My linear algebra book says that a vector is a one-column matrix. However, in the context of what we are studying (linear equations) it would make more sense if a vector was of the form of the augmented matrix:
$$ \left( \begin{matrix} 1 & 0 & 0 & x \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & z \end{matrix} \right) $$
This makes more sense to me because a vector is normally represented as an ordered $n-tuple$, and in the example i gave above its clear that the matrix represents the ordered triple $(x, y, z)$. Is the above the formally correct representation of a vector but for simplicity a vector is just represented as a column matrix?
One can view column vectors as $n\times 1$ matrices $$ u = \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) $$ and row vectors as $1\times n$ matrices. $$ u^T = \left( \begin{matrix} x & y & z \end{matrix} \right) $$ No problem here.
You could map your vectors one to one to your given structure, it effectively acts as matrix $(E|u)$, for the identity matrix $E$ and your vector $u$, due to the rule of multiplication of block matrices, but I fail to see the benefit for those extra components. It is a correct representation of a vector but not a minimal one. So I would not use this.
In other contexts such an extension might make sense, for example in computer graphics:
What you provided reminds a bit of a translation matrix for vectors using homogenous coordinates: $$ T = \left( \begin{matrix} 1 & 0 & 0 & x \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & z \\ 0 & 0 & 0 & 1 \end{matrix} \right) $$ Homogenous coordinates have a fourth dimension, which is (most of the time) normalized to $1$, but serve to extend the transformation matrices to $4\times 4$ matrices. Their benefit is to allow to model the important transformations, even the above mentioned translations, as matrix operations. A combined transformation can be obtained by matrix multiplication.