Algebra Linear
Författare:
Azuaite Schneider
Last Updated:
för 6 år sedan
Licens:
Creative Commons CC BY 4.0
Sammanfattning:
Algebra Linear
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Algebra Linear
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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}