Q9: Seja \beta=\left\{\vec{v}_{1}, \vec{v}_{2}, \vec{v}_{3}\right\} uma base de \mathbb{R}^{3} e A uma matriz 3 \times 3 tal que
A \vec{v}_{1}=2 \vec{v}_{1}, \quad A \vec{v}_{2}=\vec{v}_{1}-2 \vec{v}_{3} \quad \text { e } \quad A \vec{v}_{3}=\vec{v}_{1}-3 \vec{v}_{3} .
Se M=\left[\begin{array}{lll}\vec{v}_{1} & \vec{v}_{2} & \vec{v}_{3}\end{array}\right], então:
(A) M A=\left[\begin{array}{rrr}2 & 0 & 0 \\\\ 1 & 0 & -2 \\\\ 1 & 0 & -3\end{array}\right] M
(B) M A=\left[\begin{array}{rrr}2 & -2 & 3 \\\\ 0 & 0 & 0 \\\\ 0 & 1 & 1\end{array}\right] M
(C) M A=\left[\begin{array}{rrr}2 & 1 & 1 \\\\ 0 & 0 & 0 \\\\ 0 & -2 & -3\end{array}\right] M
(D) A M=M\left[\begin{array}{rrr}2 & 0 & 0 \\\\ 1 & 0 & -2 \\\\ 1 & 0 & -3\end{array}\right]
(E) M A=\left[\begin{array}{rrr}2 & 1 & 1 \\\\ 0 & -2 & -3 \\\\ 0 & 0 & 0\end{array}\right] M
(F) A M=M\left[\begin{array}{rrr}2 & -2 & 3 \\\\ 0 & 0 & 0 \\\\ 0 & 1 & 1\end{array}\right]
(G) A M=M\left[\begin{array}{rrr}2 & 1 & 1 \\\\ 0 & -2 & -3 \\\\ 0 & 0 & 0\end{array}\right]
(H) A M=M\left[\begin{array}{rrr}2 & 1 & 1 \\\\ 0 & 0 & 0 \\\\ 0 & -2 & -3\end{array}\right]

Passo 1

O enunciado fornece a base \beta=\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}, a matriz
M = [\,\vec{v}_1\;\vec{v}_2\;\vec{v}_3\,]
e as imagens
\displaystyle A\vec{v}_1, A\vec{v}_2 e A\vec{v}_3. Queremos achar a matriz P tal que
A\,M \;=\; M\,P.

Passo 2

Note que
A\,M \;=\;\bigl[\,A\vec{v}_1,\;A\vec{v}_2,\;A\vec{v}_3\bigr].

Passo 3

Pelas hipóteses:
A\vec{v}_1 = 2\vec{v}_1,\quad A\vec{v}_2 = \vec{v}_1 - 2\vec{v}_3,\quad A\vec{v}_3 = \vec{v}_1 - 3\vec{v}_3.
Em coordenadas na base \beta, essas imagens correspondem às colunas
\begin{pmatrix}2\\\ 0\\\ 0\end{pmatrix},
\begin{pmatrix}1\\\ 0\\\ -2\end{pmatrix} e
\begin{pmatrix}1\\\ 0\\\ -3\end{pmatrix}, respectivamente.

Passo 4

Portanto, a matriz P que satisfaz A M = M P é
P = \begin{pmatrix} 2 & 1 & 1\\\\ 0 & 0 & 0\\\\ 0 & -2 & -3 \end{pmatrix}.

Passo 5

Comparando com as alternativas, vemos que a forma
A\,M \;=\; M \begin{pmatrix} 2 & 1 & 1\\\\ 0 & 0 & 0\\\\ 0 & -2 & -3 \end{pmatrix}
corresponde à opção (H).

Resposta

(H)

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