1.Let A=[2b1bb2+1b1b2]A=\begin{bmatrix} 2 & b & 1 \\ b & b^{2}+1 & b \\ 1 & b & 2 \end{bmatrix}A=2b1bb2+1b1b2 where b>0b>0b>0. Then the minimum value of det(A)b\frac{\det(A)}{b}bdet(A) is:a.232\sqrt{3}23b.−23-2\sqrt{3}−23c.3\sqrt{3}3d.−3-\sqrt{3}−3Login to continueOnly logged in users canattempt or see the solution.
