Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que
x_{3}-2 x_{4}=-3 pode ser representado por:
(A) \left[\begin{array}{r}3 \\\\ 2 \\\\ -1 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 2 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 5 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 2\end{array}\right]\right\}
(B) \left[\begin{array}{r}1 \\\\ -3 \\\\ -7 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 2\end{array}\right],\left[\begin{array}{l}0 \\\\ 2 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}-4 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(C) \left[\begin{array}{r}-1 \\\\ 3 \\\\ 1 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 2 \\\\ 1\end{array}\right],\left[\begin{array}{l}5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -5 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}-1 \\\\ 3 \\\\ 1 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 2\end{array}\right],\left[\begin{array}{r}-2 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 4 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}2 \\\\ 1 \\\\ -9 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}2 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 6\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 7 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}2 \\\\ 1 \\\\ -9 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 2 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 3 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -7\end{array}\right]\right\}
(G) \left[\begin{array}{r}1 \\\\ -3 \\\\ -7 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 2 \\\\ 1\end{array}\right],\left[\begin{array}{r}0 \\\\ -6 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}-3 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(H) \left[\begin{array}{r}3 \\\\ 2 \\\\ -1 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}2 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -1 \\\\ 0\end{array}\right]\right\}

Passo 1

O enunciado pede a descrição do conjunto
\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}\mid x_{3}-2x_{4}=-3\}
Esse é um hiperplano afim de dimensão 3. Vamos achar uma solução particular p e a base do subespaço direção V=\{v\mid v_{3}-2v_{4}=0\}, e então comparar com as opções.

Passo 2

Para parametrizar, note que podemos escolher x_{1},x_{2},x_{4} livres e escrever
x_{3}=-3+2x_{4}.
Logo todo vetor na solução é
(x_{1},\,x_{2},\,-3+2x_{4},\,x_{4}) = (0,0,-3,0)+x_{1}(1,0,0,0)+x_{2}(0,1,0,0)+x_{4}(0,0,2,1).
Aqui
p=(0,0,-3,0)
é uma solução particular e
V=\span\{(1,0,0,0),\,(0,1,0,0),\,(0,0,2,1)\}.

Passo 3

Qualquer representação correta deve ter vetores de direção que satisfaçam v_{3}-2v_{4}=0. Isso ocorre exatamente quando os três geradores são combinações lineares de (1,0,0,0), (0,1,0,0) e (0,0,2,1). Das opções A–H só as alternativas C e G têm essa propriedade.

Passo 4

Falta verificar se o ponto inicial p de cada alternativa satisfaz p_{3}-2p_{4}=-3.
Na C temos p=(-1,3,1,-2) e
1-2\cdot(-2)=1+4=5\neq -3,
logo C está errada.
Na G temos p=(1,-3,-7,-2) e
-7-2\cdot(-2)=-7+4=-3,
logo G cumpre a equação e, junto com seus geradores, descreve o mesmo conjunto.

Resposta

(G)

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