The value of x for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$, is

±2\(\sqrt{2}\)

$\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=3(1)-x(x)$

$= 3 - x^2$ ...(i)

$\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|=3(1)-4(2)$

$=3-8=-5$  ...(ii)

Given, $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$

Putting values 

$⇒ 3 - x^2 = - 5$ (from eq. (i) & (ii))

$-x^2=-5-3$

$-x^2=-8⇒x^2=8⇒x=±\sqrt{8}⇒±\sqrt{2×2×2}$

$=±2\sqrt{2}$ 

∴ value of x is $±2\sqrt{2}$