If A(θ)=sin⁡θicos⁡θicos⁡θsin⁡θ, then which of the following is not true?

We have, |A(θ)|=1Hence, A is invertible.A(π+θ)=A(π+θ)=sin⁡(π+θ)icos⁡(π+θ)icos⁡(π+θ)sin⁡(π+θ)=−sin⁡θ−icos⁡θ−icos⁡θ−sin⁡θ=−A(θ)adj⁡(A(θ))=sin⁡θ−icos⁡θ−icos⁡θsin⁡θ⇒ A(θ)−1=sin⁡θ−icos⁡θ−icos⁡θsin⁡θ=A(π−θ)