If $\text{det}\begin{pmatrix}2x&5\\8&x\end{pmatrix} =\text{det}\begin{pmatrix}6&-2\\1&1\end{pmatrix}$, then the value of $x$ is

$±2\sqrt{6}$

The correct answer is Option (1) → $±2\sqrt{6}$

Given:

$\det\begin{pmatrix}2x & 5 \\ 8 & x\end{pmatrix} = \det\begin{pmatrix}6 & -2 \\ 1 & 1\end{pmatrix}$

Compute the determinant on both sides:

$\det\begin{pmatrix}2x & 5 \\ 8 & x\end{pmatrix} = 2x \cdot x - 8 \cdot 5 = 2x^2 - 40$

$\det\begin{pmatrix}6 & -2 \\ 1 & 1\end{pmatrix} = 6 \cdot 1 - (-2) \cdot 1 = 6 + 2 = 8$

Equating:

$2x^2 - 40 = 8$

$2x^2 = 48 \Rightarrow x^2 = 24$

$x = \pm \sqrt{24} = \pm 2\sqrt{6}$