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- #1

^{2}.

- A reflection about the line y=x, followed by a rotation counterclockwise of 60

^{o}.

This is how I proceeded.

y=x

$\begin{bmatrix}0&1\\1&0 \end{bmatrix}$

counter clockwise 60degs.

$\begin{bmatrix}1/2&-sqrt(3)/2\\sqrt(3)/2&1/2 \end{bmatrix}$

...then I multiplied them. Causing a dilemma. What order?

$\begin{bmatrix}0&1\\1&0 \end{bmatrix}$$\begin{bmatrix}1/2&-sqrt(3)/2\\sqrt(3)/2&1/2 \end{bmatrix}$

yields

$\begin{bmatrix}sqrt(3)/2&1/2\\1/2&-sqrt(3)/2 \end{bmatrix}$

The reverse obviously gave me a different answer, but I couldn't really visualize it.

$\begin{bmatrix}1/2&-sqrt(3)/2\\sqrt(3)/2&1/2 \end{bmatrix}$$\begin{bmatrix}0&1\\1&0 \end{bmatrix}$

Was my initial step correct?