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 $Q =PAP \phi$ then $P \phi Q ^{2007} P$ is equal to

• 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. $\begin{bmatrix}\sqrt{3} / 2 & -1 / 2 \\ 1 & 2007\end{bmatrix}$

Answer 1

Answer: (A)

Note that $P'= P ^{-1}$ Now, $Q = PAP'= PAP ^{-1}$
$\Rightarrow Q^{2007}=P A^{2007} P^{-1}$
$\therefore P' Q ^{2007} P = P ^{-1}\left( PA ^{2007} P ^{-1}\right) P = A ^{2007}=( I + B )^{2007}$
where $B = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$.
As $B ^{2}=0$, we get $B ^{ r }=0 \forall r \geq 2$.
Thus, by binomial theorem,
$A ^{2007}= I +2007 B = \begin{bmatrix}1 & 2007 \\ 0 & 1\end{bmatrix}$