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If `A= [a_(ij)] _(nxxn)` and f is a function, we define `f (A) = [f(a_(ij))] _(nxxn)."Let"A = [[pi/2-theta,theta],[-theta,pi/2-theta]]` thenA. sin A is invertibleB. sin A = cos AC. sin A is orthogonalD. sin 2 A=2 sin A cos A |
Answer» Correct Answer - A::C `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` |
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