[tex3]M=\begin{pmatrix}
\sen k\theta & \cos \theta & 0 \\
-\sen \theta & \cos \theta & 0 \\
0 & 0 & 1 \\
\end{pmatrix}[/tex3] , [tex3]\begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix}[/tex3] e [tex3]\begin{pmatrix}
1 \\
0 \\
3 \\
\end{pmatrix}[/tex3]
A) Calcule o determinante de M e a matriz inversa de M.
B) Resolva o sistema MX=Y.
A) [tex3]\det M=\cos^2\theta -(-\sen^2\theta )\rightarrow \det M=1[/tex3]
[tex3]\det M^{-1}=\frac{1}{\det M}\rightarrow \det M^{-1}=\frac{1}{1}\rightarrow \det M^{-1}=1[/tex3]
B)
[tex3]MX=Y\\\\\begin{pmatrix}
\cos \theta \cdot x+\sen \theta \cdot y \\
-\sen \theta \cdot x+\cos \theta \cdot y \\
z \\
\end{pmatrix}=\begin{pmatrix}
1 \\
0 \\
3 \\
\end{pmatrix}\\\\\begin{cases}
\cos \theta \cdot x+\sen \theta \cdot y=1 \\
-\sen \theta \cdot x+\cos \theta \cdot y=0
\end{cases}[/tex3]
Como resolver o sistema acima?
Resposta
[tex3]S=(z=3,x=\cos \theta ,y=\sen \theta )[/tex3]