Determinants Matrices Question 121
Question: If $ P \begin{bmatrix} \cos (\pi /6) & \sin (\pi /6) \\ -\sin (\pi /6) & \cos (\pi /6) \\ \end{bmatrix} ,A= \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} $ and $ Q=PAP’ $ then $ P’Q^{2007}P $ is equal to
Options:
A) $ \begin{bmatrix} 1 & 2007 \\ 0 & 1 \\ \end{bmatrix} $
B) $ \begin{bmatrix} 1 & \sqrt{3}/2 \\ 0 & 2007 \\ \end{bmatrix} $
C) $ \begin{bmatrix} \sqrt{3}/2 & 2007 \\ 0 & 1 \\ \end{bmatrix} $
D) $ \begin{bmatrix} \sqrt{3}/2 & -1/2 \\ 1 & 2007 \\ \end{bmatrix} $
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Answer:
Correct Answer: A
Solution:
- [a] Note that $ P’={P^{-1}} $ Now, $ Q=PAP’=PA{P^{-1}} $
$ \Rightarrow Q^{2007}=PA^{2007}{P^{-1}} $
$ \therefore P’Q^{2007}P={P^{-1}}(PA^{2007}{P^{-}}^{1})P $
$ =A^{2007}={{(I+B)}^{2007}} $ Where $ B= \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} $ . As $ B^{2}=0 $ , we get $ B^{r}=0\forall r\ge 2 $ . Thus, by binomial theorem, $ A^{2007}=I+2007B= \begin{bmatrix} 1 & 2007 \\ 0 & 1 \\ \end{bmatrix} $
BETA
