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Question

If A=211121112, verify that A36A2+9A4I=0. Hence find A1

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Solution

A=211121112

To Prove : A36A2+9A4I=O

Consider, A2=211121112211121112

=4+1+12212+1+22211+4+11222+1+21221+1+4

A2=655565556

Now, A3=A2A
=655565556211121112

=12+5+561056+5+1010655+12+5561010+5+651065+5+12

A3=222121212221212122

Consider, A36A2+9A4I

=2221212122212121226655565556+92111211124100010001

=222121212221212122363030303630303036+189991899918400040004

=2236+18421+3092130+921+3092236+18421+3092130+921+3092236+184

=000000000
A36A2+9A4I=O .....(1)

A1 :
Multiplying eqn (1) by A1, we get
A3A16A2A1+9AA14IA1=OA1
A26A+9I4A1=O
4A1=A26A+9I
Substituting the values in above eqn
4A1=6555655566211121112+9100010001

4A1=655565556126661266612+900090009

4A1=612+95+6565+6612+95+6565+6612+9

A1=14311131113

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