Dadas as matrizes A=\left(\begin{array}{cc}2 & 0 \\\\ -1 & 3\end{array}\right) e B=\left(\begin{array}{cc}2 & -1 / 2 \\\\ 3 & 1\end{array}\right), então a matriz -2AB é igual a:
\left(\begin{array}{cc}8 & -2 \\\\ 14 & 7\end{array}\right)
b) \left(\begin{array}{cc}-8 & -2 \\\\ 14 & 7\end{array}\right)
c) \left(\begin{array}{ll}-8 & -2 \\\\ -14 & -7\end{array}\right)
d) \left(\begin{array}{cc}8 & 2 \\\\ 14 & 7\end{array}\right)
е) \left(\begin{array}{cc}-8 & 2 \\\\ -14 & -7\end{array}\right)

Passo 1

O enunciado dá as matrizes
A = \begin{pmatrix}2 & 0 \\[6pt] -1 & 3\end{pmatrix} \quad\text{e}\quad B = \begin{pmatrix}2 & -\tfrac12 \\[4pt] 3 & 1\end{pmatrix}
e pede para calcular -2AB. O conteúdo principal envolve produto de matrizes e multiplicação escalar.]

Passo 2

Primeiro calculamos o produto AB usando a regra linha × coluna:
AB = \begin{pmatrix}2 & 0 \\[4pt] -1 & 3\end{pmatrix} \begin{pmatrix}2 & -\tfrac12 \\[4pt] 3 & 1\end{pmatrix} = \begin{pmatrix} 2\cdot2 + 0\cdot3 & 2\cdot(-\tfrac12) + 0\cdot1 \\[6pt] (-1)\cdot2 + 3\cdot3 & (-1)\cdot(-\tfrac12) + 3\cdot1 \end{pmatrix} = \begin{pmatrix}4 & -1 \\[4pt] 7 & \tfrac{7}{2}\end{pmatrix}.]

Passo 3

Agora multiplicamos a matriz AB pelo escalar -2:
-2AB = -2 \begin{pmatrix}4 & -1 \\[4pt] 7 & \tfrac{7}{2}\end{pmatrix} = \begin{pmatrix}-8 & 2 \\[4pt] -14 & -7\end{pmatrix}.]

Resposta

e) \displaystyle \begin{pmatrix}-8 & 2 \\\\ -14 & -7\end{pmatrix}

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