Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que -5 x_{1}+x_{2}=-2 pode ser representado por:
(A) \left[\begin{array}{r}-2 \\\\ -12 \\\\ 3 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 5 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 4 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}-1 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(B) \left[\begin{array}{r}-1 \\\\ -7 \\\\ 1 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 6 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 2\end{array}\right]\right\}
(C) \left[\begin{array}{r}-1 \\\\ 13 \\\\ -3 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 4\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}-2 \\\\ -12 \\\\ 3 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 5\end{array}\right],\left[\begin{array}{r}-6 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -4 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}1 \\\\ 3 \\\\ -2 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 5 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}1 \\\\ 3 \\\\ -2 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 5\end{array}\right],\left[\begin{array}{l}3 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 5 \\\\ 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 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -7\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -3 \\\\ 0\end{array}\right]\right\}
(H) \left[\begin{array}{r}-1 \\\\ -7 \\\\ 1 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}5 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -1 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 7\end{array}\right]\right\}

Passo 1

O enunciado pede o conjunto de (x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4} que satisfaz
-5x_{1}+x_{2}=-2.
Isso é uma hipersuperfície afim de dimensão 3. Vamos achar primeiro o subespaço homogêneo -5x_{1}+x_{2}=0 e depois uma solução particular.

Passo 2

Para o homogêneo resolvemos -5x_{1}+x_{2}=0. Daí x_{2}=5x_{1} e x_{3},x_{4} são livres. Logo qualquer solução homogênea tem forma (x_{1},x_{2},x_{3},x_{4})=(t,5t,s,r) com t,s,r\in\mathbb{R}. Assim o subespaço homogêneo é
\{(t,5t,s,r)\mid t,s,r\in\mathbb{R}\}
e pode ser escrito como
\mathrm{span}\{(1,5,0,0),\,(0,0,1,0),\,(0,0,0,1)\}.

Passo 3

Para achar uma solução particular de -5x_{1}+x_{2}=-2, escolhemos x_{3}=x_{4}=0 e x_{1}=1. Então
-5\cdot1 + x_{2} = -2 \quad\Longrightarrow\quad x_{2}=3.
Logo uma solução particular é
p = (1,3,0,0),
pois -5\cdot1 + 3 = -2.

Passo 4

A solução geral da equação afim é
(1,3,0,0) + \mathrm{span}\{(1,5,0,0),\,(0,0,1,0),\,(0,0,0,1)\}.

Passo 5

Agora comparamos com as opções. Na alternativa (B) o vetor particular é (\\ -1,-7,1,-2), que satisfaz -5(-1)+(-7)=5-7=-2, e os vetores do span são (1,5,0,0), (0,0,6,0) e (0,0,0,2). Cada um deles satisfaz -5x_{1}+x_{2}=0 (o segundo e o terceiro são múltiplos dos nossos (0,0,1,0) e (0,0,0,1)), de modo que esse span coincide com o subespaço homogêneo. As demais opções ou não satisfazem a parte homogênea ou o ponto particular não satisfaz a equação.

Resposta

(B)

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