If \[P = \left[ {\begin{array}{*{20}{c}}{\dfrac{{\sqrt 3 }}{2}}&{\dfrac{1}{2}}\\{ - \dfrac{1}{2}}&{\dfrac{{\sqrt 3 }}{2}}\end{array}} \right]\] , \[A = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\] and \[Q = PA{P^T}\], then find the value of \[{P^T}\left( {{Q^{2005}}} \right)P\].
A. \[\left[ {\begin{array}{*{20}{c}}1&{2005}\\0&1\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}{\dfrac{{\sqrt 3 }}{2}}&{2005}\\1&0\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}1&{2005}\\{\dfrac{{\sqrt 3 }}{2}}&1\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}1&{\dfrac{{\sqrt 3 }}{2}}\\0&{2005}\end{array}} \right]\]

Hint: Here, 2 matrices are given. First, simplify the required equation by using the given equation and basic identities of matrices. In the end, substitute the values in the equation and solve it to get the required answer.

Formula used:
The transposition of a matrix is found by interchanging its rows into columns or columns into rows.
For a matrix \[A\], the transpose matrix is denoted by \[{A^T}\]

Complete step by step solution:
The given matrices are \[P = \left[ {\begin{array}{*{20}{c}}{\dfrac{{\sqrt 3 }}{2}}&{\dfrac{1}{2}}\\{ - \dfrac{1}{2}}&{\dfrac{{\sqrt 3 }}{2}}\end{array}} \right]\], and \[A = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\].
The given equation is \[Q = PA{P^T}\].

Let’s calculate the matrix multiplication of \[{P^T}\] and \[P\].
\[{P^T}P = \left[ {\begin{array}{*{20}{c}}{\dfrac{{\sqrt 3 }}{2}}&{\dfrac{1}{2}}\\{ - \dfrac{1}{2}}&{\dfrac{{\sqrt 3 }}{2}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\dfrac{{\sqrt 3 }}{2}}&{ - \dfrac{1}{2}}\\{\dfrac{1}{2}}&{\dfrac{{\sqrt 3 }}{2}}\end{array}} \right]\]
\[ \Rightarrow {P^T}P = \left[ {\begin{array}{*{20}{c}}{\dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2} \times \dfrac{1}{2}}&{\dfrac{{\sqrt 3 }}{2} \times \left( { - \dfrac{1}{2}} \right) + \dfrac{1}{2} \times \dfrac{{\sqrt 3 }}{2}}\\{ - \dfrac{1}{2} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{2}}&{\left( { - \dfrac{1}{2}} \right) \times \left( { - \dfrac{1}{2}} \right) + \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2}}\end{array}} \right]\]
\[ \Rightarrow {P^T}P = \left[ {\begin{array}{*{20}{c}}{\dfrac{3}{4} + \dfrac{1}{4}}&{ - \dfrac{{\sqrt 3 }}{4} + \dfrac{{\sqrt 3 }}{4}}\\{ - \dfrac{{\sqrt 3 }}{4} + \dfrac{{\sqrt 3 }}{4}}&{\dfrac{1}{4} + \dfrac{3}{4}}\end{array}} \right]\]
\[ \Rightarrow {P^T}P = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\]
\[ \Rightarrow {P^T}P = I\] \[.....\left( 1 \right)\]

Now solve the required equation \[{P^T}\left( {{Q^{2005}}} \right)P\] by using the given equation \[Q = PA{P^T}\].
\[{P^T}\left( {{Q^{2005}}} \right)P = {P^T}{\left( {PA{P^T}} \right)^{2005}}P\]
\[ \Rightarrow {P^T}\left( {{Q^{2005}}} \right)P = {P^T}\left( {PA{P^T}} \right)\left( {PA{P^T}} \right)....2005\, times P\]
\[ \Rightarrow {P^T}\left( {{Q^{2005}}} \right)P = {P^T}PA\left( {{P^T}P} \right)A\left( {{P^T}P} \right)....2005 \, times \left( {{P^T}P} \right)\]
Use the equation \[\left( 1 \right)\].
\[ \Rightarrow {P^T}\left( {{Q^{2005}}} \right)P = IA\left( I \right)A\left( I \right)....2005\, times \left( I \right)\]
\[ \Rightarrow {P^T}\left( {{Q^{2005}}} \right)P = AA....2005 \, times\]
\[ \Rightarrow {P^T}\left( {{Q^{2005}}} \right)P = {A^{2005}}\]

Here \[A = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\]
\[{A^2} = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]\]
\[{A^3} = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right]\]
After observing the above pattern, we get that
\[{A^n} = \left[ {\begin{array}{*{20}{c}}1&n\\0&1\end{array}} \right]\]

Therefore, \[{P^T}\left( {{Q^{2005}}} \right)P = {A^{2005}} = \left[ {\begin{array}{*{20}{c}}1&{2005}\\0&1\end{array}} \right]\]
Hence the correct option is A.

Note: Students should keep in mind that the product of two matrices is defined only if the number of columns of the first matrix is equal to the number of rows of the second matrix. Also, the transpose of a matrix is obtained by changing rows to columns and columns to rows. In other words, transpose of A[N][M] is obtained by changing A[i][j] to A[j][i].