If `A=[[cos theta , sin theta],[sin theta,-costheta]], B = [[1,0],[-1,1]], C=ABA^(T),` then
`A^(T) C^(n) A, n in I^(+)` equals to

Options

A. `[[-n,1],[1,0]]`
B. `[[1,-n],[0,1]]`
C. `[[0,1],[1,-n]]`
D. `[[1,0],[-n,1]]`

Answer

D

Explanation

Here, `A= [[cos theta, sin theta ],[sin theta,-cos theta]]`
`rArr A A^(T) = I`
`because C=ABA^(T)rArr A^(T) C = BA^(T)`
`A^(T) C^(n) A=A^(T) Ccdot C^(n-1) A`
` = BA^(T)C^(n-1)A= BA^(T) C C^(n-2) A`
`= B^(2) A^(T) C^(n-2)A`
`..." "..." "...`
`=B^(n-1) A^(T) CA=B^(n-1)(BA^(T))A`
`= B^(n) A^(T) A=B^(n)I = B^(n)= [[1,0],[-n,1]]`