1
Question
Using elementary transformations, Find the inverse of matrix [1−123] if it exists.
Using elementary transformations, Find the inverse of matrix [1−123] if it exists.
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Solution
Let A=[1−123]
We know that A=IA
⇒[1−123]=[1001]A
Applying R2→R2−2R1
⇒[1−12−2(1)3−2(−1)]=[100−2(1)1−2(0)]A
⇒[1−105]=[10−21]A
Applying R2→15R2
⇒⎡⎣1−10555⎤⎦=⎡⎣10−2515⎤⎦A
⇒[1−101]=⎡⎣10−2515⎤⎦A
Applying R1→R1+R2
⇒[1+0−1+101]=⎡⎢
⎢⎣1−250+15−2515⎤⎥
⎥⎦A
⇒[1001]=⎡⎢
⎢⎣3515−2515⎤⎥
⎥⎦A
⇒I=⎡⎢
⎢⎣3515−2515⎤⎥
⎥⎦A
This is similar to I=A−1A
Thus, A−1⎡⎢
⎢⎣3515−2515⎤⎥
⎥⎦
We know that A=IA
⇒[1−123]=[1001]A
Applying R2→R2−2R1
⇒[1−12−2(1)3−2(−1)]=[100−2(1)1−2(0)]A
⇒[1−105]=[10−21]A
Applying R2→15R2
⇒⎡⎣1−10555⎤⎦=⎡⎣10−2515⎤⎦A
⇒[1−101]=⎡⎣10−2515⎤⎦A
Applying R1→R1+R2
⇒[1+0−1+101]=⎡⎢ ⎢⎣1−250+15−2515⎤⎥ ⎥⎦A
⇒[1001]=⎡⎢ ⎢⎣3515−2515⎤⎥ ⎥⎦A
⇒I=⎡⎢ ⎢⎣3515−2515⎤⎥ ⎥⎦A
This is similar to I=A−1A
Thus, A−1⎡⎢ ⎢⎣3515−2515⎤⎥ ⎥⎦
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