Algebra Linear
Forfatter:
Azuaite Schneider
Sidst opdateret:
6 år siden
Licens:
Creative Commons CC BY 4.0
Resumé:
Algebra Linear
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
Algebra Linear
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
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\title[\'Algebra Linear]{\'{A}lgebra Linear } % The short title appears at the bottom of every slide, the full title is only on the title page
\author{Azuaite A. Schneider} % Your name
\institute[IFPR] % Your institution as it will appear on the bottom of every slide, may be shorthand to save space
{
Instituto Federal do Paran\'a - Campus de Paranava\'i \\ % Your institution for the title page
\medskip
\textit{azuaite.schneider@ifpr.edu.br} % Your email address
}
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\newtheorem{defi}{Defini\c{c}\~{a}o}
\newtheorem{teo}{Teorema}
\newtheorem{lema}{Lema}
\newtheorem{prop}{Proposi\c{c}\~{a}o}
\newtheorem{exemplo}{Exemplo}
\newtheorem{corolario}{Corol\'{a}rio}
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\begin{document}
\begin{frame}
\titlepage % Print the title page as the first slide
\end{frame}
\section{C\'{a}lculo do Determinante}
\frame{
\frametitle{C\'{a}lculo do Determinante}
\begin{itemize}
\item<1-> Ordem 1:
$$ \mathbf{A}=\left(\begin{array}{c}
a_{11}
\end{array}\right)_{1\times 1} \quad\Rightarrow\quad\quad det(A)=a_{11} $$
\item<2-> Ordem 2:
$$ \mathbf{A}=\left(\begin{array}{cc}
a_{11}&a_{12}\\
a_{21}&a_{22}
\end{array}\right)_{2\times 2} \quad \Rightarrow\quad\quad det(A)=a_{11}a_{22}-a_{12}
a_{21} $$
\item<3-> Ordem 3:
$$\mathbf{A}=
\left( \begin{array}{ccc}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{array} \right)_{3\times 3} \quad \Rightarrow \quad det(A)=\left| \begin{array}{ccc}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{array} \right|$$
$$=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32} $$
\end{itemize}
}
%-------------------------------------------------%
\begin{frame}
\frametitle{Exemplos}
\begin{block}{Exemplo 1}
$$ \mathbf{A}=\left(\begin{array}{c}
-6
\end{array}\right)_{1\times 1}$$
\end{block}
$$\Rightarrow\quad\quad det(A)=-6 $$
\begin{block}{Exemplo 2}
$$ \mathbf{A}=\left(\begin{array}{cc}
1&3\\
-2&-4
\end{array}\right)_{2\times 2}$$
\end{block}
$$\quad \Rightarrow\quad\quad det(A)=1(-4)-(-2)
3 = -4+6=2$$
\end{frame}
\frame{
\frametitle{}
\begin{block}{Exemplo 3}
$$\mathbf{A}=
\left( \begin{array}{ccc}
1 & 3 & -1\\
5 & 2 & 0\\
4 & -2 & -3
\end{array} \right)_{3\times 3} \quad $$
\end{block}
$$\Rightarrow \quad \quad det(A)=\left| \begin{array}{ccc}
1 & 3 & -1\\
5 & 2 & 0\\
4 & -2 & -3
\end{array} \right|\begin{array}{cc}
1 & 3\\
5 & 2\\
4 & -2
\end{array}=63-6=57.$$
}
%-------------------------------------------------%
\begin{frame}
\frametitle{Desenvolvimento de Laplace}
Vimos que
$$ |A|=\left| \begin{array}{ccc}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{array} \right| $$
$$ \quad $$
$$=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31} $$
$$=a_{11}(a_{22}a_{33}-a_{23}a_{32})+a_{12}(a_{23}a_{31}-a_{21}a_{33})+a_{13}(a_{21}a_{32}-a_{22}a_{31}) $$
$$ \quad $$
$$=a_{11}\left|\begin{array}{cc}
a_{22}&a_{23}\\
a_{32}&a_{33}
\end{array}\right|+a_{12}\left|\begin{array}{cc}
a_{21}&a_{23}\\
a_{31}&a_{33}
\end{array}\right|+a_{13}\left|\begin{array}{cc}
a_{21}&a_{22}\\
a_{31}&a_{32}
\end{array}\right| $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Observe que o determinante da matriz inicial $ 3\times 3 $ pode ser expressa em fun\c{c}\~{a}o dos determinantes de submatrizes $ 2\times 2 $, isto \'{e}\\
$ \quad $\\
$ detA=a_{11}|A_{11}|-a_{12}|A_{12}|+a_{13}|A_{13}|$, onde $ A_{ij} $ \'{e} a submatriz da inicial, de onde a i-\'{e}sima linha e a j-\'{e}sima coluna foram retiradas. Al\'{e}m disso, se chamarmos
\begin{center}
$ \Delta_{ij}=(-1)^{i+j}|A_{ij}| $, obtemos
\end{center}
$$ detA=a_{11}\Delta_{11}+a_{12}\Delta_{12}+a_{13}\Delta_{13}$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{Caso Geral $ n\times n $}
Se $ A $ \'{e} uma matriz quadrada $ n\times n $, ent\~{a}o
$$ detA=a_{i1}\Delta_{i1}+a_{i2}\Delta_{i2}+\cdots+a_{in}\Delta_{in}=\sum_{j=1}^{n}{a_{ij}\Delta_{ij}}$$
onde $ \Delta_{ij} $\'{e} chamado cofator de $ a_{ij} $.
\begin{block}{Exemplo}
Calcule o determinante:
$$ \mathbf{|A|}=
\left| \begin{array}{cccc}
-1 & 2 & 3 & -4\\
4 & 2 & 0 & 0\\
-1 &2 & -3 &0\\
2& 0&5&1
\end{array} \right| $$
\end{block}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{Solu\c{c}\~{a}o:}
$ detA= a_{21}\Delta_{21}+a_{22}\Delta_{22}+a_{23}\Delta_{23}+a_{24}\Delta_{24}= 4\Delta_{21}+2\Delta_{22}+0\Delta_{23}+0\Delta_{24} $\\
$ \quad $\\
$ \Delta_{21}=(-1)^{2+1} \left| \begin{array}{ccc}
2 & 3 & -4\\
2 & -3 &0\\
0&5&1
\end{array} \right|=(-1)^3.(-52)=52$
$$ \quad $$
$ \Delta_{22}=(-1)^{2+2}\left| \begin{array}{ccc}
-1 & 3 & -4\\
-1 & -3 &0\\
2&5&1
\end{array} \right|=(-1)^4.2=2 $\\
$$ \therefore detA=4.52+2.2=212 $$
\end{frame}
%-------------------------------------------------%
\section{Matriz Adjunta e Matriz Inversa}
\begin{frame}
\frametitle{Matriz Adjunta e Matriz Inversa}
\begin{defi}
Matriz dos cofatores de $ A $ \'{e} a formada pelos cofatores $ \Delta_{ij} $ da matriz $ A $.
$$ \underbrace{\bar{\mathrm{A}}}_{\mathsf{Matriz \:dos\: cofatores}}=(\Delta_{ij})_{\mathsf{ordem\,igual\,a\,de\,A}} $$
\end{defi}
\begin{block}{Exemplo}
Determine a matriz dos cofatores da matriz A.
$$\mathbf{A}=
\left( \begin{array}{ccc}
2 & 1 & 0\\
-3 & 1 & 4\\
1 & 6 & 5
\end{array} \right)$$
\end{block}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{Solu\c{c}\~{a}o}
$ \bar{\mathrm{A}}=(\Delta_{ij})_{3\times 3}=
\left( \begin{array}{ccc}
\Delta_{11} & \Delta_{12} & \Delta_{13}\\
\Delta_{21} & \Delta_{22} & \Delta_{23}\\
\Delta_{31} & \Delta_{32} & \Delta_{33}
\end{array} \right) $
$$ \quad $$
$ \Delta_{11}=(-1)^{1+1}\left|\begin{array}{cc}
1 & 4\\
6 & 5
\end{array}\right|=1.(5-24)=-19$
$$ \quad $$
$ \Delta_{12}=(-1)^{1+2}\left|\begin{array}{cc}
-3 &1 \\
4&5
\end{array}\right|=-1.(-15-4)=19$
$$ \quad $$
$ \Delta_{13}=(-1)^{1+3}\left|\begin{array}{cc}
-3 & 1\\
1 & 6
\end{array}\right|=1.(-18-1)=-19$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
$ \Delta_{21}=(-1)^{2+1}\left|\begin{array}{cc}
1& 0\\
6& 5
\end{array}\right|=-1.(5)=5$
$$ \quad $$
$ \Delta_{22}=(-1)^{2+2}\left|\begin{array}{cc}
2 &0 \\
1& 5
\end{array}\right|=1.(10)=10$
$$ \quad $$
$ \Delta_{23}=(-1)^{2+3}\left|\begin{array}{cc}
2& 1\\
1& 6
\end{array}\right|=-1.(12-1)=-11$
$$ \quad $$
$ \Delta_{31}=(-1)^{3+1}\left|\begin{array}{cc}
1& 0\\
1& 4
\end{array}\right|=1.(4)=4$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
$ \Delta_{32}=(-1)^{3+2}\left|\begin{array}{cc}
2 & 0\\
-3 &4
\end{array}\right|=-1.(8)=-8$
$$ \quad $$
$ \Delta_{33}=(-1)^{3+3}\left|\begin{array}{cc}
2 & 1\\
-3 & 1
\end{array}\right|=1.(2+3)=5$
$$ \quad $$
$ \bar{\mathrm{A}}=(\Delta_{ij})_{3\times 3}=
\left( \begin{array}{ccc}
-19 & 19 & -19\\
-5 & 10 & -11\\
4 & -8 & 5
\end{array} \right) $
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{defi}
Dada uma matriz quadrada A, chamamos de matriz adjunta de A \`{a} transposta da matriz dos cofatores de A.
$$ Adj (A)=( \bar{\mathrm{A}} )^t $$
\end{defi}
\begin{block}{Exemplo}
Pelo exemplo anterior, temos que a matriz adjunta de A \'{e} dada por:\\
$$ Adj (A)=( \bar{\mathrm{A}} )^t=
\left( \begin{array}{ccc}
-19 &-5 & 4\\
19& 10 &-8 \\
-19& -11 & 5
\end{array} \right)_{3\times 3}$$
\end{block}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Calculando agora $ A. adj(A) temos:$
\begin{center}
$$\left( \begin{array}{ccc}
2 &1 & 0\\
-3& 1 &4 \\
1& 6 & 5
\end{array} \right)
\left( \begin{array}{ccc}
-19 &-5 & 4\\
19& 10 &-8 \\
-19& -11 & 5
\end{array} \right)$$
$$= \left( \begin{array}{ccc}
-38+19 &-10+10 & 8-8\\
57+19-76& 15+10-44 &-12-8+20 \\
-19+144-95& -5+60-55 & 4-48+25
\end{array} \right)$$
$$=\left( \begin{array}{ccc}
-19 &0 & 0\\
0& -19 &0\\
0& 0 & -19
\end{array} \right)=(-19).I_3$$
\end{center}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Como\\
$ detA=\left| \begin{array}{ccc}
2 &1 & 0\\
-3& 1 &4 \\
1& 6 & 5
\end{array} \right|=-19 $\\
Temos que\\
$ \quad $\\
$ A. adj (A)= det(A). I $
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{teo}
Se A \'{e} uma matriz quadrada, ent\~{a}o:
$$ A. adj (A)= det(A). I_n $$
\end{teo}
\textbf{Demonstra\c{c}\~{a}o:} Sendo
$$ A. adj(A)=(c_{uv})_{n\times n} $$
temos que
$$ c_{uv}=\sum_{k=1}^{n}{a_{uk}.\Delta_{vk}}\footnote{A $ adj(A) $ \'{e} a transposta da matriz dos cofatores, portanto, temos $ \Delta_{vk} $ e n\~{a}o $ \Delta_{kv} $.} $$
Assim, se
$$ u=v \Rightarrow c_{uu}=\sum_{k=1}^{n}{a_{uk}.\Delta_{uk}} $$
$$ c_{uu}=a_{u1}\Delta_{u1}+a_{u2}\Delta_{u2}+\cdots+a_{un}\Delta_{un}=det A\footnote{Pelo desenvolvimento de Laplace.} $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Se $ u\neq v $
$$ \Rightarrow c_{uv}=\sum_{k=1}^{n}{a_{uk}.\Delta_{vk}} $$
$$ = a_{u1}\Delta_{v1}+a_{u2}\Delta_{v2}+\cdots+a_{un}\Delta_{vn}=det B=0 $$
onde B \'{e} uma matriz obtida a partir da substitui\c{c}\~{a}o da linha $ v $ pela linha $ u $ da matriz A. Ent\~{a}o as linhas $ u $ e $ v $ da matriz B s\~{a}o id\^{e}nticas, logo pela propriedade P5 da aula anterior temos que $ det B=0 $.\\
$ \quad $\\
Assim,\\
$ c_{uv}=det A $, se $ u=v $ e\\
$ c_{uv}=0 $, se $ u\neq v $.
$$ \therefore A. adj(A)=(c_{uv})_n=(det A).I_n\;\;\;\;\quad\qquad \blacksquare$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{defi}
Dada uma matriz quadrada A de ordem $ n $, chamamos de inversa de A a uma matriz B tal que
$$ AB=BA=I_n $$
onde $ I_n $ \'{e} a matriz identidade de ordem $ n $. Escrevemos $ A^{-1} $ para a inversa de A.
\end{defi}
\begin{block}{Observa\c{c}\~{a}o}
Vimos anteriormente, que
$$ A. adj(A)=(det A).I_n$$
Ent\~{a}o $ A.\dfrac{adjA}{detA}=I_n $ se $ detA\neq 0 $.Deste modo,
$$ \dfrac{1}{det A}.adjA=A^{-1}. $$
\end{block}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{teo}
Uma matriz quadrada A admite inversa se, e somente se, $ detA\neq 0 $.
\end{teo}
\textbf{Demonstra\c{c}\~{a}o:} \\
$$ \quad $$
$(\Leftarrow)$ Observa\c{c}\~{a}o anterior.
$$ \quad $$
$ (\Rightarrow) $ Se A admite inversa, ent\~{a}o existe uma matriz B tal que,
$$ A.B=I_n $$
$$\Rightarrow det(AB)=det(I_n) $$
$$ detA.detB=1 $$
$$ \qquad\qquad\quad\quad\qquad detA\neq 0 \qquad\quad\quad\quad\quad\blacksquare$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{defi}
Dizemos que uma matriz A \'{e} singular se $ detA=0 $ e \'{e} n\~{a}o-singular se $ detA\neq 0 $.
\end{defi}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{block}{Exemplo}
Vimos no \'{u}ltimo exemplo que a adjunta da matriz $A= \left( \begin{array}{ccc}
2 &1 & 0\\
-3& 1 &4 \\
1& 6 & 5
\end{array} \right)$ \'{e} a matriz $ adj A=\left( \begin{array}{ccc}
-19 &-5 & 4\\
19& 10 &-8 \\
-19& -11 & 5
\end{array} \right) $ e que o determinante de A \'{e} $detA=-19 $. Assim pela \'{u}ltima observa\c{c}\~{a}o temos que a inversa de A existe e \'{e} dada por
$$A^{-1}=\dfrac{1}{det A}.adjA = \left( \begin{array}{ccc}
1 &5/19 & -4/19\\
-1& -10/19 &8/19 \\
1& 11/19 & -5/19
\end{array} \right)$$
\end{block}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{block}{Exemplo}
Calcule a inversa da matriz $ A=\left(\begin{array}{cc}
a&b\\
c&d
\end{array}\right) $
\end{block}
\textbf{Solu\c{c}\~{a}o}\\
$$ detA=ad-bc \textsf{ e } \bar{A}=\left(\begin{array}{cc}
\Delta_{11}&\Delta_{12}\\
\Delta_{21}&\Delta_{22}
\end{array}\right) \textsf{, logo }$$
$$ \Delta_{11}=(-1)^{1+1}|d|=d$$
$$\Delta_{12}=(-1)^{1+2}|c|=-c$$
$$\Delta_{21}=(-1)^{2+1}|b|=-b$$
$$\Delta_{22}=(-1)^{2+2}|a|=a $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
$$ \therefore \bar{A} = \left(\begin{array}{cc}
d&-c\\
-b&a
\end{array}\right)$$
Assim,
$$ adjA=(\bar{A})^t= \left(\begin{array}{cc}
d&-b\\
-c&a
\end{array}\right)$$
$$\quad$$
$$ A^{-1}=\dfrac{1}{detA}.adjA$$
$$\quad$$
$$ A^{-1}=\left(\begin{array}{cc}
\dfrac{d}{detA}&\dfrac{-b}{detA}\\
&\\
\dfrac{-c}{detA}&\dfrac{a}{detA}
\end{array} \right) $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{block}{Sub-exemplo}
Determine a inversa da matriz
$$ A=\left(\begin{array}{cc}
1&-2\\
3&-1
\end{array} \right) $$
\end{block}
\textbf{Solu\c{c}\~{a}o:}\\
$$ A^{-1}=\left(\begin{array}{cc}
-1/5&2/5\\
-3/5&1/5
\end{array} \right) $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{Propriedades}
\begin{block}{Propriedade 1}
Se A e B s\~{a}o matrizes quadradas de mesma ordem, ambas invers\'{i}veis, ent\~{a}o AB tamb\'{e}m \'{e} invers\'{i}vel.
\end{block}
De fato,\\
$$ AB.X=I_n $$
$$ (A^{-1}).ABX=A^{-1}\cdot I_n $$
$$ BX=A^{-1} $$
$$ B^{-1}BX= B^{-1}A^{-1 } $$
$$ X= B^{-1}A^{-1 } $$
$$\Rightarrow (AB)^{-1}=B^{-1}A^{-1}$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{block}{Propriedade 2}
Nem toda matriz tem inversa (basta ser singular, ou seja, $ detA=0 $).
\end{block}
\begin{block}{Exemplo}
$$ A=\left(\begin{array}{cc}
1&2\\
2&4
\end{array} \right) \Rightarrow detA=4-4=0$$
$$\therefore A^{-1} \textsf{ n\~{a}o existe!}$$
\end{block}
Observe que se $ \left(\begin{array}{cc}
1&2\\
2&4
\end{array} \right)
\left(\begin{array}{cc}
x&y\\
z&w
\end{array} \right)
= \left(\begin{array}{cc}
1&0\\
0&1
\end{array} \right)$\\
$ \quad $\\
Ter\'{i}amos que $\left\{\begin{array}{ccc}
x+2z=1&\quad\quad& 2(x+2z=0)\\
2x+4z=0&\quad\quad& \quad \quad\quad \Rightarrow 2.1=0\quad (Absurdo!)\\
y+2w=0&\quad\quad&2(y+2w)=1\\
2y+4w=1&\quad\quad&\quad\quad\quad\Rightarrow2.0=1 \quad (Absurdo!)
\end{array}\right.$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{block}{Propriedade 3}
Se A tem inversa, ent\~{a}o $ detA^{-1}=\dfrac{1}{detA}$\\
\end{block}
De fato,\\
$$ A.A^{-1}=I_n $$
$$ det(A.A^{-1})=det(I_n) $$
$$ detA.detA^{-1} =1$$
$$ detA^{-1} =\dfrac{1}{detA} \quad \quad pois \quad detA\neq 0 $$
\end{frame}
%-------------------------------------------------%
\section{Regra de Cramer}
\begin{frame}
\frametitle{Regra de Cramer}
O c\'{a}lculo da inversa de uma matriz fornece um outro m\'{e}todo de resolu\c{c}\~{a}o de sistemas lineares de equa\c{c}\~{o}es.\\
Este s\'{o} se aplica a sistemas lineares em que o n\'{u}mero de equa\c{c}\~{o}es \'{e} igual ao n\'{u}mero de inc\'{o}gnitas.\\
Suponha que queremos resolver o seguinte sistema linear.\\
$$(*)=\left\{\begin{array}{ccccccccc}
a_{11}x_{1}&+&a_{12}x_{2}&+& \cdots&+&a_{1n}x_{n}&=&b_{1}\\
a_{21}x_{1}&+&a_{22}x_{2}&+&\cdots &+&a_{2n}x_{n}&=&b_{2}\\
\vdots&\quad&\vdots&\quad&\quad&\quad&\vdots&\quad&\vdots\\
a_{m1}x_{1}&+&a_{m2}x_{2}&+&\cdots &+&a_{mn}x_{n}&=&b_{m}
\end{array}\right.$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Na forma matricial temos:
\begin{equation}
AX=B \nonumber
\end{equation}
Suponha que $detA\neq 0$\\
Ent\~{a}o existe $A^{-1}$ tal que $A^{-1}.A=I_n$.\\
Assim,
$$ X=A^{-1}B \quad \quad \Rightarrow \quad X=\dfrac{adjA}{detA}B $$
Logo,\\
$$ \left(\begin{array}{c}
x_1\\
x_2\\
\vdots\\
x_n
\end{array}\right) = \dfrac{1}{detA}.\left(\begin{array}{cccc}
\Delta_{11}&\Delta_{21}&\cdots&\Delta_{n1}\\
\Delta_{12}&\Delta_{22}&\cdots&\Delta_{n2}\\
\vdots&\vdots&\ddots&\vdots\\
\Delta_{1n}&\Delta_{2n}&\cdots&\Delta_{nn}
\end{array}\right).\left(\begin{array}{c}
b_1\\
b_2\\
\vdots\\
b_n
\end{array}\right)$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Desse modo,
$$ x_1=\dfrac{b_1\Delta_{11}+b_2\Delta_{21}+\cdots b_n\Delta_{n1}}{detA} $$
$$ x_2=\dfrac{b_1\Delta_{12}+b_2\Delta_{22}+\cdots b_n\Delta_{n2}}{detA} $$
$$\quad\quad \vdots $$
$$ x_n=\dfrac{b_1\Delta_{1n}+b_2\Delta_{2n}+\cdots b_n\Delta_{nn}}{detA} $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Ou seja,
$$ x_1=\left|\begin{array}{cccc}
b_{1}&a_{12}&\cdots&a_{1n}\\
b_{2}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
b_{n}&a_{n2}&\cdots&a_{nn}
\end{array}\right|.\frac{1}{detA} $$
$$ x_2=\left|\begin{array}{cccc}
a{11}&b_{1}&\cdots&a_{1n}\\
a_{21}&b_{2}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{n1}&b_{n}&\cdots&a_{nn}
\end{array}\right|.\dfrac{1}{detA} $$
$$\quad\quad \vdots $$
$$ x_n=\left|\begin{array}{cccc}
a_{11}&a_{12}&\cdots&b_{1}\\
a_{21}&a_{22}&\cdots&b_{2}\\
\vdots&\vdots&\ddots&\vdots\\
a_{n1}&a_{n2}&\cdots&b_{n}
\end{array}\right|.\dfrac{1}{detA} $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Portanto, podemos generalizar a solu\c{c}\~{a}o de cada inc\'{o}gnita atrav\'{e}s da express\~{a}o:
\begin{center}
\fbox{$ x_i=\dfrac{detA_i}{detA}, 1\leq i\leq n $}
\end{center}
Onde $A_i$ \'{e} a matriz obtida pela troca da coluna $i$ da matriz A pela coluna 1 da matriz B.
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{block}{Exemplo}
Determine a solu\c{c}\~{a}o do sistema
$ \left\{\begin{array}{ccccccc}
2x&-&3y&+&7z&=&1\\
x&+&&&3z&=&5\\
&&2y&-&z&=&0
\end{array}\right. $
\end{block}
\textbf{Solu\c{c}\~{a}o:}\\
$$A= \left( \begin{array}{ccc}
2 & -3 & 7 \\
1 & 0 & 3 \\
0 & 2 & -1
\end{array} \right)\textsf{ e } B=\left(\begin{array}{c}
1\\
5\\
0
\end{array}\right) $$
$$ detA=\left|\begin{array}{ccc}
2 & -3 & 7 \\
1 & 0& 3 \\
0 & 2 & -1
\end{array} \right| = -1 \quad\quad\quad detA_1=\left|\begin{array}{ccc}
1 & -3 & 7 \\
5 & 0 & 3 \\
0 & 2 & -1
\end{array} \right|=49 $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
$$ detA_2=\left|\begin{array}{ccc}
2&1&7\\
1&5&3\\
0&0&-1
\end{array}\right|=-9 \quad \quad \quad det A_3=\left|\begin{array}{ccc}
2&-3&1\\
1&0&5\\
0&2&0
\end{array}\right|=-18$$
$$ \quad $$
$$ x=\dfrac{detA_1}{detA}=\dfrac{49}{-1}=-49 $$
$$y=\dfrac{detA_2}{detA}=\dfrac{-9}{-1}=9 $$
$$z=\dfrac{detA_3}{detA}=\dfrac{-18}{-1}=18.$$
\end{frame}
%-------------------------------------------------%
\section{Matrizes Elementares}
\begin{frame}
\frametitle{Matrizes Elementares}
\begin{block}{Exemplo}
Considere a matriz A, onde $ A=\left(\begin{array}{ccc}
1 & 0 & 2 \\
2 & -1 & 3 \\
5 & 1 & 4
\end{array} \right) $.
\end{block}
Temos que
$$ A'=\begin{array}{c}
L_1\\
2L_2\\
L_3
\end{array}\left(\begin{array}{ccc}
1 & 0 & 2 \\
4 & -2 & 6 \\
5 & 1 & 4
\end{array} \right)=\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 1
\end{array} \right)\left(\begin{array}{ccc}
1 & 0 & 2 \\
2 & -1 & 3 \\
5 & 1 & 4
\end{array} \right)$$
%---------------------------------------------------------------
$$ A''=\begin{array}{c}
L_1\\
L_3\\
L_2
\end{array}\left(\begin{array}{ccc}
1 & 0 & 2 \\
5 & 1 & 4 \\
2 & -1 & 3
\end{array} \right)=\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array} \right)\left(\begin{array}{ccc}
1 & 0 & 2 \\
2 & -1 & 3 \\
5 & 1 & 4
\end{array} \right)$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
$$ A'''=\begin{array}{c}
L_1\\
L_2\\
2L_1-L_3
\end{array}\left(\begin{array}{ccc}
1 & 0 & 2 \\
2 & -1 & 3 \\
-3 & -1 & 0
\end{array} \right)=\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
2 & 0 & -1
\end{array} \right)\left(\begin{array}{ccc}
1 & 0 & 2 \\
2 & -1 & 3 \\
5 & 1 & 4
\end{array} \right)$$
\begin{defi}
Uma matriz elementar \'{e} uma matriz obtida a partir da identidade, atrav\'{e}s da aplica\c{c}\~{a}o de uma opera\c{c}\~{a}o elementar com linhas.
\end{defi}
\begin{teo}
Seja A uma matriz qualquer, o resultado da aplica\c{c}\~{a}o de uma opera\c{c}\~{a}o elementar com as linhas de A \'{e} o mesmo que o resultado da multiplica\c{c}\~{a}o da matriz elementar E correspondente \`{a} opera\c{c}\~{a}o com as linhas pela matriz A.
\end{teo}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{corolario}
Uma matriz elementar $E_1$ \'{e} invers\'{i}vel e sua inversa \'{e} a matriz elementar $E_2$ que corresponde \`{a} opera\c{c}\~{a}o com linhas inversa da opera\c{c}\~{a}o efetuada por $E_1$.
\end{corolario}
\begin{block}{Exemplo}
Determine a matriz elementar $ E_2 $, inversa da matriz elementar
$$ E_1=\left(\begin{array}{ccc}
1&0&0\\
0&2&0\\
0&0&1\\
\end{array}\right) $$
\end{block}
Ent\~{a}o, $ detE_1=2\neq 0 \quad \Rightarrow \quad E_1$ admite inversa (n\~{a}o-singular).
\end{frame}
%-------------------------------------------------%
%-------------------------------------------------%
\begin{frame}
\frametitle{}
Assim,
$$ E_1.E_2=I_n \Leftrightarrow \left(\begin{array}{ccc}
1&0&0\\
0&2&0\\
0&0&1
\end{array}\right)\left(\begin{array}{ccc}
a&b&c\\
d&e&f\\
g&h&i
\end{array}\right)=\left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}\right) $$
$$ \Rightarrow \left(\begin{array}{ccc}
a&b&c\\
2d&2e&2f\\
g&h&i
\end{array}\right)=\left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}\right) \Rightarrow \begin{array}{ccc}
a=1&d=0&g=0\\
b=0&e=1/2&h=0\\
c=0&f=0&i=1
\end{array} $$
$$ \therefore E_2=\left(\begin{array}{ccc}
1&0&0\\
0&1/2&0\\
0&0&1
\end{array}\right) $$
\end{frame}
\begin{frame}
\frametitle{}
\begin{block}{Exemplo}
Determine a matriz elementar $ E_2 $, inversa da matriz elementar
$$ E_1=\left(\begin{array}{ccc}
1&0&0\\
0&0&1\\
0&1&0
\end{array}\right) $$
\end{block}
$ detE_1=-1\neq 0 \Rightarrow E_1 $ admite inversa.\\
Assim,
$$ E_1.E_2=I_n \Leftrightarrow \left(\begin{array}{ccc}
1&0&0\\
0&0&1\\
0&1&0
\end{array}\right)\left(\begin{array}{ccc}
a&b&c\\
d&e&f\\
g&h&i
\end{array}\right)=\left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}\right) $$
$$ \Rightarrow \left(\begin{array}{ccc}
a&b&c\\
g&h&i\\
d&e&f
\end{array}\right)=\left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}\right) \Rightarrow \begin{array}{c}
a=f=h=1\\
b=c=g=i=d=e=0
\end{array}$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
$$ \therefore E_2=\left(\begin{array}{ccc}
1&0&0\\
0&0&1\\
0&1&0
\end{array}\right) $$
\begin{block}{Exemplo}
Determine a matriz elementar $ E_2 $, inversa da matriz elementar
$$ E_1=\left(\begin{array}{ccc}
1&0&0\\
x&-y&0\\
0&0&1
\end{array}\right)\begin{array}{c}
\quad\\
xL_1-yL_2\\
\quad
\end{array} $$
\end{block}
$ detE_1=-y\neq 0, \Rightarrow E_1 $ admite inversa.\\
Assim,
$$ E_1.E_2=I_n \Leftrightarrow \left(\begin{array}{ccc}
1&0&0\\
x&-y&0\\
0&0&1
\end{array}\right)\left(\begin{array}{ccc}
a&b&c\\
d&e&f\\
g&h&i
\end{array}\right)=\left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}\right) $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
$$ \Rightarrow \left(\begin{array}{ccc}
a&b&c\\
ax-dy&bx-ey&cx-fy\\
g&h&i
\end{array}\right)=\left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}\right)$$
$$ \Rightarrow \begin{array}{ccc}
a=1&ax-dy=0 \Rightarrow d=x/y&g=0\\
b=0& bx-ey=1 \Rightarrow e=-1/y &h=0\\
c=0& cx-fy=0 \Rightarrow f=0 &i=1
\end{array} $$
$$ \therefore E_2=\left(\begin{array}{ccc}
1&0&0\\
x/y&-1/y&0\\
0&0&1
\end{array}\right) \begin{array}{c}
\quad\\
(xL_1-L_2)/y\\
\,
\end{array}$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{block}{$L'_i=kL_i$}
$$L'_i=kL_i \quad\quad \begin{array}{c}
\textsf{Inv} \\
\rightsquigarrow
\end{array} \quad \quad L_i=kL'_i \quad \quad \Rightarrow \quad \quad L'_i=\dfrac{L_i}{k}$$
\end{block}
\begin{block}{$L'_i=L_j$}
$$ \begin{array}{ccccc}
L'_i=L_j&\quad\quad&\mathsf{Inv}&\quad\quad&L_i=L'_j\\
L'_j=L_i&\quad\quad&\rightsquigarrow&\quad\quad&L_j=L'_i
\end{array} $$
\end{block}
\begin{block}{$L'_i=xL_j-yL_i$}
$$ \begin{array}{ccccc}
L'_i=xL_j-yL_i & \quad & \mathsf{Inv}& \quad & L_i=xL_j-yL'_i\\
\quad & \quad & \rightsquigarrow & \quad & yL'_i=xL_j-L_i \Rightarrow L'_i=(xL_j-L_i)/y
\end{array} $$
\end{block}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{block}{Exemplo}
Determine a inversa da matriz elementar dada
$$ E_1=\left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 3 & 0 & -4
\end{array} \right) $$
\end{block}
\textbf{Solu\c{c}\~{a}o:}
$$ L'_4=3L_2-4L_4 \rightsquigarrow L_4=3L_2-4L'_4 \Rightarrow -4L'_4=L_4-3L_2 \Rightarrow L'_4=\dfrac{3L_2-L_4}{4} $$
$$ E_2= \left(\begin{array}{cccc}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&3/4&0&-1/4
\end{array}\right)$$
\end{frame}
%-------------------------------------------------%
\section{Procedimento para a invers\~{a}o de uma Matriz}
\begin{frame}
\frametitle{Procedimento para a invers\~{a}o de uma Matriz}
\begin{teo}
Se A \'{e} uma matriz invers\'{i}vel, sua matriz-linha reduzida \`{a} forma escalonada, \'{e} a identidade. Al\'{e}m disso, \'{e} dada por um produto de matrizes elementares.\\
\end{teo}
De acordo com o teorema acima temos que
$$ I_n=A.E_1E_2\dots E_{n-1}E_n $$
$$\Rightarrow E_1E_2\dots E_n=A^{-1} $$
$$\Rightarrow E_nE_{n-1}\dots E_2 E_1.I=A^{-1} $$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\begin{teo}
Se uma matriz A pode ser reduzida \`{a} matriz identidade, por uma sequência de opera\c{c}\~{o}es elementares com linhas, ent\~{a}o A \'{e} invers\'{i}vel e a matriz inversa de A \'{e} obtida a partir da matriz identidade, aplicando-se a mesma sequ\^{e}ncia de opera\c{c}\~{o}es com linhas.
\end{teo}
\begin{block}{Exemplo}
Determine, se existir, a inversa da matriz:
$ A= \left(\begin{array}{cccc}
2&1&0&0\\
1&0&-1&1\\
0&1&1&1\\
-1&0&0&3
\end{array}\right)$
\end{block}
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
\textbf{Solu\c{c}\~{a}o:}
$$ \begin{array}{c}
L_1\\
L_2\\
L_3\\
L_4
\end{array} \left(\begin{array}{cccc}
2&1&0&0\\
1&0&-1&1\\
0&1&1&1\\
-1&0&0&3
\end{array}\begin{array}{c}
\vdots\\
\vdots\\
\vdots
\end{array} \begin{array}{cccc}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{array}\right)$$
$$ \quad $$
1a Etapa (Piv\^{o} $ a_{11} $)
$$ \begin{array}{c}
L_1\\
L_1-2L_2\\
L_3\\
L_1+2L_4
\end{array} \left(\begin{array}{cccc}
2&1&0&0\\
0&1&2&-2\\
0&1&1&1\\
0&1&0&6
\end{array}\begin{array}{c}
\vdots\\
\vdots\\
\vdots
\end{array} \begin{array}{cccc}
1&0&0&0\\
1&-2&0&0\\
0&0&1&0\\
1&0&0&2
\end{array}\right)$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
2a Etapa (Piv\^{o} $a_{22} $)
$$ \begin{array}{c}
L_2-L_1\\
L_2\\
L_2-L_3\\
L_2-L_4
\end{array} \left(\begin{array}{cccc}
-2&0&2&-2\\
0&1&2&-2\\
0&0&1&-3\\
0&0&2&-8
\end{array}\begin{array}{c}
\vdots\\
\vdots\\
\vdots
\end{array} \begin{array}{cccc}
0&-2&0&0\\
1&-2&0&0\\
1&-2&-1&0\\
0&-2&0&-2
\end{array}\right)$$
3a Etapa (Piv\^{o} $ a_{33} $)
$$ \begin{array}{c}
2L_3-L_1\\
2L_3-L_2\\
L_3\\
2L_3-L_4
\end{array} \left(\begin{array}{cccc}
2&0&0&-4\\
0&1&0&-4\\
0&0&1&-3\\
0&0&0&2
\end{array}\begin{array}{c}
\vdots\\
\vdots\\
\vdots
\end{array} \begin{array}{cccc}
2&-2&-2&0\\
1&-2&-2&0\\
1&-2&-1&0\\
2&-2&-2&2
\end{array}\right)$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
4a Etapa (Piv\^{o} $ a_{44} $)
$$ \begin{array}{c}
2L_4+L_1\\
2L_4+L_2\\
3L_4+2L_3\\
L_4
\end{array} \left(\begin{array}{cccc}
2&0&0&0\\
0&-1&0&0\\
0&0&2&0\\
0&0&0&2
\end{array}\begin{array}{c}
\vdots\\
\vdots\\
\vdots
\end{array} \begin{array}{cccc}
6&-6&-6&4\\
5&-6&-6&4\\
8&-10&-8&6\\
2&-2&-2&2
\end{array}\right)$$
5a Etapa ($ I_n $)
$$ \begin{array}{c}
L_1/2\\
L_2/-1\\
L_3/2\\
L_4/2
\end{array} \left(\begin{array}{cccc}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{array}\begin{array}{c}
\vdots\\
\vdots\\
\vdots
\end{array} \begin{array}{cccc}
3&-3&-3&2\\
-5&6&6&-4\\
4&-5&-4&3\\
1&-1&-1&1
\end{array}\right)$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{}
$$ A^{-1=}\left(\begin{array}{cccc}
3&-3&-3&2\\
-5&6&6&-4\\
4&-5&-4&3\\
1&-1&-1&1
\end{array}\right)$$
\end{frame}
%-------------------------------------------------%
\begin{frame}
\frametitle{Exerc\'{i}cio}
\begin{block}{Exerc\'{i}cio}
Determine, se existir, a inversa da matriz
$$ A= \left(\begin{array}{ccc}
1&2&0\\
-2&3&4\\
2&0&2
\end{array}\right)$$
\end{block}
\textbf{Solu\c{c}\~{a}o:}
$$ A^{-1}= \left(\begin{array}{cccc}
1/5&-2/15&4/15\\
2/5&1/15&-2/15\\
-1/5&2/15&7/30
\end{array}\right)$$
\end{frame}
\end{document}