Question

`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.

Answer 1

Since, AX =U has infinetly many solutions,
`rArr |A| =0 rArr [{:(a,0, 1),(1,c, b), (1, d,b):}] = 0`
`rArr a(bc-bd)+1(d-c) = 0 rArr (d-c)(ab-1) =0`
`therefore ab =1 "or" d= c`
Again, `|A_(3)| = [{:(a,0, f),(1,c, g), (1, d,h):}] = 0 rArr g = h`
`rArr |A_(2)| = [{:(a,f, 1),(1,g, b), (1, h,b):}] = 0 rArr g = h`
`"and "|A_(1)| = [{:(f,0, 1),(g,c, b), (h, d,b):}] = 0 rArr g = h`
`therefore g =h, c=d " and "ab= 1 " "...(i)`
Now, BX=V
`|B| = [{:(a,1, 1),(0,d, c), (f, g,h):}] = 0 " "["From Eq. (i)"]`
`["since", C_(2)" and"C_(3) "are equal"]`
`therefore` BX =V has no solution.
`|B_(1)| = [{:(a^(2),1, 1),(0,d, c), (0, g,h):}] = 0 " "["from Eq. (i)"]`
[since, c =d and g=h]
`|B_(2)| = [{:(a,a^(2), 1),(0,0, c), (f, 0,h):}] =a^(2) cf =a^(2)df " " [because c=d]`
Since, `adf ne 0 rArr |B_(2)|ne 0`
`|B| =0 " and " |B_(2)|ne 0`
BX = V has no solution.