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

We are given the symmetric matrix

$$A=\begin{bmatrix}2 & b & 1\\ b & b^{2}+1 & b\\ 1 & b & 2\end{bmatrix},\qquad b\gt 0.$$

To minimise $$\dfrac{\det(A)}{b}$$ we first need the explicit value of $$\det(A).$$

For a $$3\times3$$ matrix $$\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix},$$ the expansion along the first row is

$$\det=\;a_{11}\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix} \;-\;a_{12}\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix} \;+\;a_{13}\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}.$$

Applying this formula to $$A$$, where

$$a_{11}=2,\;a_{12}=b,\;a_{13}=1,\quad a_{21}=b,\;a_{22}=b^{2}+1,\;a_{23}=b,\quad a_{31}=1,\;a_{32}=b,\;a_{33}=2,$$

we have

$$\det(A)=2\begin{vmatrix}b^{2}+1 & b\\ b & 2\end{vmatrix} \;-\;b\begin{vmatrix}b & b\\ 1 & 2\end{vmatrix} \;+\;1\begin{vmatrix}b & b^{2}+1\\ 1 & b\end{vmatrix}.$$

Now we evaluate each $$2\times2$$ determinant one by one.

Firstly,

$$\begin{vmatrix}b^{2}+1 & b\\ b & 2\end{vmatrix} =(b^{2}+1)(2)-b\cdot b =2b^{2}+2-b^{2} =b^{2}+2.$$

Secondly,

$$\begin{vmatrix}b & b\\ 1 & 2\end{vmatrix} =b\cdot2-b\cdot1 =2b-b =b.$$

Thirdly,

$$\begin{vmatrix}b & b^{2}+1\\ 1 & b\end{vmatrix} =b\cdot b-(b^{2}+1)\cdot1 =b^{2}-b^{2}-1 =-1.$$

Substituting these values back, we obtain

$$\det(A)=2(b^{2}+2)-b(b)+1(-1).$$

Simplifying step by step,

$$2(b^{2}+2)=2b^{2}+4,$$

$$-b(b)=-b^{2},$$

and $$1(-1)=-1.$$

Adding them,

$$\det(A)=\left(2b^{2}+4\right)+\left(-b^{2}\right)+\left(-1\right) =\;b^{2}+3.$$

So we have obtained the neat expression

$$\det(A)=b^{2}+3.$$

We now form the required ratio:

$$\frac{\det(A)}{b}=\frac{b^{2}+3}{b}=b+\frac{3}{b},\qquad b\gt 0.$$

To find the minimum of $$f(b)=b+\dfrac{3}{b}$$ for positive $$b,$$ we recall the Arithmetic-Geometric Mean Inequality which states that for any two positive numbers $$x$$ and $$y,$$

$$\frac{x+y}{2}\ge \sqrt{xy},$$

with equality when $$x=y.$$

Here we treat $$x=b$$ and $$y=\dfrac{3}{b}.$$ Then

$$b+\frac{3}{b}\;\ge\;2\sqrt{b\cdot\frac{3}{b}} =2\sqrt{3}.$$

Equality holds when

$$b=\frac{3}{b}\;\Longrightarrow\;b^{2}=3\;\Longrightarrow\;b=\sqrt{3},$$

which is admissible since $$b\gt 0.$$"

Therefore the minimum value of $$\dfrac{\det(A)}{b}$$ is $$2\sqrt{3}.$$

Hence, the correct answer is Option A.