Let A=[1-21 -231 115]verify that
(i)[adjA]-1=adj(A-1)
(ii)(A-1)-1=A
(i)[adjA]-1=adj(A-1)
(ii)(A-1)-1=A
Solution and Explanation
A=[1-21 -231 115]
∴|A|=1(15-1)+2(-10-1)+1(-2-3)=14-22-5=-13
Now,A11=14,A12=11,A13=-5
A21=11,A22=4,A23=-3
A31=-5,A32=-3,A33=-1
∴adjA=[14 11 -5 11 4 -3 -5 -3 -1]
∴A-1=1/|A|(adjA)
=-1/13[14 11 -5 11 4 -3 -5 -3 -1]
=1/13[-14 -11 5 -11 -4 3 5 3 1]
(i)|adjA|=14(-4-9)-11(-11-15)-5(-33+20)
=14(-13)-11(-26)-5(-13)
=-182+286+65=169
we have,
adj(adjA)=[-13 26 -13 26 -39 -13 -13 -13 65]
∴[adjA]-1=1/|adjA|(adj(adjA))
=1/169[-13 26 -13 26 -39 -13 -13 -13 65]
=1/13[-12-1 2-3-1 -1-1-5]
Now,A-1=1/13[-14 -11 5 -11 -4 3 5 3 1]=[-14/13 -11/13 5/13 -11/13 -4/13 3/13 5/13 3/13 1/13]
∴adj(A-1)=[-4/169-9/169 -(-11/169-15/169) -33/169+20/169 -(-11/169-15/169) -14/169-25/169 -(-42/169+55/169) -33/169+20/169 -(-42/169+55/169) 56/169-121/169]
=1//169[-13 26 -13 26 -39 -13 -13 -13 65]
=1/13[-12-1 2-3-1 -1-1-5]
Hence,[adjA]-1=adj(A-1).
(ii)
we have shown that
A-1=1/13[-14 -11 5 -11 -4 3 5 3 1]
And,adjA-1=1/13[-12-1 2-3-1 -1-1-5]
Now,
|A-1|=(1/13)3[-14x(-13)+11x(-26)+5x(-13)]=(1/13)3x(-169)=-1/13
∴(A-1)-1=adjA-1/|A-1|=1/(-1/13)x1/13[-12-1 2-3-1 -1-1-5]=[1-21 -231 115]=A
∴(A-1)-1=A
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