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
AB

Passo 1

O 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}
e pede AB. Para multiplicar matrizes usamos a definição
(AB)_{ij} = \sum_{k=1}^{3} A_{ik}\,B_{kj}.

Passo 2

Calculamos a primeira linha de AB usando a primeira linha de A, que é (0,1,3):
(AB)_{11} = 0\cdot2 + 1\cdot3 + 3\cdot1 = 6,
(AB)_{12} = 0\cdot4 + 1\cdot0 + 3\cdot2 = 6,
(AB)_{13} = 0\cdot(-1) + 1\cdot1 + 3\cdot5 = 16.

Passo 3

Calculamos a segunda linha de AB usando (4,-2,5):
(AB)_{21} = 4\cdot2 + (-2)\cdot3 + 5\cdot1 = 7,
(AB)_{22} = 4\cdot4 + (-2)\cdot0 + 5\cdot2 = 26,
(AB)_{23} = 4\cdot(-1) + (-2)\cdot1 + 5\cdot5 = 19.

Passo 4

Calculamos a terceira linha de AB usando (-1,2,7):
(AB)_{31} = (-1)\cdot2 + 2\cdot3 + 7\cdot1 = 11,
(AB)_{32} = (-1)\cdot4 + 2\cdot0 + 7\cdot2 = 10,
(AB)_{33} = (-1)\cdot(-1) + 2\cdot1 + 7\cdot5 = 38.

Resposta

AB = \begin{pmatrix} 6 & 6 & 16\\\\ 7 & 26 & 19\\\\ 11 & 10 & 38 \end{pmatrix}

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