<s>[INST] Finding the basis for the nullspace, the row space, and the column space of the given matrix. 
Find the basis for the nullspace, the row space, and the column space of the given matrix.$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 3 & 2 & 5 & 2 \\ -1 & 2 & 1 & 3 \\ 1 & 1 & 2 & 2 \end{bmatrix}$$

My Try
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 3 & 2 & 5 & 2 \\ -1 & 2 & 1 & 3 \\ 1 & 1 & 2 & 2 \end{bmatrix}_{R_2\rightarrow R_2-3R_1\\R_3\rightarrow R_3+R_1\\R_4\rightarrow R_4-R_1}$$
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & -4 & -4 & -1 \\ 0 & 4 & 4 & 4 \\ 0 & -1 & -1 & 1 \end{bmatrix}_{R_2\rightarrow \frac{R_2}{-1}\\R_3\rightarrow \frac{R_3}{4}}$$
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & -1 & -1 & 1 \end{bmatrix}_{R_3\rightarrow 4R_3-R_2\\R_4\rightarrow4R_4+R_2}$$
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
I got solution as inconsistent. How to proceed further?
I also referred to $1^{st}$ question of This
 [/INST] We don't say that the RREF is not consistent in that case. 
We need another step to obtain
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix}\to \begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
therefore a basis for the nullspace is given by solving $Ax=0$ that is $$(1,1,-1,0)$$
For the column space, a basis is formed by the columns of the original matrix containing the pivots in the RREF.
For the row space, as a basis we can select the first three rows in the RREF.
can you see why?
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