How can the "vector of the unknowns" in matrix be a true vector? How can the "vector of the unknowns" in matrix be a true vector?
In linear algebra,if
$$\begin{cases}
x-3y=-1 \\
x+y=2\end{cases}$$
then the system can be written as $Av=b$, where $A= \begin{bmatrix}1 & -3\\1 & 1\end{bmatrix}$ , $v = \begin{bmatrix}x \\ y\end{bmatrix}$ and 
$b = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$. 
And this $v$ is called the vector of the unknowns.
But a vector should have length and a size, right? So how can this be a true "vector"? $\begin{bmatrix}x \\ y\end{bmatrix}$ does not have a set length or a size; rather, it is inherently undefined, or incomplete, in my opinion. (It is not possible to draw $\begin{bmatrix}x \\ y\end{bmatrix}$). 
 A: The length of the vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is $\sqrt{x^2 + y^2}$.  It's a very specific quantity.
... And, like every other expression containing variables, has a definite value when you bind values to the variables.  (And the resulting length is the same as the length of the likewise bound vector.)

In comments, the OP asks a follow-on quetsion about the direction of the vector $\begin{pmatrix} x \\ y \end{pmatrix}$.  Elaborating on the implicit solution...
From $\sin \theta = \frac{y}{\sqrt{x^2 + y^2}}$, we get $\theta = \sin^{-1}\frac{y}{\sqrt{x^2 + y^2}} + 2 \pi k$ or $\theta = \pi - \sin^{-1} \frac{y}{\sqrt{x^2 + y^2}} + 2\pi k$ for any integer $k$.  This gives two infinite families of solutions, $U$.  From $\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}$, we get $\theta = \pm \cos^{-1} \frac{x}{\sqrt{x^2 + y^2}} + 2\pi k$, for any integer $k$.  This gives the two infinite families of solutions, $V$.  Then the infinite family of solutions $U \cap V = \theta + 2\pi k$, for any integer $k$, gives every angle coterminal with the direction of the vector.
A: The vector of unknowns is not "undefined." The length of the vector is equal to the number of unknowns in the systems of equations. In other words, if A is $m x n$, then $v$ is of dimension $nx1$. The only things that are "incomplete" are the entries of the vector. (But those can be found via multiple methods e.g. elimination, inversion, etc). 
Remember that for matrix multiplication to work the columns of the first matrix should equal the rows of the second. In other words, for matrices $A$, $B$ with dimensions $mxn$ and $cxd$ respectively, the matrix multiplication $A*B$ is defined if and only if $n = c$. (The resulting matrix would be of dimension $mxd$).
EDIT: Direction of a vector: There is no "formula" or a "mathematical" way to express the direction of a vector besides giving the vector's unit vector. Consider the following example: $$v =\begin{bmatrix}6 \\ 8\\\end{bmatrix}$$The length of $v$ would be:$$\sqrt{6^2 + 8^2} = 10$$To find the unit vector of $v$, you would simply divide each entry of $v$ by $10$: $$v_u = \frac {1}{10}* \begin{bmatrix}6 \\ 8\\\end{bmatrix} = \begin{bmatrix}0.6 \\ 0.8\\\end{bmatrix}$$
A: I came to my own conclusion that perhaps the correct answer to my question is that vector is not defined as an entity with a length and a width. Rather, it is defined as an element of a vector space that satisfies the ten vector space axioms. If I understood that was how a vector is defined, then it would have made so much more sense that one does not have to worry about an undetermined length and direction. Ex, doesn’t make sense to say, is v=sinx a vector? What is the direction or length of the vector? But make sense v=sinx is a vector as it fits as an element within a vector space.
A: One way to think about the problem is in term of linear transformations (transformation that keep the origin fixed, and keep parallel lines parallel). Matrices are a convenient way of packaging the information necessary to describe a particular linear transformation -- the first column describes the location that the unit vector in the one direction is sent, and the second column describes where the unit vector in the orthogonal direction is sent; this information is enough to uniquely describe the transformation. So when you look at an equation of the form: $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} =\begin{bmatrix} u \\ v \end{bmatrix} $$ you should read it as asking the question "considering that the linear transformation sends the vector [1,0] to [a,c] and [0,1], what vector [x,y] under the same transformation, will be sent to [u,v]." That way, it's easy to imagine the vector [x,y] as a "roaming" vector, searching for the right length and direction so that the linear transformation will send it to [u,v]. Watch 3blue1browns videos on linear algebra for excellent visualizations of this idea. 
