If A is a square matrix of order 3 such ...

If A is a square matrix of order 3 such that `abs(A)=2,` then
`abs((adjA^(-1))^(-1))` is

A

1

B

2

C

4

D

8

Verified by Experts

The correct Answer is:

C

`because abs(adj A^(-1)) = abs(A^(-1))^(2) = 1/(abs(A))^(2)`
`therefore abs((adj A^(-1) )^(-1) ) = 1/(abs(adjA^(-1)))=abs(A)^(2) = 2^(2) = 4`

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