Question

if $p=\begin {bmatrix} \sqrt 3 /2 & 1/2 \\ -1/2 & \sqrt 3/2 \end {bmatrix} , A=\begin {bmatrix} 1 & 1 \\ 0 & 1 \end {bmatrix}$ and $ Q=PAP^T,\ then$ $P^TQ^{2005}P $ is

• A. $\begin {bmatrix} 1 & 2005 \\ 0 & 1 \end {bmatrix}$
• B. $\begin {bmatrix} 1 & 2005 \\ 2005 & 1 \end {bmatrix}$
• C. $\begin {bmatrix} 1 & 0 \\ 2005 & 1 \end {bmatrix}$
• D. $\begin {bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix}$

Answer 1

The Correct Option is A

Now, $ \ \ \ \ \ \ \ P^T P =\begin {bmatrix}
\sqrt 3 /2 & -1/2 \\
1/2 & \sqrt 3/2 \end {bmatrix} \begin {bmatrix}
\sqrt 3 /2 & 1/2 \\
-1/2 & \sqrt 3/2 \end {bmatrix} $
$\Rightarrow \ \ \ P^TP=\begin {bmatrix}
1 & 1 \\
0 & 1 \end {bmatrix} \Rightarrow \ \ P^T P= I \ \ \Rightarrow \ \ P^T=P^{-1}$
Since $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q=PAP^T$
$\therefore \ \ \ P^TQ^{2005}P=P^T[(PAP^T)(PAP^T).....2005 \ times ]P$
$=\frac{(PAP^T)A(PAP^T)A(PAP^T)..........(PAP^T)A(PAP^T)}{2005 \ times}$
$=IA^{2005}=A^{2005}$
$\therefore \ \ A^1 \begin {bmatrix}
1 & 1 \\
0 & 1 \end {bmatrix} $
$A^2 \begin {bmatrix}
1 & 1 \\
0 & 1 \end {bmatrix} \begin {bmatrix}
1 & 1 \\
0 & 1 \end {bmatrix}=\begin {bmatrix}
1 & 2 \\
0 & 1 \end {bmatrix}$
$A^3=\begin {bmatrix}
1 & 2 \\
0 & 1 \end {bmatrix} \begin {bmatrix}
1 & 1 \\
0 & 1 \end {bmatrix} =\begin {bmatrix}
1 & 3 \\
0 & 1 \end {bmatrix}$
.... ..... ..... ........ . ........
.... ...... ..... ....... .......
$A^{2005} =\begin {bmatrix}
1 & 2005 \\
0 & 1 \end {bmatrix}$
$\therefore \ \ \ \ P^TQ^{2005}P=\begin {bmatrix}
1 & 2005 \\
0 & 1 \end {bmatrix}$