Let A = [ 2 b 1 b b² + 1 b 1 b 2 ] where b > 0. Then the minimum value of (det (A))/b is -

Matrices and Determinants: JEE Main 2019 Mathematics question

JEE Main 2019 · 10 Jan, Evening shift · Q4455

JEE Main 201910 Jan, Evening shiftMathematicsMediumMatrices and Determinants245× since 2002

Let A =

\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]

where b > 0. Then the minimum value of

{{\det \left( A \right)} \over b}

is -

JEE Main 2019 · 10 Jan, Evening shift · Q4455

JEE Main 201910 Jan, Evening shiftMathematicsMediumMatrices and Determinants245× since 2002

Let A =

\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]

where b > 0. Then the minimum value of

{{\det \left( A \right)} \over b}

is -