Let $A=\begin{bmatrix}2 &b  & 1\\b & b^2+1 & b\\1 & b & 2\end{bmatrix},$ when b > 0. Then the minimum value of $\frac{det(A)}{b},$ is.

$2\sqrt{3}$

The correct answer is option (4) : $2\sqrt{3}$

We have,

$A= \begin{bmatrix}2 &b  & 1\\b & b^2+1 & b\\1 & b & 2\end{bmatrix}$

$∴|A|=\begin{bmatrix}2 &b  & 1\\b & b^2+1 & b\\1 & b & 2\end{bmatrix}=\begin{bmatrix}0 & 0  & 1\\-b & 1 & b\\-3 & -b & 2\end{bmatrix}$    Applying $C_1→C_1-2C_3,C_2→C_2 -bC_3$

$⇒|A|=b^2 + 3.$

$⇒\frac{|A|}{b}=b +\frac{3}{b}≥2\sqrt{b×\frac{3}{b}}=2\sqrt{3}$    [Using AM ≥ GM]