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}=7 pode ser representado por:
(A) \left[\begin{array}{r}-1 \\\\ 3 \\\\ 17 \\\\ 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 \\\\ 4\end{array}\right],\left[\begin{array}{l}7 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(B) \left[\begin{array}{r}-2 \\\\ -3 \\\\ 22 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ 0 \\\\ 0 \\\\ -5\end{array}\right],\left[\begin{array}{l}0 \\\\ 6 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -1 \\\\ 0\end{array}\right]\right\}
(C) \left[\begin{array}{r}-2 \\\\ -3 \\\\ 22 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 2 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 5 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}-1 \\\\ 3 \\\\ 17 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 1 \\\\ -5 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 2\end{array}\right],\left[\begin{array}{r}-3 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}1 \\\\ 3 \\\\ -8 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 1 \\\\ -5 \\\\ 0\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -6\end{array}\right]\right\}
(F) \left[\begin{array}{r}-3 \\\\ 1 \\\\ 2 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -7 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}-3 \\\\ 1 \\\\ 2 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ 0 \\\\ 0 \\\\ -5\end{array}\right],\left[\begin{array}{l}0 \\\\ 3 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right]\right\}
(H) \left[\begin{array}{r}1 \\\\ 3 \\\\ -8 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ -5 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{l}5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -4\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}\mid5x_{2}+x_{3}=7\}
como um ponto particular p mais a span de três vetores v_{1},v_{2},v_{3}. Para isso, precisamos encontrar
  1. uma solução particular p que satisfaça \,5p_{2}+p_{3}=7,
  2. uma base do núcleo da aplicação linear A(x)=5x_{2}+x_{3}, i.e.\ o espaço \{x\mid5x_{2}+x_{3}=0\} de dimensão 3.

Passo 2

Caracterizamos o núcleo resolvendo
5x_{2}+x_{3}=0\implies x_{3}=-5x_{2}.
Tomando parâmetros livres t=x_{1}, s=x_{2}, r=x_{4}, qualquer vetor do núcleo é
(x_{1},x_{2},x_{3},x_{4})=(t,s,-5s,r) =t\,(1,0,0,0)+s\,(0,1,-5,0)+r\,(0,0,0,1).
Logo uma base para o núcleo é
(1,0,0,0),\;(0,1,-5,0),\;(0,0,0,1).

Passo 3

Agora examinamos cada alternativa, verificando:
  1. se o ponto p satisfaz 5p_{2}+p_{3}=7,
  2. se cada vetor v_{i} satisfaz 5v_{i,2}+v_{i,3}=0.
(A) \;p=(-1,3,17,2) falha pois 5\cdot3+17=32\neq7.
(B) \;p=(-2,-3,22,-1) satisfaz -15+22=7, mas v_{2}=(0,6,0,0) 5\cdot6+0=30\neq0.
(C) mesmo p=(-2,-3,22,-1) ok, porém v_{2}=(0,0,2,0) 5\cdot0+2=2\neq0.
(D) ponto p não satisfaz a equação.
(E) \;p=(1,3,-8,-3) satisfaz 5\cdot3+(-8)=7, e
v_{1}=(0,1,-5,0),\quad v_{2}=(-5,0,0,0),\quad v_{3}=(0,0,0,-6)
dão em cada caso 5v_{i,2}+v_{i,3}=0 e são linearmente independentes.
(F) ou (G) algum v_{i} falha em 5v_{i,2}+v_{i,3}=0.
(H) v_{1}=(0,-5,1,0) falha pois 5\cdot(-5)+1=-24\neq0.
Portanto somente a alternativa (E) satisfaz ambas as condições.

Resposta

(E)

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