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If A=[[cos theta , sin theta],[sin theta...

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
a. `[[-n,1],[1,0]]`
b. `[[1,-n],[0,1]]`
c. `[[0,1],[1,-n]]`
d. `[[1,0],[-n,1]]`

A

`[[-n,1],[1,0]]`

B

`[[1,-n],[0,1]]`

C

`[[0,1],[1,-n]]`

D

`[[1,0],[-n,1]]`

Text Solution

Verified by Experts

The correct Answer is:
D
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]]`
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