Questão 4: Considere as seguintes matrizes:
\begin{aligned} A & =\left[\begin{array}{ll} 2 & 0 \\\\ 6 & 7 \end{array}\right] & D & =\left[\begin{array}{ccc} -6 & 4 & 0 \\\\ 1 & 1 & 4 \\\\ -6 & 0 & 6 \end{array}\right] \\\ B & =\left[\begin{array}{cc} 0 & 4 \\\\ 2 & -8 \end{array}\right] & & E=\left[\begin{array}{ccc} 6 & 9 & -9 \\\\ -1 & 0 & -4 \\\\ -6 & 0 & -1 \end{array}\right] \\\ C & =\left[\begin{array}{ccc} -6 & 9 & -7 \\\\ 7 & -3 & -2 \end{array}\right] & & \end{aligned}
Se for possível, calcule:
\left(2 D^{\top}-3 E^{\top}\right)^{\top}

Passo 1

O enunciado pede para calcular
(2D^{\top}-3E^{\top})^{\top}
onde
D=\begin{pmatrix}-6 & 4 & 0\\\ 1 & 1 & 4\\\ -6 & 0 & 6\end{pmatrix} \quad\text{e}\quad E=\begin{pmatrix}6 & 9 & -9\\\ -1 & 0 & -4\\\ -6 & 0 & -1\end{pmatrix}.
O conteúdo necessário é a propriedade da transposição de soma e escalar, e a dupla transposição.

Passo 2

Usando as propriedades
(aM + bN)^{\top} = a\,M^{\top} + b\,N^{\top} \quad\text{e}\quad (M^{\top})^{\top}=M,
temos
(2D^{\top}-3E^{\top})^{\top} =2\,(D^{\top})^{\top}-3\,(E^{\top})^{\top} =2D-3E.

Passo 3

Calculemos primeiro 2D e 3E:
2D =\begin{pmatrix} 2\cdot(-6) & 2\cdot4 & 2\cdot0\\\\ 2\cdot1 & 2\cdot1 & 2\cdot4\\\\ 2\cdot(-6) & 2\cdot0 & 2\cdot6 \end{pmatrix} = \begin{pmatrix} -12 & 8 & 0\\\\ 2 & 2 & 8\\\\ -12 & 0 & 12 \end{pmatrix},
3E =\begin{pmatrix} 3\cdot6 & 3\cdot9 & 3\cdot(-9)\\\\ 3\cdot(-1) & 3\cdot0 & 3\cdot(-4)\\\\ 3\cdot(-6) & 3\cdot0 & 3\cdot(-1) \end{pmatrix} = \begin{pmatrix} 18 & 27 & -27\\\\ -3 & 0 & -12\\\\ -18 & 0 & -3 \end{pmatrix}.

Passo 4

Agora subtraímos termo a termo:
2D-3E = \begin{pmatrix} -12-18 & 8-27 & 0-(-27)\\\\ 2-(-3) & 2-0 & 8-(-12)\\\\ -12-(-18) & 0-0 & 12-(-3) \end{pmatrix} = \begin{pmatrix} -30 & -19 & 27\\\\ 5 & 2 & 20\\\\ 6 & 0 & 15 \end{pmatrix}.

Resposta

(2D^{\top}-3E^{\top})^{\top} = \begin{pmatrix} -30 & -19 & 27\\\\ 5 & 2 & 20\\\\ 6 & 0 & 15 \end{pmatrix}.

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