Most linear algebra books do an OK job of showing the connection. This leads me to believe that you may be treating this connection with more reverence than it merits. If you don't mind here's a quick explanation that may show the beginnings of the connection. Please forgive me if I've misunderstood you and this is too elementary for what you need:
If there's one common thread in linear algebra it's the solving of this equation:
$A \mathbf{x} = \mathbf{b}$
where $\mathbf{x} \in \mathbb{R}^n$ and $\mathbf{b} \in \mathbb{R}^m$ are vectors and $A$ is a linear function with domain in $\mathbb{R}^n$ and range in $\mathbb{R}^m$. Much of your time in linear algebra is spent exploring various ways of solving this equation. Many of these methods depend on simplifying $A$. This is very algeraic in nature. So where is the geometry?
Nestled somewhere in the linear algebra curriculum you always learn that the function $A$ can always be represented by a matrix and once you've done this you can write our equation above in much more detail. It's here that the geometry becomes apparent. Here's a quick and dirty example:
Suppose we've already figured out a matrix for $A$ and we have the following:
$\left( \begin{array}{cc} 1 & 0 \\ 2 & 1 \\ \end{array} \right) \left( \begin{array}{c} x \\ y \\ \end{array} \right) = \left( \begin{array}{c} 4 \\ 0 \\ \end{array} \right)$
Somethings going on with my TEX here but I hope you can make out what that's supposed to be. Once we do the matrix multiplication we get: $ x + y = 4$ $ 2x + y = 0$
These are both equations for lines. A solution of the original problem $\mathbf{x} = (x,y)$ is then a point $(x,y)$ satisfying both equations. We know this to be the intersection of the two lines. And there's your geometry plain and simple. For larger matrices and vectors the geometric picture is more complicated. For instance if we were to try a 3 by 3 matrix A the corresponding problem viewed geometrically would be the intersection of 3 planes.
Hope this helps.