Variables cancelling after matrix rotation

Question

Matrix Rotation Question

The above question is from a past exam paper. Unfortunately, I am struggling with the arithmetic as follows:

Using the standard 2D matrix rotation transformation, i obtained the following equations: $$ \frac{a}{2} - \frac{3^{1/2}b}{2} = b $$ $$ \frac{3^{1/2}a}{2} + \frac{b}{2} = a $$ However when I try to solve these, the variables cancel. Is this intended or is my approach/procedure incorrect?

EDIT: here is my solution: (from the standard 2D rotation matrix) $$ cos\frac{\pi}{3}a - sin\frac{\pi}{3}b = b $$ $$ sin\frac{\pi}{3}a + cos\frac{\pi}{3}b = a $$ I treated the rotation matrix as a transformation of (a,b) to (b,a)

EDIT 2: Seeing as there are infinite solutions to this problem, does that indicate that it is the span of vectors that follow the above relationship that satisfy the conditions?

Answer 1

The variables does not ''cancel'', but we have: $a=(2+\sqrt{3})b \qquad \forall b \in \mathbb{R} $.

Do you understand what does this means?

Answer 2

For any $b$, $a$ is $$a = (\sqrt{3}+2) b$$

Another way to look at it, would be if the magnitude $s = \sqrt{a^2+b^2}$ is to be preserved then $$\begin{aligned} a & = s \frac{\sqrt{3}+1}{2^{3/2}} \\ b & = s \frac{\sqrt{3}-1}{2^{3/2}} \end{aligned} $$