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Mathematics Question on Matrices

if p=[3/21/2 −1/23/2],A=[11 01]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=PAPT, then Q=PAP^T,\ then PTQ2005PP^TQ^{2005}P is

A

[12005 01]\begin {bmatrix} 1 & 2005 \\\ 0 & 1 \end {bmatrix}

B

[12005 20051]\begin {bmatrix} 1 & 2005 \\\ 2005 & 1 \end {bmatrix}

C

[10 20051]\begin {bmatrix} 1 & 0 \\\ 2005 & 1 \end {bmatrix}

D

[10 01]\begin {bmatrix} 1 & 0 \\\ 0 & 1 \end {bmatrix}

Answer

[12005 01]\begin {bmatrix} 1 & 2005 \\\ 0 & 1 \end {bmatrix}

Explanation

Solution

Now,        PTP=[3/2−1/2 1/23/2][3/21/2 −1/23/2]\ \ \ \ \ \ \ 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}
⇒   PTP=[11 01]⇒  PTP=I  ⇒  PT=P−1\Rightarrow \ \ \ P^TP=\begin {bmatrix} 1 & 1 \\\ 0 & 1 \end {bmatrix} \Rightarrow \ \ P^T P= I \ \ \Rightarrow \ \ P^T=P^{-1}
Since                Q=PAPT \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q=PAP^T
∴   PTQ2005P=PT[(PAPT)(PAPT).....2005 times]P\therefore \ \ \ P^TQ^{2005}P=P^T[(PAP^T)(PAP^T).....2005 \ times ]P
=(PAPT)A(PAPT)A(PAPT)..........(PAPT)A(PAPT)2005 times=\frac{(PAP^T)A(PAP^T)A(PAP^T)..........(PAP^T)A(PAP^T)}{2005 \ times}
=IA2005=A2005=IA^{2005}=A^{2005}
∴  A1[11 01]\therefore \ \ A^1 \begin {bmatrix} 1 & 1 \\\ 0 & 1 \end {bmatrix}
A2[11 01][11 01]=[12 01]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}
A3=[12 01][11 01]=[13 01]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}
.... ..... ..... ........ . ........
.... ...... ..... ....... .......
A2005=[12005 01]A^{2005} =\begin {bmatrix} 1 & 2005 \\\ 0 & 1 \end {bmatrix}
∴    PTQ2005P=[12005 01]\therefore \ \ \ \ P^TQ^{2005}P=\begin {bmatrix} 1 & 2005 \\\ 0 & 1 \end {bmatrix}