1
Question
Using elementary transformations find the inverse of
⎡⎢⎣3212432−12⎤⎥⎦
⎡⎢⎣3212432−12⎤⎥⎦
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Solution
Let A=⎡⎢⎣3212432−12⎤⎥⎦I=⎡⎢⎣100010001⎤⎥⎦IA=AA⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3212432−12⎤⎥⎦Applying R1→R13A⎡⎢
⎢⎣1300010001⎤⎥
⎥⎦=⎡⎢
⎢⎣123132432−12⎤⎥
⎥⎦Applying R2→R2−2R1 and R3→R3−2R1A⎡⎢
⎢
⎢⎣1300−2310−2301⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣12313083730−7343⎤⎥
⎥
⎥⎦Applying R2→38R2A⎡⎢
⎢
⎢⎣1300−14380−2301⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1231301780−7343⎤⎥
⎥
⎥⎦Applying R1→R1−23R2 and R3→R3+73R2A⎡⎢
⎢
⎢⎣12−140−14380−54781⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣10−14017800278⎤⎥
⎥
⎥⎦
Applying R3→827R3A⎡⎢
⎢
⎢⎣12−140−14380−1027727827⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣10−140178001⎤⎥
⎥
⎥⎦
Applying R1→R1+14R3 and R2→R2−78R3A⎡⎢
⎢
⎢⎣1127−527227227427−727−1027727827⎤⎥
⎥
⎥⎦=⎡⎢⎣100010001⎤⎥⎦
Therefore,A−1=⎡⎢
⎢
⎢⎣1127−527227227427−727−1027727827⎤⎥
⎥
⎥⎦
Let A=⎡⎢⎣3212432−12⎤⎥⎦
I=⎡⎢⎣100010001⎤⎥⎦
IA=A
A⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3212432−12⎤⎥⎦
Applying R1→R13
A⎡⎢
⎢⎣1300010001⎤⎥
⎥⎦=⎡⎢
⎢⎣123132432−12⎤⎥
⎥⎦
Applying R2→R2−2R1 and R3→R3−2R1
A⎡⎢
⎢
⎢⎣1300−2310−2301⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣12313083730−7343⎤⎥
⎥
⎥⎦
Applying R2→38R2
A⎡⎢
⎢
⎢⎣1300−14380−2301⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1231301780−7343⎤⎥
⎥
⎥⎦
Applying R1→R1−23R2 and R3→R3+73R2
A⎡⎢
⎢
⎢⎣12−140−14380−54781⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣10−14017800278⎤⎥
⎥
⎥⎦
Applying R3→827R3
A⎡⎢
⎢
⎢⎣12−140−14380−1027727827⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣10−140178001⎤⎥
⎥
⎥⎦
Applying R1→R1+14R3 and R2→R2−78R3
A⎡⎢
⎢
⎢⎣1127−527227227427−727−1027727827⎤⎥
⎥
⎥⎦=⎡⎢⎣100010001⎤⎥⎦
Therefore,
A−1=⎡⎢
⎢
⎢⎣1127−527227227427−727−1027727827⎤⎥
⎥
⎥⎦
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