If P = [ * 20 c 3 2 & 1 2 - 1 2 & 3 2 ], ,A = [ * 20 c 1&1 0&1 ] … - Vidyadip

MCQ

If $P = \left[ {\begin{array}{*{20}{c}}{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\end{array}} \right],\,A = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]$ and $Q = PA{P^T}$, then ${P^T}({Q^{2005}})P$ equal to

✓
$\left[ {\begin{array}{*{20}{c}}1&{2005}\\0&1\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{2005}\\1&0\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}1&{2005}\\{\sqrt 3 /2}&1\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}1&{\sqrt 3 /2}\\0&{2005}\end{array}} \right]$

✓

Answer

Correct option: A.

$\left[ {\begin{array}{*{20}{c}}1&{2005}\\0&1\end{array}} \right]$

a
(a) If $Q = PA{P^T}$

${P^T}Q = A{P^T}$, $({\rm{as}}\,\,P{P^T} = I)$

${P^T}{Q^{2005}}P = A{P^T}{Q^{2004}}P$

$ = {A^2}{P^T}{Q^{2003}}P$ $ = {A^3}{P^T}{Q^{2002}}P$ $ = {A^{2004}}{P^T}(QP)$

$ = {A^{2004}}{P^T}(PA)$ $(Q = PA{P^T} \Rightarrow QP = PA)$ $ = {A^{2005}}$

==> ${A^{2005}} = \left[ {\begin{array}{*{20}{c}}1&{2005}\\0&1\end{array}} \right]$.

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