If a transformation has matrix M=(cosθ−sinθsinθcosθ)M = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}M=(cosθsinθ−sinθcosθ), is it an isometry? Justify.
Yes; det(M)=1\det(M) = 1det(M)=1 and MTM=IM^TM = IMTM=I
No; det(M)=0\det(M) = 0det(M)=0
Only if θ=90∘\theta = 90^\circθ=90∘
Yes; rotation matrices preserve distances by definition