`If A=[[costheta,sintheta],[-sintheta,costheta]],then Lim_(ntooo)1/nA^n` is

Options

A. (A) an identity matrix
B. (B) [0 10 -1 0 ]
C. (C) a null matrix
D. (D) none of these

Answer

Zero Matrix

Explanation

`A=[(cos theta,sin theta),(-sin theta,cos theta)]`
`A^(2)=[(cos theta,sin theta),(-sin theta,cos theta)][(cos theta,sin theta),(-sin theta,cos theta)]`
`=[(cos^(2) theta-sin^(2) theta,2 sin theta cos theta),(-2 sin theta cos theta,cos^(2) theta - sin^(2) theta)]`
`=[(cos 2 theta,sin 2 theta),(-sin 2 theta,cos 2 theta)]`
`A^(3)=A^(2) A=[(cos 2 theta,sin 2 theta),(-sin 2 theta,cos 2 theta)][(cos theta,sin theta),(-sin theta,cos theta)]`
`=[(cos 3 theta,sin 3 theta),(-sin 3 theta,cos 3 theta)]`
Hence, `A^(n)=[(cos n theta,sin n theta),(-sin n theta,cos n theta)]`
or `A^(n)/n = [((cos n theta)/n,(sin n theta)/n),((-sin n theta)/n,(cos n theta)/n)]`
or `lim_(n rarr oo) A^(n)/n =[(lim_(n rarr oo) (cos n theta)/n,lim_(n rarr oo) (sin n theta)/n),(-lim_(n rarr oo) (sin n theta)/n,lim_(n rarr oo) (cos n theta)/n)]`
`=[(0,0),(0,0)]=` Zero matrix