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 $\frac{\det(A)}{b}$ is

  • A
    $2\sqrt{3}$
  • B
    $-2\sqrt{3}$
  • C
    $-\sqrt{3}$
  • D
    $\sqrt{3}$

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