I've got a solution for this problem. I just wanna confirm this as my text book
doesn't provide answers for all questions.

Let A be a 5x4 matrix with real entries such that the space of all solutions of the linear system

AXt = $\displaystyle \begin {bmatrix} 1\\2\\3\\4\\5 \end{bmatrix} $ is given by

{$\displaystyle \begin {bmatrix} 1+2s\\2+3s\\3+4s\\4+5s\end{bmatrix} : s \epsilon R $}.

Then the rank of the matrix A is
a) 4 b) 3 c) 2 d)1


My argument is as follows.

i) The rank of the matrix A is 4 atmost.
ii) Since the solution matrix has one free variable, the rank cannot be 4 and has to be 4-1 = 3.

Let me know if there is a flaw in my argument.