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}=-7 pode ser representado por:
(A) \left[\begin{array}{l}-2 \\\\ -2 \\\\ -1 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -3 \\\\ 0\end{array}\right]\right\}
(B) \left[\begin{array}{r}3 \\\\ 8 \\\\ 1 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 5\end{array}\right],\left[\begin{array}{r}-2 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -6 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(C) \left[\begin{array}{l}-2 \\\\ -2 \\\\ -1 \\\\ -3\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 \\\\ -5 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}-2 \\\\ -17 \\\\ 2 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 5 \\\\ 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\}
(E) \left[\begin{array}{l}2 \\\\ 3 \\\\ 3 \\\\ 1\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\}
(F) \left[\begin{array}{r}3 \\\\ 8 \\\\ 1 \\\\ -1\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}-4 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}-2 \\\\ -17 \\\\ 2 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}5 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -4 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 2\end{array}\right]\right\}
(H) \left[\begin{array}{l}2 \\\\ 3 \\\\ 3 \\\\ 1\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}{l}6 \\\\ 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 -5x_{1}+x_{2}=-7\}
como um afim (vetor particular mais span). O conteúdo é resolver uma equação linear em \mathbb{R}^{4} e escrever a solução geral como soma de uma solução particular com o espaço nulo da forma homogênea.

Passo 2

Para resolver
-5x_{1}+x_{2}=-7
tomamos x_{1},x_{3},x_{4} como parâmetros livres t_{1},t_{2},t_{3}. Da equação vem
x_{2}=5t_{1}-7.
Então
(x_{1},x_{2},x_{3},x_{4}) =(t_{1},5t_{1}-7,t_{2},t_{3}) =(0,-7,0,0) +t_{1}(1,5,0,0) +t_{2}(0,0,1,0) +t_{3}(0,0,0,1).
Aqui (0,-7,0,0) é uma solução particular p e os três vetores são uma base do espaço nulo da equação homogênea -5x_{1}+x_{2}=0.

Passo 3

Para cada opção devemos ter
  1. O vetor particular p satisfaz -5p_{1}+p_{2}=-7.
  2. Cada vetor da span satisfaz -5v_{1}+v_{2}=0.
    Verifica-se que, de todas as alternativas, apenas a (D) escolhe
p=(-2,-17,2,3),\quad v_{1}=(1,5,0,0),\; v_{2}=(0,0,-1,0),\; v_{3}=(0,0,0,7),
onde de fato -5(-2)+(-17)=10-17=-7 e -5\cdot1+5=0,\,-5\cdot0+0=0,\,-5\cdot0+0=0.

Resposta

(D)

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