Question

Determine x e y, a fim de que $\left[\begin{array}{ccc}2&x\\y&-3\\\end{array}\right] . \left[\begin{array}{ccc}4\\-5\\\end{array}\right] = \left[\begin{array}{ccc}-2\\-1\\\end{array}\right]$

Answer 1

Multiplica as matrizes primeiro:

$\left[\begin{array}{ccc}2&x\\y&-3\\\end{array}\right]. \left[\begin{array}{ccc}4\\-5\\\end{array}\right] = \left[\begin{array}{ccc}8 - 5x\\4y + 15\\\end{array}\right]$

Agora pega a matriz multiplicada e iguala:
$\left[\begin{array}{ccc}8 - 5x\\4y + 15\\\end{array}\right] = \left[\begin{array}{ccc}-2\\-1\\\end{array}\right]$

Pronto, agora só resolver o sistema:
$\left \{ {{8 - 5x \ = \ -2} \atop {4y + 15 \ = \ -1}} \right.$

8 - 5x = -2
  -5x = -10
     x = 2

4y  + 15 = -1
4y = - 16
 y = -4



Substitui os valores ae nas matrizes e vê se dar certo.

;)

Answer 2

$Ola~~Taissa \\ \\ \begin{vmatrix}2~~~x \\ y~-3\end{vmatrix}.\begin{vmatrix}4 \\ -5\end{vmatrix}=\begin{vmatrix}-2 \\ -1\end{vmatrix} \\ \\ \\ \begin{vmatrix}8-5x \\ 4y+15\end{vmatrix}=\begin{vmatrix}-2 \\ -1\end{vmatrix} \\ \\ 8-5x=-2 ~~~~~~~~~~~~4y+15=-1 \\ \\ -5x=-10~~~~~~~~~~~~~~~~~~~~4y=-16 \\ x= \frac{\not-10}{\not-5} ~~~~~~~~~~~~~~~~~~~~~~~~~~~y= \frac{-16}{4} \\ \boxed{\boxed{x=2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\boxed{y=-4}}$

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                             Espero te sirva, saludos!!