Question

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

Answer 1

Correct Answer - 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`