			0&0&1&-10&-3&6
%		\end{array} 	
%		\right)\\[0.5em]				
%		&{\widesim{R_2\to \frac{1}{2}R_2}}	
%		\left(
%		\begin{array}{ccc|ccc}
%			1&0&0&17&5&-10\\
%			0&1&0&5&2&-3\\
%			0&0&1&-10&-3&6
%		\end{array} 	
%		\right)\\						
%\end{align*}
%ìëï 
%$A^{-1}=\bm
%17&5&-10\\
%5&2&-3\\
%-10&-3&6
%\em$.

\item 
çùáå àú 
$(A^2)^{-1}$.

%\answer
%
%ìôé ùàìä 
%\ref{inverse-of-products}
% ñòéó ä' îú÷ééí 
%$(A^2)^{-1}=(A^{-1})^2$.
%ìôé äñòéó ä÷åãí ð÷áì 
%\begin{align*}
%(A^2)^{-1}&=(A^{-1})^2\\
%&=A^{-1}A^{-1}\\
%&=\bm
%17&5&-10\\
%5&2&-3\\
%-10&-3&6
%\em\bm
%17&5&-10\\
%5&2&-3\\
%-10&-3&6
%\em\\
%&=\bm
%414&125&-245\\
%125&38&-74\\
%-245&-74&145
%\em
%\end{align*}

\end{enumerate}
\end{enumerate}



\textbf{úæëåøú:}
òáåø îèøéöä äôéëä
$\bm p&q\\ r&s \em$
 îñãø 
$2\times 2$
äîèøéöä ääåôëéú ðúåðä òì éãé 
\[\bm p&q\\ r&s \em^{-1}=\frac{1}{ps-qr}\bm s&-q\\ -r&p \em.\]

\item 

éäéå
	$a, b\in \setR$ 
	ëê ùîú÷ééí 
	$a^2-b^2=1$.
	ðâãéø îèøéöåú 
	\[C=\bm a&b\\ b&a \em, D=\bm a&b\\ a&b \em \in \setR^{2\times 2}.\]

\begin{enumerate}
	\item 
	äøàå ù $D$ àéðä äôéëä.

%\answer 
%
%\textbf{ùéèä 1:}	
%$\det D=ab-ba=0$ 
%ìëï $D$ àéðä äôéëä.
%
%\textbf{ùéèä 2:}
%äòîåãåú ùì $D$ î÷áéìåú ìëï äï úìåéåú ìéðàøéú. ðåáò ëé 
%$D$ 
%àéðä äôéëä.
%
%
%\textbf{ùéèä 3:}
%äùåøåú ùì $D$ æäåú ìëï 
%$\r (D)<2$ 
%)áòöí 
%$\r (D)=1$ 
%ëé 
%$a^2-b^2=1$ 
%ìëï 
%$(a, b)\neq (0,0)$.(
%ðåáò ëé 
%$D$ 
%àéðä äôéëä.
%
%\textbf{ùéèä 4:}
%$D\bm b\\-a \em=\bm 0\\0 \em$
%ìëï 
%$\bm b\\ -a\em\in \Nu(D)$.
%àáì 
%$\bm b\\-a \em\neq \bm 0\\0 \em$
%ëé 
%$a^2-b^2=1$.
%ðåáò ëé 
%$\Nu(D)\neq \{\0\}$ 
%ìëï 
%$D$ 
%àéðä äôéëä.
	
	\item 
	äøàå ëé 
	$C$ 
	äôéëä åîöàå àú äîèøéöä ääåôëéú 
	$C^{-1}$.
	
%\answer	
%	
%ðçùá: 
%$\det C=a^2-b^2=1\neq 0$ 
%ìëï 
%$C$ 
%äôéëä. ìôé èòðä ääåôëéú ðúåðä òì éãé 
%$C^{-1}=\frac{1}{\det C}\bm a&-b\\ -b&a \em=\bm a&-b\\ -b&a \em$.
	
	\item 
	îöàå àú ëì äòøëéí äòöîééí ùì 
	$C$
)éù ìëúåá àú äò"ò áàîöòåú äôøîèøéí
$a, b$.(	
	òáåø ëì òøê òöîé îöàå å÷èåø òöîé îúàéí.
	
%\textbf{ùéèä 1:}
%ìîèøéöä 
%$C$ 
%éù äøáä ñéîèøéä. áôøè, ñëåí äî÷ãîéí áëì ùåøä ùååä. ðåáò ù 
%$\bm 1\\ 1\em$ 
%å÷èåø òöîé ùì 
%$C$ 
%àí òøê òöîé 
%$a+b$:
%
%\[C\bm 1\\ 1\em=\bm a&b\\ b&a \em\bm 1\\1 \em=\bm a+b\\ a+b \em=(a+b)\bm 1\\1 \em.\]	
%áàåôï ãåîä, äôøù äî÷ãîéí áëì ùåøä ãåîä ìëï ðáãå÷ àú äåå÷èåø 
%$\bm 1\\ -1\em$:
%
%\[C\bm 1\\ -1\em=\bm a&b\\ b&a \em\bm 1\\-1 \em=\bm a-b\\ b-a \em=(a-b)\bm 1\\-1 \em.\]
%ìîèøéöä îñãø 
%$2\times 2$ 
%éù ìëì äéåúø ùðé ùåøùéí ìôåìéðåí äàåôééðé ìëï éù ìëì äéåúø ùðé òøëéí òöîééí. ðéúï ìäñé÷ ùîöàðå àú ëì äòøëéí äòöîééí ùì $C$.
%\textbf{ùéèä 2:}
%ðçùá àú äôåìéðåí äàåôééðé ùì $C$:
%\begin{align*}
%p_C(\lambda)&=\bd a-\lambda&b\\ b&a-\lambda \ed\\
%&=(a-\lambda)^2-b^2\\
%&=\lambda^2-2a\lambda+a^2-b^2\\
%&=(\lambda-(a+b))(\lambda-(a-b))	
%\end{align*}	
%ìëï éù òøëéí òöîééí 
%$a+b,\ \ a-b$.
%òì îðú ìîöåà å÷èåø òöîé òáåø òøê äòöîé 
%$a+b$ 
%ðîöà å÷èåø ìà àôñ á 
%$\Nu(A-(a+b)I)$:
%\begin{align*}
%(A-(a+b)I|\0)&=\bse{cc} -b&b&0\\ b&-b&0 \ese{\widesim{R_2\to R_2+R_1}}\bse{cc} -b&b&0\\ 0&0&0 \ese
%\end{align*}	
%àæ ðéúï ì÷çú å"ò
%$\v_1=\bm1\\1 \em$.\\
%
%òì îðú ìîöåà å÷èåø òöîé òáåø òøê äòöîé 
%$a-b$ 
%ðîöà å÷èåø ìà àôñ á 
%$\Nu(A-(a-b)I)$:
%\begin{align*}
%	(A-(a-b)I|\0)&=\bse{cc} b&b&0\\ b&b&0 \ese{\widesim{R_2\to R_2-R_1}}\bse{cc} b&b&0\\ 0&0&0 \ese
%\end{align*}	
%àæ ðéúï ì÷çú å"ò
%$\v_2=\bm1\\-1 \em$.
		\item 
		îöàå àú ëì äòøëéí äòöîééí ùì 
		$D$
		)éù ìëúåá àú äò"ò áàîöòåú äôøîèøéí
		$a, b$.(
		òáåø ëì òøê òöîé îöàå å÷èåø òöîé îúàéí.
		
%\textbf{ùéèä 1:}
%ëîå áñòéó ä÷åãí, ìîèøéöä 
%$D$ 
%éù äøáä ñéîèøéä. áôøè, ñëåí äî÷ãîéí áëì ùåøä ùååä. ðåáò ù 
%$\bm 1\\ 1\em$ 
%å÷èåø òöîé ùì 
%$D$ 
%àí òøê òöîé 
%$a+b$:
%
%\[D\bm 1\\ 1\em=\bm a&b\\ a&b \em\bm 1\\1 \em=\bm a+b\\ a+b \em=(a+b)\bm 1\\1 \em.\]	
%ëòú, ðéúï ìàôñ àú ùúé äùåøåú àí ðëôéì áåå÷èåø
%$\bm b\\ -a\em$
%ìëï äåà å÷èåø òöîé òí òøê òöîé 
%$0$:
%
%\[D\bm b\\ -a\em=\bm a&b\\ a&b \em\bm b\\-a \em=\bm ab-ba\\ ab-ba \em=\bm 0\\0 \em=0\bm b\\-a \em.\]
%ìîèøéöä îñãø 
%$2\times 2$ 
%éù ìëì äéåúø ùðé ùåøùéí ìôåìéðåí äàåôééðé ìëï éù ìëì äéåúø ùðé òøëéí òöîééí. ðéúï ìäñé÷ ùîöàðå àú ëì äòøëéí äòöîééí ùì $D$.
%\textbf{ùéèä 2:}
%ðçùá àú äôåìéðåí äàåôééðé ùì $D$:
%\begin{align*}
%	p_D(\lambda)&=\bd a-\lambda&b\\ a&b-\lambda \ed\\
%	&=(a-\lambda)(b-\lambda)-ab\\
%	&=\lambda^2-(a+b)\lambda\\
%	&=\lambda(\lambda-(a+b))
%\end{align*}	
%ìëï éù òøëéí òöîééí 
%$a+b,\ \ 0$.
%òì îðú ìîöåà å÷èåø òöîé òáåø òøê äòöîé 
%$a+b$ 
%ðîöà å÷èåø ìà àôñ á 
%$\Nu(A-(a+b)I)$:
%\begin{align*}
%	(A-(a+b)I|\0)&=\bse{cc} -b&b&0\\ a&-a&0 \ese{\widesim{}}\bse{cc} 1&-1&0\\ 0&0&0 \ese
%\end{align*}	
%îëéååï ù 
%$a^2-b^2=1$ 
%ìëï 
%$a\neq 0$ 
%àå 
%$b\neq 0$ 
%åäùåøåú î÷áéìåú.
%àæ ðéúï ì÷çú å"ò
%$\v_1=\bm1\\1 \em$.\\
%
%òì îðú ìîöåà å÷èåø òöîé òáåø òøê äòöîé 
%$0$ 
%ðîöà å÷èåø ìà àôñ á 
%${\Nu(A-(0)I)}$:
%\begin{align*}
%	(A-(0)I|\0)=(A|\0)&=\bse{cc} a&b&0\\ a&b&0 \ese{\widesim{R_2\to R_2-R_1}}\bse{cc} a&b&0\\ 0&0&0 \ese
%\end{align*}	
%àæ ðéúï ì÷çú å"ò
%$\v_2=\bm b\\-a \em$
%åäåà àéðå å÷èåø äàôñ ëé 
%$a\neq 0$ 
%àå 
%$b\neq 0$ 
%ëôé ùëáø öééðå.		
\item 
	äøàå ëé éù ì 
	$C$ 
	å 
	$C^{-1}$ 
	àåúí òøëéí òöîééí.
	
%	\answer
%	
%	ìôé ñòéó á' ùì äùàìä äðåëçéú äîèøéöä 
%	$C$ 
%	äôéëä.
%	ìôé ùàìä 
%	\ref{eigen-of-inverse}
%	ñòéó á' ìëì òøê òöîé 
%	$\lambda$ 
%	ùì $C$ éù ì
%	$C^{-1}$ 
%	éùðí òøëéí òöîééí 
%	 $\frac{1}{\lambda}$.
%	 ìôé ñòéó â' ùì äùàìä äðåëçéú äòøëéí äòöîééí ùì $C$ äí 
%	 $a+b,\ a-b$. 
%	 ðåáò ùäòøëéí äòöîééí ùì 
%	 $C^{-1}$ 
%	 äí 
%	 $\frac{1}{a+b}, \ \frac{1}{a-b}$.
%	 ðúåï ëé 
%	 $1=a^2-b^2=(a+b)(a-b)$
%	 ìëï
%	 $\frac{1}{a+b}=a-b$
%	 å 
%	 $\frac{1}{a-b}=a+b$.
%	 ìëï ì 
%	 $C^{-1}$ 
%éù àåúí òøëéí òöîééí 
%	 $a+b, \ a-b$ 
%	 ëîå 
%	 $C$.
%	
	\item 
	úðå ãåâîä ìîèøéöä äôéëä $Y$ îñãø 
	$2\times 2$ 
	ëê ùäòøëéí òöîééí ùì 
	$Y$ 
	ùåðéí îäòøëéí äòöîééí ùì 
	$Y^{-1}$.
	
	
%	\answer
%	
%ð÷ç ìîùì
%$Y=\bm 2&0\\ 0&3 \em$
%åàæ 
%$Y^{-1}=\bm \frac{1}{2}&0\\ 0&\frac{1}{3} \em$.
%äòøëéí äòöîééí ùì 
%$Y$ 
%äí 
%$2, 3$
%)ðéúï ì÷çú ìîùì å÷èåøéí òöîééí 
%$\bm 1\\ 0\em, \bm 0\\ 1\em$.(
%áàåôï ãåîä, ÷ì ìøàåú ùäòøëéí äòöîééí ùì 
%$Y^{-1}$ 
%äí 
%$\frac{1}{2}, \frac{1}{3}$. 
%ìëï ì 
%$Y, Y^{-1}$
%éù òøëéí òöîééí ùåðéí.
	
\end{enumerate}

\item 
ðâãéø ÷áåöú îèøéöåú òì éãé 
\[\S=\left\{ \bm a&-b\\ b&a  \em\mid a,b\in \setR, a^2+b^2=1 \right\}.\]
\begin{enumerate}
	\item 
	äøàå ùëì îèøéöä á 
	$\S$ 
	äôéëä.
	
%	\answer 
%	
%àí 
%$A=\bm a&-b\\ b&a\em\in \S$ 
%àæ 
%$\det A=a^2-(-b^2)=a^2+b^2=1\neq 0$ 
%ìëï 
%$A$ 
%äôéëä.

	\item 
	äåëéçå ùìëì 
	$A\in \S$ 
	îú÷ééí 
	$A^{-1}\in \S$.
	
%\answer	
%	
%àí 
%$A=\bm a&-b\\ b&a\em\in \S$ 	
%	àæ ìôé èòðä îú÷ééí 
%\[A^{-1}=\frac{1}{\det A}\bm a&b\\ -b&a \em=\frac{1}{a^2+b^2}\bm a&b\\ -b&a \em=\frac{1}{1}\bm a&b\\ -b&a \em=\bm a&-(-b)\\ b&a \em\in \S\]
%)äúðàé 
%$a^2+b^2=1$
%òãééï îú÷ééí.(	
	
	\item 
	äåëéçå ùìëì 
	$A, B\in \S$
	îú÷ééí 
	$AB\in \S$.
	
%\answer
%
%úééðä 
%	$A=
%	\bm a&-b\\ b&a \em, B=
%	\bm c&-d\\ d&c \em.
%	$
%	
%	
%	ëòú, 
%	\begin{align*}
%	AB&=\bm a&-b\\ b&a \em
%	\bm c&-d\\ d&c \em=\bm ac-bd&-ad-bc \\bc+ad&ac-bd  \em	
%	\end{align*}
	
	\item 
	îöàå ëì äôúøåðåú ìîùååàä 
	$A^2=\bm 1&0\\ 0&1 \em$
	ëàùø 
	$A\in \S$.
	
%\answer 
%
%ð÷ç 
%$a=-1, b=0$ 
%åð÷áì 	
%	$A=\bm a&-b\\b&a \em=\bm -1&0\\0&-1 \em$.
%	îëàï ð÷áì 
%	\[A^2=\bm -1&0\\0&-1 \em^2=\bm -1&0\\0&-1 \em\bm -1&0\\0&-1 \em=\bm 1&0\\0&1 \em\]
%	
\end{enumerate}

%\item 
%
%\begin{enumerate}
%	\item 
%	\item 
%\end{enumerate}




%\item 
%
%\begin{enumerate}
%	\item 
%	\item 
%\end{enumerate}





\end{enumerate}



\textbf{äåøàåú:}

\begin{itemize}
	\item
îèìä æå àéðä ìäâùä.
	\item 
ôúøåðåú éôåøñîå ìëì äîàåçø á 11 áéåìé.

\end{itemize}













\end{document}