<s>[INST] Limitations of Cramer's rule? I'm trying to find the line about which this matrix reflects a vector
$\frac{1}{2}\left(\begin{array}{cc}
   -1 & -\sqrt{3} \\
   -\sqrt{3} & 1 \\
  \end{array}\right)$ $\left(\begin{array}{c}
   x \\
 y \\
  \end{array}\right)$ = $\left(\begin{array}{c}
   x \\
 y \\
  \end{array}\right)$
using cramer's rule i.e
$x =  \frac{(\frac{1}{2})^2 \left|\begin{array}{cc}
   x & -\sqrt{3} \\
   y & 1 \\
  \end{array}\right|}{(\frac{1}{2})^2 \left|\begin{array}{cc}
   -1 & -\sqrt{3} \\
   -\sqrt{3} & 1 \\
  \end{array}\right|}$ and $y = \frac{(\frac{1}{2})^2 \left|\begin{array}{cc}
   x & -\sqrt{3} \\
   y & 1 \\
  \end{array}\right|}{(\frac{1}{2})^2 \left|\begin{array}{cc}
   -1 & -\sqrt{3} \\
   -\sqrt{3} & 1 \\
  \end{array}\right|}$
But for some reason this gives me this $y=\frac{-5}{\sqrt{3}}x$ (incorrect) when I use the x-component and $y=-\sqrt{3}x$ when I use the y-component. Is there some obvious assumption I'm missing? Can Cramer's rule not be used like this? Any help would be deeply appreciated.
 [/INST] Note, that the matrix has a factor $\frac{1}{2}$ in front.
So, you made a small mistake when substituting $\begin{pmatrix} x \\ y \end{pmatrix}$ into the determinant:
$$x =  \frac{(\frac{1}{2})^2 \left|\begin{array}{cc}
   \color{blue}{2}x & -\sqrt{3} \\
   \color{blue}{2}y & 1 \\
  \end{array}\right|}{(\frac{1}{2})^2 \left|\begin{array}{cc}
   -1 & -\sqrt{3} \\
   -\sqrt{3} & 1 \\
  \end{array}\right|}$$
Btw., it is much easier to solve
$$\left[ \frac{1}{2}\left(\begin{array}{cc}
   -1 & -\sqrt{3} \\
   -\sqrt{3} & 1 \\
  \end{array}\right) - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right]\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{2}\left(\begin{array}{cc}
   -3 & -\sqrt{3} \\
   -\sqrt{3} & -1 \\
  \end{array}\right)\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
</s>