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

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.
