Determining whether an orthogonal matrix represents a rotation or reflection The exercise asks us to determine whether the given orthogonal matrix represents a rotation or a reflection. If it is a rotation, give the angle of rotation; if it is a reflection, give the line of reflection.
$$
A = \begin{bmatrix}
       -\frac{3}{5} & -\frac{4}{5}\\[0.3em]
       -\frac{4}{5} & \frac{3}{5}\\[0.3em]
     \end{bmatrix}
$$
I know you can check whether it is a reflection or rotation by calculating the determinant. So for example for the matrix above
$$det(A) = -\frac{3}{5}\cdot\frac{3}{5}-(-\frac{4}{5})\cdot(-\frac{4}{5}) = -1$$
And so that means that matrix $A$ corresponds to a reflection in $R^2$, but how do you get the line of reflection from this?
$$
B = \begin{bmatrix}
       -\frac{1}{2} & \frac{\sqrt{3}}{2}\\[0.3em]
       -\frac{\sqrt{3}}{2} & -\frac{1}{2}\\[0.3em]
     \end{bmatrix}
$$
And for a rotation, so for example matrix $B$ given above, can you simply say that it corresponds to the rotation matrix $R$
$$ 
  R =   \begin{bmatrix}
       cos(\theta) & -sin(\theta)\\[0.3em]
       sin(\theta) & cos(\theta)\\[0.3em]
     \end{bmatrix}
$$
So this gives
$$cos(\theta) = -\frac{1}{2}, \theta = cos^{-1}(-\frac{1}{2}) = 120^{\circ}\\
  sin(\theta) = -\frac{\sqrt{3}}{2}, \theta = sin^{-1}(-\frac{\sqrt{3}}{2}) = -60^{\circ}
$$
Which means that matrix $B$ corresponds to a counterclockwise rotation of $120^{\circ}$, right?

for the line of reflection look for a mirror lines that stays fixed under the transformation. for example, in the case of $A = \begin{bmatrix}
       -\frac{3}{5} & -\frac{4}{5}\\[0.3em]
       -\frac{4}{5} & \frac{3}{5}\\[0.3em]
     \end{bmatrix}$  try the line $x = \pmatrix{5\\5a}.$  we want $Ax =  x$ that is 
$$\pmatrix{-3 -4a\\-4  + 3a } =  \pmatrix{5\\5a}$$ which gives you $a= -2.$  therefore the line $y = -2x$ is the mirror.
