If A = [ 0 - tan ((θ)/2) tan ((θ)/2) 0 ] and (I₂ + A)(I₂ - A)⁻¹ = [ a - b b a ], then 13(a² + b²) is equal to

Matrices and Determinants: JEE Main 2021 Mathematics question

JEE Main 2021 · 25 Feb, Morning shift · Q36

JEE Main 202125 Feb, Morning shiftMathematicsMediumMatrices and Determinants245× since 2002

A = \left[ {\matrix{ 0 & { - \tan \left( {{\theta \over 2}} \right)} \cr {\tan \left( {{\theta \over 2}} \right)} & 0 \cr } } \right]

and

({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]

, then

13({a^2} + {b^2})

is equal to

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2021 · 25 Feb, Morning shift · Q36

JEE Main 202125 Feb, Morning shiftMathematicsMediumMatrices and Determinants245× since 2002

A = \left[ {\matrix{ 0 & { - \tan \left( {{\theta \over 2}} \right)} \cr {\tan \left( {{\theta \over 2}} \right)} & 0 \cr } } \right]

and

({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]

, then

13({a^2} + {b^2})

is equal to

Your answerType your answer. Integers and decimals are accepted.