If [A = [ pmatrix & i i & pmatrix ]], [( = 24 )] and [A^5 = [ pmatrix a & b c & d pmatrix …

Question

If $$A = \left[ {\begin{pmatrix} {\cos \theta } & {i\sin \theta }  \
   {i\sin \theta } & {\cos \theta }  \ \end{pmatrix} } \right]$$, $$\left( {\theta  = {\pi  \over {24}}} \right)$$

and $${A^5} = \left[ {\begin{pmatrix} a & b  \
   c & d  \ \end{pmatrix} } \right]$$, where $$i = \sqrt { - 1} $$  then which one of the following is
not true?

Accepted Answer

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

\end{pmatrix} } \right]$$

$$ \therefore $$ $${A^n} = \left[ {\begin{pmatrix} {\cos \,n\theta } & {i\sin \,n\theta }  \
   {i\sin \,n\theta } & {\cos \,n\theta }  \

\end{pmatrix} } \right],n \in N$$

$$ \therefore $$ ${A^5} = \left[ {\begin{pmatrix} {\cos \,5\theta } & {i\sin \,5\theta }  \
   {i\sin \,5\theta } & {\cos \,5\theta }  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} a & b  \
   c & d  \

\end{pmatrix} } \right]$$ \therefore $$a = \cos 5\theta ,\,b = i\sin 5\theta  = c,\,d = \cos 5\theta $$ \therefore $${a^2} - {b^2} = {\cos ^2}5\theta  + {\sin ^2}5\theta  = 1$${a^2} - {c^2} = {\cos ^2}5\theta  + {\sin ^2}5\theta  = 1$${a^2} - {d^2} = {\cos ^2}5\theta  - {\cos ^2}5\theta  = 1$${a^2} + {b^2} = {\cos ^2}5\theta  - {\sin ^2}5\theta  = \cos 10\theta  = \cos {{10\pi } \over {24}}$and$0 < \cos {{5\pi } \over {12}} < 1 $$\Rightarrow 0 \le {a^2} + {b^2} \le 1$

Solutions · 1