JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(A = \left[ {\begin{array}{*{20}{c}}
2&b&1 \\
b&{{b^2} + 1}&b \\
1&b&2
\end{array}} \right]\) where \(b > 0\). Then the minimum value of \(\frac{{\det \left( A \right)}}{b}\) is
- A \(2\sqrt 3\)
- B \(-2\sqrt 3\)
- C \(-\sqrt 3\)
- D \(\sqrt 3\)
Answer & Solution
Correct Answer
(A) \(2\sqrt 3\)
Step-by-step Solution
Detailed explanation
Det \(A = {b^2} + 3\) \(\frac{{\det \,A}}{b} = b + \frac{3}{b}\) \(\therefore \) Least value \( = 2\sqrt 3 \)
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