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Question

The inverse of the matrix ⎡⎢⎣20−1510013⎤⎥⎦ is

A
3111565522
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B
3111565522
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C
3111565522
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D
None of these
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Solution

The correct option is D None of these
Let A=201510013
We known than A=IA. Therefore,
201510013=100010001A
Applying R112R1, we have
⎢ ⎢ ⎢1012510013⎥ ⎥ ⎥=⎢ ⎢ ⎢1200010001⎥ ⎥ ⎥A
Applying R2R25R1, we have
⎢ ⎢ ⎢ ⎢ ⎢10120152013⎥ ⎥ ⎥ ⎥ ⎥=⎢ ⎢ ⎢ ⎢ ⎢12005210001⎥ ⎥ ⎥ ⎥ ⎥
Applying R3R3R2, we have
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢101201520012⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢120052105211⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ A
Applying R32R3, we have
⎢ ⎢ ⎢ ⎢ ⎢10120152001⎥ ⎥ ⎥ ⎥ ⎥=⎢ ⎢ ⎢ ⎢ ⎢12005210522⎥ ⎥ ⎥ ⎥ ⎥ A
Applying R1R1+12R3 and R2R252R3, we have
100010001=3111565522A
A1=3111565522

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