Question
If \[P = \left[ {\begin{array}{*{20}{c}}
{\cos \left( {\frac{\pi }{6}} \right)}&{\sin \left( {\frac{\pi }{6}} \right)}\\
{ - \sin \left( {\frac{\pi }{6}} \right)}&{\cos \left( {\frac{\pi }{6}} \right)}
\end{array}} \right],A = \left[ {\begin{array}{*{20}{c}}
0&1\\
0&0
\end{array}} \right]\] and $$Q = PAP'$$ then $$P'Q^{2007}P$$ is equal to
A.
\[\left[ {\begin{array}{*{20}{c}}
1&{2007}\\
0&1
\end{array}} \right]\]
B.
\[\left[ {\begin{array}{*{20}{c}}
1&{\frac{{\sqrt 3 }}{2}}\\
0&{2007}
\end{array}} \right]\]
C.
\[\left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{2007}\\
0&1
\end{array}} \right]\]
D.
\[\left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{ - \frac{1}{2}}\\
1&{2007}
\end{array}} \right]\]
Answer :
\[\left[ {\begin{array}{*{20}{c}}
1&{2007}\\
0&1
\end{array}} \right]\]
Solution :
Note that, $$P' = {P^{ - 1}}$$
$$\eqalign{
& {\text{Now, }}Q = PAP' = PA{P^{ - 1}} \cr
& \Rightarrow {Q^{2007}} = P{A^{2007}}{P^{ - 1}} \cr
& \therefore P'{Q^{2007}}P = {P^{ - 1}}\left( {P{A^{2007}}{P^{ - 1}}} \right)P = {A^{2007}} = {\left( {I + B} \right)^{2007}} \cr} $$
where, \[B = \left[ {\begin{array}{*{20}{c}}
0&1\\
0&0
\end{array}} \right].\]
As, $$B^2 = 0,$$ we get $${B^r} = 0\forall r \geqslant 2.$$
Thus, by binomial theorem,
\[{A^{2007}} = I + 2007B = \left[ {\begin{array}{*{20}{c}}
1&{2007}\\
0&1
\end{array}} \right]\]