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Question
Using elementary row transformation, find the inverse of the matrix A=⎡⎢⎣123257−2−4−5⎤⎥⎦.
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Solution
⎡⎢⎣123257−2−4−5⎤⎥⎦
⎡⎢⎣123|100257|010−2−4−5|001⎤⎥⎦
R1↔R2
⎡⎢⎣257|010123|100−2−4−5|001⎤⎥⎦
R2→R2−12R1,R3→R3+(1)R1
⎡⎢
⎢⎣257|0100−12−12|1−120012|011⎤⎥
⎥⎦
R2↔R3
⎡⎢
⎢⎣257|010012|0110−12−12|1−120⎤⎥
⎥⎦
R3→R3+12R2
⎡⎢
⎢⎣257|010012|0110012|1012⎤⎥
⎥⎦
R3→2R3,R2→R2−2R3,R1→R1−7R3
⎡⎢⎣250|−141−7010|−41−1001|201⎤⎥⎦
R1→R1−5R2
⎡⎢⎣200|6−4−2010|−41−1001|201⎤⎥⎦
R1→12R1
⎡⎢⎣100|3−2−1010|−41−1001|201⎤⎥⎦
A−1=⎡⎢⎣3−2−1−41−1201⎤⎥⎦
⎡⎢⎣123257−2−4−5⎤⎥⎦
⎡⎢⎣123|100257|010−2−4−5|001⎤⎥⎦
R1↔R2
⎡⎢⎣257|010123|100−2−4−5|001⎤⎥⎦
R2→R2−12R1,R3→R3+(1)R1
⎡⎢
⎢⎣257|0100−12−12|1−120012|011⎤⎥
⎥⎦
R2↔R3
⎡⎢
⎢⎣257|010012|0110−12−12|1−120⎤⎥
⎥⎦
R3→R3+12R2
⎡⎢
⎢⎣257|010012|0110012|1012⎤⎥
⎥⎦
R3→2R3,R2→R2−2R3,R1→R1−7R3
⎡⎢⎣250|−141−7010|−41−1001|201⎤⎥⎦
R1→R1−5R2
⎡⎢⎣200|6−4−2010|−41−1001|201⎤⎥⎦
R1→12R1
⎡⎢⎣100|3−2−1010|−41−1001|201⎤⎥⎦
A−1=⎡⎢⎣3−2−1−41−1201⎤⎥⎦
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