Find a vector $z$ so that system of linear equations $A^{T} \cdot y=d$ has solution if and only if $d^{T} \cdot z=0 \\$ Matrix $A= \begin{bmatrix} 1 & 0 & -1 & 1 \\ 2 & 1 & -3 &4 \\ 1 & -2 & 1&-4 \\ \end{bmatrix}$
I tried to create augmented matrix with $A^{T}$ because of $A^{T} \cdot y=d\space$ (maybe this is wrong ?) $$[A^{T}|d]= \left[ \begin{array}{ccc|c} 1&2&1&d_1\\ 0&1&-2&d_2\\ -1&-3&1&d_3\\ 1&4&-4&d_4\\ \end{array} \right]$$ for some $d=(d_1,d_2,d_3,d_4 \in \Bbb{R})$ $$-$$
Ref of this augmented matrix is $$\left[ \begin{array}{ccc|c} 1&2&1&d_1\\ 0&1&-2&d_2\\ 0&0&-1&d_4-d_1-2d_2\\ 0&0&0&d_1+d_2+d_3\\ \end{array} \right] $$ For sure $d_1+d_2+d_3=0$, because system has to be solvable, and $d_1=-d_2-d_3$ $$-$$ Can I now do multicipation $\space d^{T} \cdot z=0 \\$ like $$\begin{bmatrix} -d_2-d_3 & d_2 & d_3 \\ \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}=0 \\$$ $$-$$ So, my solution will be, $$ x_1=(1/d_3)\cdot x_2 + (1/d_2)\cdot x_3 \\[1ex] \Rightarrow z= \begin{bmatrix} (1/d_3)\cdot x_2 + (1/d_2)\cdot x_3 \\ x_2 \\ x_3 \\ \end{bmatrix}\\$$ If this is correct, can someone help me to solve this but in another dirrection..
Note that $$ A^Ty=d \ \ \text {is solvable}\Longleftrightarrow d\in\mathcal R(A^T) \stackrel{\text{assumption}} {\color{blue}{\Longleftrightarrow}} d\perp z $$ where $\mathcal R(T)$ denotes the range of $T$ and $\mathcal N(T)$ the null space. Thus $\mathcal R(A^T) =\langle z \rangle^\perp$ and this implies that $$ \mathcal N(A) =\mathcal R(A^T)^\perp = \langle z \rangle. $$ We find that $Ax=0$ is solved by $$ x=c \left(\begin{array}{c}1\\1\\1\\0 \end{array} \right). $$ Therefore $z$ can be any vector of the form $c(1,1,1,0)^T$ where $c\ne 0$.