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 Also find A–1 Solution We have A  = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0
Now A11 = 7,  A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0,
A33 = 1
Therefore
adj A =
7
3
3
1
1
0
1
0
1
−
−




−



−

Rationalised 2023-24
DETERMINANTS     91
Now
A (adj A) =
1
3
3
7
3
3
1
4
3
1
1
0
1
3
4
1
0
1
−
−

 


 

−

 


 

−

 

=
7
3
3
3
3
0
3
0
3
7
4
3
3
4
0
3
0
3
7
3
4
3
3
0
3
0
4
−
−
− +
+
− +
+




−
−
− +
+
− +
+




−
−
− +
+
− +
+


=
1
0
0
0
1
0
0
0
1










 = (1)   
1
0
0
0
1
0
0
0
1










 = A I
Also
A–1
1
A
A
a d j
=
 =
7
3
3
1
1
1
0
1
1
0
1
−
−




−



−

 = 
7
3
3
1
1
0
1
0
1
−
−




−



−

Example 14 If A = 
2
3
1
2
and B
1
4
1
−3




=




−
−




, then verify that (AB)–1 = B–1A–1