If $x\begin{bmatrix} 2 \ 1 \end{bmatrix} + y\begin{bmatrix} 3 \ 5 \end{bmatrix} + \begin{bmatrix} -8 \ -11 \end{bmatrix}$ = 0, then the value of (2x – 3y) is:

1. 1

2. -4

3. 3

4. -3

$\Rightarrow x\begin{bmatrix} 2 \ 1 \end{bmatrix} + y\begin{bmatrix} 3 \ 5 \end{bmatrix} + \begin{bmatrix} -8 \ -11 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \ x \end{bmatrix} + \begin{bmatrix} 3y \ 5y \end{bmatrix} + \begin{bmatrix} -8 \ -11 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x + 3y - 8 \ x + 5y - 11 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.$

∴ 2x + 3y - 8 = 0

⇒ 2x + 3y = 8 ….(1)

∴ x + 5y - 11 = 0

⇒ x = 11 - 5y ….(2)

Substituting value of x from equation (2) in (1), we get :

⇒ 2(11 - 5y) + 3y = 8

⇒ 22 - 10y + 3y = 8

⇒ -7y = 8 - 22

⇒ -7y = -14

⇒ y = $\dfrac{-14}{-7}$

⇒ y = 2.

Substituting value of y in equation (2) we get:

⇒ x = 11 - 5(2)

⇒ x = 11 - 10

⇒ x = 1.

∴ 2x - 3y = 2(1) - 3(2) = 2 - 6 = -4.

Hence, option 2 is the correct option.