Exercício 4. Sejam A=\left(\begin{array}{cc} 1 & 2 \\\\ 3 & -4 \end{array}\right), \quad B=\left(\begin{array}{cc} 5 & 0 \\\\ -6 & 7 \end{array}\right), \quad C=\left(\begin{array}{ccc} 1 & -3 & 4 \\\\ 2 & 6 & -5 \end{array}\right), \quad D=\left(\begin{array}{ccc} 3 & 7 & -1 \\\\ 4 & -8 & 9 \end{array}\right) Encontre (a) A^{T}, (b) B^{T}, (c) (A B)^{T}, (d) A^{T} B^{T}. Passo 1O enunciado nos dá as matrizes A = \begin{pmatrix}1 & 2 \\[6pt] 3 & -4\end{pmatrix}, \quad B = \begin{pmatrix}5 & 0 \\[6pt] -6 & 7\end{pmatrix} e pede: (a) A^{T}, (b) B^{T}, (c) (AB)^{T}, (d) A^{T}B^{T}. O conteúdo principal é definição de transposta e multiplicação de matrizes.] Passo 2Para achar a transposta de A, basta trocar linhas por colunas: A^{T} = \begin{pmatrix} 1 & 3 \\\\ 2 & -4 \end{pmatrix}. Passo 3De modo análogo, a transposta de B é B^{T} = \begin{pmatrix} 5 & -6 \\\\ 0 & 7 \end{pmatrix}. Passo 4Primeiro calculamos o produto AB. Seja C=AB; cada entrada c_{ij} é c_{ij} = \sum_{k=1}^{2} a_{ik}\,b_{kj}. Isso dá AB = \begin{pmatrix} 1\cdot5 + 2\cdot(-6) & 1\cdot0 + 2\cdot7 \\[4pt] 3\cdot5 + (-4)\cdot(-6) & 3\cdot0 + (-4)\cdot7 \end{pmatrix} = \begin{pmatrix} -7 & 14 \\\\ 39 & -28 \end{pmatrix}. Agora, usando a definição de transposta, trocamos linhas por colunas em AB: (AB)^{T} = \begin{pmatrix} -7 & 39 \\\\ 14 & -28 \end{pmatrix}.] Passo 5Para A^{T}B^{T}, multiplicamos \displaystyle A^{T}=\begin{pmatrix}1&3\\\ 2&-4\end{pmatrix} por \displaystyle B^{T}=\begin{pmatrix}5&-6\\\ 0&7\end{pmatrix}. Cada entrada é o produto escalar de linha de A^{T} por coluna de B^{T}: A^{T}B^{T} = \begin{pmatrix} 1\cdot5 + 3\cdot0 & 1\cdot(-6) + 3\cdot7 \\[4pt] 2\cdot5 + (-4)\cdot0 & 2\cdot(-6) + (-4)\cdot7 \end{pmatrix} = \begin{pmatrix} 5 & 15 \\\\ 10 & -40 \end{pmatrix}.] Resposta(a) A^{T} = \begin{pmatrix}1 & 3 \\\\ 2 & -4\end{pmatrix} (b) B^{T} = \begin{pmatrix}5 & -6 \\\\ 0 & 7\end{pmatrix} (c) (AB)^{T} = \begin{pmatrix}-7 & 39 \\\\ 14 & -28\end{pmatrix} (d) A^{T}B^{T} = \begin{pmatrix}5 & 15 \\\\ 10 & -40\end{pmatrix} MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosExercício 3. Sejam A=\left(\begin{array}{cc} 1 & 2 \\\\ 3 & -4 \end{array}\right), \quad B=\left(\begin{array}{cc} 5 & 0 \\\\ -6 & 7 \end{array}\right), \quad C=\left(\begin{array}{ccc} 1 & -3 & 4 \\\\ 2 & 6 & -5 \end{array}\right), \quad D=\left(\begin{array}{ccc} 3 & 7 & -1 \\\\ 4 & -8 & 9 \end{array}\right) Encontre C D Exercício 3. Sejam A=\left(\begin{array}{cc} 1 & 2 \\\\ 3 & -4 \end{array}\right), \quad B=\left(\begin{array}{cc} 5 & 0 \\\\ -6 & 7 \end{array}\right), \quad C=\left(\begin{array}{ccc} 1 & -3 & 4 \\\\ 2 & 6 & -5 \end{array}\right), \quad D=\left(\begin{array}{ccc} 3 & 7 & -1 \\\\ 4 & -8 & 9 \end{array}\right) Encontre C D Exercício 4. Sejam A=\left(\begin{array}{cc} 1 & 2 \\\\ 3 & -4 \end{array}\right), \quad B=\left(\begin{array}{cc} 5 & 0 \\\\ -6 & 7 \end{array}\right), \quad C=\left(\begin{array}{ccc} 1 & -3 & 4 \\\\ 2 & 6 & -5 \end{array}\right), \quad D=\left(\begin{array}{ccc} 3 & 7 & -1 \\\\ 4 & -8 & 9 \end{array}\right) Encontre (a) A^{T}, (b) B^{T}, (c) (A B)^{T}, (d) A^{T} B^{T}. Exercício 5. Sejam A=\left(\begin{array}{ccc} 1 & -1 & 2 \\\\ 0 & 3 & 4 \end{array}\right), \quad B=\left(\begin{array}{ccc} 4 & 0 & -3 \\\\ -1 & -2 & 3 \end{array}\right), \quad C=\left(\begin{array}{cccc} 2 & -3 & 0 & 1 \\\\ 5 & -1 & -4 & 2 \\\\ -1 & 0 & 0 & 3 \end{array}\right), \quad D=\left(\begin{array}{c} 2 \\\\ -1 \\\\ 3 \end{array}\right) Encontre B C