If A−1=⎡⎢⎣3−11−156−55−22⎤⎥⎦ and B=⎡⎢⎣12−2−1300−21⎤⎥⎦, find (AB)−1
We know that (AB)−1=B−1A−1 .....(1)
Given A−1=⎡⎢⎣3−11−156−55−22⎤⎥⎦
B=⎡⎢⎣12−2−1300−21⎤⎥⎦
Here, |B|=1(3−0)+1(2−4)+0
⇒|B|=3−2=1
Since, |B|≠0
Hence, B−1 exists.
B−1=adjB|A| and adjB=CT
C11=(−1)1+1∣∣∣30−21∣∣∣
⇒C11=3−0=3
C12=(−1)1+2∣∣∣−1001∣∣∣
⇒C12=−(−1−0)=1
C13=(−1)1+3∣∣∣−130−2∣∣∣
⇒C13=2−0=2
C21=(−1)2+1∣∣∣2−2−21∣∣∣
⇒C21=−(2−4)=2
C22=(−1)2+2∣∣∣1−201∣∣∣
⇒C22=1−0=1
C23=(−1)2+3∣∣∣120−2∣∣∣
⇒C23=−(−2−0)=2
C31=(−1)3+1∣∣∣2−230∣∣∣
⇒C31=0+6=6
C32=(−1)3+2∣∣∣1−2−10∣∣∣
⇒C32=−(0−2)=2
C33=(−1)3+3∣∣∣12−13∣∣∣
⇒C33=3+2=5
Hence, the co-factor matrix is C=⎡⎢⎣312212625⎤⎥⎦
⇒adjB=CT=⎡⎢⎣326112225⎤⎥⎦
⇒B−1=adjB|B|=⎡⎢⎣326112225⎤⎥⎦
Substituting the values in (1), we get
(AB)−1=⎡⎢⎣326112225⎤⎥⎦⎡⎢⎣3−11−156−55−22⎤⎥⎦
(AB)−1=⎡⎢⎣9−30+30−3+12−123−10+123−15+10−1+6+41−5+46−30+25−2+12−102−10+10⎤⎥⎦
(AB)−1=⎡⎢⎣9−35−290102⎤⎥⎦