Linear equations and matrices

I have seen a system of linear equations represented in both these ways.

(1) $$a_{1}x + b_{1}y = c_{1}$$ $$a_{2}x + b_{2}y = c_{2}$$

(2) $$\underbrace{\begin{bmatrix}a_{1} & b_{1} \\a_{2} & b_{2} \end{bmatrix}}_{A}\underbrace{\begin{bmatrix}x\\y \end{bmatrix}}_{\overrightarrow{x}}=\underbrace{\begin{bmatrix}c_{1} \\c_{2} \end{bmatrix}}_{\overrightarrow{b}}$$

I can see that (1) and (2) are equivalent, since if I carry out the matrix multiplication in (2) I get the system of equations in (1). But I'm not able to get an intuition for why this is so.

The geometric interpretation of (1) is that each equation represents a line in the xy plane and the solution to the system of equations is the point at which the lines intersect. The geometric interpretation of (2) is some vector $\overrightarrow{x}$ in the xy plane such that when the linear transformation A is applied to $\overrightarrow{x}$ it transforms into $\overrightarrow{b}$. I'm not able to visualize why these two interpretations are equivalent. Can somebody please help me out?

Here's one way to think of it:

The points on the line $a_1 x + a_2 y = c_1$ are all points $(x,y)$ that, when acted upon by the transformation, have an $x$-coordinate of $c_1$. In other words, the points that solve the first equation are exactly those that end up on the line $x = c_1$, after the transformation. In other words: if we were to take the line $x = c_1$ and play the transformation in reverse, we would get the line $a_1 x + a_2y = c_1$.

Let's focus on that last perspective. Our transformation takes points from our "starting space" $\Bbb R^2$ to our "target space" $\Bbb R^2$. Begin with the lines $x = c_1$ and $y = c_2$ in the target space of the transformation. Of course, the intersection of these lines is $(c_1,c_2)$. Playing the transformation in reverse brings each line to a line in the "starting space", and the intersection of the lines in the starting space must come from the intersection of the corresponding lines in the target space. In other words, the intersection of the lines in the starting space is the point that gets mapped to $(c_1,c_2)$.

Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be defined as $T\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$, where $a, b, c$, and $d$ are real constants. For a fixed $\begin{pmatrix}c_1 \\ c_2 \end{pmatrix}$, there's only one specific vector $\begin{pmatrix}x \\ y \end{pmatrix}$ such that, under the transformation $T$, $\begin{pmatrix}x\\y\end{pmatrix} \mapsto \begin{pmatrix}c_1\\c_2\end{pmatrix}$.

That is to say that if I fix my constants $c_1$ and $c_2$ and write the linear equations $ax + by = c_1$ and $cx + dy = c_2$, then there's only one specific coordinate where those lines can intersect. Depending on which values of $c_1$ and $c_2$ I choose, the lines will intersect at different places.

Alternatively, if I don't fix $c_1$ and $c_2$, I can generate different values of $c_1$ and $c_2$ by plugging different $(x, y)$ coordinates into the equations of my lines (i.e. choosing different vectors $\begin{pmatrix}x\\y\end{pmatrix}$ on which $T$ will act).

What this means is that, if I choose a specific vector in $\mathbb{R}^2$, then -- given a specific transformation $T$ -- there exists a vector in $\mathbb{R}^2$ that can be transformed into that vector I chose. That is, for any $u \in \mathbb{R}^2$, there exists a $v \in \mathbb{R}^2$ such that $T(v) = u$. This indicates that $T$ is surjective, in which case $T$ maps $\mathbb{R}^2$ onto itself. Likewise, when we choose different $(x, y)$ coordinates, we generate different values of $c_1$ and $c_2$. That is, for any $v_1 \in \mathbb{R}^2$ and $v_2 \in \mathbb{R}^2$ where $v_1 \neq v_2$, one has that $T(v_1) \neq T(v_2)$. This indicates that $T$ is injective.

What this indicates is that $\mathbb{R}^2$ is isomorphic to itself -- that is, we can take any point in $\mathbb{R}^2$ and scale its coordinates, and we'll still be in $\mathbb{R}^2$.

Warning: Of course, these notions generally depend on the entries of the matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ -- I'm making the assumption they're "nice" and give rise to these properties. For a counter-example to what I was talking about, consider the transformation $T\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$. How does $T$ act on $\mathbb{R}^2$? Well, under this transformation, $\begin{pmatrix}x\\y\end{pmatrix} \mapsto \begin{pmatrix}0\\y\end{pmatrix}$. This map isn't a surjection, since I can't generate any vector in $\mathbb{R}^2$ under this transformation -- I can only generate the ones with an $x$-coordinate of zero. In other words, it sends $\mathbb{R}^2$ to the $y$-axis.