1
Question
Find the inverse of the matrix, if it exists.
⎡⎢⎣250011−103⎤⎥⎦
⎡⎢⎣250011−103⎤⎥⎦
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Solution
Let A=⎡⎢⎣250011−103⎤⎥⎦
We know that A=IA
∴⎡⎢⎣250011−103⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R1→12R1, we have:
⎡⎢
⎢⎣150011−1203⎤⎥
⎥⎦=⎡⎢
⎢⎣1200010001⎤⎥
⎥⎦A
Applying R2→R2−5R1, we have:
⎡⎢
⎢⎣100011−12523⎤⎥
⎥⎦=⎡⎢
⎢⎣12−520010001⎤⎥
⎥⎦A
Applying R3→R3−R2, we have:
⎡⎢
⎢
⎢⎣100010−125212⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣12−525201−1001⎤⎥
⎥
⎥⎦A
Applying R3→2R3, we have:
⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−155−16−21−52⎤⎥⎦A
∴A−1=⎡⎢⎣3−155−16−21−52⎤⎥⎦
We know that A=IA
∴⎡⎢⎣250011−103⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R1→12R1, we have:
⎡⎢ ⎢⎣150011−1203⎤⎥ ⎥⎦=⎡⎢ ⎢⎣1200010001⎤⎥ ⎥⎦A
Applying R2→R2−5R1, we have:
⎡⎢ ⎢⎣100011−12523⎤⎥ ⎥⎦=⎡⎢ ⎢⎣12−520010001⎤⎥ ⎥⎦A
Applying R3→R3−R2, we have:
⎡⎢ ⎢ ⎢⎣100010−125212⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣12−525201−1001⎤⎥ ⎥ ⎥⎦A
Applying R3→2R3, we have:
⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−155−16−21−52⎤⎥⎦A
∴A−1=⎡⎢⎣3−155−16−21−52⎤⎥⎦
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