Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que 5 x_{1}+x_{3}=3 pode ser representado por:
(A) \left[\begin{array}{l}-1 \\\\ -2 \\\\ -2 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-5 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -6 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -7\end{array}\right]\right\}
(B) \left[\begin{array}{r}-1 \\\\ 3 \\\\ 8 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ 0 \\\\ -5 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -3\end{array}\right],\left[\begin{array}{l}0 \\\\ 6 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(C) \left[\begin{array}{r}3 \\\\ -3 \\\\ -12 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ -5 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}2 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -1 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}2 \\\\ 1 \\\\ -7 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ -5 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}1 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 5 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}3 \\\\ -3 \\\\ -12 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 1 \\\\ 0 \\\\ -5\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 3 \\\\ 0\end{array}\right],\left[\begin{array}{r}-4 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}-1 \\\\ 3 \\\\ 8 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-5 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -4\end{array}\right],\left[\begin{array}{r}0 \\\\ -2 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{l}-1 \\\\ -2 \\\\ -2 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ 0 \\\\ -5 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 3 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 4\end{array}\right]\right\}
(H) \left[\begin{array}{r}2 \\\\ 1 \\\\ -7 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 1 \\\\ 0 \\\\ -5\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -5 \\\\ 0\end{array}\right],\left[\begin{array}{l}7 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}

Passo 1

O enunciado pede a descrição do conjunto
S=\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}\mid 5x_{1}+x_{3}=3\}
como um ponto particular mais o span de três vetores. O conteúdo é a solução de uma equação linear em \mathbb{R}^{4}, que define um hiperplano afim de dimensão 3.

Passo 2

Para achar a forma geral, resolvemos
5x_{1}+x_{3}=3
de onde
x_{1}=\frac{3-x_{3}}{5}.
Tomamos x_{2},x_{3},x_{4} livres. Assim
(x_{1},x_{2},x_{3},x_{4}) =\Bigl(\tfrac{3-x_{3}}{5},\,x_{2},\,x_{3},\,x_{4}\Bigr).

Passo 3

Escolhemos um ponto particular. Por exemplo, fixando
x_{2}=3,\quad x_{3}=8,\quad x_{4}=2
temos
x_{1}=\frac{3-8}{5}=-1,
logo um particular é
p=(-1,3,8,2),
e de fato 5(-1)+8=3.

Passo 4

A solução homogênea de 5x_{1}+x_{3}=0 parametriza-se por
x_{2}=s,\quad x_{3}=t,\quad x_{4}=u,\quad x_{1}=-\frac{t}{5}.
Vetores independentes geradores do subespaço homogêneo podem ser escolhidos como
v_{1}=(1,0,-5,0),\quad v_{2}=(0,1,0,0),\quad v_{3}=(0,0,0,1).
De fato cada um satisfaz 5x_{1}+x_{3}=0.

Passo 5

Portanto
S =\{\,p + \alpha\,v_{1}+\beta\,v_{2}+\gamma\,v_{3}\mid \alpha,\beta,\gamma\in\mathbb{R}\}.

Passo 6

Na lista de alternativas, a opção (B) apresenta exatamente
p=(-1,3,8,2), \quad v_{1}=(1,0,-5,0), \quad v_{2}=(0,6,0,0)=6\,(0,1,0,0), \quad v_{3}=(0,0,0,-3)=-3\,(0,0,0,1),
que geram o mesmo afim e o mesmo subespaço homogêneo, só mudando escala dos vetores. Logo a resposta é a alternativa (B).

Resposta

(B)

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