If A=0 1 21 2 33 x 1 and A−1=1/2−1/21/2−43y5/2−3/21/2, then
x=1, y=-1
x=-1 , y=1
x=2 , y=-1/2
x=1/2 , y=1/2
We have 1 0 00 1 00 0 1=AA−1
=0121233x11/2−1/21/2−43y5/2−3/21/2=10y+1012(y+1)41−x3x−12+xy
⇒1−x=0,x−1=0,y+1=0,2+xy=1⇒x=1,y=−1.