Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que -7 x_{3}+x_{4}=-5 pode ser representado por:
(A) \left[\begin{array}{r}1 \\\\ -3 \\\\ -1 \\\\ -12\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}7 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -1\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -6 \\\\ 0\end{array}\right]\right\}
(B) \left[\begin{array}{r}1 \\\\ -3 \\\\ -1 \\\\ -12\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 7 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -3 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -4\end{array}\right]\right\}
(C) \left[\begin{array}{r}-3 \\\\ -2 \\\\ 1 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 7 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 4\end{array}\right]\right\}
(D) \left[\begin{array}{r}-1 \\\\ 1 \\\\ 2 \\\\ 9\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 7 \\\\ 1\end{array}\right],\left[\begin{array}{l}2 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 3 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}-3 \\\\ -2 \\\\ 1 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}7 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -7\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 2 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}-2 \\\\ 3 \\\\ 2 \\\\ -26\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 7 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 7 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}-1 \\\\ 1 \\\\ 2 \\\\ 9\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 7\end{array}\right],\left[\begin{array}{l}5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 6 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(H) \left[\begin{array}{r}-2 \\\\ 3 \\\\ 2 \\\\ -26\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 7\end{array}\right],\left[\begin{array}{r}0 \\\\ -7 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}1 \\\\ 0 \\\\ 0 \\\\ 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 -7x_{3}+x_{4}=-5\}
como uma reta afim de dimensão 3, isto é, um ponto particular mais a combinação linear de três vetores. Vamos usar parametrização e depois comparar com as alternativas.

Passo 2

Como apenas x_{3} e x_{4} entram na equação, escolhemos
x_{1}=s,\quad x_{2}=t,\quad x_{3}=u,
e da equação -7x_{3}+x_{4}=-5 vem
x_{4}=-5+7u.
Logo o vetor geral é
(x_{1},x_{2},x_{3},x_{4}) =(s,t,u,-5+7u) =(0,0,0,-5)+s\,(1,0,0,0)+t\,(0,1,0,0)+u\,(0,0,1,7).

Passo 3

Para que uma alternativa esteja certa, o ponto particular p deve satisfazer
-7p_{3}+p_{4}=-5,
e cada vetor da base \{v_{1},v_{2},v_{3}\} deve satisfazer
-7v_{i,3}+v_{i,4}=0
(pois são soluções da parte homogênea).

Passo 4

Testando rapidamente cada opção:
A a F e H ou têm o ponto particular errado (não satisfazem -7p_{3}+p_{4}=-5) ou algum vetor da base não satisfaz -7v_{3}+v_{4}=0.
Só em G temos
p=(-1,1,2,9),\quad -7\cdot2+9=-14+9=-5,
e os vetores
v_{1}=(0,0,1,7),\;-7\cdot1+7=0,\quad v_{2}=(5,0,0,0),\;v_{3}=(0,6,0,0)
satisfazendo todos -7v_{i,3}+v_{i,4}=0. Além disso são três vetores independentes, gerando o espaço das soluções homogêneas.

Resposta

(G)

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