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Let $$\theta = {\pi \over 5}$$ and $$A = \left[ {\begin{matrix} {\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \\ \end{matrix} } \right]$$. If B = A + A4 , then det (B) :

JEE · Math · previous-year question

  1. A.lies in (1, 2)correct
  2. B.lies in (2, 3).
  3. C.is zero.
  4. D.is one.

Answer

A. lies in (1, 2)

Explanation

$$A = \left[ {\begin{matrix} {\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \\ \end{matrix} } \right]$$ A2 = $$\left[ {\begin{matrix} {\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \\ \end{matrix} } \right]$$$$\left[ {\begin{matrix} {\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \\ \end{matrix} } \right]$$ $$ \Rightarrow $$ A2 = $$\left[ {\begin{matrix} {\cos 2\theta } & {\sin 2\theta } \\ { - \sin 2\theta } & {\cos 2\theta } \\ \end{matrix} } \right]$$ Similarly, An = $$\left[ {\begin{matrix} {\cos n\theta } & {\sin n\theta } \\ { - \sin n\theta } & {\cos n\theta } \\ \end{matrix} } \right]$$ $$ \therefore $$ B = A + A4 = $$\left[ {\begin{matrix} {\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \\ \end{matrix} } \right]$$ + $$\left[ {\begin{matrix} {\cos 4\theta } & {\sin 4\theta } \\ { - \sin 4\theta } & {\cos 4\theta } \\ \end{matrix} } \right]$$ = $$\left[ {\begin{matrix} {\cos 4\theta + \cos \theta } & {\sin 4\theta + \sin \theta } \\ { - \sin 4\theta - \sin \theta } & {\cos 4\theta + \cos \theta } \\ \end{matrix} } \right]$$ detB = (cos4$$\theta $$ + cos$$\theta $$)2 + (sin4$$\theta $$ + sin$$\theta $$)2 = cos24$$\theta $$ + cos2$$\theta $$ + 2cos4$$\theta $$ cos$$\theta $$ + sin24$$\theta $$ + sin2$$\theta $$ + 2sin4$$\theta $$ –sin$$\theta $$ = 2 + 2 ( cos4$$\theta $$ cos$$\theta $$ + sin4$$\theta $$ sin$$\theta $$) $$ \Rightarrow $$ detB = 2 + 2 cos3$$\theta $$ at $$\theta $$ = $${\pi \over 5}$$ detB = 2 + 2cos $${{3\pi } \over 5}$$ = 2(1 - sin18) = 2(1 - $${{\sqrt 5 - 1} \over 4}$$) = 2$$\left( {{{5 - \sqrt 5 } \over 4}} \right)$$ = $${{{5 - \sqrt 5 } \over 2}}$$ $$ \simeq $$ 1.385 $$ \therefore $$ detB $$ \in $$ (1, 2)

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