Let $z \in \mathbb R^{4}$. Say I am given:
$A=\begin{pmatrix}−1 & −1 & 0 & −6 \\ −6 &−2& 3& −11 \\ 1& 7& −10& −2 \\ 3& 1& −1& 12\end{pmatrix}$
and $c:=\begin{pmatrix} 47\\ 93\\ 17\\ −93 \end{pmatrix}$
Let $y:=\begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \\ y_{4}\end{pmatrix}$ be the solution to $Ay=c$
It follows that $y=\begin{pmatrix} -3 \\ -2 \\ -2 \\ -7\end{pmatrix}$.
I am told that $c^{T}z=\sum_{i = 1}^{4}y_{i}A_{\{i\}\cdot}z$ but I do not see why this is the case, note that $A_{\{i\}\cdot}$ denotes the $i-$th row.
I know that $y$ signfies the vector containing the scalar quantities that form a linear combination of the columns(!) of $A$, meaning:
$y_{1}\begin{pmatrix}-1\\-6\\1\\3\end{pmatrix}+y_{2}\begin{pmatrix}-1\\-2\\7\\1\end{pmatrix}+y_{3}\begin{pmatrix}0\\3\\-10\\-1\end{pmatrix}+y_{4}\begin{pmatrix}-6\\-11\\-2\\12\end{pmatrix}=\begin{pmatrix}47\\93\\17\\-93\end{pmatrix}$
But $A_{\{i\}\cdot}$ represents the $i-$th row, so how does $c^{T}z=\sum_{i = 1}^{4}y_{i}A_{\{i\}\cdot}z$ make any sense at all?
Write $$ Ay = c. $$ Transpose to get $$ c^t = y^t A^t $$ Check to see that this means that $$ c^{t} = \sum_{i = 1}^{4}y_{i} A_{\{i\}} $$ Now noting that both $c^t$ and the expression on the right hand side are $1 \times 4$ row vectors, take a matrix product with an arbitrary $4 \times 1$ column vector $z$ to get \begin{align} c^{t}z &= \biggl(\sum_{i = 1}^{4}y_{i} A_{\{i\}}\biggr) z\\ &= \sum_{i = 1}^{4}\biggl(y_{i} A_{\{i\}}z \biggr) & \text{by linearity of matrix mult'n}\\ &= \sum_{i = 1}^{4}y_{i} \biggl(A_{\{i\}}z \biggr) & \text{by linearity of matrix mult'n again} \end{align}
Your text has perhaps misused the centered dot notation, i.e., $\cdot$, to indicate a matrix product rather than a dot-product, or maybe it defines dot-products between row and column vectors. Either way, this is what's really meant.