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

Answer 1

Answer: (D)

$2\sqrt{3}$

$A=\left[\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right](b>0)$

$|A|=2(2b^2+2-b^2)-b(2b-b)+1(b^2-b^2-1)$

$|A|=2(b^2+2)-b^2-1$

$|A|=b^2+3$
$\frac{|A|}{b}=b+\frac{3}{b}\Rightarrow \frac{b+\frac{3}{b}}{2}\ge \sqrt{3}$
$b+\frac{3}{b}\ge 2\sqrt{3}$

So, the minimum value is $2\sqrt{3}$