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Let $A$ be a $5×4$ matrix with real enries such that the space of all solutions of the linear system $AX^t=[1,2,3,4,5]^t$ is given by $\{[1+2s,2+3s,3+4s,4+5s]^t:s\in\mathbb{R}\}$. Then the rank of $A$ is equal to

  • $4$
  • $3$
  • $2$
  • $1$

I am completely stuck on it. Can anyone help me please.

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Do you know the Rank-Nullity Theorem? – JohnD Dec 16 '12 at 6:11
yes .but how can i apply this? – ketu Dec 16 '12 at 6:18
Apply it to the matrix $A$. You want to solve for the rank of $A$, and you know the number of columns of $A$. So all that remains is to calculate the nullity of $A$. Can you do that given the information about the linear system? Another way of asking this is: what is the connection between the nullity of a matrix and the solution space of a linear system with that coefficient matrix? – user108903 Dec 16 '12 at 15:17
You have $A:\mathbb{R^4}\rightarrow \mathbb{R^5}$ and $X\in \mathbb{R^5}$, so how you can apply $A$ to $X$? – Mhenni Benghorbal Dec 18 '12 at 14:22

Since the solution space is of dimension $1$, consider the reduced row echelon form of the matrix $A$. It must have exactly $1$ free variable. So $\operatorname{rank}(\operatorname{Null}(A))=1$. Hence $\operatorname{rank}(A)=4-1=3$.

Another way to look at this is there has to be $3$ pivot columns in the ref of $A$, and the corresponding columns of $A$ form a basis for the column space of $A$. Hence $\operatorname{rank}(A)=3$.

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