Sendo \underset{\sim}{C}=\left[\begin{array}{cc}1 & 0 \\\\ 5 & -2\end{array}\right], \underset{\sim}{D}=\left[\begin{array}{cc}-3 & 4 \\\\ 1 & 1\end{array}\right] e o escalar \alpha=5 :
Verifique que (\underset{\sim}{C}+\underset{\sim}{D})^{T}={\underset{\sim}{C}}^{T}+{\underset{\sim}{D}}^{T}.

Passo 1

O enunciado dá as matrizes
\underset{\sim}{C} = \begin{pmatrix}1 & 0 \\[6pt]5 & -2\end{pmatrix}, \quad \underset{\sim}{D} = \begin{pmatrix}-3 & 4 \\[6pt]1 & 1\end{pmatrix}
e pede para verificar a propriedade de transposta da soma, ou seja
(\underset{\sim}{C} + \underset{\sim}{D})^{T} \;=\; \underset{\sim}{C}^{T} \;+\;\underset{\sim}{D}^{T}.
Para isso vamos calcular separadamente \underset{\sim}{C} + \underset{\sim}{D}, sua transposta, e depois as transpostas individuais somadas.]

Passo 2

Calcular a soma das matrizes:
\underset{\sim}{C} + \underset{\sim}{D} = \begin{pmatrix}1 + (-3) & 0 + 4 \\[4pt]5 + 1 & -2 + 1\end{pmatrix} = \begin{pmatrix}-2 & 4 \\[4pt]6 & -1\end{pmatrix}.]

Passo 3

Transpor a soma obtida:
(\underset{\sim}{C} + \underset{\sim}{D})^{T} = \begin{pmatrix}-2 & 4 \\[4pt]6 & -1\end{pmatrix}^{T} = \begin{pmatrix}-2 & 6 \\[4pt]4 & -1\end{pmatrix}.]

Passo 4

Calcular as transpostas individuais:
\underset{\sim}{C}^{T} = \begin{pmatrix}1 & 0 \\[4pt]5 & -2\end{pmatrix}^{T} = \begin{pmatrix}1 & 5 \\[4pt]0 & -2\end{pmatrix}, \quad \underset{\sim}{D}^{T} = \begin{pmatrix}-3 & 4 \\[4pt]1 & 1\end{pmatrix}^{T} = \begin{pmatrix}-3 & 1 \\[4pt]4 & 1\end{pmatrix}.]

Passo 5

Somar as transpostas:
\underset{\sim}{C}^{T} + \underset{\sim}{D}^{T} = \begin{pmatrix}1 + (-3) & 5 + 1 \\[4pt]0 + 4 & -2 + 1\end{pmatrix} = \begin{pmatrix}-2 & 6 \\[4pt]4 & -1\end{pmatrix}.]

Resposta

Como
(\underset{\sim}{C} + \underset{\sim}{D})^{T} = \begin{pmatrix}-2 & 6 \\[4pt]4 & -1\end{pmatrix} \quad\text{e}\quad \underset{\sim}{C}^{T} + \underset{\sim}{D}^{T} = \begin{pmatrix}-2 & 6 \\[4pt]4 & -1\end{pmatrix},
vemos que
(\underset{\sim}{C} + \underset{\sim}{D})^{T} = \underset{\sim}{C}^{T} + \underset{\sim}{D}^{T}.]

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