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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

Question

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

Solution

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4 (170 Votos)
Rosana Especialista · Tutor por 3 anos

Resposta

opção correta é a letra c) . Para verificar isso, vamos calcular os dois lados da equação. Primeiro, vamos calcular : (\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