If $(\begin{matrix} 3 & y x & 1 \end{matrix} ∣ ∣ ∣ = (\begin{matrix} 3 & 2 4 & 1 \end{matrix} ∣ ∣ ∣$ , where x and y are natural numbers, then possible values of x and y are:

x = 1, y = 8

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x = 2 , y = 4

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x = 4 , y = 2

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x = 8 , y = 1

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Solution

The correct option is D x = 8 , y = 1
If
$(\begin{matrix} 3 & y x & 1 \end{matrix} ∣ ∣ ∣ = (\begin{matrix} 3 & 2 4 & 1 \end{matrix} ∣ ∣ ∣$
$⟹ 3 - x y = 3 - 8$
$⟹ x y = 8$
$⟹ Substituting the values given in the options$
We get $x = 1, y = 8; x = 2, y = 4; x = 4, y = 2; x = 8, y = 1$
all satisfies the condition.

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