![\begin{pmatrix}
\frac{1}{\sqrt[]{2}} & 0 & - \frac{1}{\sqrt[]{2}} \\
0 & 1 & 0 \\
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix}](/latexrender/pictures/3f5211781b318534ca75dde864f5c425.png "\begin{pmatrix}
\frac{1}{\sqrt[]{2}} & 0 & - \frac{1}{\sqrt[]{2}} \\
0 & 1 & 0 \\
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix}")
![\begin{pmatrix}
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
0 & 1 & 0 \\
-\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix}](/latexrender/pictures/d8c1a86504c7f5605bcf4ed8790f1435.png "\begin{pmatrix}
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
0 & 1 & 0 \\
-\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix}")
e encontrei como resultado a seguinte matriz:
![\begin{pmatrix}
\frac{1}{2} & \frac{1}{\sqrt[]{2}} & - \frac{1}{2} \\
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
-\frac{1}{2} & - \frac{1}{\sqrt[]{2}} & \frac{1}{2} \\
\end{pmatrix}](/latexrender/pictures/a7bcf011c1c7453692fb11c747658162.png "\begin{pmatrix}
\frac{1}{2} & \frac{1}{\sqrt[]{2}} & - \frac{1}{2} \\
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
-\frac{1}{2} & - \frac{1}{\sqrt[]{2}} & \frac{1}{2} \\
\end{pmatrix}")
Está correto?
por barbara-rabello » Qui Abr 25, 2013 22:35
por e8group » Sex Abr 26, 2013 21:11
![A = \begin{pmatrix}\frac{1}{\sqrt[]{2}} & 0 & - \frac{1}{\sqrt[]{2}} \\
0 & 1 &0 \\
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix}](/latexrender/pictures/1b6ae74181c712bf8d0790ef3f4acc54.png "A = \begin{pmatrix}\frac{1}{\sqrt[]{2}} & 0 & - \frac{1}{\sqrt[]{2}} \\
0 & 1 &0 \\
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix}")
![C = \begin{pmatrix}
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
0 & 1 & 0 \\
-\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix}](/latexrender/pictures/e132ff10642800148d111754bcee8210.png "C = \begin{pmatrix}
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
0 & 1 & 0 \\
-\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix}")
. ![BC= \begin{pmatrix}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{pmatrix} \cdot \begin{pmatrix}
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
0 & 1 & 0 \\
-\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix} = \begin{pmatrix}
\0 & -1 & 0\\
\frac{1}{\sqrt{2}} &0 & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}}& 0 & \frac{1}{\sqrt{2}}
\end{pmatrix}](/latexrender/pictures/7052b39282a36573c053919de88b09b7.png "BC= \begin{pmatrix}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{pmatrix} \cdot \begin{pmatrix}
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
0 & 1 & 0 \\
-\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix} = \begin{pmatrix}
\0 & -1 & 0\\
\frac{1}{\sqrt{2}} &0 & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}}& 0 & \frac{1}{\sqrt{2}}
\end{pmatrix}")
![ABC = \begin{pmatrix}\frac{1}{\sqrt[]{2}} & 0 & - \frac{1}{\sqrt[]{2}} \\
0 & 1 &0 \\
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix} \cdot \begin{pmatrix}
\0 & -1 & 0\\
\frac{1}{\sqrt{2}} &0 & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}}& 0 & \frac{1}{\sqrt{2}}
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2} & -\frac{1}{\sqrt{2}} & - \frac{1}{2}\\
\frac{1}{\sqrt{2}}& 0 &\frac{1}{\sqrt{2}} \\
- \frac{1}{2}& -\frac{1}{\sqrt{2}} & \frac{1}{2}
\end{pmatrix} = \frac{1}{2}\cdot \begin{pmatrix}
1&-\sqrt{2} & -1 \\
\sqrt{2}& 0 & \sqrt{2} \\
-1& -\sqrt{2} &1
\end{pmatrix}](/latexrender/pictures/7fcc236c5dfe2119cc502d4bf07317da.png "ABC = \begin{pmatrix}\frac{1}{\sqrt[]{2}} & 0 & - \frac{1}{\sqrt[]{2}} \\
0 & 1 &0 \\
\frac{1}{\sqrt[]{2}} & 0 & \frac{1}{\sqrt[]{2}} \\
\end{pmatrix} \cdot \begin{pmatrix}
\0 & -1 & 0\\
\frac{1}{\sqrt{2}} &0 & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}}& 0 & \frac{1}{\sqrt{2}}
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2} & -\frac{1}{\sqrt{2}} & - \frac{1}{2}\\
\frac{1}{\sqrt{2}}& 0 &\frac{1}{\sqrt{2}} \\
- \frac{1}{2}& -\frac{1}{\sqrt{2}} & \frac{1}{2}
\end{pmatrix} = \frac{1}{2}\cdot \begin{pmatrix}
1&-\sqrt{2} & -1 \\
\sqrt{2}& 0 & \sqrt{2} \\
-1& -\sqrt{2} &1
\end{pmatrix}")
Voltar para Matrizes e Determinantes
por weverton » Sáb Jul 24, 2010 06:14
Usuários navegando neste fórum: Nenhum usuário registrado e 14 visitantes
em um sistema de coordenadas cartesianas xOy. Determine o número complexo b , de módulo igual a 1 , cujo afixo M pertence ao quarto quadrante e é tal que o ângulo LÔM é reto.
o ângulo entre o eixo horizontal e o afixo 
. O triângulo é retângulo com catetos 
e 
, tal que 
. Seja 
o ângulo complementar. Então 
. Como 
, o ângulo que o afixo 
formará com a horizontal será , mas negativo pois tem de ser no quarto quadrante. Se 
, então 
. Como módulo é um: 
.
.