If $\left|\begin{array}{cc}3 & -4 \\ 2 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 5 \\ 1 & x\end{array}\right|$ then $|x|$ is equal to

$\sqrt{\frac{5}{2}}$ 4 $2 \sqrt{2}$ 2

$2 \sqrt{2}$

$\left|\begin{array}{cc}3 & -4 \\ 2 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 5 \\ 1 & x\end{array}\right|$ $\Rightarrow 3-(-4) \times 2=2 x^2-5$ $\Rightarrow 3+8=2 x^2-5$ so $2 x^2=16$ ⇒ so $x^2=8$ $|x| = 2 \sqrt{2}$