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
Answer & explanation
Correct answer: option 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 $±2\sqrt{2}$