If ${∣ ∣_{2}^{4}_{1}^{1} ∣ ∣}^{2} = ∣ ∣_{1}^{3}_{x}^{2} ∣ ∣ - ∣ ∣_{- 2}^{x}_{1}^{3} ∣ ∣,$ then $x$ =

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Solution

The correct option is A 6
${∣ ∣ ∣ \begin{matrix} 4 & 1 2 & 1 \end{matrix} ∣ ∣ ∣}^{2} = ∣ ∣ ∣ \begin{matrix} 3 & 2 1 & x \end{matrix} ∣ ∣ ∣ - ∣ ∣ ∣ \begin{matrix} x & 3 - 2 & 1 \end{matrix} ∣ ∣ ∣$
$(4 - 2)^{2} = (3 x - 2) - (x + 6)$
$4 = (3 x - x) - 8$
$12 = 2 x$
$\Rightarrow x = 6$

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