Let A=⎡⎢⎣1−21−231115⎤⎥⎦, verify that (a) [adjA]−1=adj(A−1)
(b) (A−1)−1=A
Let A=⎡⎢⎣1−21−231115⎤⎥⎦, verify that (a) [adjA]−1=adj(A−1)
(b) (A−1)−1=A
Given A=⎡⎢⎣1−21−231115⎤⎥⎦,⇒|A|=∣∣
∣∣1−21−231115∣∣
∣∣
=1(15−1)−(−2)(−10−1)+1(−2−3)=14−22−5=−13≠0
A11=15−1=14A12=−(−10−1)=11,A13=−2−3=−5A21=−(−10−1)=11,A22=5−1=4A23=−(1+2)=−3A31=(−2−3)=−5A32=−(1+2)=−3A33=3−4=−1
adj(A)=⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦T=⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦
∴ A−1=1|A|(adj(A))=1−13⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦=⎡⎢
⎢
⎢⎣−1413−1113513−1113−413313513313113⎤⎥
⎥
⎥⎦
Here, adj(A)=⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦=B (let)
∴ |B|=|adj A|=⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦=14(−4−9)−11(−11−15)−5(−33+20)=−182+286+65=169≠0
Cofactors of B are
B11=(−4−9)=−13,B12=−(−11−15)=26,B13=(−33+20)=−13B21=−(−11−15)=26,B22=−14−25=−39,B23=−(−42+55)=−13B31=(−33+20)=−13,B32=−(−42+55)=−13,B33=56−121=−65
adj(B)=adj(adjA)=⎡⎢⎣−1326−1326−39−13−13−13−65⎤⎥⎦T=⎡⎢⎣−1326−1326−39−13−13−13−65⎤⎥⎦
∴ B−1=[adj A]−1=1|adj A(adj(adj A))⇒[adjA]−1=1169⎡⎢⎣−1326−1326−39−13−13−13−65⎤⎥⎦=113⎡⎢⎣−12−12−3−1−1−1−5⎤⎥⎦ …(i)
Cofactors of A−1 are
A11=−13169,A1226169,A13=−13169,A21=26169,A22=−39169,A23=−13169A31=−13169,A32=−13169,A33=−65169
Now, adj(A−1)=⎡⎢
⎢
⎢⎣−1316926169−1316926169−39169−13169−13169−13169−65169⎤⎥
⎥
⎥⎦T=⎡⎢
⎢
⎢⎣−1316926169−1316926169−39169−13169−13169−13169−65169⎤⎥
⎥
⎥⎦
⇒adj(A−1)=113⎡⎢⎣−12−12−3−1−1−1−5⎤⎥⎦ …(ii)
From Eqs. (i) and (ii), we get that adj(A−1)=(adjA)−1
|A−1|=−1413(−4169−9169)+1113(−11169−15169)+513(−33169+20169)=−1413(−13169)+1113(−26169)+513(−13169)=14169−22169−5169=−13169=−113
and adj(A−1)=113⎡⎢⎣−12−12−3−1−1−1−5⎤⎥⎦
∴ (A−1)=1|A−1|=(adj A−1)=1−113×113⎡⎢⎣−12−12−3−1−1−1−5⎤⎥⎦=⎡⎢⎣1−21−231115⎤⎥⎦=A
Given A=⎡⎢⎣1−21−231115⎤⎥⎦,⇒|A|=∣∣
∣∣1−21−231115∣∣
∣∣
=1(15−1)−(−2)(−10−1)+1(−2−3)=14−22−5=−13≠0
A11=15−1=14A12=−(−10−1)=11,A13=−2−3=−5A21=−(−10−1)=11,A22=5−1=4A23=−(1+2)=−3A31=(−2−3)=−5A32=−(1+2)=−3A33=3−4=−1
adj(A)=⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦T=⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦
∴ A−1=1|A|(adj(A))=1−13⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦=⎡⎢
⎢
⎢⎣−1413−1113513−1113−413313513313113⎤⎥
⎥
⎥⎦
Here, adj(A)=⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦=B (let)
∴ |B|=|adj A|=⎡⎢⎣1411−5114−3−5−3−1⎤⎥⎦=14(−4−9)−11(−11−15)−5(−33+20)=−182+286+65=169≠0
Cofactors of B are
B11=(−4−9)=−13,B12=−(−11−15)=26,B13=(−33+20)=−13B21=−(−11−15)=26,B22=−14−25=−39,B23=−(−42+55)=−13B31=(−33+20)=−13,B32=−(−42+55)=−13,B33=56−121=−65
adj(B)=adj(adjA)=⎡⎢⎣−1326−1326−39−13−13−13−65⎤⎥⎦T=⎡⎢⎣−1326−1326−39−13−13−13−65⎤⎥⎦
∴ B−1=[adj A]−1=1|adj A(adj(adj A))⇒[adjA]−1=1169⎡⎢⎣−1326−1326−39−13−13−13−65⎤⎥⎦=113⎡⎢⎣−12−12−3−1−1−1−5⎤⎥⎦ …(i)
Cofactors of A−1 are
A11=−13169,A1226169,A13=−13169,A21=26169,A22=−39169,A23=−13169A31=−13169,A32=−13169,A33=−65169
Now, adj(A−1)=⎡⎢
⎢
⎢⎣−1316926169−1316926169−39169−13169−13169−13169−65169⎤⎥
⎥
⎥⎦T=⎡⎢
⎢
⎢⎣−1316926169−1316926169−39169−13169−13169−13169−65169⎤⎥
⎥
⎥⎦
⇒adj(A−1)=113⎡⎢⎣−12−12−3−1−1−1−5⎤⎥⎦ …(ii)
From Eqs. (i) and (ii), we get that adj(A−1)=(adjA)−1
|A−1|=−1413(−4169−9169)+1113(−11169−15169)+513(−33169+20169)=−1413(−13169)+1113(−26169)+513(−13169)=14169−22169−5169=−13169=−113
and adj(A−1)=113⎡⎢⎣−12−12−3−1−1−1−5⎤⎥⎦
∴ (A−1)=1|A−1|=(adj A−1)=1−113×113⎡⎢⎣−12−12−3−1−1−1−5⎤⎥⎦=⎡⎢⎣1−21−231115⎤⎥⎦=A
verify
that
