
Find matrix B
B is a 2x2 matrix with the property that there is an invertible 2x2 matrix R such that
R^1BR = [(sqrt 3) 0 ] [ 1.2 (sqrt 3)/2]
[ 0 (sqrt 3) ] [(sqrt 3)/2 1/2 ]
I know the first part is the scaling matrix and the second on in the rotation matrix.
I'm suppose to find B^6 which is equal to
B^6 = R^1D^6R
But I dont know how I would find the matrix for R or D.

Are you saying that $\displaystyle B$ is a matrix such that there exists invertible $\displaystyle R$ such that
$\displaystyle R^{1}BR=\begin{bmatrix}\sqrt{3}&0\\0&\sqrt{3}\end{bma trix}
\begin{bmatrix}1/2&\sqrt{3}/2\\\sqrt{3}/2&1/2\end{bmatrix}?$
And you're asked to find $\displaystyle B^{6}?$

Yea, that is what we were given and then asked to find B^6

Hmm. Interesting problem. Well, the nice thing about diagonal matrices is that it's easy to find their nth power: just take the nth power of the elements on the main diagonal. Now the problem you've been given is not a typical diagonalization problem. Presumably, you've seen those already, so you know how you can always do this:
If
$\displaystyle B=R^{1}DR,$ then
$\displaystyle B^{n}=(R^{1}DR)^{n}=\underbrace{R^{1}DRR^{1}DR\dots R^{1}DR}_{n\times}=R^{1}\underbrace{DD\dots D}_{n\times}R=R^{1}D^{n}R.$
I played some "what if" games with your situation. Let's say you've got
$\displaystyle R^{1}BR=\begin{bmatrix}\sqrt{3}&0\\0&\sqrt{3}\end{bma trix}
\begin{bmatrix}1/2&\sqrt{3}/2\\\sqrt{3}/2&1/2\end{bmatrix}=C.$
So, $\displaystyle C$ is known. Then
$\displaystyle B=RCR^{1},$
and the $\displaystyle R^{1}$ is on the wrong side of $\displaystyle C.$ We want to have
$\displaystyle B=R^{1}DR.$ If I compare these two, it leads me to ask this question: what if I could find a diagonal $\displaystyle D$ such that
$\displaystyle C=R^{2}DR^{2}?$ Then
$\displaystyle B=RCR^{1}=RR^{2}DR^{2}R^{1}=R^{1}DR,$ as desired.
So, can you diagonalize $\displaystyle C=Q^{1}DQ$ but then, when you find the invertible matrix $\displaystyle Q,$ can you take its square root? There are ways to do that, I think, that aren't overly difficult, especially if you orthogonally diagonalize $\displaystyle C.$
So, give this program a shot: orthogonally diagonalize $\displaystyle C=Q^{1}DQ.$ What do you get?

Are you sure about that? I recommend you go back and reread the problem. I am inclined to suspect that it is R itself that is equal to $\displaystyle \begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}$ and that then $\displaystyle R^{1}BR= \begin{bmatrix}\sqrt{3} & 0 \\ 0 & \sqrt{3}\end{bmatrix}$
which is then the same as saying that $\displaystyle \begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}B= \bmatrix{\sqrt{3} & 0 \\ 0 & \sqrt{3}\end{bmatrix}\begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}$.
If that is the case, that you have $\displaystyle R^{1}BR= D$, where I am using "D" to stand for that diagonal matrix, then you also have $\displaystyle B= RDR^{1}$. Now notice that $\displaystyle B^2= (RDR^{1})(RDR^{1})= (RD)(R^{1}R)(DR^{1})= (RD)(DR^{1})= RD^2R^{1}$
$\displaystyle B^3= B(B^2)= (RDR^{=1})(RD^2R^{1})= RD^3R^{1}$, etc.
Further, since D is a diagonal matrix, it is easy to find powers of D:
$\displaystyle \begin{bmatrix}d & 0 \\ 0 & d\end{bmatrix}^n= \begin{bmatrix}d^n & 0 \\ 0 & d^n\end{bmatrix}$.