Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que x_{2}+7 x_{4}=7 pode ser representado por:
(A) \left[\begin{array}{r}-1 \\\\ 0 \\\\ 3 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ 0 \\\\ -7 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 3 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 6\end{array}\right]\right\}
(B) \left[\begin{array}{r}3 \\\\ 21 \\\\ 1 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 1 \\\\ 0 \\\\ -7\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 4 \\\\ 0\end{array}\right],\left[\begin{array}{r}-2 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(C) \left[\begin{array}{r}2 \\\\ 14 \\\\ -3 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ -7 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{r}-5 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -7 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}-3 \\\\ -14 \\\\ -1 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-7 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 5\end{array}\right],\left[\begin{array}{r}0 \\\\ -6 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}3 \\\\ 21 \\\\ 1 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ -7 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 2 \\\\ 0\end{array}\right],\left[\begin{array}{l}7 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(F) \left[\begin{array}{r}2 \\\\ 14 \\\\ -3 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}0 \\\\ 1 \\\\ 0 \\\\ -7\end{array}\right],\left[\begin{array}{l}1 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -4 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}-3 \\\\ -14 \\\\ -1 \\\\ 3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}1 \\\\ 0 \\\\ -7 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 4\end{array}\right]\right\}
(H) \left[\begin{array}{r}-1 \\\\ 0 \\\\ 3 \\\\ 1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{r}-7 \\\\ 0 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -1\end{array}\right],\left[\begin{array}{r}0 \\\\ -3 \\\\ 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 x_{2}+7x_{4}=7\}
em forma afim “vetor particular + span de três vetores”. É um hiperplano de dimensão 3, pois impomos uma única equação linear em \mathbb{R}^{4}.

Passo 2

Para encontrar a solução geral, isolamos x_{2}:
x_{2}=7-7x_{4}.
Seja s=x_{1}, t=x_{3}, u=x_{4}. Então
(x_{1},x_{2},x_{3},x_{4}) =(s,\;7-7u,\;t,\;u) =(0,7,0,0)+s\,(1,0,0,0)+t\,(0,0,1,0)+u\,(0,-7,0,1).
Logo uma representação canônica é
\begin{pmatrix}0\\\ 7\\\ 0\\\ 0\end{pmatrix} +\operatorname{span}\Bigl\{\begin{pmatrix}1\\\ 0\\\ 0\\\ 0\end{pmatrix}, \begin{pmatrix}0\\\ 0\\\ 1\\\ 0\end{pmatrix}, \begin{pmatrix}0\\\ -7\\\ 0\\\ 1\end{pmatrix}\Bigr\}.

Passo 3

Uma condição para cada opção ser correta é:
  1. O vetor particular p deve satisfazer p_{2}+7p_{4}=7.
  2. Cada vetor da base da direção deve satisfazer a versão homogênea x_{2}+7x_{4}=0.
Testando rapidamente as opções, verifica-se que apenas na alternativa (E) esses dois requisitos valem simultaneamente:
– Vetor particular
p=\begin{pmatrix}3\\\ 21\\\ 1\\\ -2\end{pmatrix},
tem 21+7\cdot(-2)=21-14=7.
– Vetores da direção
v_{1}=\begin{pmatrix}0\\\ -7\\\ 0\\\ 1\end{pmatrix},\quad v_{2}=\begin{pmatrix}0\\\ 0\\\ 2\\\ 0\end{pmatrix},\quad v_{3}=\begin{pmatrix}7\\\ 0\\\ 0\\\ 0\end{pmatrix}
satisfazem
(-7)+7\cdot1=0,\quad 0+7\cdot0=0,\quad 0+7\cdot0=0,
respectivamente.

Resposta

(E) \displaystyle \begin{pmatrix}3\\\ 21\\\ 1\\\ -2\end{pmatrix} +\operatorname{span}\Bigl\{\begin{pmatrix}0\\\ -7\\\ 0\\\ 1\end{pmatrix}, \begin{pmatrix}0\\\ 0\\\ 2\\\ 0\end{pmatrix}, \begin{pmatrix}7\\\ 0\\\ 0\\\ 0\end{pmatrix}\Bigr\}

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