If $\begin{bmatrix$}[r] x + 2y & -y \ 3x & 4 \end{bmatrix} = \begin{bmatrix}[r] -4 & 3 \ 6 & 4 \end{bmatrix}, then the values of x and y are

1. x = 2, y = 3
2. x = 2, y = -3
3. x = -2, y = 3
4. x = 3, y = 2

Given,

$\begin{bmatrix$}[r] x + 2y & -y \ 3x & 4 \end{bmatrix} = \begin{bmatrix}[r] -4 & 3 \ 6 & 4 \end{bmatrix} \\[0.5em]

By definition of equality of matrices we get,

⇒ x + 2y = -4     (…Eq 1)

⇒ -y = 3 or y = -3

⇒ 3x = 6 or x = 2

Putting, x = 2 and y = -3 in Eq 1,

⇒ x + 2y = -4
⇒ L.H.S. = 2 + 2(-3) = 2 - 6 = -4 = R.H.S.

Since, x = 2 and y = -3 satisfies the equation x + 2y = -4,

∴ x = 2 and y = -3.

∴ Option 2 is the correct option.