Find x and y, if :

$\begin{bmatrix$}[r] 3 & -2 \ -1 & 4 \end{bmatrix}\begin{bmatrix}[r] 2x \ 1 \end{bmatrix} + 2\begin{bmatrix}[r] -4 \ 5 \end{bmatrix} = 4\begin{bmatrix}[r] 2 \ y \end{bmatrix}

Given,

$\Rightarrow \begin{bmatrix$}[r] 3 & -2 \ -1 & 4 \end{bmatrix}\begin{bmatrix}[r] 2x \ 1 \end{bmatrix} + 2\begin{bmatrix}[r] -4 \ 5 \end{bmatrix} = 4\begin{bmatrix}[r] 2 \ y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 3 \times 2x + (-2) \times 1 \ -1 \times 2x + 4 \times 1 \end{bmatrix} + \begin{bmatrix}[r] -8 \ 10 \end{bmatrix} = \begin{bmatrix}[r] 8 \ 4y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 6x - 2 \ -2x + 4 \end{bmatrix} + \begin{bmatrix}[r] -8 \ 10 \end{bmatrix} = \begin{bmatrix}[r] 8 \ 4y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 6x - 2 + (-8) \ -2x + 4 + 10 \end{bmatrix} = \begin{bmatrix}[r] 8 \ 4y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 6x - 10 \ -2x + 14 \end{bmatrix} = \begin{bmatrix}[r] 8 \ 4y \end{bmatrix}

By definition of equality of matrices we get,

6x - 10 = 8
⇒ 6x = 18
⇒ x = 3

-2x + 14 = 4y
⇒ -2(3) + 14 = 4y
⇒ -6 + 14 = 4y
⇒ 8 = 4y
⇒ y = 2.

Hence, x = 3 and y = 2.