Question

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

Answer 1

$\left| A\right| =\begin{vmatrix}2&b&1\\ b&b^{2}+1&b\\ 1&b&2\end{vmatrix}$
$=2\left(2b^{2} +2 -b^{2}\right) -b\left(2b -b\right)+1 \left(b^{2} -b^{2} -1\right)$
$= 2b^{2} +4 -b^{2} -1 =b^{2} +3$
$\frac{\left| A\right|}{b} = b+\frac{3}{b}$
$\because\frac{b+\frac{3}{b}}{2}\ge\left(b \frac{3}{b}\right)^{\frac{1}{2}}$
$\Rightarrow b +\frac{3}{b} \ge2\sqrt{3}$
$\therefore \frac{\left|A\right|}{b} \ge2\sqrt{3}$
Minimum value of$\frac{\left| A\right|}{b}$ is $2\sqrt{3} = 2 \times1.73 = 3.46$