Question:

**Step 1: Calculate $\det(A)$.** The determinant of the matrix $A = \begin{bmatrix} 2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2 \end{bmatrix}$ is calculated using cofactor expansion along the first row: $$ \det(A) = 2 \begin{vmatrix} b^2+1 & b \\ b & 2 \end{vmatrix} - b \begin{vmatrix} b & b \\ 1 & 2 \end{vmatrix} + 1 \begin{vmatrix} b & b^2+1 \\ 1 & b \end{vmatrix} $$ $$ \det(A) = 2((b^2+1)(2) - b \cdot b) - b(b \cdot 2 - b \cdot 1) + 1(b \cdot b - (b^2+1) \cdot 1) $$ $$ \det(A) = 2(2b^2+2 - b^2) - b(2b - b) + 1(b^2 - b^2 - 1) $$ $$ \det(A) = 2(b^2+2) - b(b) + 1(-1) $$ $$ \det(A) = 2b^2+4 - b^2 - 1 $$ $$ \det(A) = b^2+3 $$ **Step 2: Form the expression to minimize.** The expression to minimize is $\frac{\det(A)}{b}$. Given $b > 0$: $$ f(b) = \frac{b^2+3}{b} = b + \frac{3}{b} $$ **Step 3: Find the critical point.** To find the minimum value, we take the first derivative of $f(b)$ with respect to $b$ and set it to zero: $$ f'(b) = \frac{d}{db}\left(b + \frac{3}{b}\right) = 1 - \frac{3}{b^2} $$ Setting $f'(b) = 0$: $$ 1 - \frac{3}{b^2} = 0 $$ $$ b^2 = 3 $$ Since $b > 0$, we have $b = \sqrt{3}$. **Step 4: Verify this is a minimum.** We use the second derivative test to confirm that $b = \sqrt{3}$ corresponds to a minimum: $$ f''(b) = \frac{d}{db}\left(1 - \frac{3}{b^2}\right) = \frac{6}{b^3} $$ For $b = \sqrt{3}$, $f''(\sqrt{3}) = \frac{6}{(\sqrt{3})^3} = \frac{6}{3\sqrt{3}} = \frac{2}{\sqrt{3}}$. Since $f''(\sqrt{3}) > 0$, the critical point $b = \sqrt{3}$ corresponds to a local minimum. **Step 5: Calculate the minimum value.** Substitute $b = \sqrt{3}$ into the expression for $f(b)$: $$ f(\sqrt{3}) = \sqrt{3} + \frac{3}{\sqrt{3}} $$ $$ f(\sqrt{3}) = \sqrt{3} + \sqrt{3} $$ $$ f(\sqrt{3}) = 2\sqrt{3} $$