I know that the rank of the matrix must be $3$, since there is exactly one free variable (since exactly one vector spans the null space).
But I am not sure how to reconstruct the exact $RREF$ that would have this vector spanning its null space. All I know is that the last column would be a free variable, and that the other three columns would be linearly independent, since again there is only one variable in the basis of the null space.
I actually would not even think that it is possible to get the $RREF$ with just this information. We can know that $2\times$column$1$ $+ 3 \times $column$2$ $+$column$3$ equals $0$ for all three rows, but how does that uniquely determine a matrix?
Best Answer
Observe that the null space and row space of the matrix are orthogonal complements. So $v=[x,y,z,w]^T$ is a vector in the row-space of the matrix iff$$\langle[2,3,1,0]^T,v\rangle=0\iff 2x+3y+z=0\\v=[x,y,z,w]^T=x[1,0,-2,0]^T+y[0,1,-3,0]^T+w[0,0,0,1]^T$$We want to select three linearly independent row vectors so one possible option is$$\begin{bmatrix}1&0&-2&0\\0&1&-3&0\\0&0&0&1\end{bmatrix}$$Note that we cannot determine the matrix uniquely. We may change the order of the row vectors or take some other linearly independent vectors in the row-space and row-reduce the matrix to obtain another valid choice.