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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 ${P^T}{Q^{2015}}P$
(A) $\left[ {\begin{array}{*{20}{c}}
  0&{2015} \\
  0&0
\end{array}} \right]$
(B) $\left[ {\begin{array}{*{20}{c}}
  {2015}&0 \\
  1&{2015}
\end{array}} \right]$
(C) $\left[ {\begin{array}{*{20}{c}}
  {2015}&1 \\
  0&{2015}
\end{array}} \right]$
(D) $\left[ {\begin{array}{*{20}{c}}
  1&{2015} \\
  0&1
\end{array}} \right]$

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Last updated date: 14th Jun 2024
Total views: 411.6k
Views today: 10.11k
Answer
VerifiedVerified
411.6k+ views
Hint: Firstly, multiply the matrix P and transpose of P together. Then using that and the value of matrix Q, transform the given expression into a simpler form. Multiply matrix A with itself to figure out the pattern in the higher power.

Complete step-by-step answer:
In the question, two matrices P and A are given and matrix Q is defined as $Q = PA{P^T}$, i.e. it is a multiplication of matrix P, A and transpose of matrix P. With all these information, we need to find the value of ${P^T}{Q^{2015}}P$.
Let’s go step by step and first solve the transpose of P. In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal; that is, it switches the row and column indices of the matrix by producing another matrix.
$P = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{\sqrt 3 }}{2}}&{\dfrac{1}{2}} \\
  { - \dfrac{1}{2}}&{\dfrac{{\sqrt 3 }}{2}}
\end{array}} \right] \Rightarrow {P^T} = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{\sqrt 3 }}{2}}&{ - \dfrac{1}{2}} \\
  {\dfrac{1}{2}}&{\dfrac{{\sqrt 3 }}{2}}
\end{array}} \right]$
Let’s multiply these two matrices
$ \Rightarrow {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] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{3 - 1}}{2}}&0 \\
  0&{\dfrac{{3 - 1}}{2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = I$
Therefore, we got the product of matrix P and its transpose as an identity matrix.
We also have $Q = PA{P^T}$, let’s substitute it into the expression for which we need to find the value.
${P^T}{Q^{2015}}P = {P^T}{\left( {PA{P^T}} \right)^{2005}}P = {P^T}\left( {PA{P^T}} \right)\left( {PA{P^T}} \right)\left( {PA{P^T}} \right).....\left( {PA{P^T}} \right)P$
But this can be rewritten similarly, but this time it will be more useful to us:
$ \Rightarrow {P^T}{Q^{2015}}P = \left( {{P^T}P} \right)A\left( {{P^T}P} \right)A\left( {{P^T}P} \right)A.........\left( {{P^T}P} \right)A\left( {{P^T}P} \right)$
But, we already calculated that ${P^T}P = I$ which gives us:
$ \Rightarrow {P^T}{Q^{2015}}P = \left( {{P^T}P} \right)A\left( {{P^T}P} \right)A\left( {{P^T}P} \right)A.........\left( {{P^T}P} \right)A\left( {{P^T}P} \right) = I{A^{2005}} = {A^{2005}}$
Now, we just got to evaluate ${A^{2005}}$and we already have the value $A = \left[ {\begin{array}{*{20}{c}}
  1&1 \\
  0&1
\end{array}} \right]$.
\[ \Rightarrow A \times A = {A^2} = \left[ {\begin{array}{*{20}{c}}
  1&1 \\
  0&1
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  1&1 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {1 - 0}&{1 + 1} \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&1
\end{array}} \right]\]
And, ${A^3} = {A^2} \times A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&1 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {1 - 0}&{2 + 1} \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  0&1
\end{array}} \right]$
We can clearly see the pattern here when we are multiplying with A
Therefore, we can say: ${A^{2005}} = \left[ {\begin{array}{*{20}{c}}
  1&{2005} \\
  0&1
\end{array}} \right]$
Hence, we get the required product of matrices as: ${P^T}{Q^{2015}}P = {A^{2005}} = \left[ {\begin{array}{*{20}{c}}
  1&{2005} \\
  0&1
\end{array}} \right]$

So, the correct answer is “Option D”.

Note:Follow a step by step procedure for a question involving multiplication of matrices. You can multiply two matrices if, and only if, the number of columns in the first matrix equals the number of rows in the second matrix. An alternative approach to the same problem is by finding ${A^{2005}}$ before the transformation of ${P^T}{Q^{2015}}P$.