Finding the rank of a matrix Let $A$ be a $5\times 4$ matrix with real entries such that the space of all solutions of the linear system $AX^t = (1,2,3,4,5)^t$ is given by$\{(1+2s,  2+3s, 3+4s, 4+5s)^t :s\in \mathbb{R}\}$ where $t$ denotes the transpose of a matrix. Then what would be the rank of $A$?
Here is my attempt
Number of linearly independent solution of a non homogeneous system of linear equations is given by $n-r+1$ where n refers to number of unknowns and $r$ denotes rank of the coefficient matrix $A$.
Based on this fact, we may write $n-r+1 =1  $. Since there seems only one linearly independent solution (Here i am confused).
Am i right? or How can i do it right?
thanks
 A: Let $b=(1,2,3,4,5)^T$, $x_0 = (1,2,3,4)^T$, and $v_0 = (2,3,4,5)^T$. We know $A x_0 = b$, and that $x'$ is another solution iff $x'-x_0 \in \mathbb{sp}\{v_0\}$. Consequently we have $\ker A  = \mathbb{sp}\{v_0\}$, and hence $\dim \ker A = 1$. Since $A : \mathbb{R}^4 \rightarrow \mathbb{R}^5$, we have
$$\dim \ker A + \mathrm{rank} A = 4.$$
Hence the rank is 3.
A: The number of leading variables plus the number of parameters in a consistent system equals the number of unknowns. Here you have a system with $4$ unknowns and one parameter, so the number of "leading variables" is $3$. The number of leading variables equals the rank of the matrix, so the rank is $3$.
(It makes no sense to say "there is only one linearly independent solution"; every nonzero solution, by itself, is linearly independent; what you have "only one" of is parameters.) 
A: Rank theorem says that if $A$ be the coefficient matrix of a consistent system of linear equations with $n$ variables then
number of free variables (parameters) = $n$ - $rank(A)$
By using this we have $1 = 4 - rank(A)$
Thus $rank(A) = 3$
thanks to Dr Arturo sir for clearing my doubt.
