Once a matrix is put into RREF, it can be re-arranged as R=[I F], the identity variables followed by free variables. The null space matrix is that which, multiplying R, gives the zero vector. In the MIT open courseware lecture, Dr. Strang says this can be expressed as $N=\begin{bmatrix} -F \cr I \end{bmatrix}$, making RN=0. I understand how this works in theory, but in practice I'm having difficulty setting up N once I figure out R. He says the columns of N are the special solutions. I think I'm missing a conceptual intuition. For example, after Gauss-Jordan Elimination, I have the matrix R:

$\begin{bmatrix} 1\text{ }3\text{ }0\text{ }0\text{ }3\cr 0\text{ }0\text{ }1\text{ }0\text{ }0\text{ }9\cr0\text{ }0\text{ }0\text{ }1\text{ -}4 \end{bmatrix}$

If I rearrange columns so the pivot columns are followed by the free columns, what does N look like?

Once a matrix is put into RREF, it can be re-arranged as R=[I F], the identity variables followed by free variables. The null space matrix is that which, multiplying R, gives the zero vector. In the MIT open courseware lecture, Dr. Strang says this can be expressed as $N=\begin{bmatrix} -F \cr I \end{bmatrix}$, making RN=0. I understand how this works in theory, but in practice I'm having difficulty setting up N once I figure out R. He says the columns of N are the special solutions. I think I'm missing a conceptual intuition. For example, after Gauss-Jordan Elimination, I have the matrix R:

$\begin{bmatrix} 1\text{ }3\text{ }0\text{ }0\text{ }3\cr 0\text{ }0\text{ }1\text{ }0\text{ }9\cr0\text{ }0\text{ }0\text{ }1\text{ -}4 \end{bmatrix}$

If I rearrange columns so the pivot columns are followed by the free columns, what does N look like?