Let `A=[{:(1,-2,1),(-2,3,1),(1,1,5):}]`. Verify that ltbtgt (i) `[adjA]^(-1)=adj (A^(-1))`
(ii) `A^((-1)^(-1))=A`

Explanation

`[[1,-2,1],[-2,3,-1],[1,1,5]]`
`|A|=1xx14+2xx-11+1xx-5=-13ne0`
Co-factor of A=`[[14,11,-5],[11,4,-3],[-5,-3,-1]]`
Adj.A=`[[14,11,-5],[11,4,-3],[-5,-3,-1]]`
`A^-1=1/|A|xxAdj.A=-1/13[[14,11,-5],[11,4,-3],[-5,-3,-1]]`
`|Adj.A|=14(14-9)-11(-11-15)-5(-33+20)=169ne0`
Co-factor (Adj.A)=`[[-13,-26,-13],[-26-,39,-13],[-13-,13,-65]]`
`(Adj.A)^-1=1/|Adj.A|xxAdj.(Adj.A)=1/69[[-13,-26,-13],[-26-,39,-13],[-13-,13,-65]]=1/13[[-1,2,-1],[-2,3,-1],[-1,-1,-5]]`......(1)
`Adj(A^-1)=Adj.-1/13[[14,11,-5],[11,4,-3],[-5,-3,-1]]`
`=-[[-14/13,-11/13,5/13],[-11/13-,4/13,3/13],[5/13,3/13,1/13]]`
`1/13[[-1,2,-1],[-2,3,-1],[-1,-1,-5]]`.....(2)
from `(1)` and `(2)` `(Adj.A)^-1=Adj(A^-1)`
`|A|^-1=|-1/13[[14,-11,5],[-11,-4,3],[5,3,1]]|`
=`1/13xx1/13xx1/13|[-14,-11,5],[-11,-4,-3],[5,3,1]|`
`=1/13^3xx(-14xx-13+11xx-26+5xx-13)`
`|A^-1|=1/13^3xx-169=-1/13ne0`
Cofactor of `A^-1=1/13^2[[-13,26,-13],[26,-39,-13],[-13,-13,-65]]`
`Adj.of (A^-1)=1/13^2[[-13,26,-13],[26,-39,-13],[-13,-13,-65]]=1/13[[-1,2,-2],[2,-3,-1],[-1,-1,-5]]`
`(A^-1)^-1=1/|A^-1|xxAdj.(A^-1)`
`Adj.(A^-1)=1/(-1/13)xx1/13[[-1,2,-1],[2,-3,-1],[-1,-1,-5]]`
`=[[-1,2,-1],[2,-3,-1],[-1,-1,-5]]=A`
Hence `(A^-1)^-1=A`