Question 1021823
{{{abs(adjA) = abs(A)^(n-1)}}} by a property of the adjoint operator.

==> {{{abs(adjA) = abs(A)^(n-1) = abs(A)^2 = (-3)^2 = 9}}}, where n = 3.

Proof.

For an invertible nxn matrix, {{{A*(adjA) = abs(A)I[n])}}}

==> {{{abs(A)abs(adjA) = (abs(A))^n}}}, 
the last equation arising from the determinant of a scalar matrix and the identity matrix.

==> {{{abs(adjA) = (abs(A))^(n-1)}}}, since {{{abs(A) <>0}}}