Let A = [ cases 2 & b & 1 b & b ^ 2 + 1 & b 1 & b & 2 cases ],...

\left|A\right| = \left|$\begin{cases}2 & b & 1 \\ b & {b}^{2} + 1 & b \\ 1 & b & 2\end{cases}$\right|$<br /><br />$= 2\left(2{b}^{2} + 2 - {b}^{2}\right) - b\left(2b - b\right) + 1\left({b}^{2} - {b}^{2} - 1\right)$<br /><br />$= 2\left({b}^{2} + 2\right) - {b}^{2} - 1$<br /><br />$= 2{b}^{2} + 4 - {b}^{2} - 1$<br /><br />$= {b}^{2} + 3$<br /><br />$\dfrac{\left|A\right|}{b} = b + \dfrac{3}{b} = {\left(\sqrt{b} - \sqrt{\dfrac{3}{b}}\right)}^{2} + 2\sqrt{3}$<br /><br />Hence, the minimum value of$\dfrac{\left|A\right|}{b}$is$2\sqrt{3}$. \left(∵ {\left(\sqrt{b} - \sqrt{\dfrac{3}{b}}\right)}^{2} \ge 0\right)$