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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?

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    You have not reflected the exam question adequately. It could have been phrased: find all points that will move the same under an anticlockwise rotation of $\pi/2$ around the origin moves them to the same place as a reflection in the diagonal line given by $a=b$. – 2017-01-08
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    Is that still a vector span also? – 2017-01-08
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    I'm not entirely sure how you interpret that. The solutions vectors (a,b) to the equation form a subspace that is spanned by, in this case, one vector. The answers below also imply this. – 2017-01-09

2 Answers 2

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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?

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    Oh, does it mean that any vectors that follow this relationship fit the above conditions? – 2017-01-08
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    Yes: any vector of the form $[(2+\sqrt{3})b,b]$ is a solution. – 2017-01-08
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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 $$\begin{aligned} a & = s \frac{\sqrt{3}+1}{2^{3/2}} \\ b & = s \frac{\sqrt{3}-1}{2^{3/2}} \end{aligned} $$