Equivalence of two different matrix multiplications I have a vector $\vec{x}$ and a symmetric, invertible matrix $A$.
I define a vector $\vec{y}$ such that the $i$th element, $\vec{y}_i$, is equal to $$\vec{y}_i = \frac{\sum_j A^{-1}_{ji}}{\sum_{ij}{A_{ij}^{-1}}}$$ which is  the sum of one column of the inverse of $A$, divided by the sum of all elements of the inverse of $A$.
I also define a vector $\vec{z}$ such that the following is true:
$$\begin{bmatrix} z_1  \\ \vdots \\ z_N \\ q\end{bmatrix} = \begin{bmatrix} A_{11} & \cdots & A_{1N} & 1  \\ \vdots  & \ddots & \vdots & \vdots\\ A_{N1} & \cdots & A_{NN} & 1  \\ 1 & \cdots & 1 & 0\end{bmatrix}^{-1}\begin{bmatrix} 0  \\ \vdots \\ 0 \\ 1\end{bmatrix}$$
It turns out that $\vec{y}$ and $\vec{z}$ are equal. How can I see that that must be the case?
For example, if 
$A = \left[ \begin{array}{}
3 & 2 & 0 \\
2 & 4 & 1 \\
0 & 1 & 5
\end{array} \right]$
then $\vec{y} = \vec{z} = \left[ 0.55, 0.1,  0.35 \right]$.
 A: $$\begin{bmatrix} z_1  \\ \vdots \\ z_N \\ q\end{bmatrix} = \begin{bmatrix} A_{11} & \cdots & A_{1N} & 1  \\ \vdots  & \ddots & \vdots & \vdots\\ A_{N1} & \cdots & A_{NN} & 1  \\ 1 & \cdots & 1 & 0\end{bmatrix}^{-1}\begin{bmatrix} 0  \\ \vdots \\ 0 \\ 1\end{bmatrix} $$
is equivalent to $$\begin{bmatrix} A_{11} & \cdots & A_{1N} & 1  \\ \vdots  & \ddots & \vdots & \vdots\\ A_{N1} & \cdots & A_{NN} & 1  \\ 1 & \cdots & 1 & 0\end{bmatrix}\begin{bmatrix} z_1  \\ \vdots \\ z_M \\ q\end{bmatrix} = \begin{bmatrix} 0  \\ \vdots \\ 0 \\ 1\end{bmatrix}$$
which can be rewritten as $z_1 + \cdots + z_N = 1$ and $$Az = -q\begin{bmatrix} 1   \\ \vdots  \\ 1\end{bmatrix}$$
Multiplying by $A^{-1}$ gives $$z = -qA^{-1}\begin{bmatrix} 1  \\ \vdots  \\ 1\end{bmatrix} = -q\begin{bmatrix} A_{11}^{-1} + \cdots + A_{1N}^{-1} \\ \vdots  \\ A_{N1}^{-1} + \cdots + A_{NN}^{-1}\end{bmatrix} = -q\begin{bmatrix} \sum_{j=1}^N A_{1j}^{-1}\\ \vdots  \\ \sum_{j=1}^N A_{Nj}^{-1}\end{bmatrix}$$
and $$1 = \sum_{i=1}^n z_i = -q \sum_{i=1}^N \sum_{j=1}^N A_{ij}^{-1} \implies q = -\frac1{\sum_{i,j=1}^NA_{ij}^{-1}}$$
Therefore $$z = \frac1{\sum_{i,j=1}^NA_{ij}^{-1}} \begin{bmatrix} \sum_{j=1}^N A_{1j}^{-1}\\ \vdots  \\ \sum_{j=1}^N A_{Nj}^{-1}\end{bmatrix} = \frac1{\sum_{i,j=1}^NA_{ij}^{-1}} \begin{bmatrix} \sum_{i=1}^N A_{i1}^{-1}\\ \vdots  \\ \sum_{i=1}^N A_{iN}^{-1}\end{bmatrix} = y$$
We used that $A$ is symmetric to get $\sum_{j=1}^N A_{kj}^{-1} = \sum_{i=1}^N A_{ik}^{-1}$.
