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If $A =$ $\left[ {\,\begin{array}{*{20}{c}}{{{\cos }^2}\alpha }&{\sin \alpha \,\,\cos \alpha }\\{\sin \alpha \,\,\cos \alpha }&{{{\sin }^2}\alpha }\end{array}\,} \right]$; $B =$ $\left[ {\,\begin{array}{*{20}{c}}{{{\cos }^2}\beta }&{\sin \beta \,\,\cos \beta }\\{\sin \beta \,\,\cos \beta }&{{{\sin }^2}\beta }\end{array}\,} \right]$ are such that, $AB$ is a null matrix, then which of the following should necessarily be an odd integral multiple of $\frac{\pi }{2}$.
$\alpha$
$\beta$
$\alpha - \beta$
$\alpha + \beta$
Solution
$AB =$ $\left( {\begin{array}{*{20}{c}}{{{\cos }^2}\alpha }&{\sin \alpha \,\,\cos \alpha }\\{\sin \alpha \,\cos \alpha }&{{{\sin }^2}\alpha }\end{array}} \right)$ $\left( {\begin{array}{*{20}{c}}{{{\cos }^2}\beta }&{\sin \beta \,\,\cos \beta }\\ {\sin \beta \,\cos \beta }&{{{\sin }^2}\beta }\end{array}} \right)$
$=$ $\left( {\begin{array}{*{20}{c}}{{{\cos }^2}\alpha \,\,{{\cos }^2}\beta \,\, + \,\,\sin \alpha \,\,\cos \alpha \,\,\sin \beta \,\,\cos \beta \,}&{\,{{\cos }^2}\alpha \,\,\sin \beta \,\,\cos \beta \,\, + \,\,\sin \alpha \,\cos \alpha \,\,{{\sin }^2}\beta }\\{{{\cos }^2}\beta \,\,\sin \alpha \,\,\cos \alpha \,\, + \,\,{{\sin }^2}\alpha \,\,\sin \beta \,\,\cos \beta }&{\sin \alpha \,\,\cos \alpha \,\,\sin \beta \,\,\cos \beta \,\, + \,\,{{\sin }^2}\alpha \,\,{{\sin }^2}\beta }\end{array}} \right)$
$=$ $\left( {\begin{array}{*{20}{c}}{\cos \alpha \,\,\cos \beta \,\,\cos (\alpha – \beta )}&{\cos \alpha \,\,\sin \beta \,\,\cos (\alpha – \beta )}\\{\sin \alpha \,\,\cos \beta \,\,\cos (\alpha – \beta )}&{\sin \alpha \,\,\sin \beta \,\,\cos (\alpha – \beta )}\end{array}} \right)$
==>$\alpha – \beta$ must be an odd integral multiple of $\pi /2$
