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Question
Using elementary transformations, find the inverse of matrix [3152], if it exists.
Using elementary transformations, find the inverse of matrix [3152], if it exists.
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Solution
Let A=[3152]
We know that A=IA
⇒[3152]=[1001]A
Applying R1→R1−25R2
⇒⎡⎣3−25(5)1−25(2)52⎤⎦=⎡⎣1−25(0)0−25(1)01⎤⎦A
⇒⎡⎣3−21−4552⎤⎦=⎡⎣1−00−2501⎤⎦A
⇒⎡⎣11552⎤⎦=⎡⎣1−2501⎤⎦A
Applying R2→R2−5R1
⇒⎡⎢
⎢
⎢⎣1155−5(1)2−5(15)⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1−250−5(1)1−5(−25)⎤⎥
⎥
⎥⎦A
⇒⎡⎣1155−52−1⎤⎦=⎡⎢⎣1−250−51−(−2)⎤⎥⎦A
⇒⎡⎣11501⎤⎦=⎡⎣1−25−53⎤⎦A
Applying R1→R1−15R2
⇒⎡⎣1−15(0)15−15(1)01⎤⎦=⎡⎣1−15(−5)−25−15(3)−53⎤⎦A
⇒⎡⎣1−015−1501⎤⎦=⎡⎣1−(−1)−25−35−53⎤⎦A
⇒[1001]=⎡⎣1+1−55−53⎤⎦A
⇒[1001]=[2−1−53]A
⇒I=[2−1−53]A
This is similar to I=A−1A
Thus, A−1=[2−1−53]
We know that A=IA
⇒[3152]=[1001]A
Applying R1→R1−25R2
⇒⎡⎣3−25(5)1−25(2)52⎤⎦=⎡⎣1−25(0)0−25(1)01⎤⎦A
⇒⎡⎣3−21−4552⎤⎦=⎡⎣1−00−2501⎤⎦A
⇒⎡⎣11552⎤⎦=⎡⎣1−2501⎤⎦A
Applying R2→R2−5R1
⇒⎡⎢ ⎢ ⎢⎣1155−5(1)2−5(15)⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣1−250−5(1)1−5(−25)⎤⎥ ⎥ ⎥⎦A
⇒⎡⎣1155−52−1⎤⎦=⎡⎢⎣1−250−51−(−2)⎤⎥⎦A
⇒⎡⎣11501⎤⎦=⎡⎣1−25−53⎤⎦A
Applying R1→R1−15R2
⇒⎡⎣1−15(0)15−15(1)01⎤⎦=⎡⎣1−15(−5)−25−15(3)−53⎤⎦A
⇒⎡⎣1−015−1501⎤⎦=⎡⎣1−(−1)−25−35−53⎤⎦A
⇒[1001]=⎡⎣1+1−55−53⎤⎦A
⇒[1001]=[2−1−53]A
⇒I=[2−1−53]A
This is similar to I=A−1A
Thus, A−1=[2−1−53]
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