4
$\begingroup$

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.

  • 0
    You have $A:\mathbb{R^4}\rightarrow \mathbb{R^5}$ and $X\in \mathbb{R^5}$, so how you can apply $A$ to $X$?2012-12-18

1 Answers 1

5

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 $\operatorname{rank}(A)=3$.