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\headline={\setfonts[/10]\rm\hfill [\the\pageno]}
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\def\Q{{\bf Q}}
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\def\A{{\bf A}}
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% ===============================================================

\vglue 2cm

\gg Soustavy

\gg line�rn�ch rovnic

\bigskip\bigskip

* vlastnosti mno�in �e�en�

* metody hled�n� �e�en�

* nejednozna�nost z�pisu �e�en�


\n ---------------------------------------------------------------

\g Terminologie

{\bf Definice:} Nech� $\A=(a_{i,j})\in\R^{m,n}$ je matice,
${\bf b}\in\R^{m,1}$ je jednosloupcov� matice.
Maticov� rovnost
$$
  \A\cdot{\bf x} = {\bf b}
$$
s nezn�mou jednosloupcovou matic� ${\bf x}\in\R^{n,1}$ naz�v�me {\em soustavou
line�rn�ch rovnic\/} ($m$ rovnic, $n$ nezn�m�ch). 
$\A$ je {\em matice soustavy}, ${\bf b}$ je {\em sloupec prav�ch stran},
$(\A\,|\,{\bf b})$ je {\em roz���en� matice soustavy}.

Je-li ${\bf o}\in\R^{m,1}$ nulov� sloupcov� vektor, 
pak $\A\cdot{\bf x} = {\bf o}$ je 
{\em homogenn� soustava line�rn�ch rovnic}.

{\em �e�en� soustavy} je takov� vektor z $\R^n$, kter�, zapsan� do
sloupce m�sto nezn�m� matice ${\bf x}$, spl�uje danou maticovou
rovnost.

{\bf �loha:} Naj�t v�echna �e�en�, tj. vymezit podmno�inu
$M\subseteq\R^n$ v�ech �e�en� soustavy.

\n ---------------------------------------------------------------

\g Dva pohledy na soustavu lin. rovnic

{\bf Pohled po ��dc�ch}, tedy po jednotliv�ch rovnic�ch. Ka�d� rovnice
sama vymezuje podmno�inu v�ech sv�ch �e�en� $M_i\subseteq\R^n$, 
$i\in\{1,2,\ldots,m\}$. Geometricky
je $M_i$ nadrovinou (podprostorem dimenze $n-1$ posunut�m 
z po��tku do n�jak�ho jin�ho bodu). V�echny rovnice maj� b�t spln�ny
sou�asn�, hled�me tedy spole�n� {\em pr�nik\/} v�ech t�chto nadrovin $M_i$.

{\bf Pohled po sloupc�ch}. Rozepi�me matici soustavy $\A$ do sloupc�:
$\A=(\A_1,\A_2,\ldots,\A_n)$. Soustava $\A\cdot{\bf x}={\bf b}$
p�ech�z� na:
$$
  \A\cdot{\bf x}\ = \
  (\A_1,\A_2,\ldots,\A_n)\cdot\pmatrix{x_1\cr x_2\cr\vdots\cr x_n} = \
  x_1\A_1 + x_2\A_2 + \cdots + x_n\A_n \ = \ {\bf b}
$$
Hled�me tedy koeficienty line�rn� kombinace sloupc� matice $\A$, kter�
zaru��, �e se dan� kombinace rovn� prav� stran� ${\bf b}$.

\n -----------------------------------------------------------------

\g K �emu slou�� elimina�n� metoda

M�-li soustava $\A\cdot{\bf x}={\bf b}$ mno�inu �e�en� $M$ a je
$$
  (\A\,|\,{\bf b}) \sim (\C\,|\,{\bf d})
$$
pak soustava $\C\cdot{\bf x}={\bf d}$ m� stejnou mno�inu �e�en� $M$.

Kdy� eliminujeme na schodovitou matici $\C$, pak p�jde u soustavy  
$\C\cdot{\bf x}={\bf d}$ hledan� mno�ina �e�en� $M$ l�pe naj�t.


\n -----------------------------------------------------------------

\g Frobeniova v�ta, �e�itelnost soustavy

{\bf V�ta}: Soustava $\A\cdot{\bf x}={\bf b}$ m� aspo� jedno �e�en�
pr�v� tehdy, kdy� $\hod\A = \hod(\A\,|\,{\bf b})$.

D�kaz: (sloupcov� pohled): soustava m� �e�en� pr�v� kdy� vektor~${\bf b}$ 
le�� v line�rn�m obalu sloupcov�ch vektor� $\A_1,\A_2,\ldots,\A_n$, co�
je pr�v� tehdy, kdy� $\hod\A = \hod(\A\,|\,{\bf b})$. Vyu�ijeme tak�
toho, �e hodnost matice nen� jen dimenze lin. obalu ��dk�, ale
je to tak� dimenze line�rn�ho obalu sloupc� matice, nebo� \ $\hod\A=\hod\A^T$.

D�kaz: (elimina�n� pohled): Po eliminaci na schodovitou matici m�me
soustavu $\C\cdot{\bf x}={\bf d}$ kter� nem� �e�en� pr�v� tehdy, kdy�
existuje nulov� ��dek v matici $\C$ s nenulov�m ��slem na prav�
stran�. Tj. pr�v� tehdy kdy� $\hod\C\not=\hod(\C\,|\,{\bf
d})$. Elimina�n� metoda ov�em nem�n� hodnost.


\n -----------------------------------------------------------------

\g Homogenn� soustava Ax = o

* Homogenn� soustava line�rn�ch rovnic m� v�dy nulov� �e�en�.

* Mno�inou �e�en� $M_0$ homogenn� soustavy je v�dy podprostor\thinspace:

$$ 
  \eqalign{
   {\bf u}\in M_0, \ {\bf v}\in M_0,& \quad \hbox{tj.}\quad
   \A{\bf u}={\bf o}, \quad \A{\bf v}={\bf o}. \cr
   &\A({\bf u}+{\bf v}) = \A{\bf u}+\A{\bf v} = {\bf o}+{\bf o} = {\bf o},
   \quad\hbox{tj.}\ {\bf u}+{\bf v}\in M_0. \cr
   {\bf u}\in M_0, \ \alpha\in\R,& \quad \hbox{tj.}\quad
   \A{\bf u}={\bf o}.\cr
   &\A(\alpha{\bf u}) = \alpha\A({\bf u}) = \alpha\,{\bf o} = {\bf o},
   \quad\hbox{tj.}\quad \alpha\,{\bf u}\in M_0.
}
$$

\n ----------------------------------------------------------------

\g Jak vy�e�it homogenn� soustavu

* Nejprve p�evedeme eliminac� na soustavu se stejnou mno�inou �e�en�,
  ale se schodovitou matic� soustavy:
$$
  (\A\,|\,{\bf o}) \sim (\C\,|\,{\bf o})
$$
* Ka�d� nenulov� rovnice v soustav� $\C{\bf x}={\bf o}$ umo�n� spo��tat
jednu nezn�mou (p�i zp�tn� substituci zespoda nahoru). T�mto prom�nn�m
��k�me {\em v�zan� prom�nn�}. Ostatn�
(takto nespo��tan�) prom�nn� jsou {\em voln� prom�nn�}, neboli parametry.
Nech� $t_1,t_2,\ldots,t_k$ jsou v�echny voln� prom�nn�.
M��eme volit tyto hodnoty za $(t_1,t_2,\ldots,t_k)$
$$
  (1,0,\ldots,0), \quad(0,1,\ldots,0),\quad \ldots ,\quad(0,0,\ldots,1)
$$
a pro ka�dou tuto volbu voln�ch prom�nn�ch dopo��t�me prom�nn� v�zan�.
Dost�v�me tak line�rn� nez�vislou mno�inu �e�en�, kter� je b�z�
podprostoru $M_0$ v�ech �e�en�.


\n ------------------------------------------------------------------

\g P��klad (homogenn� soustava)

�e�me soustavu $\A\cdot{\bf x}={\bf o}$ s matic�:

$$
  \def\\#1{\vbox to0pt{\vss\hbox to0pt{\hbox to#1mm{$\scriptstyle
       x_1,\hss x_2,\hss x_3,\hss x_4,\hss x_5$}\hss}\bigskip\vskip7pt}}
  \A = \pmatrix {\\{49}1&\ 3\ &2&0&3\cr 1&1&1&-1&5\cr 2&8&5&3&7\cr 3&9&6&2&12}
  \sim \pmatrix {\\{42}1&3&2&0&3\cr 0&2&1&3&1\cr 0&0&0&2&3}
$$
V�zan� prom�nn�: $x_1,\ x_2,\ x_4$, voln� prom�nn�: $x_3,\ x_5$.

P�i volb� $x_3=1,\ x_5=0$ vych�z�: $x_4=0,\ x_2=-1/2,\ x_1=-1/2$, \hb
p�i volb� $x_3=0,\ x_5=1$ vych�z�: $x_4=-3/2,\ x_2=7/4,\ x_1=-33/4$.

Tak�e m�m dv� line�rn� nez�visl� �e�en�:
$$
  (-1/2,-1/2,1,0,0),\ (-33/4, 7/4, 0, -3/2, 1).
$$
V�echna �e�en� tvo�� line�rn� obal t�chto dvou �e�en�:
$$
  M_0 = \lob<(-1,-1,2,0,0),\ (-33,7,0,-6,4)>
$$

\n -------------------------------------------------------------------

\g Dimenze prostoru �e�en�

* Dimenze prostoru �e�en� homogenn� soustavy je rovna po�tu voln�ch
  prom�nn�ch, 

* co� je rovno po�tu v�ech prom�nn�ch minus po�tu v�zan�ch prom�nn�ch, 

* co� je rovno po�tu v�ech prom�nn�ch minus po�tu nenulov�ch rovnic
  soustavy $\C{\bf x}={\bf o}$ se schodovitou matic�,

* co� je rovno po�tu v�ech rovnic minus hodnost matice soustavy.

{\bf Z�v�r}: Nech� $M_0$ je mno�ina �e�en� soustavy $\A{\bf x} = {\bf o}$
s $m$ rovnicemi a $n$ nezn�m�mi. Pak $$\dim M_0 = n - \hod\A.$$


\n ----------------------------------------------------------------

\g Dva podprostory v $\setmath[//]\R^n$ vymezen� matic� A

Nech� je d�na matice $\A\in\R^{m,n}$

* Ozna�me $R$ line�rn� obal ��dk� matice $\A$. Je to podprostor v $\R^n$.

* Ozna�me $M_0$ mno�inu v�ech �e�en� homogenn� soustavy $\A{\bf x} = {\bf o}$.
  Je to rovn� podprostor v $\R^n$. Naz�v� se {\em nulov�m prostorem
  matice $\A$}.

* Do ��dk� matice $\B$ napi�me n�jakou b�zi prostoru $M_0$.

{\bf Plat�}:

* Ka�d� vektor z $M_0$ �e�� soustavu $\A{\bf x} = {\bf o}$.

* Ka�d� vektor z $R$ �e�� soustavu $\B{\bf x} = {\bf o}$.

* Je-li $\vec u\in R$ a $\vec v\in M_0$, pak $\vec u\cdot\vec v^T = 0$.

* $\dim R+\dim M_0 = n = \dim\R^n$


\n ----------------------------------------------------------

\g Algoritmus hled�n� b�ze nulov�ho prostoru

{\bf Algoritmus}: Nech� $\A\sim(\E\,|\,\C)$, kde $\E$ je jednotkov� matice. Pak ��dky
matice $(-\C^T\,|\,\E')$ obsahuj� b�zi �e�en� soustavy $\A{\bf x}={\bf o}$.

Pozn�mka: $\E'$ je zde tak� jednotkov� matice, ov�em obecn� jin�ho typu
ne� matice $\E$.

D�kaz: ��dky matice $(-\C^T\,|\,\E')$ jsou line�rn� nez�visl� a jejich
po�et je roven $n-\hod\A$. Tak�e lin. obal t�chto ��dk� m� stejnou
dimenzi, jako prostor �e�en� $M_0$. Sta�� uk�zat, �e ka�d� ��dek matice
$(-\C^T\,|\,\E')$ �e�� soustavu $\A{\bf x}={\bf o}$:
$$
  (\E\,|\,\C)\cdot \pmatrix{-\C\cr\noalign{\smallskip\hrule\smallskip}\E'} =
  \E\cdot(-\C) + \C\cdot\E' = -\C+\C = {\bf O}.
$$
Pozn�mka: nelze-li prov�st $\A\sim(\E\,|\,\C)$, pak je mo�n� dostat $(\E\,|\,\C)$
po vhodn� permutaci sloupc� (zm�na po�ad� nezn�m�ch). Zp�tnou permutaci
sloupc� pak provedeme na matici $(-\C^T\,|\,\E')$ a m�me hledanou b�zi
prostoru �e�en�.

\n -----------------------------------------------------------

\g P��klad

Metodou ze sl�du [11] vy�e��me soustavu ze sl�du [8]. Matici p�evedeme
Gauss-Jordanovou eliminac�:
$$
  \def\\#1{\vbox to0pt{\vss\hbox to0pt{\hbox to#1mm{$\scriptstyle
       x_1,\hss x_2,\hss x_3,\hss x_4,\hss x_5$}\hss}\bigskip\vskip7pt}}
  \A = \pmatrix {1&\ 3\ &2&0&3\cr 1&1&1&-1&5\cr 2&8&5&3&7\cr 3&9&6&2&12}
  \sim \pmatrix {\\{52}1&\ 0\ &1/2&0&33/4\cr 0&1&1/2&0&-7/4\cr 0&0&0&1&3/2}
$$
Prohod�me sloupce a p�ejdeme od matice $(\E\,|\,\C)$ k matici $(-\C^T\,|\,\E')$:

$$
  \def\\#1{\vbox to0pt{\vss\hbox to0pt{\hbox to#1mm{$\scriptstyle
       x_1,\hss x_2,\hss x_4,\hss x_3,\hss x_5$}\hss}\bigskip\vskip7pt}}
     \pmatrix {\\{50}1&\ 0\ &0&1/2&33/4\cr 0&1&0&1/2&-7/4\cr
       0&0&1&0&3/2}, \quad
     \pmatrix {-1\\{58}/2&\!\!-1/2&0&1&\ 0\cr -33/4&7/4&\!\!\!-3/2&0&\ 1}
$$
Po zp�tn�m prohozen� sloupc� dost�v�me v ��dc�ch b�zi mno�iny �e�en� $M_0$:
$$
  \def\\#1{\vbox to0pt{\vss\hbox to0pt{\hbox to#1mm{$\scriptstyle
       x_1,\hss x_2,\hss x_3,\hss x_4,\hss x_5$}\hss}\bigskip\vskip7pt}}
\pmatrix {-1\\{58}/2&\!\!-1/2&1&0&\ 0\cr -33/4&7/4&0&\!\!\!-3/2&\ 1}
$$

\n ------------------------------------------------------------

\g Nehomogenn� soustava line�rn�ch rovnic

{\bf Terminologie}: Jak�koli �e�en� soustavy $\A{\bf x} = {\bf b}$
naz�v�me {\em partikul�rn� �e�en�} t�to soustavy. 

Soustava $\A{\bf x} = {\bf o}$ se naz�v� {\em p�idru�en� homogenn� soustava}
k~soustav� $\A{\bf x} = {\bf b}$.

{\bf V�ta}: Mno�ina $M$ v�ech �e�en� soustavy $\A{\bf x} = {\bf b}$
je bu� pr�zdn�, nebo je tvaru
$$
  M = {\bf v} + M_0
$$
kde ${\bf v}$ je partikul�rn� �e�en� soustavy $\A{\bf x} = {\bf b}$
a $M_0$ je mno�ina v�ech �e�en� p�idru�en� homogenn� soustavy
$\A{\bf x} = {\bf o}$.

D�kaz: Ozna�me ${\bf v}$ partikul�rn� �e�en� a nech� ${\bf u}\in M_0$. 
Sta�� ov��it, �e ${\bf v}+{\bf u}\in M$.
D�le mus�me ov��it, �e pro ka�d� ${\bf w}\in M$ existuje ${\bf u}\in M_0$ tak,
�e ${\bf v}+{\bf u} = {\bf w}$. 

Pozn�mka: V�hodn� je geometrick� p�edstava, ud�lejte si n��rtek.

\n ------------------------------------------------------------------

\g Jak vy�e�it nehomogenn� soustavu

* Naj�t jedno partikul�rn� �e�en� ${\bf v}$.

* Vy�e�it p�idru�enou homogenn� soustavu, naj�t $M_0$.

* V�echna �e�en� napsat ve tvaru $M = {\bf v} + M_0$.

Jedin� probl�m: naj�t partikul�rn� �e�en� ${\bf v}$.
Typick� postup:

* Eliminovat $(\A\,|\,{\bf b}) \sim (\C\,|\,{\bf d})$, na soustavu
  se schodovitou matic�. 

* Sloupce s voln�mi prom�nn�mi odstranit
  (tj. dosadit za voln� prom�nn� nuly). Vznik� soustava s regul�rn�
  �tvercovou matic�. 

* Do�e�it tuto soustavu zp�tn�m chodem eliminace.
  
* K �e�en� p�ipsat nuly na m�sta voln�ch prom�nn�ch.

\n -----------------------------------------------------------------

\g P��klad (nehomogenn� soustava)

Soustava ze sl�du [8] je dopln�na o pravou stranou. Gauss-Jorda\-novou
eliminac� uprav�m roz���enou matici soustavy:
$$
  \def\+{\phantom0}\def\|{\hfill\kern8pt\vrule height1.1em depth.3em\kern-5pt}
  \def\c{\cr\noalign{\vskip-2pt}}
  \pmatrix {1&\ 3\ &2&0&3\|&3\c 1&1&1&-1&5\|&-2\c 2&8&5&3&7\|&13\c 3&9&6&2&\!\!12\|&11}
  \sim \pmatrix {1&\ 0\ &1/2&0&33/4\|&-3\c 0&1&1/2&0&-7/4\|&2\c 0&0&0&1&\ 3/2\|&1}
$$
Odstran�m sloupce odpov�daj�c� voln�m prom�n�m:
\vskip-5pt
$$
   \def\\#1{\vbox to0pt{\vss\hbox to0pt{\hbox to#1mm{$\scriptstyle
       x_1,\hss x_2,\hss x_4$}\hss}\bigskip\vskip7pt}}
 \def\+{\phantom0}\def\|{\hfill\kern8pt\vrule height1.1em depth.3em\kern-5pt}
  \def\c{\cr\noalign{\vskip-2pt}}
  \pmatrix {\\{26}1&\ 0\ &0\|&-3\c 0&1&0\|&2\c 0&0&1\|&1}
$$

Partikul�rn� �e�en� je $(-3,2,0,1,0)$ a mno�ina v�ech �e�en� je
$$
  M \ =\ (-3,2,0,1,0) \ + \ \lob<(-1,-1,2,0,0),\ (-33,7,0,-6,4)>.
$$


\n ----------------------------------------------------------------

\g Probl�m nejednozna�nosti z�pisu �e�en�

Stejn� mno�ina �e�en� soustavy $\A{\bf x} = {\bf b}$ se d� vyj�d�it
r�zn�mi vektory b�ze �e�en� p�idru�en� homogenn� soustavy a r�zn�mi
partikul�rn�mi �e�en�mi. Jak poznat, �e:
$$
  \vec v + \lob<\vecc u_k> \ = \ \vec w + \lob<\vecc z_k>\ ?
$$
Sta�� porovnat hodnosti n�sleduj�c�ch matic:
$$
  \hod\pmatrix{\vec u_1\cr\vdots\cr\vec u_k\cr\vec z_1\cr\vdots\cr\vec z_k\cr
               \vec w - \vec v} =
  \hod\pmatrix{\vec u_1\cr\vdots\cr\vec u_k} =
  \hod\pmatrix{\vec z_1\cr\vdots\cr\vec z_k}
$$
Viz t� stranu [9] k t�matu \uv{matice}.  
               
\n -----------------------------------------------------------------

\g Soustavy se �tvercovou matic� A

* Je-li matice $\A$ {\bf regul�rn�}, pak soustava m� jedin� �e�en�.

* Je-li matice $\A$ {\bf regul�rn�}, pak lze soustavu $\A\,{\bf x}={\bf b}$
  �e�it vyn�soben�m t�to rovnosti inverzn� matic� $\A^{-1}$ zleva:
  $$ 
      \A^{-1}\A\,{\bf x} = \A^{-1}\,{\bf b}, \quad
      \hbox{tj.}\quad   {\bf x} = \A^{-1}\,{\bf b}.
  $$
\vskip-.5\baselineskip
%* Je-li matice $\A$ {\bf regul�rn�}, je mo�n� tak� prov�st LU rozklad t�to matice a �e�it
%  jednu soustavu dop�ednou substituc� a dal�� zp�tnou substituc�.
%  Viz stranu [3] k t�matu \uv{LU rozklad}. Je to nepatrn� numericky
%  v�hodn�j�� ne� po��tat inverzn� matici.

* Je-li matice $\A$ {\bf regul�rn�} a zaj�maj� n�s jen n�kter� slo�ky �e�en�, je
  mo�n� pou��t {\em Cramerovo pravidlo}, viz n�sleduj�c� stranu. 

* Je-li $\A$ {\bf singul�rn�}, pak po eliminaci $(\A\,|\,{\bf b}) \sim  (\C\,|\,{\bf d})$
  dost�v�me soustavu s matic� $\C$, kter� nen� �tvercov�. D�le je
  nutn� pou��t postupy uveden� na p�edchoz�ch stran�ch.


\n -----------------------------------------------------------------

\g Cramerovo pravidlo

Nech� $\A$ je regul�rn� �tvercov� matice. Pak pro $i$-tou slo�ku �e�en�
soustavy $\A\,{\bf x} = {\bf b}$ plat�
$$
  x_i = {\det \B_i \over \det\A}\,,
$$
kde matice $\B_i$ je shodn� s matic� $\A$ a� na $i$-t� sloupec, kter�
je zam�n�n za sloupec prav�ch stran.

D�kaz: Vyu�ijeme vztah ${\bf x}=\A^{-1}\cdot{\bf b}$ a zam���me se v
maticov�m sou�inu na v�po�et $i$-t�ho ��dku v matici ${\bf x}$.
P�itom matici $\A^{-1}$ zap��eme pomoc� dopl�k�.
Viz stranu [19] k t�matu \uv{determinant}.

\n -----------------------------------------------------------------

\g P��klad

Vy�e��me soustavu s parametrem $p\in\R$, kter� m� roz���enou matici:
$$
  \def\+{\phantom0}\def\|{\hfill\kern8pt\vrule height1.1em depth.3em\kern-5pt}
  \def\c{\cr\noalign{\vskip-2pt}}
  (\A\,|\,{\bf b}) =
  \pmatrix{2&-p&-1\|&3\c 1&-7&-5\|&0\c -1&3&p\|&-1}
$$
Je $\det\A = (p-2)\,(p-17)$. Tak�e pro $p=17$ nebo $p=2$ je matice
soustavy singul�rn�:
$$
  \def\+{\phantom0}\def\|{\hfill\kern8pt\vrule height1.1em depth.3em\kern-5pt}
  \def\c{\cr\noalign{\vskip-2pt}}
  \eqalign{
  p=17:& \   \pmatrix{2&-17&-1\|&3\c 1&-7&-5\|&0\c -1&3&17\|&-1}
        \sim \pmatrix{1&-7&-5\|&0\c 0&-1&3\|&1\c 0&0&0\|&1}\ 
        \ldots \ M=\emptyset
\cr\noalign{\bigskip}
  p=2:& \   \pmatrix{2&-2&-1\|&3\c 1&-7&-5\|&0\c -1&3&2\|&-1} 
       \sim \pmatrix {1&-7&-5\|&0\c 0&4&3\|&1}\ \ldots
}
$$
\hfill$\ldots\ M \ = \ (5/3,0,1/3)\ +\ \lob<(1,3,-4)>$

\n ---------------------------------------------------------

\g P��klad (pokra�ov�n�)

Pro $p\not=17$ a $p\not=2$ je matice soustavy regul�rn� a soustava m�
jedin� �e�en�. Najdeme toto �e�en� pomoc� Cramerova pravidla.
$$
  \eqalign{
  \det\pmatrix{3&-p&-1\cr 0&-7&-5\cr -1&3&p}\! &= -26(p-2), \quad \!
  \det\pmatrix{2&3&-1\cr 1&0&-5\cr -1&-1&p}\! = -3(p-2) \cr\noalign{\bigskip}
  \det\pmatrix{2&-p&3\cr 1&-7&0\cr -1&3&-1}\! &= -(p-2). \quad\!
  \hbox{Proto�e $\det\A=(p-2)\,(p-17)$, je:}
}
$$
$$
  x_1 = {-26(p-2)\over(p-2)\,(p-17)} = {26\over 17-p}, \quad
  x_2 = {3\over 17-p}, \quad
  x_3 = {1\over 17-p}.
$$

Pro p��pad $p\not=17$ a $p\not=2$ obsahuje mno�ina $M$ jedin� �e�en�:
$$
  M=\left\{\left({26\over17-p},\ {3\over17-p},\ {1\over17-p}\right)\right\}.$$


\n ---------------------------------------------------------------

\g Maticov� rovnice AX = B

* Je-li $\A$ regul�rn� matice, pak rovnice m� jedin� �e�en�
  $\X = \A^{-1}\,\B$.

* Jinak sta�� matice $\X$ a $\B$ rozepsat do sloupc�:
  $$
    \X = (\X_1\ \X_2\ \ldots\ \X_k), \quad
    \B = (\B_1\ \B_2\ \ldots\ \B_k),
  $$
  tak�e maticov� rovnice p�ech�z� na $k$ soustav line�rn�ch rovnic
  $$
    \A\,\X_1=\B_1, \quad \A\,\X_2 = \B_2, \quad \ldots \A\,\X_k=\B_k.
  $$
  Tyto soustavy maj� spole�nou matici soustavy, tj. spole�nou
  p�idru�enou homogenn� soustavu, tj. spole�nou mno�inu $M_0$ v�ech
  �e�en� p�idru�en� homogenn� soustavy. Pro ka�dou soustavu zvlṻ je
  t�eba spo��tat partikul�rn� �e�en�.

* Mno�ina v�ech �e�en� je mno�na v�ech matic $\X$, kter� maj� ve sloupc�ch
  odpov�daj�c� partikul�rn� �e�en�, ke kter�m je v ka�d�m sloupci
  (nez�visle) p�i�tena mno�ina �e�en� $M_0$.

\end
