Let $\begin{vmatrix}3&y\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$ and x, y are natural numbers, then the number of solutions for the system is:

4

The correct answer is Option (3) → 4

Given determinant: $\begin{vmatrix}3 & y \\ x & 1\end{vmatrix} = \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix}$

Compute RHS determinant: $3*1 - 2*4 = 3 - 8 = -5$

LHS determinant: $3*1 - x*y = 3 - xy$

Equate: $3 - xy = -5 \Rightarrow xy = 8$

Natural numbers $(x,y)$ such that $xy = 8$:

$(1,8), (2,4), (4,2), (8,1)$

Number of solutions = 4