Exercício 2. 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
B C e A(B C)

Passo 1

O enunciado dá as matrizes
A = \begin{pmatrix}1 & 2 \\\\ 3 & -4\end{pmatrix},\quad B = \begin{pmatrix}5 & 0 \\\\ -6 & 7\end{pmatrix},\quad C = \begin{pmatrix}1 & -3 & 4 \\\\ 2 & 6 & -5\end{pmatrix},
e pede primeiro B\,C (produto de uma 2\times2 por 2\times3) e depois A\,(B\,C). Vamos usar a definição de multiplicação de matrizes, em que
(BC)_{ij} = \sum_{k=1}^{2} B_{ik}\,C_{kj}.

Passo 2

Calculamos B\,C entrada a entrada:
(BC)_{11} = 5\cdot1 + 0\cdot2 = 5, \quad (BC)_{12} = 5\cdot(-3) + 0\cdot6 = -15, \quad (BC)_{13} = 5\cdot4 + 0\cdot(-5) = 20;
(BC)_{21} = (-6)\cdot1 + 7\cdot2 = -6 + 14 = 8, \quad (BC)_{22} = (-6)\cdot(-3) + 7\cdot6 = 18 + 42 = 60, \quad (BC)_{23} = (-6)\cdot4 + 7\cdot(-5) = -24 - 35 = -59.
Portanto,
BC = \begin{pmatrix} 5 & -15 & 20 \\\\ 8 & 60 & -59 \end{pmatrix}.

Passo 3

Agora multiplicamos A pela matriz BC. Para cada entrada de A(BC):
(A\,(BC))_{1j} = 1\cdot(BC)_{1j} + 2\cdot(BC)_{2j}, \quad (A\,(BC))_{2j} = 3\cdot(BC)_{1j} + (-4)\cdot(BC)_{2j}.
Fazendo j=1,2,3:
(A\,BC)_{11} = 1\cdot5 + 2\cdot8 = 21, \quad (A\,BC)_{12} = 1\cdot(-15) + 2\cdot60 = 105, \quad (A\,BC)_{13} = 1\cdot20 + 2\cdot(-59) = -98;
(A\,BC)_{21} = 3\cdot5 + (-4)\cdot8 = -17, \quad (A\,BC)_{22} = 3\cdot(-15) + (-4)\cdot60 = -285, \quad (A\,BC)_{23} = 3\cdot20 + (-4)\cdot(-59) = 296.

Resposta

BC = \displaystyle\begin{pmatrix} 5 & -15 & 20 \\\\ 8 & 60 & -59 \end{pmatrix} A\,(BC) = \displaystyle\begin{pmatrix} 21 & 105 & -98 \\\\ -17 & -285 & 296 \end{pmatrix}

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