If `{:P=[(sqrt3/2,1/2),(-1/2,sqrt3/2)],A=[(1,1),(0,1)]:}and Q=PAP^T," then" P^TQ^2015P`, is

Options

A. `{:[(2015,1),(1,2015)]:}`
B. `{:[(1,2015),(0,1)]:}`
C. `{:[(0,2015),(0,0)]:}`
D. `{:[(2015,0),(1,2015)]:}`

Answer

B

Explanation

We have,
`{:P=[(sqrt3/2,1/2),(-1/2,sqrt3/2)]rArr P^T=[(sqrt3/2,-1/2),(1/2,sqrt3/2)]:}`
Clearly, `P^T=P^(-1)`
`:. Q=PAP^T=PAP^(-1)`
`rArr Q^2015=(PAP^(-1))^2015=PA^2015P^(-1)`
`:. P^TQ^2015P=P^(-1)(PA^2015P^(-1))P=A^2015`
Now, `A={:[(1,1),(0,1)]:}`
`:.{:A^2=[(1,1),(0,1)][(1,1),(0,1)]=[(1,2),(0,1)]:}`
`{:A^3=A^2A=[(1,2),(0,1)][(1,1),(0,1)]=[(1,3),(0,1)]:}`and so no.
In general `{:A^2015=[(1,2015),(0,1)]:}`
Hence, `P^TQ^2015P=A^2015={:[(1,2015),(0,1)]:}`.