Using Cramer's Rule to Derive Dot Product I'm hoping to derive an equation for the dot product using Cramer's rule.  Here I'm going to try in $\mathbb{R}^2$ and will generalize once I get this first issue cleared. 
I'm hoping to arrive at an expression for the dot product by first asking how to describe some vector $\mathbf{r}$ in terms of two basis vectors $\mathbf{\hat{m}}, \mathbf{\hat{n}}$. If you can, then the coefficients of the linear combination ought to be equal to the projection of $\mathbf{r}$ onto one of its basis vectors. 
Thus by solving the system
$$\mathbf{r} = \alpha\mathbf{\hat{m}} + \beta\mathbf{\hat{n}}$$
in the unknowns $\alpha, \beta$, I want to see that $\alpha = \mathbf{r}\cdot \mathbf{\hat{m}}$ and $\beta = \mathbf{r} \cdot \mathbf{\hat{n}}$
Using Cramer's rule and the fact that $\det A = \det A^T$, we see that
$$\alpha = 
\frac{
\begin{array}{|cc|}
r_x & r_y \\
n_x & n_y
\end{array}
}
{
\begin{array}{|cc|}
m_x&m_y\\
n_x&n_y
\end{array}
} \qquad 
\beta = 
\frac{
\begin{array}{|cc|}
m_x & m_y \\
r_x & r_y
\end{array}
}
{
\begin{array}{|cc|}
m_x&m_y\\
n_x&n_y
\end{array}
} 
$$
Now it seems that $\alpha \ne \mathbf{r}\cdot\mathbf{\hat{m}} = r_xm_x + r_ym_y = 
\begin{array}{|cc|}
r_x & r_y \\
-m_y & m_x
\end{array}
$ and similarly our expectations did not hold for $\beta$. 
Can someone please explain what I am misunderstanding that leads to this unexpected conclusion

This only works if the two basis vectors are orthonormal. Putting that in your assumption presupposes the conclusion. However, there is another way to see it: this should define a new inner product, not necessarily the dot product, in which they are orthonormal.
