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If A(0) = [[2 ,-2,-4],[-1,3,4],[1,-2,-3]...

If `A_(0) = [[2 ,-2,-4],[-1,3,4],[1,-2,-3]] and B_(0) [[-4,-4,-4],[1,0,1],[4,4,3]]` and
`B_(n) = adj (B_(n-1)),n inN` and `I` is an dientity matrix of order 3.
det `(A_(0) + A_(0)^(2) B_(0) ^(2) + A_(0)^(2) + A_(0) ^(4) B_(0)^(4) + ... ` upto 12 terms ) is
equal to

A

1200

B

-960

C

0

D

-9600

Text Solution

Verified by Experts

The correct Answer is:
C
`because A_(0) = [[2,-2,-4],[-1,3,4],[1,-2,-3]] rArr abs(A_(0)) = 0`
and abj `B_(0) = [[-4,1,4],[-3,0,4],[-3,1,-3]]^(T) = [[-4,-3,-3],[1,0,1],[4,4,3]]=B_(0) `
`because B_(n) = adj (B_(n-1)), n in N `
`therefore B_(1) = adj (B_(0) )=B_(0)`
`rArr B_(2) = adj (B_(1)) = adj (B_(0)) = B_(0),`
`B_(3) = B_(0) , B_(4) = B_(0) , ...`
`therefore B_(n) = B_(0) AA n in N`
det `(A_(0) + A_(0)^(2) B_(0)^(2) + A_(0)^(3) + A_(0)^(4) B_(0)^(4) + ...` upto 12 terms ) = det
`{A_(0) (I + A_(0) B_(0) ^(2) +A_(0)^(2) + A_(0) ^(3) B_(0)^(4)+...` upto 12 terms ) }
`=abs(A_(0)) (I + A_(0) B_(0) ^(2) +A_(0)^(2) + A_(0) ^(3) B_(0)^(4)+...` upto 12 terms)
= 0 `[because abs(A_(0)) = 0]`
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