Reference Request: Matrix Notation Role Swap It is customary to use matrix notation to find the intersection of two lines in $\mathbb{R}^2$. Consider the special case,
$$ax + by = c$$
$$bx + ay = d$$
$$x,y \in \mathbb{R}; a,b,c,d > 0$$
One would write,
$$
\begin{bmatrix} a & b \\ b & a \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c \\ d\end{bmatrix}
$$
Due to the pattern in the coefficients of x and y, one could also write:
$$
\begin{bmatrix} x & y \\ y & x \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} c \\ d \end{bmatrix}
$$
Applying Cramer's rule to the second matrix (barring some mistake), one obtains the following equations:
$$
\left(x-\frac{c}{2a}\right)^{2}-\left(y-\frac{d}{2a}\right)^{2}=\frac{\left(c^{2}-d^{2}\right)}{4a^{2}}
$$
$$
\left(x-\frac{d}{2b}\right)^{2}-\left(y-\frac{c}{2b}\right)^{2}=\frac{\left(d^{2}-c^{2}\right)}{4b^{2}}
$$
A plot of the resulting hyperbolas, as well as the original lines, is available on this interactive graphing calculator. As you can see from the plot:

*

*Each line in the original system has two axis-intercepts and it shares both intercepts with both hyperbolas in an alternating manner. That is, the first line (containing $ax$) shares its x-intercept with the first hyperbola (containing $c/2a$), and it's y-intercept with the second hyperbola (containing $d/2b$). The second line has the reverse relationship.

*The solution to the original equation (the point where the lines intersect) is also one of the points where the hyperbolas intersect.

For those that are curious, I extended this approach to a different system (the second equation is replaced with $ax^2 + by^2 = d$), and obtained other intersecting geometries. These curves are also available at the above link, but are turned off by default.
Clearly there is a relationship between the geometries of these two representations, and this approach can be used to generate families of curves with very specific properties. Question: has anyone published an exploration of the consequences of re-writing the system this way, and extending it to other problems (e.g. starting from hyperbolas and working backwards)?
Note: this approach does not appear to have obvious applications for solving the original system (i.e. I have not yet found a use for the eigenvalues of the alternative representation, except that they are closely related to the determinant), and a couple of mathematician friends have informed me that this might explain why I haven't found other examples of this approach just yet.
-------- Edit: Example derivations with Cramer's Rule --------
Cramer's Rule enables one to solve for each component of the unknown vector by taking the ratio of a modified matrix and the original coefficient matrix.
Original System:
In order to solve for the first term in the vector of unknowns, the numerator of the ratio of determinants uses a matrix where the first column in the coefficient matrix is replaced by the resultant vector.
$$
x = \frac{ 
det \begin{vmatrix} c & b \\ d & a \end{vmatrix}
}{
det \begin{vmatrix} a & b \\ b & a \end{vmatrix}
}
= \frac {ca - db}{a^2 - b^2}
$$
Since the coefficients are known, one obtains the value of x.
Alternate System:
Similarly, for the first value of the vector (which is known in this case),
$$
a = \frac{ 
det \begin{vmatrix} c & y \\ d & x \end{vmatrix}
}{
det \begin{vmatrix} x & y \\ y & x \end{vmatrix}
}
= \frac {cx - dy}{x^2 - y^2}
$$
From there,
$$
ax^2 - ay^2 = cx - dy
$$
Re-arrange terms,
$$
ax^2 - cx = ay^2 - dy
$$
Divide across by $a$,
$$
x^2 - \frac{c}{a}x = y^2 - \frac{d}{a}y
$$
Then complete the squares,
$$
x^2 - \frac{c}{a}x + \frac{c^2}{4a^2} - \frac{c^2}{4a^2} = y^2 - \frac{d}{a}y + \frac{d^2}{4a^2} - \frac{d^2}{4a^2}
$$
Simplify,
$$
\left(x - \frac{c}{2a} \right) ^2 - \frac{c^2}{4a^2} = \left(y - \frac{d}{2a} \right) ^2 - \frac{d^2}{4a^2}
$$
Re-arrange terms,
$$
\left(x - \frac{c}{2a} \right) ^2 - \left(y - \frac{d}{2a} \right) ^2 = \frac{c^2}{4a^2} - \frac{d^2}{4a^2}
$$
-------- Edit 2: Derivations without Cramer's Rule 2D --------
The way to solve the 2D system without matrix algebra is by substitution. The first step is to multiply each equation by the "other" uncommon constant:
$$
\left( ax+by \right) d =\left( c \right) d
$$
$$
\left( bx+ay \right) c =\left( d \right) c
$$
The equations can be set equal to each other, and after re-arranging, one gets,
$$
\left( ad - bc \right) x = \left( ac - bd \right) y
$$
Since we have the benefit of knowing what we're looking for, we can perform the following substitutions,
$$
x = \frac{\left( ac - bd \right)}{D}, y = \frac{\left( ad - bc \right)}{D}
$$
Substituting $D$ into either equation of the original system and solving for $D$, one quickly finds that $D = a^2 - b^2$ as expected. Thus, we have re-invented the wheel.
Similarly, to generate the hyperbolas, rather than multiplying by the "other" constant, multiply by the variables as follows:
$$
\left( ax+by \right) x =\left( c \right) x
$$
$$
\left( bx+ay \right) y =\left( d \right) y
$$
This is one connection to convolution (i.e. multiplying by polynomials). Since the coefficients on the left-hand side are the same, we will obtain terms that enable substitution on the left hand side as shown in this slightly re-arranged result:
$$
bxy = cx - ax^2
$$
$$
bxy = dy - ay^2
$$
Clearly, by substitution we obtain the intermediate step seen earlier in the Cramer's rule derivations:
$$
cx - ax^2 = dy - ay^2
$$
If I proceed with completing the squares, etc., I obtain one of the hyperbola equations I presented earlier. Alternatively, if I "solve for $a$", then I obtain,
$$
a = \frac{cx - dy}{x^2 - y^2}
$$
which is Cramer's rule for the alternate formulation. To obtain the second hyperbola equation, I multiply the system this way instead:
$$
\left( ax+by \right) y =\left( c \right) y
$$
$$
\left( bx+ay \right) x =\left( d \right) x
$$
That is, I simply reverse the multiplication (1st equation times y rather than x).
NOTE: this also explains why the hyperbolas share their intercepts with the lines in the original system (the linear term is preserved in the multiplication and becomes a factor of the hyperbola).
 A: Let us consider the following $3 \times 3$ system on which it is easier to see the possible generalization :
$$\begin{cases}ax+by+cz&=&u\\
cx+ay+bz&=&v\\
bx+cy+az&=&w\end{cases}\tag{1}$$
It can be written under two different matricial forms :
$$\text{Either }\begin{pmatrix}a&b&c\\c&a&b\\b&c&a\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}u\\v\\w\end{pmatrix}$$
$$\text{or} \ \ \ \ \ \begin{pmatrix}x&y&z\\y&z&x\\z&x&y\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}u\\v\\w\end{pmatrix}\tag{2'}$$
The hidden mathematical property behind these two equivalent matrix representations is the fact that these "forward vs. backwards" circulant matrices correspond to discrete (circular) convolution $\star$, here :
$$\text{Either} \ \underbrace{(a, b, c)}_{\to \text{matrix}} \star \underbrace{(x,y,z)}_{\to \text{vector}} = \underbrace{(u,v,w)}_{\to \text{vector}} \ \text{or} \ \ \underbrace{(x, y, z)}_{\to \text{matrix}} \star \underbrace{(a,b,c)}_{\to \text{vector}} = \underbrace{(u,v,w)}_{\to \text{vector}} $$
Indeed, convolution corresponds to the first signal $abc$ (periodized) sliding to the right over the second one $xyz$ with, at each step, a totalization of products of corresponding terms.
Dualy, one can consider that it is the second signal, $xyz$ which slides to the left in front of the first one, explaining the structure of the second matrix in (2).
$$\begin{array}{cccccc}a&b&c&a&b&c\\ \hline&&&x&y&z\end{array} \ \ \ \text{see equ. (1)i)}$$
$$\begin{array}{cccccc}a&b&c&a&b&c\\ \hline&&x&y&z&\end{array}  \ \ \ \text{see equ. (1)ii)}$$
$$\begin{array}{cccccc}a&b&c&a&b&c\\ \hline&x&y&z&&\end{array}  \ \ \ \text{see equ. (1)iii)}$$
For more, see here.
Remark : I don't address in this analysis the hyperbolas representation, just because I have no idea of the way one the $2 \times 2$ case could be generalized.
