I don't understand a step in a solution of the following problem:
Let $A\in Mat_{4x3}(\mathbb{R})$ and suppose all solutions of the equation $xA= 0$ are linear combinations of $x_1= (1,1,1,0)$ and $x_2= (−2,−1,0,1)$. Find its reduced column echelon form and the dimensions of the subspaces ker($T_A$) and im($T_A$).
Solution Since $x_1$ and $x_2$ are linearly independent we have that dim(ker($T_A$)) = 2, hence $rank(A) = 4−2 = 2$. Denote with $a_{ij}$ the entry of the matrix A at row i and column j. Since $x_1$ and $x_2$ are solutions it must hold that:
$$a_{11}+a_{21}+a_{31}=0$$ $$a_{12}+a_{22}+a_{32}=0$$ $$a_{13}+a_{23}+a_{33}=0$$ $$-2a_{11}-a_{21}+a_{41}=0$$ $$-2a_{12}-a_{22}+a_{42}=0$$ $$-2a_{13}-a_{23}+a_{43}=0$$ From this, we can compute the reduced echelon form given by: \begin{equation*} \mathbf{A} = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 0 \\ -2 & -1 & 0 \end{array} \right] \end{equation*} Now, it's not clear to me how we obtain this matrix from the equations above. How do we do this?