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Let ∣3yx1∣=∣3241∣\begin{vmatrix} 3 & y \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix} and x,yx, y are natural numbers, then the number of solutions for the system is:

Solution

✅ Correct Option: 3

The value of a 2×22 \times 2 determinant ∣abcd∣\begin{vmatrix} a & b \\ c & d \end{vmatrix} is ad−bcad - bc.

Left Side: 3(1)−x(y)=3−xy3(1) - x(y) = 3 - xy

Right Side: 3(1)−4(2)=3−8=−53(1) - 4(2) = 3 - 8 = -5

3−xy=−53 - xy = -5

−xy=−5−3-xy = -5 - 3

−xy=−8-xy = -8

xy=8xy = 8

We are looking for pairs of natural numbers (x,y)(x, y) whose product is 88. Natural numbers are positive integers (1,2,3,…1, 2, 3, \dots).

The pairs of factors of 88 are:

  • x=1,y=8⇒(1,8)x = 1, y = 8 \Rightarrow (1, 8)
  • x=2,y=4⇒(2,4)x = 2, y = 4 \Rightarrow (2, 4)
  • x=4,y=2⇒(4,2)x = 4, y = 2 \Rightarrow (4, 2)
  • x=8,y=1⇒(8,1)x = 8, y = 1 \Rightarrow (8, 1)

There are 4 distinct solutions for the system.

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