If $A = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\;\alpha }&{\cos \;\alpha \;\sin \;\alpha }\\ {\cos \;\alpha \;\sin \;\alpha }&{{\rm{\;\;\;si}}{{\rm{n}}^2}\alpha } \end{array}} \right]\;$ and $A = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\;\beta }&{\cos \;\alpha \;\sin \;\beta }\\ {\cos \;\beta \;\sin \;\beta }&{{\rm{\;\;\;si}}{{\rm{n}}^2}\beta } \end{array}} \right]$ are two matrices such that the product $AB$ is null matrix, then $\alpha - \beta $ is

JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}\;\alpha }&{\cos \;\alpha \;\sin \;\alpha }\\ {\cos \;\alpha \;\sin \;\alpha }&{\mathrm{\;\;\;si}{\mathrm{n}^2}\alpha } \end{array}} \right]\;

and

A = \left[ {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}\;\beta }&{\cos \;\alpha \;\sin \;\beta }\\ {\cos \;\beta \;\sin \;\beta }&{\mathrm{\;\;\;si}{\mathrm{n}^2}\beta } \end{array}} \right]

are two matrices such that the product

AB

is null matrix, then

\alpha - \beta

View answer and explanation

Correct option: C

An odd multiple of

\pi /2

Option C (An odd multiple of

\pi /2

) is correct because the product AB being a null matrix implies that

\cos(\alpha - \beta) = 0

, which means

\alpha - \beta

is an odd multiple of

\pi /2