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Question
Using elementary row transformations, find the inverse of the matrix A=⎡⎢⎣123257−2−4−5⎤⎥⎦.
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Solution
Let A=⎡⎢⎣123257−2−4−5⎤⎥⎦
We Know that A=IA. Therefore,
⎡⎢⎣123257−2−4−5⎤⎥⎦ =⎡⎢⎣100010001⎤⎥⎦A
Applying R3→R3+R2,we have
⎡⎢⎣123257012⎤⎥⎦ =⎡⎢⎣100010011⎤⎥⎦A
Applying R2→R2−2R1
⎡⎢⎣123011012⎤⎥⎦ =⎡⎢⎣100−110011⎤⎥⎦A
Applying R1→R1−(R3+R2)
⎡⎢⎣100011012⎤⎥⎦ =⎡⎢⎣2−2−1−110011⎤⎥⎦A
Applying C3→C3−C2
⎡⎢⎣100010011⎤⎥⎦ =⎡⎢⎣2−21−11−1010⎤⎥⎦A
Applying C2→C2−C3
⎡⎢⎣100010001⎤⎥⎦ =⎡⎢⎣2−31−12−1010⎤⎥⎦A
Therefore A−1=⎡⎢⎣2−31−12−1010⎤⎥⎦
Let A=⎡⎢⎣123257−2−4−5⎤⎥⎦
We Know that A=IA. Therefore,
⎡⎢⎣123257−2−4−5⎤⎥⎦ =⎡⎢⎣100010001⎤⎥⎦A
Applying R3→R3+R2,we have
⎡⎢⎣123257012⎤⎥⎦ =⎡⎢⎣100010011⎤⎥⎦A
Applying R2→R2−2R1
⎡⎢⎣123011012⎤⎥⎦ =⎡⎢⎣100−110011⎤⎥⎦A
⎡⎢⎣100011012⎤⎥⎦ =⎡⎢⎣2−2−1−110011⎤⎥⎦A
Applying C3→C3−C2
⎡⎢⎣100010011⎤⎥⎦ =⎡⎢⎣2−21−11−1010⎤⎥⎦A
Applying C2→C2−C3
⎡⎢⎣100010001⎤⎥⎦ =⎡⎢⎣2−31−12−1010⎤⎥⎦A
Therefore A−1=⎡⎢⎣2−31−12−1010⎤⎥⎦
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