If matrix $\begin{bmatrix$}[r] 2 & a + b \ b - a & 3 \end{bmatrix} = \begin{bmatrix}[r] 2 & 8 \ 2 & 3 \end{bmatrix}, then :

1. a = 4 = b

2. a = 2 and b = 6

3. a = 1 and b = 7

4. a = 3 and b = 5

Given,

⇒ $\begin{bmatrix$}[r] 2 & a + b \ b - a & 3 \end{bmatrix} = \begin{bmatrix}[r] 2 & 8 \ 2 & 3 \end{bmatrix}

⇒ a + b = 8 ………(1)

⇒ b - a = 2 ………(2)

Adding equations (1) and (2), we get :

⇒ (a + b) + (b - a) = 8 + 2

⇒ 2b = 10

⇒ b = $\dfrac{10}{2}$

⇒ b = 5.

Substituting value of b in equation (1), we get :

⇒ a + 5 = 8

⇒ a = 8 - 5 = 3.

Hence, Option 4 is the correct option.