Dadas as matrizes A=\left[\begin{array}{ccc}0 & 1 & 3 \\\\ 4 & -2 & 5 \\\\ -1 & 2 & 7\end{array}\right], \quad B=\left[\begin{array}{ccc}2 & 4 & -1 \\\\ 3 & 0 & 1 \\\\ 1 & 2 & 5\end{array}\right] e C=\left[\begin{array}{ccc}-1 & 2 & 0 \\\\ 3 & 4 & 6 \\\\ 1 & 0 & -1\end{array}\right] Obtenha A(B+C) Passo 1O enunciado dá as matrizes A=\begin{pmatrix}0 & 1 & 3 \\\\ 4 & -2 & 5 \\\\ -1 & 2 & 7\end{pmatrix},\quad B=\begin{pmatrix}2 & 4 & -1 \\\\ 3 & 0 & 1 \\\\ 1 & 2 & 5\end{pmatrix},\quad C=\begin{pmatrix}-1 & 2 & 0 \\\\ 3 & 4 & 6 \\\\ 1 & 0 & -1\end{pmatrix} e pede o produto A(B+C). O conteúdo é soma de matrizes e multiplicação de matrizes, usando a regra de que cada elemento do produto é o produto escalar da linha de A com a coluna de B+C. Passo 2Primeiro calculamos a soma B+C somando termo a termo: B+C =\begin{pmatrix} 2+(-1) & 4+2 & -1+0 \\\\ 3+3 & 0+4 & 1+6 \\\\ 1+1 & 2+0 & 5+(-1) \end{pmatrix} =\begin{pmatrix} 1 & 6 & -1 \\\\ 6 & 4 & 7 \\\\ 2 & 2 & 4 \end{pmatrix}. Passo 3Agora definimos D = B + C e calculamos A\,D. Cada elemento (A\,D)_{ij} é dado por (A\,D)_{ij} =\sum_{k=1}^{3}A_{ik}\,D_{kj}, isto é, o produto escalar da i-ésima linha de A com a j-ésima coluna de D. Passo 4Calculando linha a linha: 1) Primeira linha de A é (0,1,3). (A\,D)_{11}=0\cdot1+1\cdot6+3\cdot2=12,\quad (A\,D)_{12}=0\cdot6+1\cdot4+3\cdot2=10,\quad (A\,D)_{13}=0\cdot(-1)+1\cdot7+3\cdot4=19. 2) Segunda linha de A é (4,-2,5). (A\,D)_{21}=4\cdot1+(-2)\cdot6+5\cdot2=2,\quad (A\,D)_{22}=4\cdot6+(-2)\cdot4+5\cdot2=26,\quad (A\,D)_{23}=4\cdot(-1)+(-2)\cdot7+5\cdot4=2. 3) Terceira linha de A é (-1,2,7). (A\,D)_{31}=(-1)\cdot1+2\cdot6+7\cdot2=25,\quad (A\,D)_{32}=(-1)\cdot6+2\cdot4+7\cdot2=16,\quad (A\,D)_{33}=(-1)\cdot(-1)+2\cdot7+7\cdot4=43. Resposta A(B+C)=\begin{pmatrix} 12 & 10 & 19 \\\\ 2 & 26 & 2 \\\\ 25 & 16 & 43 \end{pmatrix}. MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosDadas as matrizes A=\left[\begin{array}{ccc}0 & 1 & 3 \\\\ 4 & -2 & 5 \\\\ -1 & 2 & 7\end{array}\right], \quad B=\left[\begin{array}{ccc}2 & 4 & -1 \\\\ 3 & 0 & 1 \\\\ 1 & 2 & 5\end{array}\right] e C=\left[\begin{array}{ccc}-1 & 2 & 0 \\\\ 3 & 4 & 6 \\\\ 1 & 0 & -1\end{array}\right] Obtenha A(B+C) Dadas as matrizes A=\left[\begin{array}{ccc}0 & 1 & 3 \\\\ 4 & -2 & 5 \\\\ -1 & 2 & 7\end{array}\right], \quad B=\left[\begin{array}{ccc}2 & 4 & -1 \\\\ 3 & 0 & 1 \\\\ 1 & 2 & 5\end{array}\right] e C=\left[\begin{array}{ccc}-1 & 2 & 0 \\\\ 3 & 4 & 6 \\\\ 1 & 0 & -1\end{array}\right] Obtenha BC Dadas as matrizes A=\left[\begin{array}{ccc}0 & 1 & 3 \\\\ 4 & -2 & 5 \\\\ -1 & 2 & 7\end{array}\right], \quad B=\left[\begin{array}{ccc}2 & 4 & -1 \\\\ 3 & 0 & 1 \\\\ 1 & 2 & 5\end{array}\right] e C=\left[\begin{array}{ccc}-1 & 2 & 0 \\\\ 3 & 4 & 6 \\\\ 1 & 0 & -1\end{array}\right] Obtenha B(A-B) Dadas as matrizes A=\left[\begin{array}{ccc}0 & 1 & 3 \\\\ 4 & -2 & 5 \\\\ -1 & 2 & 7\end{array}\right], \quad B=\left[\begin{array}{ccc}2 & 4 & -1 \\\\ 3 & 0 & 1 \\\\ 1 & 2 & 5\end{array}\right] e C=\left[\begin{array}{ccc}-1 & 2 & 0 \\\\ 3 & 4 & 6 \\\\ 1 & 0 & -1\end{array}\right] Obtenha B(A-B)