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

Answer: (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$
$\therefore P^TQ^{2005}P=P^T[(PAP^T)(PAP^T).....2005 \text{ times} ]P$
$=\underbrace{\left(P^{T} P\right) A\left(P^{T} P\right) A\left(P^{T} P\right) \ldots\left(P^{T} P\right) A\left(P^{T} P\right)}_{2005 \text { times }}$
$=IA^{2005}=A^{2005}$
$\therefore A^1 \begin {bmatrix}1 & 1 \\0 & 1 \end {bmatrix} $
$A^2 \begin {bmatrix}1 & 1 \\0 & 1 \end {bmatrix} \begin {bmatrix} 1 & 1 \\0 & 1 \end {bmatrix}=\begin {bmatrix}1 & 2 \\0 & 1 \end {bmatrix}$
$A^3=\begin {bmatrix}1 & 2 \\0 & 1 \end {bmatrix} \begin {bmatrix}1 & 1 \\0 & 1 \end {bmatrix} =\begin {bmatrix}1 & 3 \\0 & 1 \end {bmatrix}$
.... ..... ..... ........ . ........
.... ...... ..... ....... .......
$A^{2005} =\begin {bmatrix}1 & 2005 \\0 & 1 \end {bmatrix}$
$\therefore P^TQ^{2005}P=\begin {bmatrix}1 & 2005 \\0 & 1 \end {bmatrix}$