If \(\begin{vmatrix}x&8\\4&x\end{vmatrix}=\begin{vmatrix}6&2\\18&6\end{vmatrix}\), then x is/are equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\pm4\sqrt{2}\)
A 2 × 2 determinant is expanded as \(\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad - bc\):
\(\displaystyle \begin{vmatrix}x&8\\4&x\end{vmatrix} = x^2 - 32\)
\(\displaystyle \begin{vmatrix}6&2\\18&6\end{vmatrix} = 36 - 36 = 0\)
Equating the two:
\(x^2 - 32 = 0\)
\(x^2 = 32\)
\(x = \pm\sqrt{32} = \pm4\sqrt{2}\)
The question asks for both values (“is/are”), so the answer is \(\pm4\sqrt{2}\).