Given `A=[(1,1,1),(2,4,1),(2,3,1)], B=[(2,3),(3,4)].` Find P such that BPA= `[(1,0,1),(0,1,0)]`

Explanation

Given, BPA=`[(1,0,1),(0,1,0)]`
` therefore " " p=b^(-1)[(1,0,1),(0,1,0)]A^(-1)`
`therefore " " B=[(2,3),(3,4)] rArr B^(-1)=(1)/((-1))[(4,-3),(-3,2)]=[(-4,3),(3,-2)]`
`therefore" " B^(-1)=[(-4,3),(3,-2)]`
and ` A= [(1,1,1),(2,4,1),(2,3,1)`
`therefore" " |A|=1(4-3)-1(2-2)+1(6-8)=-1!=0`
Now, adj `A=[(1,2,-3),(0,-1,1),(-2,-1,2)]`
`therefore" " A^(-1)=(adjA)/(|A|)=[(-1,-2,3),(0,1,-1),(2,1,-2)]`
substiruting the values of `A^(-1)` and `B^(-1)` from Eqs. (ii) and (iii) in Eq. (i), then
`P=[(-4,3),(3,-2)][(1,0,1),(0,1,0)][(-1,-2,3),(0,1,-1),(2,1,-2)]`
`P=[(-4,3),(3,-2)][(1,-1,1),(0,1,-1)][(-4,7,-7),(3,-5,5)]`