For the matrices A and B, verify that (AB)′=B′A′, where (i) `A=[{:(-1),(4),(3):}],` `B=[-1" " 2" " 1]`

Explanation

Taking L.H.S.
`(AB)'`
Finding AB first
`AB=[[1],[-4],[3]]_(3xx1)[[-1,2,1]]_(1xx3)`
`AB=[[1xx(-1),1xx2,1xx1],[-4xx(-1),-4xx2,-4xx1],[3xx(-1),3xx2,3xx1]]_(3xx3)`
`AB=[[-1,2,1],[4,-8,-4],[-3,6,3]]`
So, `(AB)'=[[-1,4,-3],[2,-8,6],[1,-4,3]]`
Taking R.H.S.
Finding `B',A'`
Given `B=[[-1 ,2, 1]]`
`therefore`
`B'=[[-1 ,2, 1]]'=[[-1],[2],[1]]`
and `A=[[1],[-4],[ 3]]`
`therefore`
`A'=[[1,-4,1]]`
`B'A'=[[-1],[2],[1]]_(3xx1)[[1,-4,1]]_(1xx3)`
`B'A'=[[-1xx1,-1xx(-4),-1xx3],[2xx1,2xx(-4),2xx3],[1xx1,1xx(-4),1xx3]]_(3xx3)`
`B'A'=[[-1,4,-3],[2,-8,6],[1,-4,3]]`= L.H.S
L.H.S= R.H.S
Hence Proved
(ii) Taking L.H.S.
`(AB)'`
Finding AB first
`AB=[[0],[1],[2]]_(3xx1)[[1,5,7]]_(1xx3)`
`AB=[[0xx1,0xx5,0xx7],[1xx1,1xx5,1xx7],[2xx1,2xx5,2xx7]]_(3xx3)`
`AB=[[0,0,0],[1,5,7],[2,10,14]]`
`(AB)'=[[0,1,2],[0,5,10],[0,7,14]]`
Taking R.H.S.
Finding `B',A'`
Given `B=[[1 ,5, 7]]`
`therefore`
`B'=[[1 ,5, 7]]'=[[1],[5],[7]]`
and `A=[[0],[1],[ 2]]`
`therefore`
`A'=[[0,1,2]]`
`B'A'=[[1],[5],[7]]_(3xx1)[[0,1,2]]_(1xx3)`
`B'A'=[[1xx0,1xx1,1xx2],[5xx0,5xx1,5xx2],[7xx0,7xx1,7xx2]]_(3xx3)`
`B'A'=[[0,1,2],[0,5,10],[0,7,14]]`= L.H.S
L.H.S= R.H.S
Hence Proved