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Question

For the matrix A=111123213. Show that A36A2+5A+11I=0. Hence, find A1

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Solution

Given, A=111123213
A2=AA=111123213111123213
=1+1+21+2113+31+261+4+316921+62232+3+9=42138147314
and A3=A2A=42138147314111123213

=4+2+24+4146+33+8283+16+143244273+2876147+9+42=871232769321358

A36A2+5A+11I

871232769321358642138147314+5111123213+11100010001

=87123276932135824126184884421884+5555101510515+110001100011

=824+5+11712+5+016+5+023+18+5+02748+10+1169+8415+03242+10+013+185+05884+15+11=000000000
Now, A36A2+5A+11I=0

(AAA)A16(AA)A1+5AA1+11IA1=0
(Pre-multiplying by A1 as |A|0)
|A|=∣ ∣111123213∣ ∣=1(63)1(3+6)+1(14)=395=110
AA(AA1)6A(AA1)+5(AA1)+11(IA1)=0
AAI6AI+5I+11A1=0
(using AA1=I and IA1=A1)
A26A+5I=11A1 (Using AAI=A2 and AI=A)
A1=111(A26A+5I)A1=111(A2+6A5I)

=11142138147314+61111232135100010001

=11142138147314+6666121812618500050005

=1114+652+601+603+608+125141807+12036+014+185=111345914531


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