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Question

Find the inverse of the matrix, if it exists.
250011103

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Solution

Let A=250011103

We know that A=IA

250011103=100010001A

Applying R112R1, we have:

⎢ ⎢1500111203⎥ ⎥=⎢ ⎢1200010001⎥ ⎥A

Applying R2R25R1, we have:

⎢ ⎢10001112523⎥ ⎥=⎢ ⎢12520010001⎥ ⎥A

Applying R3R3R2, we have:

⎢ ⎢ ⎢100010125212⎥ ⎥ ⎥=⎢ ⎢ ⎢125252011001⎥ ⎥ ⎥A

Applying R32R3, we have:

100010001=3155162152A

A1=3155162152

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