4 Dadas as matrizes, $ A=\left(\begin{array}{ll}0 & 3 \ 8 & 3\end{array}\right) $ e $ B=\left(\begin{array}{cc}-8 & -9 \ 5 & 4\end{array}\right) $, a matriz resultante de $ -6 A+3 B $, é:
(A) $ \left(\begin{array}{ll}-43 & -45 \ -33 & -25\end{array}\right) $
(B) $ \left(\begin{array}{cc}-24 & -45 \ -33 & -6\end{array}\right) $
(c) $ \left(\begin{array}{ll}-38 & -45 \ -33 & -20\end{array}\right) $
(D) $ \left(\begin{array}{cc}-15 & -45 \ -33 & 3\end{array}\right) $
(E) $ \left(\begin{array}{cc}-20 & -45 \ -33 & -2\end{array}\right) $

Q: 4 Dadas as matrizes, $ A=\left(\begin{array}{ll}0 & 3 \\ 8 & 3\end{array}\right) $ e $ B=\left(\begin{array}{cc}-8 & -9 \\ 5 & 4\end{array}\right) $, a matriz resultante de $ -6 A+3 B $, é: (A) $ \left(\begin{array}{ll}-43 & -45 \\ -33 & -25\end{array}\right) $ (B) $ \left(\begin{array}{cc}-24 & -45 \\ -33 & -6\end{array}\right) $ (c) $ \left(\begin{array}{ll}-38 & -45 \\ -33 & -20\end{array}\right) $ (D) $ \left(\begin{array}{cc}-15 & -45 \\ -33 & 3\end{array}\right) $ (E) $ \left(\begin{array}{cc}-20 & -45 \\ -33 & -2\end{array}\right) $

Para encontrar a matriz resultante de $ -6A + 3B $, primeiro precisamos calcular $ -6A $ e $ 3B $ separadamente e depois somá-las. Primeiro, calculamos $ -6A $: \[ A = \left(\begin{array}{cc}0 & 3 \\ 8 & 3\end{array}\right) \] \[ -6A = -6 \times \left(\begin{array}{cc}0 & 3 \\ 8 & 3\end{array}\right) = \left(\begin{array}{cc}-6 \cdot 0 & -6 \cdot 3 \\ -6 \cdot 8 & -6 \cdot 3\end{array}\right) = \left(\begin{array}{cc}0 & -18 \\ -48 & -18\end{array}\right) \] Agora, calculamos $ 3B $: \[ B = \left(\begin{array}{cc}-8 & -9 \\ 5 & 4\end{array}\right) \] \[ 3B = 3 \times \left(\begin{array}{cc}-8 & -9 \\ 5 & 4\end{array}\right) = \left(\begin{array}{cc}3 \cdot (-8) & 3 \cdot (-9) \\ 3 \cdot 5 & 3 \cdot 4\end{array}\right) = \left(\begin{array}{cc}-24 & -27 \\ 15 & 12\end{array}\right) \] Agora somamos as duas matrizes: \[ -6A + 3B = \left(\begin{array}{cc}0 & -18 \\ -48 & -18\end{array}\right) + \left(\begin{array}{cc}-24 & -27 \\ 15 & 12\end{array}\right) = \left(\begin{array}{cc}0 + (-24) & -18 + (-27) \\ -48 + 15 & -18 + 12\end{array}\right) = \left(\begin{array}{cc}-24 & -45 \\ -33 & -6\end{array}\right) \] **Resposta: B** A matriz resultante é $ \left(\begin{array}{cc}-24 & -45 \\ -33 & -6\end{array}\right) $.

Question

Accepted Answer

Para encontrar a matriz resultante de $ -6A + 3B $, primeiro precisamos calcular $ -6A $ e $ 3B $ separadamente e depois somá-las.

Primeiro, calculamos $ -6A $:

$$ A = \left(\begin{array}{cc}0 & 3 \ 8 & 3\end{array}\right) $$

$$ -6A = -6 \times \left(\begin{array}{cc}0 & 3 \ 8 & 3\end{array}\right) = \left(\begin{array}{cc}-6 \cdot 0 & -6 \cdot 3 \ -6 \cdot 8 & -6 \cdot 3\end{array}\right) = \left(\begin{array}{cc}0 & -18 \ -48 & -18\end{array}\right) $$

Agora, calculamos $ 3B $:

$$ B = \left(\begin{array}{cc}-8 & -9 \ 5 & 4\end{array}\right) $$

$$ 3B = 3 \times \left(\begin{array}{cc}-8 & -9 \ 5 & 4\end{array}\right) = \left(\begin{array}{cc}3 \cdot (-8) & 3 \cdot (-9) \ 3 \cdot 5 & 3 \cdot 4\end{array}\right) = \left(\begin{array}{cc}-24 & -27 \ 15 & 12\end{array}\right) $$

Agora somamos as duas matrizes:

$$ -6A + 3B = \left(\begin{array}{cc}0 & -18 \ -48 & -18\end{array}\right) + \left(\begin{array}{cc}-24 & -27 \ 15 & 12\end{array}\right) = \left(\begin{array}{cc}0 + (-24) & -18 + (-27) \ -48 + 15 & -18 + 12\end{array}\right) = \left(\begin{array}{cc}-24 & -45 \ -33 & -6\end{array}\right) $$

**Resposta: B**

A matriz resultante é $ \left(\begin{array}{cc}-24 & -45 \ -33 & -6\end{array}\right) $.

4 Dadas as matrizes,

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