Question

B = {(1,−1, 0), (1, 0, 2), (1, 1, 2)} é uma base do ℝ3 Para escrevermos o vetor v = (2,−1, 0) como combinação linear
desta base, que valores devemos multiplicar os vetores da base B ?
A) 2,-1,1
B) 2,-1,-1
C) 2,1,-1
D) 2,-1,3
E) 2,−1, 0

Answer 1

Boa noite Thais!

Solução!

Escrevendo o vetor B como combinação linear de V.


$a(1,-1,0)+b(1,0,2)+c(1,1,2)=(2,-1,0)\\\\\\\ Montando ~~um~~ sistema~~com~~ tr\^es~~variaveis!$


$\begin{cases} ~~a+b+c=2\\ -a~~~~~+c=-1\\ ~~~~~~2b+2c=0 \end{cases}~\\\\\\ Vou~~resolver~~pela~~substituic\~ao!\\\\\\\ 2b=-2c\\\\\ b=\dfrac{-2c}{2}\\\\\\ \boxed{b=-c} \\\\\\\\\ Substituindo~~na~~primeira~~equac\~ao!\\\\\\\\\ a-c+c=2\\\\\\ a+0=2\\\\ \boxed{a=2}\\\\\\\ -a+c=-1\\\\\\ -2+c=-1\\\\ c=-1+2\\\\ \boxed{c=1}$

$Finalmente!\\\\\\ b=-c\\\\ \boxed{b=-1}$

$V=(2,-1,1)$


$\boxed{Resposta:V=(2,-1,1) ~~\boxed{Alternativa~~A}}$

Boa noite!
Bons estudos!

Answer 2

Olá!
 
    Sejam  $\alpha,\;\;\beta,\;\;\gamma\in\mathbb{R}$  os coeficientes que multiplicam os vetores da base. Temos:

$\alpha(1,-1,0)+\beta(1,0,2)+\gamma(1,1,2)=(2,-1,0) \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{ll}\alpha+\beta+\gamma&=2\\ -\alpha+\gamma&=-1\;\;\sim\\ \beta+\gamma&=0\end{array}\right. \left\{ \begin{array}{ll}\alpha+\beta+\gamma&=2 \\ \beta+2\gamma&=1\;\;\sim\\ \beta+\gamma&=0 \end{array}\right. \\ \\ \\ \sim \left\{\begin{array}{ll} \alpha+\beta+\gamma&=\;\;\;2\\ \beta+2\gamma&=\;\;\;1\\-\gamma&=-1 \end{array}\right. \Rightarrow \gamma=1\Rightarrow\beta+2\cdot 1=1\Rightarrow\beta=-1$

Logo,

$\alpha+(-1)+1=2\Rightarrow\alpha=2\\ \\ \\ \therefore \\ \\ S=\{(2,-1,1)\}$

Portanto, gabarito (A).

Bons estudos!