• # question_answer Let $A=\left[ \begin{matrix} 2 & b & 1 \\ b & {{b}^{2}}+1 & b \\ 1 & b & 2 \\ \end{matrix} \right]$ 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) $\sqrt{3}$

 Det $A={{b}^{2}}+3$ $\frac{\det A}{b}=b+\frac{3}{b}$ $\therefore$      Least value $=2\sqrt{3}.$