So AB  = I     or   A  B  = 1    (since I
1, AB
A B )
=
=
This gives
A ≠ 0 Hence A is nonsingular Conversely, let A be nonsingular Then A  ≠ 0
Now
A (adj A) = (adj A) A = A I
(Theorem 1)
or
A 
1
1
A
A A
I
| A |
| A |
adj
adj




=
=








or
AB = BA = I, where B = 
1
A
| A | adj
Thus
A is invertible and A–1 = 
1
A
| A | adj
Example 13 If A = 
1
3
3
1
4
3
1
3
4








, then verify that A adj A = |A| I