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 A
1
2
1
2
4
7

 



=
=
=

 




 



Hence
2
7
12
8
12
1
0
A
4A
I
4
7
4
8
0
1






−
+ =
−
+












0
0
O
0
0


=
=




Now
A2 – 4A + I = O
Therefore
A A – 4A = – I
or
A  A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0)
or
A (A A–1) – 4I = – A–1
or
AI – 4I = – A–1
or
A–1  =  4I – A  = 
4
0
2
3
2
3
0
4
1
2
1
2
−






−
=






−






Hence
1
2
3
A
1
2
−
−


= 

−

EXERCISE 4