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 Therefore, A–1 and B–1 both exist and are given by
A–1 = −
−
−
−




= 



−
1
11
4
3
1
2
3
2
1
1
1
,B
Therefore
B A
−
− = −




−
−
−




1
1
1
11
3
2
1
1
4
3
1
2
 = −
−
−
−
−




1
11
14
5
5
1  
14
5
1
5
1
11


=




Hence (AB)–1 = B–1 A–1
Rationalised 2023-24
 92
MATHEMATICS
Example 15 Show that the matrix A = 
2
3
1
2



  satisfies the equation A2 – 4A + I = O,
where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix Using this equation, find A–1 Solution We have 
2
2
3
2
3
7
12
A
A