12. Sendo X=(} 0&1 2&3 ),alpha =2 beta =3 verifique que: Ver as Orientações para a) alpha (beta x)=(alpha beta )x o professor. alpha (X+Y)=alpha X+alpha Y c) (alpha +beta )X=alpha X+beta X

opção correta é a letra c) $(\alpha +\beta )X=\alpha X+\beta X$. Para verificar isso, vamos calcular os dois lados da equação. Primeiro, vamos calcular $(\alpha +\beta )X$:<br />$$(\alpha +\beta )X = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} ) = (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1\\ 2&3\end{matrix} )= (\alpha +\beta )(\begin{matrix} 0&1