JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 17)

Let $$\theta = {\pi \over 5}$$ and $$A = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$.

If B = A + A4 , then det (B) :
lies in (1, 2)
lies in (2, 3).
is zero.
is one.

Explanation

$$A = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$

A2 = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$$$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$

$$ \Rightarrow $$ A2 = $$\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$$

Similarly, An = $$\left[ {\matrix{ {\cos n\theta } & {\sin n\theta } \cr { - \sin n\theta } & {\cos n\theta } \cr } } \right]$$

$$ \therefore $$ B = A + A4

= $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$ + $$\left[ {\matrix{ {\cos 4\theta } & {\sin 4\theta } \cr { - \sin 4\theta } & {\cos 4\theta } \cr } } \right]$$

= $$\left[ {\matrix{ {\cos 4\theta + \cos \theta } & {\sin 4\theta + \sin \theta } \cr { - \sin 4\theta - \sin \theta } & {\cos 4\theta + \cos \theta } \cr } } \right]$$

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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