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Let $A\in \mathrm{Mat}_{5\times 4}(\mathbb R)$ be such that the space of all solutions of the linear system $AX^t=[1,2,3,4,5]^t$ is given by $\left\{[1+2s,2+3s,3+4s,4+5s]:s\in \mathbb R\right\}$. I need to find $\mathrm{Rank}(A)$.

I don't know where to start. I need some hints.

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    @ChrisLeary: Sorry. Corrected.2012-12-07

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It holds $A^{-1}\bigl((1,2,3,4,5)\bigr) = \ker A + v$ for one vector $v \in A^{-1}\bigl((1,2,3,4,5)\bigr)$ and thus you get $\dim(\ker A )= 1$. Furthermore we know

$ \dim(\ker A) + \dim( \mathrm {Im} \: A ) = \dim \: \mathbb R^4 = 4,$

following $\dim (\mathrm{Im}\: A) = 3$.