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Question
By using elementary operations, find the inverse of the matrix A=[122−1]
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Solution
Given : A=[122−1]
We know that A=IA
⇒[122−1]=[1001]A
Applying R2→R2−2R1
⇒[122−2(1)−1−2(2)]
=[100−2(1)1−2(0)]A
⇒[120−5]=[10−21]A
Applying R2→−15R2
⇒⎡⎢⎣120(−15)−5(−15)⎤⎥⎦
=⎡⎢⎣10−2(−15)1(−15)⎤⎥⎦A
⇒[1201]=⎡⎣1025−15⎤⎦A
Applying R1→R1−2R2
⇒[1−2(0)2−2(1)01]
=⎡⎢
⎢
⎢⎣1−2(25)0−2(−15)25−15⎤⎥
⎥
⎥⎦A
⇒[1−02−201]=⎡⎢
⎢
⎢⎣1−452525−15⎤⎥
⎥
⎥⎦A
⇒[1001]=⎡⎢
⎢⎣152525−15⎤⎥
⎥⎦A
⇒I=⎡⎢
⎢⎣152525−15⎤⎥
⎥⎦A
This is similar to I=A−1A
Hence, A−1=⎡⎢
⎢⎣152525−15⎤⎥
⎥⎦
We know that A=IA
⇒[122−1]=[1001]A
Applying R2→R2−2R1
⇒[122−2(1)−1−2(2)]
=[100−2(1)1−2(0)]A
⇒[120−5]=[10−21]A
Applying R2→−15R2
⇒⎡⎢⎣120(−15)−5(−15)⎤⎥⎦
=⎡⎢⎣10−2(−15)1(−15)⎤⎥⎦A
⇒[1201]=⎡⎣1025−15⎤⎦A
Applying R1→R1−2R2
⇒[1−2(0)2−2(1)01]
=⎡⎢ ⎢ ⎢⎣1−2(25)0−2(−15)25−15⎤⎥ ⎥ ⎥⎦A
⇒[1−02−201]=⎡⎢ ⎢ ⎢⎣1−452525−15⎤⎥ ⎥ ⎥⎦A
⇒[1001]=⎡⎢ ⎢⎣152525−15⎤⎥ ⎥⎦A
⇒I=⎡⎢ ⎢⎣152525−15⎤⎥ ⎥⎦A
This is similar to I=A−1A
Hence, A−1=⎡⎢ ⎢⎣152525−15⎤⎥ ⎥⎦
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