Let `A=[(1,0,0),(1,0,1),(0,1,0)]` satisfies `A^(n)=A^(n-1)+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

`A^(n)-A^(n-2)=A^(2)-I implies 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)/(A^(50)=25 A^(2)-24I)`
Here,
`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)]`
`implies A^(50)=[(25,0,0),(25,25,0),(25,0,25)]-24 [(1,0,0),(0,1,0),(0,0,1)]`
`=[(1,0,0),(25,1,0),(25,0,1)]`
`:. |A^(50)|=1`
Also, `tr(A^(50))=1+1+1=3`. Further,
`[(1,0,0),(25,1,0),(25,0,1)][(x),(y),(z)]=[(1),(25),(25)]implies [(x),(y),(z)]= uu_(1)=[(1),(0),(0)]`
Similarly,
`uu_(2)=[(0),(1),(0)]` and `uu_(3)=[(0),(0),(1)]implies uu=[(1,0,0),(0,1,0),(0,0,1)]`, i.e., `|uu|=1`