If `A=[{:(0,1,1),(1,0,1),(1,1,0):}]` show that `A^(-1)=(1)/(2)(A^(2)=3I)`

Explanation

We have,
`A[(0,1,1),(1,0,1),(1,1,0)]`
Cofactors are
`A_(11)=-1, A_(12)=1, A_(13)=1`,
`A_(21)=1, A_(22)=-1, A_(23)=1`,
`A_(31)=1, A_(31)=1, A_(32)=1 A_(33)=-1`
`:." adj A"=[(-1,1,1),(1,-1,1),(1,1,-1)]^(T)=[(-1,1,1),(1,-1,1),(1,1,-1)]`
`|A|=0-(-1)+1.1=2`
`:. A^(-1)=("adj A")/(|A|)=1/2 [(-1,1,1),(1,-1,1),(1,1,-1)]`
Now `A^(2)=[(0,1,1),(1,0,1),(1,1,0)][(0,1,1),(1,0,1),(1,1,0)]=[(2,1,1),(1,2,1),(1,1,2)]`
`:. (A^(2)-3I)/2=1/2 {[(2,1,1),(1,2,1),(1,1,2)]-[(3,0,0),(0,3,0),(0,0,3)]}`
`=1/2 |(-1,1,1),(1,-1,1),(1,1,-1)|=A^(-1)`
Hence proved.