Let $\mathrm{A}=\left[\begin{array}{ccc}2 & \mathrm{~b} & 1 \\ \mathrm{~b} & \mathrm{~b}^{2}+1 & \mathrm{~b} \\ 1 & \mathrm{~b} & 2\end{array}\right]$ where $\mathrm{b}>0$. Then the
minimum value of $\frac{\operatorname{det}(\mathrm{A})}{\mathrm{b}}$ is :
Correct Option: , 4
$A=\left[\begin{array}{ccc}2 & b & 1 \\ b & b^{2}+1 & b \\ 1 & b & 2\end{array}\right](b>0)$
$|\mathrm{A}|=2\left(2 \mathrm{~b}^{2}+2-\mathrm{b}^{2}\right)-\mathrm{b}(2 \mathrm{~b}-\mathrm{b})+1\left(\mathrm{~b}^{2}-\mathrm{b}^{2}-1\right)$
$|\mathrm{A}|=2\left(\mathrm{~b}^{2}+2\right)-\mathrm{b}^{2}-1$
$|\mathrm{~A}|=\mathrm{b}^{2}+3$
$\frac{|\mathrm{A}|}{\mathrm{b}}=\mathrm{b}+\frac{3}{\mathrm{~b}} \Rightarrow \frac{\mathrm{b}+\frac{3}{\mathrm{~b}}}{2} \geq \sqrt{3}$
$\mathrm{b}+\frac{3}{\mathrm{~b}} \geq 2 \sqrt{3}$
Option (4)
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