Let  $A=\left[\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right]$ where b>0. Then the minimum value of $\frac{\det \left(A\right)}{b}$ is

Solution:

We have,  $A=\left[\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right]$, where b>0

Now, det$\left(A\right)=2\left(2b^2+2-b^2\right)-b(2b-b)+1\left(b^2-b^2-1\right)$

$=2\left(b^2+2\right)-b^2-1=b^2+3$

Now, $\frac{\det \left(A\right)}{b}=\frac{b^2+3}{b}=b+\frac{3}{b}$

Since, $\text{ A.M. }\ge \text{ G.M.}$