Let `A=[(1,0,0),(1,0,1),(0,1,0)]` satisfies `A^(n)=A^(n-2)+A^(2)-I` for `n ge 3`. And trace of a square matrix X is equal to the sum of elements in its proncipal diagonal.
Further consider a matrix `underset(3xx3)(uu)` with its column as `uu_(1), uu_(2), uu_(3)` such that
`A^(50) uu_(1)=[(1),(25),(25)], A^(50) uu_(2)=[(0),(1),(0)], A^(50) uu_(3)=[(0),(0),(1)]`
Then answer the following question :
Trace of `A^(50)` equals

Options

A. 0
B. 1
C. 2
D. 3

Answer

D

Explanation

`because A^(n) = A^(n-2) + A^(2) - I rArr A^(50) = A^(48) + A^(2) - I`
Further, `{:(A^(48) =, A^(46) +, A^(2) ,- I) , (A^(46) =, A^(44) +, A^(2), -I),(vdots ,vdots,vdots, vdots),(A^(4)=,A^(2) +, A^(2),-I^(4)):} `
On adding all, we get
` A^(50) = 25 A^(2) - 24I` ...(i)
`because A^(2) = [[1,0,0],[1,0,1],[0,1,0]][[1,0,0],[1,0,1],[0,1,0]]= [[1,0,0],[1,1,0],[1,0,1]]`
`therefore A^(50) = 25 A^(2) - 24I = [[25,0,0],[25,25,0],[25,0,25]]-[[24,0,0],[0,24,0],[0,0,24]]` [from Eq. (i)]
`=[[1,0,0],[25,1,0],[25, 0,1]]` ...(ii)
Hence, trace of `A^(50) = 1+ 1+ 1= 3`