Find the value of x for which $|_{x}^{3}_{1}^{x} | = |_{8}^{3}_{1}^{2} |$ .

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Solution

$∣ ∣ ∣ \begin{matrix} 3 & 2 8 & 1 \end{matrix} ∣ ∣ ∣ = 3 - 16 = - 13$
$∣ ∣ ∣ \begin{matrix} 3 & x x & 1 \end{matrix} ∣ ∣ ∣ = 3 - x^{2}$
From Question $⟹ - 13 = 3 - x^{2} x^{2} = 13 + 3 x^{2} = 16 x = \pm 4$

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