Mensagem não lida por joaopcarv » Sáb 20 Abr, 2019 19:03
Mensagem não lida
por joaopcarv » Sáb 20 Abr, 2019 19:03
Esta questão é da segunda lista de Álgebra Linear 1 da Politécnica USP, mas no caso tem-se [tex3]\mathsf{\vec{w} \ = \ \dfrac{1}{\sqrt{6}} \cdot \bigg(2\cdot \hat{i} \ - \ \hat{j} \ \color{\red}+} \mathsf{\ \hat{k}\bigg).}[/tex3]
Supondo [tex3]\mathsf{E}[/tex3]
a base canônica [tex3]\mathsf{\hat{i} \ = \ (1,0,0), \hat{j} \ = \ (0,1,0), \hat{k} \ = \ (0,0,1)}[/tex3]
, as coordenadas dos vetores já estão dadas em relação a essa base [tex3]\mathsf{\hat{i}, \hat{j}, \ \hat{k}.}[/tex3]
Para provar que [tex3]\mathsf{F \ \in \ \mathbb{V}^3 \ = \ \{\vec{u}, \ \vec{v}, \vec{w} \}}[/tex3]
é uma base ortonormal, é necessário que:
[tex3]\mathsf{\hookrightarrow \ ||\vec{u}|| \ = \ ||\vec{v}|| \ = \ ||\vec{w}|| \ = 1;}[/tex3]
[tex3]\mathsf{||\vec{u}|| = \ \sqrt{\bigg(\dfrac{1}{\sqrt{3}}\bigg)^2 \ + \ \bigg(\dfrac{1}{\sqrt{3}}\bigg)^2 \ + \ \bigg(- \dfrac{1}{\sqrt{3}}\bigg)^2} \ = \ \boxed{\mathsf{1}}}[/tex3]
[tex3]\mathsf{||\vec{v}|| = \ \sqrt{0 \ + \ \bigg(\dfrac{1}{\sqrt{2}}\bigg)^2 \ + \ \bigg(\dfrac{1}{\sqrt{2}}\bigg)^2} \ = \ \boxed{\mathsf{1}}}[/tex3]
[tex3]\mathsf{||\vec{w}|| = \ \sqrt{\bigg(\dfrac{2}{\sqrt{6}}\bigg)^2 \ + \ \bigg(-\dfrac{1}{\sqrt{6}}\bigg)^2 \ + \ \bigg(\dfrac{1}{\sqrt{6}}\bigg)^2} \ = \ \boxed{\mathsf{1}}}[/tex3]
[tex3]\mathsf{\hookrightarrow \ \vec{u} \ \perp \ \vec{v} \ \perp \ \vec{v} \ \Leftrightarrow \ \vec{u} \bullet \vec{v} \ = \ \vec{v} \bullet \vec{w} \ = \ \vec{u} \bullet \vec{w} \ = \ 0.}[/tex3]
[tex3]\mathsf{ \vec{u} \bullet \vec{v} \ = \ \dfrac{1}{\sqrt{3}}\cdot 0 \ + \ \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{2}} \ - \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{2}} \ = \ \boxed{\mathsf{0}}}[/tex3]
[tex3]\mathsf{ \vec{v} \bullet \vec{w} \ = \ 0 \cdot \dfrac{2}{\sqrt{6}} \ - \ \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{6}} \ + \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{6}} \ = \ \boxed{\mathsf{0}}}[/tex3]
[tex3]\mathsf{ \vec{u} \bullet \vec{w} \ = \ \dfrac{1}{\sqrt{3}} \cdot \dfrac{2}{\sqrt{6}} \ - \ \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{6}} \ - \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{6}} \ = \ \boxed{\mathsf{0}}}[/tex3]
Então, provamos que [tex3]\mathsf{F}[/tex3]
é uma base ortonormal de [tex3]\mathbb{V}^{\mathsf{3}}[/tex3]
, sendo os três vetores geradores LI.
Agora, temos achar a matriz de mudança de base de [tex3]\mathsf{E}[/tex3]
para [tex3]\mathsf{F}[/tex3]
. Para isso, exprimindo a matriz dos vetores geradores de [tex3]\mathsf{F}[/tex3]
em relação à base [tex3]\mathsf{E}[/tex3]
, temos:
[tex3]\mathsf{M_{FE} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & 0 & \mathsf{\frac{2}{\sqrt{6}}} \\
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
-\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]
A inversa dessa matriz é [tex3]\mathsf{M_{EF}}[/tex3]
, ou seja, converte vetores construídos na base [tex3]\mathsf{E}[/tex3]
para a base [tex3]\mathsf{F}[/tex3]
. Para descobrirmos essa inversa, podemos usar escalonamento da matriz aumentada, mas usarei aqui a relação:
[tex3]\mathsf{A^{-1} \ = \ \dfrac{1}{det \ A} \ \cdot \ A_C^t}[/tex3]
, em que [tex3]\mathsf{A_C^t}[/tex3]
é a transposta da matriz dos cofatores de [tex3]\mathsf{A}[/tex3]
.
No caso, teremos: [tex3]\mathsf{M_{EF} \ = \ \dfrac{1}{det \ M_{FE}} \ \cdot \ M_{FE_{C}}^t}[/tex3]
Calculando primeiramente [tex3]\mathsf{det \ M_{FE} :}[/tex3]
[tex3]\mathsf{det \ M_{FE} \ = \ (-1)^{(1+1)} \cdot \underbrace{\begin{vmatrix} \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{- \frac{1}{\sqrt{6}}} \\ \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{6}}} \end{vmatrix}}_{menor \ associado \ de \ a_{1,1}} \cdot \dfrac{1}{\sqrt{3}} \ + \ (-1)^{(1+2)} \cdot \underbrace{\begin{vmatrix} \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{- \frac{1}{\sqrt{6}}} \\ \mathsf{- \frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{6}}} \end{vmatrix}}_{menor \ associado \ de \ a_{1,2}} \cdot \ 0 \ + \ (-1)^{(1+3)} \cdot \underbrace{\begin{vmatrix} \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{ \frac{1}{\sqrt{2}}} \\ \mathsf{-\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} \end{vmatrix}}_{menor \ associado \ de \ a_{1,3}} \cdot \dfrac{2}{\sqrt{6}}}[/tex3]
[tex3]\boxed{\mathsf{det \ M_{FE} \ = \ 1}}[/tex3]
Para calcular um cofator de um elemento [tex3]\mathsf{a_{i,j}}[/tex3]
, temos: [tex3]\mathsf{C_{a_{i,j}} \ = \ -1^{(i \ + \ j)} \ \cdot \ det \ M_{i,j}}[/tex3]
, onde a matriz [tex3]\mathsf{M_{i,j}}[/tex3]
é o "menor associado", ou seja, é a matriz formada quando eliminamos a linha e a coluna de [tex3]\mathsf{a_{i,j}.}[/tex3]
Ou seja, teremos a matriz de cofatores [tex3]\mathsf{M_{FE_{C}}}[/tex3]
:
[tex3]\mathsf{M_{FE_{C}} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & 0 & \mathsf{\frac{2}{\sqrt{6}}} \\
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{-\frac{1}{\sqrt{6}}} \\
-\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]
Transpondo essa matriz: [tex3]\mathsf{M_{FE_{C}^t} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{-\frac{1}{\sqrt{3}}} \\
0 & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{2}}} \\
\mathsf{\frac{2}{\sqrt{6}}} & \mathsf{-\frac{1}{\sqrt{6}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]
Por fim: [tex3]\mathsf{M_{EF} \ = \ \dfrac{1}{det \ M_{FE}} \ \cdot \ M_{FE_{C}}^t}[/tex3]
[tex3]\mathsf{M_{EF} \ = \ \dfrac{1}{1} \ \cdot \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{-\frac{1}{\sqrt{3}}} \\
0 & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{2}}} \\
\mathsf{\frac{2}{\sqrt{6}}} & \mathsf{-\frac{1}{\sqrt{6}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]
[tex3]\boxed{\mathsf{M_{EF}} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{-\frac{1}{\sqrt{3}}} \\
0 & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{2}}} \\
\mathsf{\frac{2}{\sqrt{6}}} & \mathsf{-\frac{1}{\sqrt{6}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]
Então, para o vetor [tex3]\mathsf{\vec{t} \ = \ \bigg(3\cdot \hat{i} \ - \ 2\cdot \hat{j} \ - \ \hat{k} \bigg)_E}[/tex3]
, suas coordenadas [tex3]\mathsf{(x, \ y, z)_F}[/tex3]
são dadas pela operação com seus vetores coluna nas bases [tex3]\mathsf{E, \ F}[/tex3]
:
[tex3]\begin{pmatrix}
\mathsf{x} \\
\mathsf{y} \\
\mathsf{z} \\
\end{pmatrix}_\mathsf{F} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{-\frac{1}{\sqrt{3}}} \\
0 & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{2}}} \\
\mathsf{\frac{2}{\sqrt{6}}} & \mathsf{-\frac{1}{\sqrt{6}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix} \ \mathsf{\cdot} \ \begin{pmatrix}
\mathsf{3} \\
\mathsf{-2} \\
\mathsf{-1} \\
\end{pmatrix}_{\mathsf{E}}[/tex3]
[tex3]\mathsf{x \ = \ 3 \cdot \dfrac{1}{\sqrt{3}} \ - \ 2 \cdot \dfrac{1}{\sqrt{3}} \ + \ 1 \cdot \dfrac{1}{\sqrt{3}} \ = \ \boxed{\mathsf{\dfrac{2}{\sqrt{3}}}}}[/tex3]
[tex3]\mathsf{y \ = \ 3 \cdot 0 \ - \ 2 \cdot \dfrac{1}{\sqrt{2}} \ - \ 1 \cdot \dfrac{1}{\sqrt{2}} \ = \ \boxed{\mathsf{-\dfrac{3}{\sqrt{2}}}}}[/tex3]
[tex3]\mathsf{z \ = \ 3 \cdot \dfrac{2}{\sqrt{6}} \ + \ 2 \cdot \dfrac{1}{\sqrt{6}} \ - \ 1 \cdot \dfrac{1}{\sqrt{6}} \ = \ \boxed{\mathsf{\dfrac{7}{\sqrt{6}}}}}[/tex3]
Logo, [tex3]\boxed{\boxed{\mathsf{\vec{t} \ = \ \bigg(\dfrac{2}{\sqrt{3}}, -\dfrac{3}{\sqrt{2}}, \dfrac{7}{\sqrt{6}}\bigg)_F}}}[/tex3]
Última edição:
joaopcarv (Dom 21 Abr, 2019 12:01). Total de 1 vez.
That's all I'd do all day. I'd just be the catcher in the rye and all.
"Last year's wishes are this year's apologies... Every last time I come home (...)"
Poli-USP