Q4: O conjunto dos \left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} tais que
-7 x_{1}+x_{2}=2 pode ser representado por:
(A) \left[\begin{array}{r}1 \\\\ 9 \\\\ 2 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 7 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 0 \\\\ 7\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right]\right\}
(B) \left[\begin{array}{r}1 \\\\ -19 \\\\ 3 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}1 \\\\ 7 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 5 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -4\end{array}\right]\right\}
(C) \left[\begin{array}{r}-1 \\\\ -5 \\\\ -3 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 7 \\\\ 1\end{array}\right],\left[\begin{array}{r}-1 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ -5 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(D) \left[\begin{array}{r}-1 \\\\ -5 \\\\ -3 \\\\ 2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 7\end{array}\right],\left[\begin{array}{l}0 \\\\ 4 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}3 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(E) \left[\begin{array}{r}1 \\\\ -19 \\\\ 3 \\\\ -2\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}7 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 0 \\\\ 2 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -7\end{array}\right]\right\}
(F) \left[\begin{array}{r}2 \\\\ 16 \\\\ -2 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 1 \\\\ 7\end{array}\right],\left[\begin{array}{l}0 \\\\ 5 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}-3 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right]\right\}
(G) \left[\begin{array}{r}1 \\\\ 9 \\\\ 2 \\\\ -1\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}7 \\\\ 1 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ 0 \\\\ -3\end{array}\right],\left[\begin{array}{r}0 \\\\ 0 \\\\ -6 \\\\ 0\end{array}\right]\right\}
(H) \left[\begin{array}{r}2 \\\\ 16 \\\\ -2 \\\\ -3\end{array}\right]+\operatorname{span}\left\{\left[\begin{array}{l}0 \\\\ 0 \\\\ 7 \\\\ 1\end{array}\right],\left[\begin{array}{l}1 \\\\ 0 \\\\ 0 \\\\ 0\end{array}\right],\left[\begin{array}{l}0 \\\\ 6 \\\\ 0 \\\\ 0\end{array}\right]\right\}

Passo 1

O enunciado pede descrever o conjunto
\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}\mid -7x_{1}+x_{2}=2\}
como uma reta afim em \mathbb{R}^4, ou seja, como
p + \span\{v_{1},v_{2},v_{3}\},
onde p é uma solução particular de -7x_{1}+x_{2}=2 e \{v_{1},v_{2},v_{3}\} é uma base do espaço solução da homogênea -7x_{1}+x_{2}=0.

Passo 2

Resolvendo a homogênea
-7x_{1}+x_{2}=0
em \mathbb{R}^4, tomamos x_{1}=u, x_{3}=v, x_{4}=w livres e então
x_{2}=7u.
Logo a solução geral da homogênea é
(u,\,7u,\,v,\,w) = u\,(1,7,0,0) + v\,(0,0,1,0) + w\,(0,0,0,1).
Então uma base do núcleo é
v_{1}=(1,7,0,0),\quad v_{2}=(0,0,1,0),\quad v_{3}=(0,0,0,1).

Passo 3

Buscamos agora uma solução particular de
-7x_{1}+x_{2}=2.
Basta escolher u=0, v=0, w=0 e impor x_{2}=2, ou seja
p=(0,2,0,0).
Portanto a solução geral da afim é
(0,2,0,0) + \span\{(1,7,0,0),\,(0,0,1,0),\,(0,0,0,1)\}.

Passo 4

Para cada alternativa, verificamos duas coisas:
  1. O vetor p satisfaz -7p_{1}+p_{2}=2.
  2. Cada vetor da base satisfaz -7v_{i1}+v_{i2}=0.
    Fazendo essas contas rapidamente, apenas a opção (A) tem
p=(1,9,2,-1),\quad -7\cdot1+9=2,
e os vetores
(1,7,0,0),\;(0,0,0,7),\;(0,0,-2,0)
satisfazem -7\cdot1+7=0, -7\cdot0+0=0, -7\cdot0+0=0.

Passo 5

As demais opções ou não cumprem -7p_{1}+p_{2}=2 ou algum vetor do span não anula -7x_{1}+x_{2}.

Resposta

(A) \displaystyle \begin{bmatrix}1\\\ 9\\\ 2\\\ -1\end{bmatrix} +\span\Bigl\{\begin{bmatrix}1\\\ 7\\\ 0\\\ 0\end{bmatrix}, \begin{bmatrix}0\\\ 0\\\ 0\\\ 7\end{bmatrix}, \begin{bmatrix}0\\\ 0\\\ -2\\\ 0\end{bmatrix}\Bigr\}

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