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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
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
  • ✓
    $2\sqrt 3$
  • B
    $-2\sqrt 3$
  • C
    $-\sqrt 3$
  • D
    $\sqrt 3$
✓

Answer

Correct option: A.
$2\sqrt 3$
a
Det $A = {b^2} + 3$

$\frac{{\det \,A}}{b} = b + \frac{3}{b}$

$\therefore $ Least value $ = 2\sqrt 3 $

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