Question

If $P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right],A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $Q=PA{P}^T$ then $P\left(Q^{2005}\right)P^T$ equal to

• A. $\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}\sqrt{3}/2& 2005\\ 1& 0\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 2005\\ \sqrt{3}/2& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}1& \sqrt{3}/2\\ 0& 2005\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$

We have,

$P^{T}P=1$

$\Rightarrow P^T=P^{-1}$

Now,

$Q=PA{P}^T$

So,

$P{Q}^{2005}{P}^T=[\left(PA{P}^T\right)\left(PA{P}^T\right)......2005\text{times}]P^T$

$=\left(P^{T}P\right)A\left(P^{T}P\right)A\left(P^{T}P\right)A.......\left(P^{T}P\right)A\left(P^{T}P\right)=I.A^{2005}$

$\Rightarrow I.A^{2005}=A^{2005}$

Now,

$A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

$A^2=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$

$A^3=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

.......

.......

$A^{2005}=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$

Hence, this is the answer.