Dadas
\mathbb{A}=\left(\begin{array}{cc} -1 & 3 \\\\ 2 & -2 \end{array}\right) \quad \text { e } \quad \mathbb{B}=\left(\begin{array}{cc} 3 & -4 \\\\ 5 & 4 \end{array}\right)
determine \mathbb{X}=\frac{1}{2} \mathbb{A}-2 \mathbb{B}^{T} e sua transposta \mathbb{X}^{T}.

Passo 1

O enunciado dá
\mathbb{A}=\begin{pmatrix}-1 & 3 \\\\ 2 & -2\end{pmatrix} \quad\text{e}\quad \mathbb{B}=\begin{pmatrix}3 & -4 \\\\ 5 & 4\end{pmatrix}
e pede \mathbb{X}=\tfrac12\mathbb{A}-2\mathbb{B}^{T} e sua transposta \mathbb{X}^{T}. Vamos usar operações de multiplicação por escalar, transposta de matriz e soma/subtração de matrizes.

Passo 2

Multiplicamos \mathbb{A} por \tfrac12:
\tfrac12\,\mathbb{A} = \tfrac12 \begin{pmatrix}-1 & 3 \\\\ 2 & -2\end{pmatrix} = \begin{pmatrix}-\tfrac12 & \tfrac32 \\\\ 1 & -1\end{pmatrix}.

Passo 3

Calculamos a transposta de \mathbb{B}:
\mathbb{B}^{T} = \begin{pmatrix}3 & -4 \\\\ 5 & 4\end{pmatrix}^{T} = \begin{pmatrix}3 & 5 \\\\ -4 & 4\end{pmatrix}.

Passo 4

Multiplicamos \mathbb{B}^{T} por -2:
-2\,\mathbb{B}^{T} = -2 \begin{pmatrix}3 & 5 \\\\ -4 & 4\end{pmatrix} = \begin{pmatrix}-6 & -10 \\\\ 8 & -8\end{pmatrix}.

Passo 5

Agora subtraímos matrizes para obter \mathbb{X}:
\mathbb{X} = \tfrac12\,\mathbb{A} \;-\; 2\,\mathbb{B}^{T} = \begin{pmatrix}-\tfrac12 & \tfrac32 \\\\ 1 & -1\end{pmatrix} + \begin{pmatrix}-6 & -10 \\\\ 8 & -8\end{pmatrix} = \begin{pmatrix} -\tfrac12-6 & \tfrac32-10 \\[6pt] 1+8 & -1-8 \end{pmatrix} = \begin{pmatrix} -\tfrac{13}{2} & -\tfrac{17}{2} \\[3pt] 9 & -9 \end{pmatrix}.]

Passo 6

Para achar \mathbb{X}^{T}, basta trocar linhas por colunas em \mathbb{X}:
\mathbb{X}^{T} = \begin{pmatrix} -\tfrac{13}{2} & -\tfrac{17}{2} \\[3pt] 9 & -9 \end{pmatrix}^{T} = \begin{pmatrix} -\tfrac{13}{2} & 9 \\[3pt] -\tfrac{17}{2} & -9 \end{pmatrix}.]

Resposta

\mathbb{X} = \begin{pmatrix} -\tfrac{13}{2} & -\tfrac{17}{2} \\[3pt] 9 & -9 \end{pmatrix}, \qquad \mathbb{X}^{T} = \begin{pmatrix} -\tfrac{13}{2} & 9 \\[3pt] -\tfrac{17}{2} & -9 \end{pmatrix}.]

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