Question

Determinar x e y de uma matriz!

Answer 1

Resposta:

Leia abaixo

Explicação passo-a-passo:

$\begin{cases}-\dfrac{2x}{3} + 7y = 2A\\5x + \dfrac{y}{2} = -3B\end{cases}$

$\begin{cases}-2x + 21y = 6A\\10x + y = -6B\end{cases}$

$\begin{cases}-10x + 105y = 30A\\10x + y = -6B\end{cases}$

$\text{Vou somar as equacoes ...}$

$106y = 30A - 6B$

$53y = 15A - 3B$

$\boxed{\boxed{y = \dfrac{15A - 3B}{53}}}$

$10x + \dfrac{15A - 3B}{53} = -6B$

$530x + 15A - 3B = -318B$

$106x = -3A - 63B$

$\boxed{\boxed{x = \dfrac{-3A-63B}{106}}}$

$y_{11} = \dfrac{15(1) - 3(8)}{53} = \dfrac{15-24}{53} = \dfrac{-9}{53}$

$y_{12} = \dfrac{15(-5)-3(7)}{53} = \dfrac{-75-21}{53}= \dfrac{-96}{53}$

$y_{13} = \dfrac{15(3) -3(-2)}{53} = \dfrac{45+6}{53} = \dfrac{51}{53}$

$x_{11} = \dfrac{-3(1) -63(8)}{106} = \dfrac{-3-504}{106} = \dfrac{-507}{106}$

$x_{12} = \dfrac{-3(-5) - 63(7)}{106} = \dfrac{15 - 441}{106} = \dfrac{-213}{53}$

$x_{13} = \dfrac{-3(3) -63(-2)}{106} = \dfrac{-9 - (-126)}{106} = \dfrac{-9 + 126}{106} = \dfrac{117}{106}$

$x = \large\mathsf{\begin{vmatrix}-507/106\\-213/53\\117/106\end{vmatrix}}$

$y = \large\mathsf{\begin{vmatrix}-9/53\\-96/53\\51/53\end{vmatrix}}$