`A=[(a,1,0),(1,b,d),(1,b,c)],B=[(a,1,1),(0,d,c),(f,g,h)],U=[(f),(g),(h)],V=[(a^2),(0),(0)]` If there is a vector matrix X, such that `AX = U` has infinitely many solutions, then prove that `BX = V` cannot have a unique solution. If `a f d != 0`. Then,prove that `BX = V` has no solution.

Explanation

Given that `A=[(a,1,0),(1,b,d),(1,b,c)], X=[(x),(y),(z)], U=[(f),(g),(h)]`.
Also, `AX=U` has infinite many solutions.
`implies |A|=0` and `|A_(1)|=|A_(2)|=|A_(3)|`
Now, `|A|=0`
`implies |(a,1,0),(1,b,d),(1,b,c)|=0`
`implies a(bc-bd)-1(c-d)=0`
`implies (ab-1) (c-d)=0`
`implies ab=1` or `c=d` (1)
and `|A_(1)|=|(f,1,0),(g,b,d),(h,b,c)|=0`
`implies f(bc-bd)-1 (gc-hd)=0`
`implies fb(c-d)=gc-hd`
`|A_(2)|=|(a,f,0),(1,g,d),(1,h,c)|=0`
`implies a(gc-hd)-f(c-d)=0`
`implies a(gc-hd)=f(c-d)`
`|A_(3)|=|(a,1,f),(1,b,g),(1,b,h)|=0`
`implies a(bh-bg)-1 (h-g)+f(b-b)=0`
`implies ab(h-g)-(h-g)=0`
`implies ab=1` or `h=g`
For `AX=U` having infinitely many solutions,
`c=d` and `h=g`
Now consider `BX=V`, where `B=[(a,1,1),(0,d,c),(f,g,h)], V=[(a^(2)),(0),(0)]`.
Then `|B|=|(a,1,1),(0,d,c),(f,g,h)|=0`
[ `:'` In view of `c=d` and `g=h, C_(2)` and `C_(3)` are identical]
So, `BX=V` has no unique solution.
Also, `|B_(1)|=|(a^(2),1,1),(0,d,c),(0,g,h)|=0" "( :' c=d, g=h)`
`|B_(2)|=|(a,a^(2),1),(0,0,c),(f,0,h)|=a^(2) cf=a^(2) df" "( :' c=d)`
`|B_(3)|=|(a,1,a^(2)),(0,d,0),(f,g,0)|=-a^(2) df`
Thus, if `adf ne 0` then `|B_(2)|=|B_(3)| ne 0`
Hence, no solution exists.