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Value(s) of xx for which, ∣x15x∣=∣8221∣\begin{vmatrix} x & 1 \\ 5 & x \end{vmatrix} = \begin{vmatrix} 8 & 2 \\ 2 & 1 \end{vmatrix} is:

Solution

✅ Correct Option: 4

Expanding both determinants:

LHS =x⋅x−1⋅5=x2−5= x \cdot x - 1 \cdot 5 = x^2 - 5

RHS =8⋅1−2⋅2=4= 8 \cdot 1 - 2 \cdot 2 = 4

So x2−5=4⇒x2=9⇒x=±3x^2 - 5 = 4 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3.

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