<p>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 \(\dfrac{\det(A)}{b}\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that the matrix has a special symmetric structure that allows the determinant to be factored cleanly, then apply AM-GM inequality to minimize the resulting expression in b.
<p><strong>Step 1:</strong> Compute det(A) by expanding along the first row:</p><p>det(A) = 2(2(b²+1) - b²) - b(2b - b) + 1(b² - (b²+1))</p><p>= 2(b² + 2) - b(b) + (-1)</p><p>= 2b² + 4 - b² - 1</p><p>= b² + 3</p><p><strong>Step 2:</strong> Form the ratio:</p><p>$\frac{\det(A)}{b} = \frac{b^2 + 3}{b} = b + \frac{3}{b}$</p><p><strong>Step 3:</strong> Apply AM-GM inequality to minimize:</p><p>By AM-GM: $b + \frac{3}{b} \geq 2\sqrt{b \cdot \frac{3}{b}} = 2\sqrt{3}$</p><p>Equality holds when $b = \frac{3}{b}$, i.e., $b = \sqrt{3}$</p><p><strong>Step 4:</strong> Verify b = √3 > 0 ✓</p><p>∴ Minimum value = $2\sqrt{3}$</p>
Correct Answer: A