Question

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

Answer 1

Answer: 1.5

$\left|A\right|=\left|\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right|$
$=2\left(2b^2+2-b^2\right)-b\left(2b-b\right)+1\left(b^2-b^2-1\right)$
$=2b^2+4-b^2-1=b^2+3$
$\frac{|A|}{b}=b+\frac{3}{b}$
$\because \frac{b+\frac{3}{b}}{2}\ge {\left(b\frac{3}{b}\right)}^{\frac{1}{2}}$
$\Rightarrow b+\frac{3}{b}\ge 2\sqrt{3}$
$\therefore \frac{|A|}{b}\ge 2\sqrt{3}$
Minimum value of$\frac{|A|}{b}$ is $2\sqrt{3}=2\times 1.73=3.46$