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Question

Using elementary row transformations, find the inverse of the matrix A=123257245

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Solution

Let A=IA
123257245=100010001A

Operate R2R22R1 and R3R3+2R1

123011001=100210201A

Operate R2R2R3

123010001=100411201A

Operate R1R13R3

120010001=503411201A

Operate R1R12R2

100010001=321411201A

Which is of the form
I=A1A

Hence, A1=321411201

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