Matrix Time

We begin by rewriting the given matrix,

$P=\begin{bmatrix} \dfrac {\sqrt{3}}{2} & \dfrac{1}{2} \\ \dfrac {-1}{2} & \dfrac {\sqrt{3}}{2} \end{bmatrix}=\begin{bmatrix} \sin \frac{\pi}{3} & \cos \frac{\pi}{3} \\ -\cos \frac{\pi}{3} & \sin \frac{\pi}{3} \end{bmatrix}$

Also, we can see that, $A=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=I$ , where $I$ is the $2\times 2$ identity matrix. We will use the identity: $\sin^2x+\cos^2x=1~\forall x\in \mathbb{R}$ while evaluating $Q$ and then the final matrix asked in the question (let's call it $X$ ).

Also, we will use the trivial identities $Y=YI_Y=I_YY$ , $(I_Y)^n=I_Y~\forall n\in \mathbb{N}$ , $(XY)^T=Y^TX^T$ and $(I_Y)^T=I_Y$ where $X,Y$ are square matrices and $I_Y$ is the corresponding identity matrix for $Y$ . Now, coming onto the calculations,

$Q=PAP^T=(PI)P^T=PP^T=\begin{bmatrix} \sin \frac{\pi}{3} & \cos \frac{\pi}{3} \\ -\cos \frac{\pi}{3} & \sin \frac{\pi}{3} \end{bmatrix}\cdot\begin{bmatrix} \sin \frac{\pi}{3} & -\cos \frac{\pi}{3} \\ \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \end{bmatrix}$

After matrix multiplication, we will obtain $Q=PP^T=I$

Now, $Q=I\implies Q^{2005}=I^{2005}=I$ . So,

$X=P^T(IP)=P^TP=(PP^T)^T=Q^T=I^T=I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

So, $X=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is our answer.

From the given data,its pretty clear that,

A=I.

Q=PAP"=PP",since A is an identity matrix. And also PP"=I,(By solving) ,hence we can conclude that P is an Orthogonal Matrix.

Hence,

P"Q^2500 P=I.

Note: P =Given matrix. P"=Transpose of P. I=Identity matrix.