Question

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 :

• A. $-2\sqrt{3}$
• B. $-\sqrt{3}$
• C. $\sqrt{3}$
• D. $2\sqrt{3}$

Answer 1

The correct option is D $2\sqrt{3}$

$A=\left[\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right]$ where $b>0$
$\Rightarrow |A|=2\left[2\right(b^2+1)-b^2]-b[2b-b]+1[b^2-b^2-1]$
$\Rightarrow |A|=2b^2+4-b^2-1$
$\Rightarrow |A|=b^2+3$
We have to calculate, $min\left(\frac{det\left(A\right)}{b}\right)$
$\Rightarrow \frac{b^2+3}{b}=b+\frac{3}{b}$
Let $f\left(b\right)=b+\frac{3}{b}$
$\Rightarrow f^{\prime }\left(b\right)=1-\frac{3}{b^2}$
$\Rightarrow f^{\prime \prime }\left(b\right)=\frac{6}{b^3}$
For maxima and minima, $f^{\prime }\left(b\right)=0$
$\therefore b=\sqrt{3}(\because b>0)$
and $f^{\prime \prime }\left(\sqrt{3}\right)>0$
Therefore, $f\left(b\right)$ is minimum at $b=\sqrt{3}$
$\therefore$ Minimum value of $f\left(b\right)=\frac{det\left(A\right)}{b}$ is
$f\left(b\right)=f\left(\sqrt{3}\right)=\sqrt{3}+\frac{3}{\sqrt{3}}.$
$\Rightarrow f\left(\sqrt{3}\right)=2\sqrt{3}$
Minimum value of $\frac{det\left(A\right)}{b}$ is $2\sqrt{3}$