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
Best Answer
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?