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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
Expanding both determinants:
\(3 - xy = 3 - 8\)
\(xy = 8\)
The natural-number pairs with product 8 come from the divisors 1, 2, 4, 8 of 8:
\((x, y) = (1, 8),\ (2, 4),\ (4, 2),\ (8, 1)\)
So there are 4 solutions.