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Let A = [[1,0,0],[1,0,1], [0,1,0]] " sat...

Let `A = [[1,0,0],[1,0,1], [0,1,0]] " satisfies " A^(n) = A^(n-2) + A^(2 ) -I` for `nge 3` and
consider matrix `underset(3xx3)(U)` with its columns as `U_(1), U_(2), U_(3),` such that
` A^(50)U_(1)=[[1],[25],[25]],A^(50) U_(2)=[[0],[1],[0]]and A^(50) U_(3)[[0],[0],[1]]`
The value of `abs(U)` equals

A

-1

B

0

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
C
`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)
Let `U_(1)=[[x],[y],[x]]`
Given, `A^(50) U_(1) = [[1],[25],[25]]rArr [[1,0,0],[25,1,0],[25,0,1]][[x],[y],[z]]= [[1],[25],[25]]`
[from Eq. (ii) ]
`rArr [[x],[25x+y],[25x+z]]=[[1],[25],[25]]`, we get x = 1, y = 20 and z = 0
`therefore U_(1) = [[1],[0],[0]], "similarly " U_(2) = [[0],[1],[0]]`
and `U_(3) = [[0],[0],[1]]rArrU = [[1,0,0],[0,1,0],[0,0,1]]=I`
`therefore abs(U)=1`
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