linear algebra help!!Adjoint?

Question

Assume that it is known that the adjoint of a matrix A is

adj A=
–6 –4 -8
-4 -4 -6
6 3 8

Then A itself is either ___ ___ ___
___ ___ ___
___ ___ ___
or
___ ____ ___
___ ____ ___
___ ____ ___

Answer 1

Brashion's answer is wrong.
Here's the theory you need:
Let A be an 3x3 matrix. A special case of a
theorem in most linear algebra books states that
A * adj A = |A|* I_3, (*)
where
|A| is the determinant of A and I_3 is the 3*3 identity matrix.
The right side of this last equation can be written
|A| 0 0
0 |A| 0
0 0 |A|
Now, taking the determinant of both sides
yields
|A|* |adj A| =|A|³
so
|adj A| = |A|².
This, and (*) yield the key equation for this problem
A = √|adj A| * (adj A)^-1.
Taking your matrix, adj A =
-6 -4 -8
-4 -4 -6
6 3 8
We find, with the aid of an online matrix calculator that
|adj A| = 4
(adj)^-1 = -7/2 2 -2
.................. -1 0 1
...................3 -3/2 2
So there are 2 possibilities for A:
-7 4 -4
-2 0 -2
6 -3 4
or
7 -4 4
2 0 2
-6 3 -4

Answer 2

The adjoint is usually defined as the conjugate transpose. Since the entries here are all real numbers, that means we just need to take the transpose:
-6 -4 6
-4 -4 3
-8 -6 8

The tranpose is well-defined, so I don't understand why there are two blanks.

*****************
Steiner's work certainly seems to fit the way the problem was asked, but I have two concerns.

1) Working out the multiplication gives
[Steiner answers] * adj(A) = 2*I(3)
but det(A) = 4, so the matrices don't quite fit the relation
A * adj(A) = det(A)*I(3).

2) I've just never seen the term "adjoint" used in this sense. The definitions I can find in textbooks, MathWorld, etc., is the one I gave initially. What seems to be intended is an interesting problem solvable by the methods below, but the terminology seems off.