Exercício 6. 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 A^{T} B Passo 1O enunciado pede para calcular o produto entre a transposta de A e a matriz B, ou seja, encontrar A^{T}B. Vamos usar a definição de matriz transposta e a regra de multiplicação de matrizes. Passo 2Primeiro obtemos a transposta de A. Como A = \begin{pmatrix} 1 & -1 & 2 \\[6pt] 0 & 3 & 4 \end{pmatrix}, a transposta A^T é A^T = \begin{pmatrix} 1 & 0 \\[4pt] -1 & 3 \\[4pt] 2 & 4 \end{pmatrix}.] Passo 3Para multiplicar A^T (de ordem 3\times2) por B (de ordem 2\times3), usamos (A^T B)_{ij}=\sum_{k=1}^{2}(A^T)_{ik}\,B_{kj}. Aplicando essa fórmula a todos os elementos, primeiro escrevemos os produtos parciais: A^T B = \begin{pmatrix} 1\cdot4 + 0\cdot(-1) & 1\cdot0 + 0\cdot(-2) & 1\cdot(-3) + 0\cdot3 \\[6pt] -1\cdot4 + 3\cdot(-1) & -1\cdot0 + 3\cdot(-2) & -1\cdot(-3) + 3\cdot3 \\[6pt] 2\cdot4 + 4\cdot(-1) & 2\cdot0 + 4\cdot(-2) & 2\cdot(-3) + 4\cdot3 \end{pmatrix} e então simplificamos cada entrada: A^T B = \begin{pmatrix} 4 & 0 & -3 \\[4pt] -7 & -6 & 12 \\[4pt] 4 & -8 & 6 \end{pmatrix}.] Resposta A^{T}B = \begin{pmatrix} 4 & 0 & -3 \\[4pt] -7 & -6 & 12 \\[4pt] 4 & -8 & 6 \end{pmatrix}] MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosExercício 6. 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 A^{T} B Exercício 6. 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 A^{T} B Exercício 6. 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 A^{T} C Exercício 6. 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 A^{T} C