A) \[\left[ \begin{matrix} 1 & 2007 \\ 0 & 1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 1 & \sqrt{3}/2 \\ 0 & 2007 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} \sqrt{3}/2 & 2007 \\ 0 & 1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} \sqrt{3}/2 & -1/2 \\ 1 & 2007 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
[a] Note that \[P'={{P}^{-1}}\] Now, \[Q=PAP'=PA{{P}^{-1}}\] \[\Rightarrow {{Q}^{2007}}=P{{A}^{2007}}{{P}^{-1}}\] \[\therefore P'\,{{Q}^{2007}}P={{P}^{-1}}(P{{A}^{2007}}{{P}^{-}}^{1})P\] \[={{A}^{2007}}={{(I+B)}^{2007}}\] Where \[B=\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right]\]. As \[{{B}^{2}}=0\], we get \[{{B}^{r}}=0\forall r\ge 2\]. Thus, by binomial theorem, \[{{A}^{2007}}=I+2007B=\left[ \begin{matrix} 1 & 2007 \\ 0 & 1 \\ \end{matrix} \right]\]You need to login to perform this action.
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