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Question

Let A=121231115, verify that (a) [adjA]1=adj(A1)

(b) (A1)1=A


Solution

Given A=121231115,|A|=∣ ∣121231115∣ ∣
=1(151)(2)(101)+1(23)=14225=130

A11=151=14A12=(101)=11,A13=23=5A21=(101)=11,A22=51=4A23=(1+2)=3A31=(23)=5A32=(1+2)=3A33=34=1
adj(A)=141151143531T=141151143531
 A1=1|A|(adj(A))=113141151143531=⎢ ⎢ ⎢141311135131113413313513313113⎥ ⎥ ⎥

Here, adj(A)=141151143531=B (let)

 |B|=|adj A|=141151143531=14(49)11(1115)5(33+20)=182+286+65=1690
Cofactors of B are

B11=(49)=13,B12=(1115)=26,B13=(33+20)=13B21=(1115)=26,B22=1425=39,B23=(42+55)=13B31=(33+20)=13,B32=(42+55)=13,B33=56121=65

adj(B)=adj(adjA)=132613263913131365T=132613263913131365
 B1=[adj A]1=1|adj A(adj(adj A))[adjA]1=1169132613263913131365=113121231115   (i)
Cofactors of A1 are
A11=13169,A1226169,A13=13169,A21=26169,A22=39169,A23=13169A31=13169,A32=13169,A33=65169

Now, adj(A1)=⎢ ⎢ ⎢131692616913169261693916913169131691316965169⎥ ⎥ ⎥T=⎢ ⎢ ⎢131692616913169261693916913169131691316965169⎥ ⎥ ⎥
adj(A1)=113121231115    (ii)
From Eqs. (i) and (ii), we get that adj(A1)=(adjA)1

|A1|=1413(41699169)+1113(1116915169)+513(33169+20169)=1413(13169)+1113(26169)+513(13169)=14169221695169=13169=113
and adj(A1)=113121231115
 (A1)=1|A1|=(adj A1)=1113×113121231115=121231115=A
 

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