Let A=⎡⎢⎣−112123311⎤⎥⎦
For applying elementary row operation we write,
A = I. A
⎡⎢⎣−112123311⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R1↔R2, we get
⎡⎢⎣123−112311⎤⎥⎦=⎡⎢⎣010100001⎤⎥⎦A
Applying R2→R2+R1 and R3→R3−3R1, we get
⎡⎢⎣1230350−5−8⎤⎥⎦=⎡⎢⎣0101100−31⎤⎥⎦A
Applying R1→R1−23R2, we get
⎡⎢
⎢⎣10−130350−5−8⎤⎥
⎥⎦=⎡⎢
⎢⎣−231301100−31⎤⎥
⎥⎦A
Applying R2→13R2, we get
⎡⎢
⎢
⎢⎣10−1301530−5−8⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣−23130131300−31⎤⎥
⎥
⎥⎦A
Applying R3→R3+5R2, we get
⎡⎢
⎢
⎢⎣10−1301530013⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢⎣−231301313053−431⎤⎥
⎥
⎥
⎥⎦A
Applying R1→R1+R3 and R2→−5R3
⎡⎢
⎢⎣1000100013⎤⎥
⎥⎦=⎡⎢
⎢⎣111−87−553−431⎤⎥
⎥⎦A
Applying R3→3R3, we get
⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣1−11−87−55−43⎤⎥⎦A
Hence A−1=⎡⎢⎣1−11−87−55−43⎤⎥⎦