Question:
Let \(A = \begin{bmatrix} 2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2 \end{bmatrix}\) where \(b \gt 0\).
Then the minimum value of \(\frac{\det(A)}{b}\) is -
Let \(A = \begin{bmatrix} 2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2 \end{bmatrix}\) where \(b \gt 0\).
Then the minimum value of \(\frac{\det(A)}{b}\) is -
Then the minimum value of \(\frac{\det(A)}{b}\) is -
Show Hint
Expand the 3x3 determinant to get \(b^2+3\), then apply AM-GM to \(b + \frac{3}{b}\).
Updated On: Nov 5, 2026
- \(\sqrt{3}\)
- \(-2\sqrt{3}\)
- \(-\sqrt{3}\)
- \(2\sqrt{3}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Expand along the first row:
\(\det A = 2[2(b^2+1) - b^2] - b[2b - b] + 1[b^2 - (b^2+1)]\)
\(= 2(b^2+2) - b^2 - 1 = b^2 + 3\).
So \(\frac{\det A}{b} = b + \frac{3}{b}\) with \(b \gt 0\).
By AM-GM, \(b + \frac3b \ge 2\sqrt{b\cdot\frac3b} = 2\sqrt3\), with equality at \(b = \sqrt3\).
Minimum value = \(2\sqrt3\).
Answer: (4) \(2\sqrt{3}\)
\(\det A = 2[2(b^2+1) - b^2] - b[2b - b] + 1[b^2 - (b^2+1)]\)
\(= 2(b^2+2) - b^2 - 1 = b^2 + 3\).
So \(\frac{\det A}{b} = b + \frac{3}{b}\) with \(b \gt 0\).
By AM-GM, \(b + \frac3b \ge 2\sqrt{b\cdot\frac3b} = 2\sqrt3\), with equality at \(b = \sqrt3\).
Minimum value = \(2\sqrt3\).
Answer: (4) \(2\sqrt{3}\)
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