Can someone please provide an intuition behind cramer's rule? See question. I usually get concepts like this very quickly (no studying required), but this one looks like Chinese. Can someone please help me understand a brief intuition behind Cramer's rule for 2x2 and 3x3 matrices? TYVM
 A: Suppose we want to solve $$x\begin{bmatrix}a_0\\a_1\end{bmatrix}+y\begin{bmatrix}b_0\\b_1\end{bmatrix}=\begin{bmatrix}c_0\\c_1\end{bmatrix}$$ for $x$ and $y$. Then we want a linear combination of vectors $\mathbf{a}$ and $\mathbf{b}$ such that they span a parallelogram with outer vertex at the point $c=(c_0,c_1)$ (draw a picture). 
If we achieve this, the parallelograms spanned by 
1) $y\mathbf{b}$ and  $x\mathbf{a}$
2) $y\mathbf{b}$ and $\mathbf{c}$ 
will have the same area because they both have base $\mathbf{b}$ and height equal (consult picture to see why this is the case - for instance http://upload.wikimedia.org/wikipedia/commons/thumb/f/f2/Cramer.jpg/400px-Cramer.jpg)
Because area is given by $2\times 2$ determinants, we get $$\det\begin{vmatrix}\mathbf{c}&y\mathbf{b}\end{vmatrix}=y\det\begin{vmatrix}\mathbf{c}&\mathbf{b}\end{vmatrix}=\det\begin{vmatrix}x\mathbf{a}&y\mathbf{b}\end{vmatrix}=xy\det\begin{vmatrix}\mathbf{a}&\mathbf{b}\end{vmatrix}$$ so that $$x=\frac{\det\begin{vmatrix}\mathbf{c}&\mathbf{b}\end{vmatrix}}{\det\begin{vmatrix}\mathbf{a}&\mathbf{b}\end{vmatrix}}$$ as required.
 This argument can be generalized to any dimension by considering the "volume" of the parallelopiped spanned by the vectors involved.
A: I'd like to provide another way of looking at the problem,
which might help building intuition.
Suppose, as in @NotNotLogical's answer, that $x$ and $y$ satisfy
\begin{equation}
x\begin{bmatrix}a_0\\a_1\end{bmatrix}+y\begin{bmatrix}b_0\\b_1\end{bmatrix}=\begin{bmatrix}c_0\\c_1\end{bmatrix}
\end{equation}
and define the following function, which describes a plane in the variables $a, b$:
\begin{equation}
    \pi(a, b) = a \, x + b \, y.
\end{equation}
By definition $\pi(0, 0) = 0$, and from the equation:
\begin{equation}
    \pi(a_0, b_0) = c_0 \quad \text{and} \quad \pi(a_1, b_1) = c_1,
\end{equation}
so we know three points contained in the plane: $(0, 0, 0)$, $(a_0, b_0, c_0)$, and $(a_1, b_1, c_1)$.
Every other point $(a, b, c)$ of the plane has to be a linear combination of the directing vectors, which can be expressed using a determinant:
\begin{equation}
    \det \begin{bmatrix}a & b & c \\ a_0 & b_0 & c_0 \\ a_1 & b_1 & c_1 \end{bmatrix} = 0.
\end{equation}
Since we know that $(1, 0, x)$ and $(0, 1, y)$ are in the plane, we can write the above equation for these vectors.
Using the method of minors, this gives, in the first case:
\begin{equation}
    \det \begin{bmatrix} b_0 & c_0 \\ b_1 & c_1 \end{bmatrix} + x \, \det \begin{bmatrix} a_0 & b_0 \\ a_1 & b_1 \end{bmatrix} = 0.
\end{equation}
This method generalizes to any dimension,
using a hyperplane instead of a plane.
In dimension two,
the use of determinants can be easily avoided using the usual vector product,
by writing the normal vector of the plane in two different ways and imposing that they be colinear:
\begin{equation}
    \begin{bmatrix} x \\ y \\ -1 \end{bmatrix} = \alpha \, \begin{bmatrix} a_0 \\ b_0 \\ c_0 \end{bmatrix} \times \begin{bmatrix} a_1 \\ b_1 \\ c_1 \end{bmatrix}
\end{equation}
From the last equation, $\alpha$ can be deduced.
