Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que
5 x_{2}+x_{3}=3 pode ser representado por:
(A) \left[\begin{array}{r}-1 \\\\ -3 \\\\ -12 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ -5 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{r}-1 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 3\end{array}\right]\right\}
(B) \left[\begin{array}{r}-1 \\\\ -3 \\\\ -12 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 1 \\\\ -5 \\\\ 0\end{array}\right],\left[\begin{array}{r}-7 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 4\end{array}\right]\right\}
(C) \left[\begin{array}{r}3 \\\\ 2 \\\\ -7 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 7 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}3 \\\\ 2 \\\\ -7 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ 0 \\\\ 0 \\\\ -5\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -4 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}-2 \\\\ 1 \\\\ -2 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{r}0 \\\\ -6 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}1 \\\\ -3 \\\\ 18 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 1 \\\\ -5 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -3\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}1 \\\\ -3 \\\\ 18 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ -5 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 2\end{array}\right],\left[\begin{array}{l}6 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(H) \left[\begin{array}{r}-2 \\\\ 1 \\\\ -2 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ 0 \\\\ 0 \\\\ -5\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -7 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 2 \\\\ 0 \\\\ 0\end{array}\right]\right\}

Passo 1

O problema pede a descrição do conjunto (x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}
sujeito à equação
5x_{2}+x_{3}=3.
Esse é um subespaço afim de dimensão 3. Para representá-lo, precisamos encontrar um ponto particular que satisfaça a equação e uma base do espaço homogêneo associado (isto é, a solução de 5x_{2}+x_{3}=0).

Passo 2

Vamos resolver a equação isolando x_{3}:
x_{3}=3-5x_{2}.
As variáveis x_{1}, x_{2} e x_{4} são livres. Denotando s=x_{1}, t=x_{2}, u=x_{4}, temos a solução geral
(x_{1},x_{2},x_{3},x_{4})=(s,\;t,\;3-5t,\;u).

Passo 3

Escrevendo em forma afim, separamos um ponto particular e a combinação linear dos vetores que geram o espaço homogêneo:
(s,t,3-5t,u) =(0,0,3,0) +s\,(1,0,0,0) +t\,(0,1,-5,0) +u\,(0,0,0,1).
Logo, um ponto particular é p=(0,0,3,0) e uma base para o núcleo de 5x_{2}+x_{3} é dada pelos vetores
v_{1}=(1,0,0,0),\quad v_{2}=(0,1,-5,0),\quad v_{3}=(0,0,0,1).

Passo 4

Agora conferimos as opções: buscamos um ponto particular que satisfaça 5x_{2}+x_{3}=3 e vetores de span que satisfaçam 5x_{2}+x_{3}=0 e cubram as direções livres em x_{1},x_{2},x_{4}. A única opção que cumpre isso é a (F), pois
– o ponto (1,-3,18,2) satisfaz 5(-3)+18=3,
– os vetores (0,1,-5,0), (0,0,0,-3) e (-5,0,0,0) estão no espaço homogêneo e são independentes.

Resposta

(F)

Exercícios relacionados