If A−1=⎡⎢⎣3−11−156−55−22⎤⎥⎦ and B=⎡⎢⎣12−2−1300−21⎤⎥⎦, find (AB)−1.
We know that, (AB)−1=B−1A−1andA−1 is known, therefore we proceed to find B−1
Here, |B|=∣∣
∣∣12−2−1300−21∣∣
∣∣=1(3−0)−2(−1−0)−2(2−0)=3+2−4=1≠0
∴B−1 exists. Cofactors of B are
B11=(3−0)=3,B12=−(−1−0)=1,B13=(2−0)=2B21=−(2−4=2),B22=1−0=1,B23=−(−2−0)=2B31=(0+6)=6B32=−(0−2)=2,B33=3+2=5
∴ adj(B)=⎡⎢⎣312212625⎤⎥⎦T=⎡⎢⎣326112225⎤⎥⎦
∴ B−1=1|B|(adj(B))=11⎡⎢⎣326112225⎤⎥⎦=⎡⎢⎣326112225⎤⎥⎦
Now, (AB)−1=B−1A−1=⎡⎢⎣326112225⎤⎥⎦=⎡⎢⎣3−11−156−55−22⎤⎥⎦
=⎡⎢⎣9−30+30−3+12−123−10+123−15+10−1+6−41−5+46−30+25−2+12−102−10+10⎤⎥⎦=⎡⎢⎣9−35−210102⎤⎥⎦