Find x and y, if :

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

Given,

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

By definition of equality of matrices we get,

x = 2

-3 + y = -2
⇒ y = -2 + 3
⇒ y = 1.

Hence, x = 2 and y = 1.