Question

Let $A = \begin{bmatrix}2&b&1\\ b&b^{2}+1&b\\ 1&b&2\end{bmatrix}$ where $b > 0$. Then the minimum value of $\frac{det (A)}{b}$ is :

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

Answer 1

Answer: (D)

$A = \begin{bmatrix}2&b&1\\ b&b^{2}+1&b\\ 1&b&2\end{bmatrix} \left(b>0\right) $
$ \left|A\right| = 2\left(2b^{2} +2 -b^{2} \right)-b\left(2b-b\right)+1\left(b^{2}-b^{2}-1\right) $
$\left|A\right| =2\left(b^{2}+2\right)-b^{2}-1$
$ \left|A\right|=b^{2}+3$
$ \frac{\left|A\right|}{b} =b+\frac{3}{b} \Rightarrow \frac{b+ \frac{3}{b}}{2} \ge\sqrt{3} $
$ b+ \frac{3}{b} \ge2\sqrt{3} $