Let A= [

2	5
1	3

]
We can write A = IA
i.e. [

2	5
1	3

] = [

1	0
0	1

] A
Applying R1 →

1

2

R1 , we get
[

1	5 2
1	3

] = [

1 2	0
0	1

] A
Applying R2 → R2−R1 , gives
[

1	5 2
0	1 2

] = [

1 2	0
− 1 2	1

] A
Applying , R1 → R1−5R2 we obtain
[

1	0
0	1 2

] = [

3	−5
− 1 2	1

] A
Applying R2 → 2R2 , gives
[

1	0
0	1

] = [able‌3,−5;−1,2] A
Therefore, A−1 = [

3	−5
−1	2

]