Dadas \underset{\sim}{A}=\left[\begin{array}{ll}1 & 2 \\\\ 3 & 4\end{array}\right] e \underset{\sim}{B}=\left[\begin{array}{ll}0 & 1 \\\\ 1 & 0\end{array}\right], verifique a propriedade (\underset{\sim}{A} \cdot \underset{\sim}{B})^{T}=\underset{\sim}{B^{T}} \cdot \underset{\sim}{A^{T}}.

Passo 1

O enunciado dá as matrizes
\underset{\sim}{A}=\begin{pmatrix}1 & 2 \\\\ 3 & 4\end{pmatrix} \quad\text{e}\quad \underset{\sim}{B}=\begin{pmatrix}0 & 1 \\\\ 1 & 0\end{pmatrix}
e pede para verificar a propriedade de transposta de produto de matrizes
(\underset{\sim}{A}\cdot \underset{\sim}{B})^{T} \;=\;\underset{\sim}{B}^{T}\!\cdot \underset{\sim}{A}^{T}.
Para isso, vamos calcular de um lado (\underset{\sim}{A}\cdot \underset{\sim}{B})^{T} e do outro \underset{\sim}{B}^{T}\!\cdot \underset{\sim}{A}^{T} e comparar.

Passo 2

Calculemos primeiro \underset{\sim}{A}\cdot \underset{\sim}{B} e depois sua transposta:
\underset{\sim}{A}\cdot \underset{\sim}{B} =\begin{pmatrix}1 & 2 \\\\ 3 & 4\end{pmatrix} \begin{pmatrix}0 & 1 \\\\ 1 & 0\end{pmatrix} =\begin{pmatrix} 1\cdot0+2\cdot1 & 1\cdot1+2\cdot0\\\\ 3\cdot0+4\cdot1 & 3\cdot1+4\cdot0 \end{pmatrix} =\begin{pmatrix}2 & 1 \\\\ 4 & 3\end{pmatrix}.
Portanto,
(\underset{\sim}{A}\cdot \underset{\sim}{B})^{T} =\begin{pmatrix}2 & 1 \\\\ 4 & 3\end{pmatrix}^{T} =\begin{pmatrix}2 & 4 \\\\ 1 & 3\end{pmatrix}.

Passo 3

Agora calculemos \underset{\sim}{B}^{T} e \underset{\sim}{A}^{T} e seu produto:
\underset{\sim}{B}^{T} =\begin{pmatrix}0 & 1 \\\\ 1 & 0\end{pmatrix}^{T} =\begin{pmatrix}0 & 1 \\\\ 1 & 0\end{pmatrix}, \quad \underset{\sim}{A}^{T} =\begin{pmatrix}1 & 2 \\\\ 3 & 4\end{pmatrix}^{T} =\begin{pmatrix}1 & 3 \\\\ 2 & 4\end{pmatrix}.
Então
\underset{\sim}{B}^{T}\!\cdot \underset{\sim}{A}^{T} =\begin{pmatrix}0 & 1 \\\\ 1 & 0\end{pmatrix} \begin{pmatrix}1 & 3 \\\\ 2 & 4\end{pmatrix} =\begin{pmatrix} 0\cdot1+1\cdot2 & 0\cdot3+1\cdot4\\\\ 1\cdot1+0\cdot2 & 1\cdot3+0\cdot4 \end{pmatrix} =\begin{pmatrix}2 & 4 \\\\ 1 & 3\end{pmatrix}.

Passo 4

Comparando os resultados obtidos em cada lado, vemos que
(\underset{\sim}{A}\cdot \underset{\sim}{B})^{T} =\begin{pmatrix}2 & 4 \\\\ 1 & 3\end{pmatrix} \quad\text{e}\quad \underset{\sim}{B}^{T}\!\cdot \underset{\sim}{A}^{T} =\begin{pmatrix}2 & 4 \\\\ 1 & 3\end{pmatrix},
logo ambos são iguais.

Resposta

A propriedade está verificada:
(\underset{\sim}{A}\cdot \underset{\sim}{B})^{T} =\underset{\sim}{B}^{T}\!\cdot \underset{\sim}{A}^{T}.

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