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Question

If A=201510013, then find A1 using elementary row operations.

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Solution

A=201510013
We know that A=IA
201510013=100010001A
Applying R112R1, we have
101/2510013=1/200010001A
Applying R2R25R1, we have
101/2015/2013=1/2005/210001A
Applying R3R3R2, we have
101/2015/2001/2=1/2005/2105/211A
Applying R32R3, we have
101/2015/2001=1/2005/210522A
Applying R1R1+12R3, and R2R252R2, we have
100010001=3111565522A
I=A1A
A1=3111565522
Hence, the answer is 3111565522.


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