Sejam A=\left[\begin{array}{ll} 1 & 2 \\\\ 4 & 5 \end{array}\right] ; B=\left[\begin{array}{ccc} -2 & 0 & 1 \\\\ 3 & 0 & 1 \end{array}\right], C=\left[\begin{array}{c} -1 \\\\ 2 \\\\ 3 \end{array}\right] Encontre, se possível: \frac{3}{5} B Passo 1O enunciado nos dá a matriz B = \begin{pmatrix} -2 & 0 & 1\\\\ 3 & 0 & 1 \end{pmatrix} e pede a multiplicação escalar \tfrac{3}{5}B. O conteúdo aqui é multiplicação escalar de matrizes, que consiste em multiplicar cada entrada de B pelo escalar \tfrac{3}{5}. Passo 2Pela definição de multiplicação escalar, se B=(b_{ij}) então \Bigl(\tfrac{3}{5}B\Bigr)_{ij} = \tfrac{3}{5}\,b_{ij}. Vamos aplicar isso a cada entrada de B. Passo 3Calculando cada elemento: \begin{aligned} \tfrac{3}{5}\cdot(-2)&=-\tfrac{6}{5},& \tfrac{3}{5}\cdot0&=0,& \tfrac{3}{5}\cdot1&=\tfrac{3}{5},\\\ \tfrac{3}{5}\cdot3&=\tfrac{9}{5},& \tfrac{3}{5}\cdot0&=0,& \tfrac{3}{5}\cdot1&=\tfrac{3}{5}. \end{aligned} Logo, \tfrac{3}{5}B =\begin{pmatrix} -\tfrac{6}{5} & 0 & \tfrac{3}{5}\\\ \tfrac{9}{5} & 0 & \tfrac{3}{5} \end{pmatrix}. Resposta \tfrac{3}{5}B = \begin{pmatrix} -\tfrac{6}{5} & 0 & \tfrac{3}{5}\\\ \tfrac{9}{5} & 0 & \tfrac{3}{5} \end{pmatrix} MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosSejam A=\left[\begin{array}{ll} 1 & 2 \\\\ 4 & 5 \end{array}\right] ; B=\left[\begin{array}{ccc} -2 & 0 & 1 \\\\ 3 & 0 & 1 \end{array}\right], C=\left[\begin{array}{c} -1 \\\\ 2 \\\\ 3 \end{array}\right] Encontre, se possível: -2 A Sejam A=\left[\begin{array}{ll} 1 & 2 \\\\ 4 & 5 \end{array}\right] ; B=\left[\begin{array}{ccc} -2 & 0 & 1 \\\\ 3 & 0 & 1 \end{array}\right], C=\left[\begin{array}{c} -1 \\\\ 2 \\\\ 3 \end{array}\right] Encontre, se possível: B \cdot C Sejam A=\left[\begin{array}{ll} 1 & 2 \\\\ 4 & 5 \end{array}\right] ; B=\left[\begin{array}{ccc} -2 & 0 & 1 \\\\ 3 & 0 & 1 \end{array}\right], C=\left[\begin{array}{c} -1 \\\\ 2 \\\\ 3 \end{array}\right] Encontre, se possível: A \cdot A Sejam A=\left[\begin{array}{ll} 1 & 2 \\\\ 4 & 5 \end{array}\right] ; B=\left[\begin{array}{ccc} -2 & 0 & 1 \\\\ 3 & 0 & 1 \end{array}\right], C=\left[\begin{array}{c} -1 \\\\ 2 \\\\ 3 \end{array}\right] Encontre, se possível: (B \cdot C)^{T}