13
Question
Using elementary row transformations, find the inverse of the matrix A=⎡⎢⎣123257−2−4−5⎤⎥⎦
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Solution
Let A=IA
⎡⎢⎣123257−2−4−5⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Operate R2→R2−2R1 and R3→R3+2R1
⇒⎡⎢⎣123011001⎤⎥⎦=⎡⎢⎣100−210201⎤⎥⎦A
Operate R2→R2−R3
⎡⎢⎣123010001⎤⎥⎦=⎡⎢⎣100−41−1201⎤⎥⎦A
Operate R1→R1−3R3
⎡⎢⎣120010001⎤⎥⎦=⎡⎢⎣−50−3−41−1201⎤⎥⎦A
Operate R1→R1−2R2
⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−2−1−41−1201⎤⎥⎦A
Which is of the form
I=A−1A
Hence, A−1=⎡⎢⎣3−2−1−41−1201⎤⎥⎦
⎡⎢⎣123257−2−4−5⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Operate R2→R2−2R1 and R3→R3+2R1
⇒⎡⎢⎣123011001⎤⎥⎦=⎡⎢⎣100−210201⎤⎥⎦A
Operate R2→R2−R3
⎡⎢⎣123010001⎤⎥⎦=⎡⎢⎣100−41−1201⎤⎥⎦A
Operate R1→R1−3R3
⎡⎢⎣120010001⎤⎥⎦=⎡⎢⎣−50−3−41−1201⎤⎥⎦A
Operate R1→R1−2R2
⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−2−1−41−1201⎤⎥⎦A
Which is of the form
I=A−1A
Hence, A−1=⎡⎢⎣3−2−1−41−1201⎤⎥⎦
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