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 Passo 1O enunciado nos dá as matrizes B=\begin{pmatrix}4 & 0 & -3\\\ -1 & -2 & 3\end{pmatrix}\in\mathbb{R}^{2\times3} \quad\text{e}\quad C=\begin{pmatrix}2 & -3 & 0 & 1\\\ 5 & -1 & -4 & 2\\\ -1 & 0 & 0 & 3\end{pmatrix}\in\mathbb{R}^{3\times4}, e pede o produto BC. Como B é 2\times3 e C é 3\times4, o produto está bem-definido e resulta numa matriz 2\times4. Vamos usar a definição: cada elemento (BC)_{ij} é o produto escalar da linha i de B pela coluna j de C. Passo 2Seja B=(b_{ik}) e C=(c_{kj}). Então (BC)_{ij} \;=\;\sum_{k=1}^{3} b_{ik}\,c_{kj}\,. Passo 3Calculo da primeira linha de BC: \begin{aligned} (BC)_{11} &= 4\cdot2 + 0\cdot5 + (-3)\cdot(-1) = 8 + 0 + 3 = 11,\\\ (BC)_{12} &= 4\cdot(-3) + 0\cdot(-1) + (-3)\cdot0 = -12 + 0 + 0 = -12,\\\ (BC)_{13} &= 4\cdot0 + 0\cdot(-4) + (-3)\cdot0 = 0 + 0 + 0 = 0,\\\ (BC)_{14} &= 4\cdot1 + 0\cdot2 + (-3)\cdot3 = 4 + 0 - 9 = -5. \end{aligned} Passo 4Calculo da segunda linha de BC: \begin{aligned} (BC)_{21} &= (-1)\cdot2 + (-2)\cdot5 + 3\cdot(-1) = -2 -10 -3 = -15,\\\ (BC)_{22} &= (-1)\cdot(-3) + (-2)\cdot(-1) + 3\cdot0 = 3 + 2 + 0 = 5,\\\ (BC)_{23} &= (-1)\cdot0 + (-2)\cdot(-4) + 3\cdot0 = 0 + 8 + 0 = 8,\\\ (BC)_{24} &= (-1)\cdot1 + (-2)\cdot2 + 3\cdot3 = -1 -4 + 9 = 4. \end{aligned} Resposta BC = \begin{pmatrix} 11 & -12 & 0 & -5\\\\ -15 & 5 & 8 & 4 \end{pmatrix}. MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosExercício 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 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 D 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 D