Let [ = 5] and [A = [ pmatrix & - & pmatrix ]].

Question

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

If B = A + A4
, then det (B) :

Accepted Answer

$$A = \left[ {\begin{pmatrix} {\cos \theta } & {\sin \theta }  \
   { - \sin \theta } & {\cos \theta }  \

\end{pmatrix} } \right]$$

A2 = $$\left[ {\begin{pmatrix} {\cos \theta } & {\sin \theta }  \
   { - \sin \theta } & {\cos \theta }  \

\end{pmatrix} } \right]$$ $\left[ {\begin{pmatrix} {\cos \theta } & {\sin \theta }  \
   { - \sin \theta } & {\cos \theta }  \

\end{pmatrix} } \right]$$ \Rightarrow $A2 =$\left[ {\begin{pmatrix} {\cos 2\theta } & {\sin 2\theta }  \
   { - \sin 2\theta } & {\cos 2\theta }  \

\end{pmatrix} } \right]$Similarly, An =$\left[ {\begin{pmatrix} {\cos n\theta } & {\sin n\theta }  \
   { - \sin n\theta } & {\cos n\theta }  \

\end{pmatrix} } \right]$$ \therefore $B = A + A4

=$\left[ {\begin{pmatrix} {\cos \theta } & {\sin \theta }  \
   { - \sin \theta } & {\cos \theta }  \

\end{pmatrix} } \right]$+$\left[ {\begin{pmatrix} {\cos 4\theta } & {\sin 4\theta }  \
   { - \sin 4\theta } & {\cos 4\theta }  \

\end{pmatrix} } \right]$=$\left[ {\begin{pmatrix} {\cos 4\theta  + \cos \theta } & {\sin 4\theta  + \sin \theta }  \
   { - \sin 4\theta  - \sin \theta } & {\cos 4\theta  + \cos \theta }  \

\end{pmatrix} } \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 $

Solutions · 1