Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que x_{3}+3 x_{4}=-2 pode ser representado por:
(A) \left[\begin{array}{r}2 \\\\ 3 \\\\ 1 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ -3 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 4\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right]\right\}
(B) \left[\begin{array}{r}-1 \\\\ 2 \\\\ -5 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 0 \\\\ 1 \\\\ -3\end{array}\right],\left[\begin{array}{r}-6 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -2 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(C) \left[\begin{array}{r}-1 \\\\ 2 \\\\ -5 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 0 \\\\ -3 \\\\ 1\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\}
(D) \left[\begin{array}{r}3 \\\\ -2 \\\\ -8 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-3 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -3 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -6\end{array}\right]\right\}
(E) \left[\begin{array}{r}-2 \\\\ -3 \\\\ -11 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 0 \\\\ -3 \\\\ 1\end{array}\right],\left[\begin{array}{r}0 \\\\ -3 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}3 \\\\ -2 \\\\ -8 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ -3 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -4\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 7 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}-2 \\\\ -3 \\\\ -11 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 0 \\\\ 1 \\\\ -3\end{array}\right],\left[\begin{array}{l}0 \\\\ 2 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}-1 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(H) \left[\begin{array}{r}2 \\\\ 3 \\\\ 1 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-3 \\\\ 1 \\\\ 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\}

Passo 1

O enunciado pede representar o conjunto de \bigl(x_{1},x_{2},x_{3},x_{4}\bigr)\in\mathbb{R}^{4} tais que
x_{3}+3x_{4}=-2
como um ponto particular mais o span da solução homogênea. Isso é típico de um hiperplano afim:
\{x\in\mathbb{R}^4\mid x_3+3x_4=-2\} = v_{0} + \ker(A),
onde A=[0\;0\;1\;3].

Passo 2

Para achar \ker(A), resolvemos a parte homogênea
x_{3}+3x_{4}=0.
Tomamos x_{1}=s, x_{2}=t, x_{4}=u livres. Então
x_{3}=-3u,
e um vetor genérico da solução homogênea é
(s,t,-3u,u) =s\,(1,0,0,0)+t\,(0,1,0,0)+u\,(0,0,-3,1).
Logo
\ker(A)=\span\bigl\{(1,0,0,0),\,(0,1,0,0),\,(0,0,-3,1)\bigr\}.

Passo 3

Agora precisamos de um ponto particular v_{0} com x_{3}+3x_{4}=-2. Uma escolha simples é
x_{1}=0,\quad x_{2}=0,\quad x_{4}=0\;\Longrightarrow\;x_{3}=-2,
portanto
v_{0}=(0,0,-2,0).
Qualquer outra escolha que satisfaça a equação também funciona (os offsets dados nas alternativas são apenas outros pontos particulares).

Passo 4

Basta comparar com as alternativas e ver qual span coincide com \ker(A) e cujo ponto particular satisfaz x_{3}+3x_{4}=-2.
Na alternativa (C) temos:
v_{0}=(-1,2,-5,1),\quad \span\{(0,0,-3,1),\,(5,0,0,0),\,(0,6,0,0)\}.
Cada vetor do span satisfaz v_{3}+3v_{4}=0, e de fato
-5+3\cdot1=-2,
logo ( -1,2,-5,1) está no hiperplano. Além disso os vetores do span são múltiplos de (1,0,0,0), (0,1,0,0) e (0,0,-3,1), exatamente a base da solução homogênea.

Resposta

(C)

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