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Let $A = \begin{bmatrix} 2 & b & 1 \ b & b^2 + 1 & b \ 1 & b & 2 \end{bmatrix}$ where $b > 0$. Then the minimum value of $\dfrac{\det(A)}{b}$ is:
(1) $2\sqrt{3}$
(2) $-2\sqrt{3}$
(3) $-\sqrt{3}$
(4) $\sqrt{3}$
BETA
