IfA=⎡⎢⎣cos xsin x0−sin xcos x0001⎤⎥⎦=f(x), then A−1 is equal to
f(-x)
A=⎡⎢⎣cos xsin x0−sin xcos x0001⎤⎥⎦|A|=1×(cos2x+sin2x)=1adj(A)=⎡⎢⎣cos x−sin x0sin xcos x0001⎤⎥⎦A−1=1|A|adj(A)=⎡⎢⎣cos x−sin x0sin xcos x0001⎤⎥⎦A−1=⎡⎢⎣cos(−x)sin(−x)0−sin(−x)cos(−x)0001⎤⎥⎦=f(−x), since cos(−x)=cos x and sin(−x)=sin x