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Question

If A=111123213 show that A36A2+5A+11I=0. Hence find A1

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Solution

A=111123213
A2=42138147314
A3=871232769321358
Now,
A36A2+5A+11I
=871232769321358+(6)42138147314+5111123213+11100010001
=871232769321358+24126184884421884+5555101510515+110001100011
=871232769321358+871232769321358
=000000000=0
Now,
A36A2+5A+11I=0
Operating A1 on both sides
A26A+5I+11A1=0 (AA1=I)
11A1=A2+6A5I
11A1=42138147314+61111232135100010001
11A1=345914531
A1=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢311411511911111411511311111⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

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