Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que
5 x_{3}+x_{4}=5 pode ser representado por:
(A) \left[\begin{array}{r}-2 \\\\ 1 \\\\ 3 \\\\ -5\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 0 \\\\ -5 \\\\ 1\end{array}\right],\left[\begin{array}{r}-1 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -4 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(B) \left[\begin{array}{r}1 \\\\ 2 \\\\ -1 \\\\ 10\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-5 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 4 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 3\end{array}\right]\right\}
(C) \left[\begin{array}{r}-1 \\\\ -3 \\\\ 3 \\\\ -10\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 0 \\\\ -5 \\\\ 1\end{array}\right],\left[\begin{array}{r}0 \\\\ -6 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}6 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}1 \\\\ 2 \\\\ -1 \\\\ 10\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ -5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -5\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -6 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}2 \\\\ -1 \\\\ -3 \\\\ 20\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-5 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 5\end{array}\right]\right\}
(F) \left[\begin{array}{r}-1 \\\\ -3 \\\\ 3 \\\\ -10\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 0 \\\\ 1 \\\\ -5\end{array}\right],\left[\begin{array}{r}0 \\\\ -7 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}7 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}2 \\\\ -1 \\\\ -3 \\\\ 20\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}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\}
(H) \left[\begin{array}{r}-2 \\\\ 1 \\\\ 3 \\\\ -5\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 0 \\\\ 1 \\\\ -5\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 2 \\\\ 0 \\\\ 0\end{array}\right]\right\}

Passo 1

O enunciado define o subconjunto
S = \{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}\mid 5x_{3}+x_{4}=5\}.
Esse é um subespaço afim de dimensão 3, pois impõe uma única equação linear. Para descrevê-lo como
p + \operatorname{span}\{v_{1},v_{2},v_{3}\}
precisamos:
  1. Encontrar uma solução particular p de 5x_{3}+x_{4}=5.
  2. Encontrar uma base \{v_{1},v_{2},v_{3}\} do núcleo da aplicação A(x)=5x_{3}+x_{4}, ou seja, solução de 5x_{3}+x_{4}=0.

Passo 2

Resolvendo o homogêneo
5x_{3}+x_{4}=0
parametrizamos por x_{1}=s, x_{2}=t, x_{3}=u livres e x_{4}=-5u. Logo
\ker A = \bigl\{(s,t,u,-5u) \mid s,t,u\in\mathbb{R}\bigr\} =\operatorname{span}\bigl\{(1,0,0,0),(0,1,0,0),(0,0,1,-5)\bigr\},
dimensão 3.

Passo 3

Bastaria achar um p tal que 5p_{3}+p_{4}=5. Por exemplo, tomando p_{1}=0, p_{2}=0, p_{3}=1, p_{4}=0, temos
5\cdot1+0=5,
logo um particular é p=(0,0,1,0).

Passo 4

Cada alternativa lista um ponto base e três vetores. Para ser correta, o ponto base deve satisfazer 5x_{3}+x_{4}=5 e cada vetor do span deve satisfazer 5x_{3}+x_{4}=0. Testando rapidamente as opções recai-mos na (F):
Ponto base (-1,-3,3,-10) 5\cdot3+(-10)=5.
Vetores \;(0,0,1,-5), (0,-7,0,0), (7,0,0,0);todos satisfazem 5x_{3}+x_{4}=0 e são linearmente independentes.

Resposta

(F)

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