Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que x_{2}-2 x_{3}=-7 pode ser representado por:
(A) \left[\begin{array}{r}2 \\\\ -11 \\\\ 1 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 1 \\\\ 2 \\\\ 0\end{array}\right],\left[\begin{array}{l}5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 2\end{array}\right]\right\}
(B) \left[\begin{array}{r}3 \\\\ -5 \\\\ 1 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 1 \\\\ 2 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -2\end{array}\right],\left[\begin{array}{l}7 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(C) \left[\begin{array}{r}-2 \\\\ -9 \\\\ -1 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}2 \\\\ 0 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -6 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}-3 \\\\ -1 \\\\ 3 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}2 \\\\ 0 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -1 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 3 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}-3 \\\\ -1 \\\\ 3 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 0 \\\\ 0 \\\\ 2\end{array}\right],\left[\begin{array}{r}0 \\\\ -5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}2 \\\\ -11 \\\\ 1 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 2 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -7\end{array}\right]\right\}
(G) \left[\begin{array}{r}3 \\\\ -5 \\\\ 1 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 2 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 6\end{array}\right],\left[\begin{array}{l}4 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(H) \left[\begin{array}{r}-2 \\\\ -9 \\\\ -1 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 0 \\\\ 0 \\\\ 2\end{array}\right],\left[\begin{array}{r}0 \\\\ -4 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -3 \\\\ 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 x_{2}-2x_{3}=-7\}
ou seja, um hiperplano afim em \mathbb{R}^{4}. Para descrever esse conjunto numérica e geometricamente, precisamos de:
  1. um vetor particular p que satisfaça p_{2}-2p_{3}=-7, e
  2. uma base do espaço homogêneo (o núcleo da forma linear x_{2}-2x_{3}), que é dado por \{v\mid v_{2}-2v_{3}=0\}.

Passo 2

Primeiro, encontramos a solução geral da equação homogênea
x_{2}-2x_{3}=0.
Tomando x_{1}=s, x_{3}=t e x_{4}=u livres, vem
x_{2}=2t,
de modo que
(x_{1},x_{2},x_{3},x_{4}) =(s,2t,t,u) =s\,(1,0,0,0)+t\,(0,2,1,0)+u\,(0,0,0,1).
Assim, o espaço homogêneo é
\operatorname{span}\bigl\{(1,0,0,0),\,(0,2,1,0),\,(0,0,0,1)\bigr\}.

Passo 3

As opções dadas usam três vetores para a parte de “span”. Devem satisfazer a condição homogênea v_{2}-2v_{3}=0. Ao verificar cada tripla, só duas opções cumprem isso:
  • (F) tem (0,2,1,0), (-5,0,0,0) e (0,0,0,-7), todas com v_{2}=2v_{3}.
  • (G) tem (0,2,1,0), (4,0,0,0) e (0,0,0,6), também todas com v_{2}=2v_{3}.
Todas as outras têm pelo menos um vetor que não satisfaz v_{2}-2v_{3}=0.

Passo 4

Resta escolher entre (F) e (G) verificando o vetor particular.
  • Em (F) o ponto é (2,-11,1,-3), para o qual
-11 - 2\cdot1 = -13 \neq -7.
  • Em (G) o ponto é (3,-5,1,-1), para o qual
-5 - 2\cdot1 = -7,
ou seja, satisfaz exatamente a condição x_{2}-2x_{3}=-7.

Passo 5

Portanto a única descrição que usa vetores de direção corretos e um ponto particular que resolve x_{2}-2x_{3}=-7 é a opção (G).

Resposta

(G) \displaystyle \begin{pmatrix}3\\\ -5\\\ 1\\\ -1\end{pmatrix} +\operatorname{span}\Bigl\{\,(0,2,1,0),\,(4,0,0,0),\,(0,0,0,6)\Bigr\}.

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