Considere as seguintes matrizes:
A=\left[\begin{array}{cc} -1 & 0 \\\\ 5 & 2 \end{array}\right], \quad B=\left[\begin{array}{cc} 3 & 7 \\\\ -2 & 1 \end{array}\right], \quad C=\left[\begin{array}{ccc} 5 & -3 & 1 \\\\ 0 & 2 & 0 \end{array}\right] \\\ D=\left[\begin{array}{ccc} 0 & 7 & 2 \\\\ 9 & -2 & -2 \\\\ 1 & 3 & 2 \end{array}\right], \quad E=\left[\begin{array}{ccc} 0 & 3 & -2 \\\\ 9 & 0 & 0 \\\\ 1 & -1 & 2 \end{array}\right]
Se for possível, calcule:
\left(2 D^{t}-3 E\right)^{t}

Passo 1

O enunciado pede para calcular
(2D^{t}-3E)^{t}
sendo D e E matrizes 3\times3. O conteúdo envolvido é a propriedade da transposta de soma e produto por escalar.

Passo 2

Usamos que para quaisquer matrizes compatíveis e escalar k vale
(M + N)^{t} = M^{t} + N^{t}, \quad (kM)^{t} = k\,M^{t}
Logo,
(2D^{t}-3E)^{t} = (2D^{t})^{t} - (3E)^{t} = 2\,D - 3\,E^{t}.

Passo 3

Primeiro calculamos
2D =2\begin{pmatrix} 0 & 7 & 2\\\\ 9 & -2 & -2\\\\ 1 & 3 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 14 & 4\\\\ 18 & -4 & -4\\\\ 2 & 6 & 4 \end{pmatrix},
depois
E^{t} =\begin{pmatrix} 0 & 3 & -2\\\\ 9 & 0 & 0\\\\ 1 & -1 & 2 \end{pmatrix}^{t} = \begin{pmatrix} 0 & 9 & 1\\\\ 3 & 0 & -1\\\\ -2 & 0 & 2 \end{pmatrix},
e então
3E^{t} =\begin{pmatrix} 0 & 27 & 3\\\\ 9 & 0 & -3\\\\ -6 & 0 & 6 \end{pmatrix}.
Finalmente,
2D - 3E^{t} = \begin{pmatrix} 0 & 14 & 4\\\\ 18 & -4 & -4\\\\ 2 & 6 & 4 \end{pmatrix} - \begin{pmatrix} 0 & 27 & 3\\\\ 9 & 0 & -3\\\\ -6 & 0 & 6 \end{pmatrix} = \begin{pmatrix} 0 & -13 & 1\\\\ 9 & -4 & -1\\\\ 8 & 6 & -2 \end{pmatrix}.

Resposta

(2D^{t}-3E)^{t} = \begin{pmatrix} 0 & -13 & 1\\\\ 9 & -4 & -1\\\\ 8 & 6 & -2 \end{pmatrix}.

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