Question

If $P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right],\hspace{0.17em}A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],$ and $Q=PA{P}^T$ then find the value of $P^T{Q}^{2005}P$ is equal to

• A. $\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 2005\\ 2005& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 2005& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Answer 1

The correct option is A

$\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$

Explanation for the correct option

Step 1: Simplification of the equation $\mathit{Q}\mathbf{=}\mathit{P}\mathit{A}{\mathit{P}}^{\mathbf{T}}$,

Since the given expression is $Q=PA{P}^T$ which can be rewritten as,

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

Where $P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$

And transpose of $P$ is

$\Rightarrow P^T=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$

Now, we know that the product of a matrix and its transpose is equal to an identity matrix,

$\therefore P{P}^T=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$

$\Rightarrow P{P}^T=\left[\begin{array}{cc}\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\times \frac{1}{2}\right)& \left(\frac{\sqrt{3}}{2}\times -\frac{1}{2}\right)+\left(\frac{1}{2}\times \frac{\sqrt{3}}{2}\right)\\ \left(-\frac{1}{2}\times \frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{3}}{2}\times \frac{1}{2}\right)& \left(\frac{1}{2}\times \frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\right)\end{array}\right]$

$\Rightarrow P{P}^T=\left[\begin{array}{cc}\frac{3}{4}+\frac{1}{4}& -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\\ -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}& \frac{1}{4}+\frac{3}{4}\end{array}\right]=\left[\begin{array}{cc}\frac{4}{4}& 0\\ 0& \frac{4}{4}\end{array}\right]$

$\Rightarrow P{P}^T=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

Where $I$ is an identity matrix

$\Rightarrow P{P}^T=I$

We know that $Q=P{P}^{T}A$

$\therefore Q=IA$

Step 2: Simplification of the required expression

Now solve the expression $P^T{Q}^{2005}P$

$\begin{array}{rcl}\therefore P^T{Q}^{2005}P& =& P^T\left[\left(PA{P}^T\right)\left(PA{P}^T\right)\left(PA{P}^T\right)\dots 2005\right]P\\ & =& \left(P{P}^T\right)A\left(P{P}^T\right)A\left(P{P}^T\right)A\dots \left(P{P}^T\right)AP{P}^T\left[\because P{P}^T=I\right]\\ & =& \left(I\right)A\left(I\right)A\left(I\right)A\dots \left(I\right)A\\ & =& I{A}^{2005}\left[\because I^n=I\right]\end{array}$

We also know that when a matrix is multiplied by an identity matrix then the result is a non-identity matrix.

$\begin{array}{rcl}{P}^T{Q}^{2005}P& =& I{A}^{2005}\\ & =& A^{2005}\end{array}$

From this, we have

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

Step 3: Calculation for the correct option

To determine $A^{2005}$, we need to multiply the matrix again and again like this

$\begin{array}{rcl}{A}^2& =& AA\\ & =& \left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\\ & =& \left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\end{array}$

Similarly,

$\begin{array}{rcl}{A}^3& =& A^{2}A\\ & =& \left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\\ & =& \left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\end{array}$

As we can see the value of $A,A^2,A^3...A^{2005}$

$\therefore A^{2005}=\left[\begin{array}{cc}1& 2005\\ 1& 0\end{array}\right]$

Finally, the value of $P^T{Q}^{2005}P=A^{2005}$

Hence option (A) is the correct answer.