A=⎡⎢⎣20−1510013⎤⎥⎦
We know that A=IA
∴⎡⎢⎣20−1510013⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R1→12R1, we have
⇒⎡⎢⎣10−1/2510013⎤⎥⎦=⎡⎢⎣1/200010001⎤⎥⎦A
Applying R2→R2−5R1, we have
⇒⎡⎢⎣10−1/2015/2013⎤⎥⎦=⎡⎢⎣1/200−5/210001⎤⎥⎦A
Applying R3→R3−R2, we have
⇒⎡⎢⎣10−1/2015/2001/2⎤⎥⎦=⎡⎢⎣1/200−5/2105/2−11⎤⎥⎦A
Applying R3→2R3, we have
⇒⎡⎢⎣10−1/2015/2001⎤⎥⎦=⎡⎢⎣1/200−5/2105−22⎤⎥⎦A
Applying R1→R1+12R3, and R2→R−2−52R2, we have
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−11−156−55−22⎤⎥⎦A
⇒I=A−1A
∴A−1=⎡⎢⎣3−11−156−55−22⎤⎥⎦
Hence, the answer is ⎡⎢⎣3−11−156−55−22⎤⎥⎦.