If `A=(a_(i j))_(nxxn)` and `f` is a function, we define `f(A)=((f(a_(i j))))_(nxxn ')` Let `A=([pi//2-theta,theta],[-theta,pi//2-theta])` . Then

Options

A. `sin A` is invertible
B. `sin A = cos A`
C. `sin A` is orthogonal
D. `sin (2A)=2 sin A cos A`

Explanation

`sin A=[(cos theta,sin theta),(-sin theta,cos theta)]` and `cos A =[(sin theta,cos theta),(cos theta,sin theta)]`
`:. |sin A|=cos^(2) theta+ sin^(2) theta=1`.
Hence, `sin A` is invertiable.
Also, `(sin A)xx(sin A)^(T)=[(cos theta,sin theta),(-sin theta,cos theta)][(cos theta,-sin theta),(sin theta,cos theta)]`
`=[(cos^(2) theta+sin^(2) theta,0),(0,cos^(2) theta+sin^(2) theta)]`
`=[(1,0),(0,1)]`
`=I`
Hence, `sin A` is orthogonal. Also,
`2 sin A cos A=2 [(cos theta,sin theta),(-sin theta,cos theta)][(sin theta,cos theta),(cos theta,sin theta)]`
`=2 [(2 sin theta cos theta,cos^(2) theta+sin^(2) theta),(cos^(2) theta-sin^(2) theta,0)]`
`=2[(sin 2 theta,1),(cos 2 theta,0)] ne sin 2A`