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 2 A=2 sin A cos A

Answer

A, C

Explanation

`sin A = [[cos theta ,sin theta],[-sin theta , cos theta]] and cos theta = [[sin theta , cos theta ],[cos theta , sin theta ]]`
`therefore abs(sin A) cos^(2) theta + sin ^(2) theta = 1 ne 0 `
Hence, sin A is incertible.
Also, `(sinA)(sin A)^(T) = [[cos theta ,sin theta],[-sin theta , cos theta]][[cos theta ,-sin theta],[sin theta , cos theta]]`
` = [[1,0],[0,1]] = I`
Hence, sin A is orthogonal.
Also ,`2sinA" " sin A = 2[[cos theta ,sin theta],[-sin theta , cos theta]][[sin theta,cos theta ],[ cos theta,sin theta]]`
`= 2 [[sin 2theta, 1 ],[cos 2 theta, 0]] ne sin 2 A`