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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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en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem is you friend –  clark Dec 7 '12 at 15:34
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You have 5 equations in 4 unknowns, but in the solution set values are specified for 3 variables only. Which of your variables is the free variable? You should be able to compute the rank using the number of variables and the number of free variables. –  Chris Leary Dec 7 '12 at 15:42
    
@ChrisLeary: Sorry. Corrected. –  Sugata Adhya Dec 7 '12 at 16:16
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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 $\dim (\mathrm{Im}\: A) = 3$.

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