If A = [cos θsin θ−sin θcos θ], then for any natural number n, find the value of Det(A").
Given A = [cos θsin θ−sin θcos θ] ⇒An=[cos nθsin nθ−sin nθcos nθ] ⇒|An|i.e.,Det(An)=[cos nθsin nθ−sin nθcos nθ]=cos2nθ+sin2nθ=1.