If $f\left( {\rm{\theta }} \right) = \left| {\begin{array}{*{20}{c}} {{{\cos }^2}{\rm{\theta }}}&{\cos {\rm{\theta }}\sin {\rm{\theta }}}&{ - {\rm{sin\theta }}}\\ {\cos {\rm{\theta }}\sin {\rm{\theta }}}&{{{\sin }^2}{\rm{\theta }}}&{{\rm{cos\theta }}}\\ {\sin {\rm{\theta }}}&{ - \cos {\rm{\theta }}}&0 \end{array}} \right|{\rm{.}}$. Then, for all $\theta $

$f\left( {\rm{\theta }} \right) = 1$

$f\left( {\rm{\theta }} \right) = 0$

$f\left( {\rm{\theta }} \right) = - 1$

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Analyzing +4 | -1

f\left( \mathrm{\theta } \right) = \left| {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}\mathrm{\theta }}&{\cos \mathrm{\theta }\sin \mathrm{\theta }}&{ - \mathrm{sin\theta }}\\ {\cos \mathrm{\theta }\sin \mathrm{\theta }}&{{{\sin }^2}\mathrm{\theta }}&{\mathrm{cos\theta }}\\ {\sin \mathrm{\theta }}&{ - \cos \mathrm{\theta }}&0 \end{array}} \right|\mathrm{.}

. Then, for all

\theta

View answer and explanation

Correct option: B

f\left( \mathrm{\theta } \right) = 1

Option B (

f\left( \mathrm{\theta } \right) = 1

) is correct because the determinant simplifies to 1 for all values of

\theta