Solve for x and y :

$\begin{bmatrix$}[r] -2 & 0 \ 3 & 1 \end{bmatrix}\begin{bmatrix}[r] -1 \ 2x \end{bmatrix} + 3\begin{bmatrix}[r] -2 \ 1 \end{bmatrix} = 2\begin{bmatrix}[r] y \ 3 \end{bmatrix}.

Given,

$\Rightarrow \begin{bmatrix$}[r] -2 & 0 \ 3 & 1 \end{bmatrix}\begin{bmatrix}[r] -1 \ 2x \end{bmatrix} + 3\begin{bmatrix}[r] -2 \ 1 \end{bmatrix} = 2\begin{bmatrix}[r] y \ 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -2 \times -1 + 0 \times 2x \ 3 \times (-1) + 1 \times 2x \end{bmatrix} + \begin{bmatrix}[r] -6 \ 3 \end{bmatrix} = \begin{bmatrix}[r] 2y \ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \ -3 + 2x \end{bmatrix} + \begin{bmatrix}[r] -6 \ 3 \end{bmatrix} = \begin{bmatrix}[r] 2y \ 6 \end{bmatrix} \\[1em] \begin{bmatrix}[r] 2 + (-6) \ -3 + 2x + 3 \end{bmatrix} = \begin{bmatrix}[r] 2y \ 6 \end{bmatrix} \\[1em] \begin{bmatrix}[r] -4 \ 2x \end{bmatrix} = \begin{bmatrix}[r] 2y \ 6 \end{bmatrix}

By definition of equality of matrices we get,

2y = -4
⇒ y = -2.

2x = 6
⇒ x = 3.

Hence, x = 3 and y = -2.