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(A^(50))` equals

(a)`-1`

(b)0

(d)25

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The correct Answer is:

`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)
`abs(A^(50)) = abs(A)^(50) = abs((1,0,0),(1,0,1),(0,1,0))^(50) = (-1)^(50) = 1`

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