Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que x_{2}-5 x_{4}=3 pode ser representado por:
(A) \left[\begin{array}{r}3 \\\\ -2 \\\\ -3 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 1 \\\\ 0 \\\\ 5\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 6 \\\\ 0\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(B) \left[\begin{array}{r}3 \\\\ -2 \\\\ -3 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 5 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 5 \\\\ 0\end{array}\right],\left[\begin{array}{l}7 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(C) \left[\begin{array}{r}1 \\\\ -7 \\\\ 2 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 5 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}3 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -6 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}1 \\\\ -7 \\\\ 2 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 1 \\\\ 0 \\\\ 5\end{array}\right],\left[\begin{array}{r}-7 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 2 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}-1 \\\\ 13 \\\\ 3 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 0 \\\\ 5 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -2\end{array}\right],\left[\begin{array}{r}0 \\\\ -1 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}2 \\\\ 8 \\\\ -1 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 0 \\\\ 5 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -3\end{array}\right],\left[\begin{array}{r}0 \\\\ -4 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}-1 \\\\ 13 \\\\ 3 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}5 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -7\end{array}\right]\right\}
(H) \left[\begin{array}{r}2 \\\\ 8 \\\\ -1 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}5 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 4\end{array}\right]\right\}

Passo 1

O enunciado pede representar o conjunto dos (x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4} satisfazendo
x_{2}-5x_{4}=3
como um subespaço afim, ou seja, uma solução particular dessa equação mais o espaço das soluções de
x_{2}-5x_{4}=0.

Passo 2

Para encontrar a forma geral, parametrize
x_{1}=u,\quad x_{3}=v,\quad x_{4}=t.
Da igualdade x_{2}-5x_{4}=3 vem x_{2}=3+5t. Logo
(x_{1},x_{2},x_{3},x_{4}) =(u,3+5t,v,t) =(0,3,0,0) +u\,(1,0,0,0) +v\,(0,0,1,0) +t\,(0,5,0,1).
Um ponto particular é (0,3,0,0) e uma base do subespaço homogêneo é \{(1,0,0,0),(0,0,1,0),(0,5,0,1)\}.

Passo 3

Cada alternativa fornece um ponto particular p=(p_{1},p_{2},p_{3},p_{4}) e três vetores de direção. Primeiro checamos em cada p se
p_{2}-5p_{4}=3.
As que satisfazem são (C), (D), (E), (F), (G) e (H).

Passo 4

Para cada opção restante, testamos se todo vetor de direção v=(v_{1},v_{2},v_{3},v_{4}) satisfaz
v_{2}-5v_{4}=0.
Somente na alternativa (C) todos os vetores (0,5,0,1), (3,0,0,0) e (0,0,-6,0) passam nesse teste. Nas demais sempre há ao menos um vetor com v_{2}-5v_{4}\neq0, eliminando-as.

Resposta

(C)

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