If `A=|[3, 2], [7, 5]|` and `B=|[6 ,7], [8, 9]|` , verify that `(A B)^(-1)=B^(-1)A^(-1)` .

Explanation

To Verify `(A B)^(-1)=B^(-1)A^(-1)`

Taking L.H.S.
`(A B)^(-1)`

First calculating `AB`

`AB=[(3, 7),( 2, 5)][(6, 8),( 7, 9)]=[[3(6)+7(7),3(8)+7(9)],[2(6)+5(7),2(8)+5(9)]]`

` " " " =[[18+49,24+63],[12+35,16+45]]=[[67,87],[47,61]]`

Now,
`(A B)^(-1)=1/|AB|adj(AB) " " " " " " "` exists if `|AB|ne0`

`|AB|=|[67,87],[47,61]|=67(61)-47(87)`
`=4087-4089=-2`

Since `|AB|ne0`
`therefore` `(A B)^(-1)` exists

Now,
`AB=[[67,87],[47,61]]`

`adj(AB)=[[61,-87],[-47,67]]`

Thus,
`(A B)^(-1)=1/|AB|adj(AB)`
Putting values

`(A B)^(-1)=1/(-2)[[61,-87],[-47,67]]`


Taking R.H.S.
`B^(-1)A^(-1)`

First calculating `B^(-1)`
`B^(-1)=1/|B|adj(B) " " "` exist if `|B|ne0`

Now,
`|B|=|(6, 8),( 7, 9)|=6(9)-7(8)=54-56=-2`

Since `|B|ne0`
`therefore B^(-1)` exists

Now,
`B=[(6, 8),( 7, 9)]`

`adj(B)=[[9,-8],[-7,6]]`

Thus, `B^(-1)=1/|B|adj(B) `

`B^(-1)=1/(-2) [[9,-8],[-7,6]]`


Calculating `A^(-1)`

`A^(-1)=1/|A|adj(A) " " "` exist if `|A|ne0`

`|A|=|(3, 7),( 2, 5)|=15-14=1`
Since `|A|ne0`, `A^(-1)` exists

`A=[(3, 7),( 2, 5)]`

`adj(A)=[(5, -7),( -2, 3)]`

`A^(-1)=1/|A|adj(A)`

`A^(-1)=1/1[[5,-7],[-2,3]]=[[5,-7],[-2,3]]`


Now,
`B^(-1)A^(-1)=(-1)/2[[9,-8],[-7,6]][[5,-7],[-2,3]]`

` " " " " =(-1)/2[[9(5)+(-8)(-2),9(-7)+(-8)3],[-7(5)+6(-2),-7(-7)+6(3)]]=(-1)/2[[45+16,-63-24],[-35-12,49+18]]`

` " " " " =[[61,-87],[-47,67]]`
` " " " " `= L.H.S.


`therefore` L.H.S. = R.H.S.

Hence Proved