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 Solution We have AB = 
2
3
1
2
1
5
1
4
1
3
5
14
−
−

 



=

 



−
−
−

 



Since,
AB  = –11 ≠ 0, (AB)–1 exists and is given by
(AB)–1 = 
14
5
1
1
(AB)
5
1
AB
11
adj
−
−


=−


−
−


 
14
5
1
5
1
11


=




Further, A  = –11 ≠ 0 and B  = 1 ≠ 0