`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

AX =U has infiinite many solutions
`rArr abs(A) = 0 = abs(A_(1))= abs(A_(2))=abs(A_(3))`
Now, `abs(A) = 0 `
`rArr abs((a,1,0),(1,b,d),(1,b,c)) = 0 rArr (ab-1)(c-d)=0`
`rArr ab = 1 or c=d" "...(i)`
and `abs(A_(1))=0`
`rArr [[f,1,0],[g,b,d],[h,b,c]]=0`
`rArr fb(c-d) - gc + hd = 0`
`rArr fb(c-d)=gc-hd` ...(i)
`rArr abs(A_(2))=0`
`rArr abs((a,f,0),(1,g,d),(1,h,c)) = 0`
`rArra (gc-dh)-f(c-d)=0`
` rArr a(gc-dh)=f(c-d)` ...(iii)
`abs(A_(3))=0`
`rArr [[a,1,f],[1,b,g],[1,h,c]]=0`
`rArr (h-g)(ab-1)=0`
`rArr h=g or ab =1 ` ...(iv)
Taking `c= d rArr h = g and ab ne 1` (from Eqs. (i). (ii) and (iv))
Now, thaing `BX=V,`
Then, `abs(B) = abs((a,1,1),(0,b,c),(f,g,h))=0`
[`because` In view of c = d and g = h, `c_(2) and c_(3)` are identical]
`rArr BX=V` has no unique solution.
and `abs(B_(1))=abs((a^(2),1,1),(0,d,c),(0,g,h))=0 " "[because c= d, g= h]`
`abs(B_(2))=abs((a,a^(2),1),(0,0,c),(f,0,h))=a^(2)fc =a^(2)df " " [because c=d]`
and `abs(B_(3))=abs((a,1,a^(2)),(0,d,0),(f,g,0))=-a^(2)df `
If `a^(2)df ne 0, ` then `abs(B_(2)) = abs(B_(3))ne 0` Hence, on solution exist.