How can I introduce "directions" $i$, $j$, and $k$ in a linear system to more "explicitly" create visualizations in Cartesian coordinates? I'm self-studying linear algebra using Professor Strang's online videos, and thought what he calls the "column interpretation" of linear systems was neat. That is, looking at
\begin{align*}
\begin{bmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
2\\
4\\
8
\end{bmatrix}
\end{align*}
as
\begin{align*}
\begin{bmatrix}
1\\
4\\
7
\end{bmatrix}x+
\begin{bmatrix}
2\\
5\\
8
\end{bmatrix}y+
\begin{bmatrix}
3\\
6\\
9
\end{bmatrix}z=
\begin{bmatrix}
2\\
4\\
8
\end{bmatrix}
\end{align*}
To visualize the above, he drew each column as a vector, e.g. $(1, 4, 7)$ drawn as an arrow from the origin to the point $(1, 4, 7)$. Then he reformulated the problem as finding a "balance" of vectors, $(1, 4, 7)$, $(2, 5, 8)$, and $(3, 6, 9)$ that "add up" (laying tails to heads) to point to the RHS column, $(2, 4, 8)$.
This must be very basic to everyone here, but it kinda blew my mind. It made $(x, y, z)$ "mere" coefficients, stripped of any "dimensionality," if that makes sense. And it was pleasing to me look at it as multiplying both sides of the system by a "directional unit vector" or whatever it's called:
\begin{align*}
\begin{bmatrix}
i & 0 & 0\\
0 & j & 0\\
0 & 0 & k
\end{bmatrix}
\begin{bmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
i & 0 & 0\\
0 & j & 0\\
0 & 0 & k
\end{bmatrix}
\begin{bmatrix}
2\\
4\\
8
\end{bmatrix}
\end{align*}
Which multiplies out to:
\begin{align*}
\begin{bmatrix}
1i & 2i & 3i\\
4j & 5j & 6j\\
7k & 8k & 9k
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
2i\\
4j\\
8k
\end{bmatrix}
\end{align*}
Which again in the "column interpretation" splits out to:
\begin{align*}
\begin{bmatrix}
1i\\
4j\\
7k
\end{bmatrix}x+
\begin{bmatrix}
2i\\
5j\\
8k
\end{bmatrix}y+
\begin{bmatrix}
3i\\
6j\\
9k
\end{bmatrix}z=
\begin{bmatrix}
2i\\
4j\\
8k
\end{bmatrix}
\end{align*}
Maybe I'm being too pedantic, but for me it's eye-opening to see that the jump from symbols to Cartesian coordinates is something I've taken for granted in my education, but only works because it was pre-selected to work (or it wouldn't be taught that way). I realize that $i$, $j$, and $k$ are just more arbitrary symbols, i.e. what makes me put faith in them following algebraic rules? I honestly don't know the answer to that. But until I'm shown I'm wrong, I feel that using $i$, $j$, $k$ is helping me to think about the translation from symbols to graphs with a little more rigor.
So, okay. Here's my question then.
How do I formulate the classic, "row interpretation" using $i$, $j$, and $k$?
The linear system breaks out to these equations:
\begin{align*}
x + 2y + 3z &= 2\\
4x + 5y + 6z &= 4\\
7x + 8y + 9z &= 8
\end{align*}
So my first puzzlement was, what "directional units" does the RHS use?
But let's take a simpler, 2-dimensional example:
$$y=2x+7$$
Intuitively, I'd insert the following:
$$y=2\frac{j}{i}x+7j$$
And this seems to align with what we're taught: 2 is a slope, rise over run, and 7 is a y-intercept.
Inverting this comes out nicely too:
$$x=\frac{1}{2}\frac{i}{j}y+\frac{7}{2}i$$
But okay, now how do I put it into a form where all the variables are on the LHS? This is the furthest I can get:
$$-2xj+yi=7ij$$
(In case it's not clear, I'm trying to work backward with algebra to a point that I can see how start from the linear system and inject some $i$, $j$, and $k$'s to make the row interpretation work.)
But the $x$ and $j$ being paired together doesn't make any intuitive sense to me. I suppose I can divide both sides by $ij$:
$$-2x\frac{1}{i}+y\frac{1}{j}=7$$
But I'm still lost on the meaning of it. (What is the reciprocal of a direction?) This also gets crazy when I add more dimensions, where each dimension's coefficient needs to be divided by a product of every other dimension.
To summarize, I would like to know if there's a way to use directional units $i$, $j$, and $k$ to more "explicitly" create the corresponding spatial representations in Cartesian space (rather than just taking it for granted and drawing them), and maybe more importantly, whether what I'm doing is actually useful toward thinking about things more precisely, or I've merely added a superficial layer of symbols that happen to match up by coincidence and I should abandon this.
 A: You could work with the elements
$$
\mathbf e_1=\begin{pmatrix}1\\0\\0\end{pmatrix}, \quad \mathbf e_2=\begin{pmatrix}0\\1\\0\end{pmatrix},\quad \mathbf e_3=\begin{pmatrix}0\\0\\1\end{pmatrix}.
$$
These are a orthonormal basis for $\mathbb R^3$. What does that mean? Orthonormality means that the inner product of $\mathbf e_i$ and $\mathbf e_j$ is $1$ if $i=j$ and $0$ otherwise. Basis of $\mathbb R^3$ means that all three dimensional vectors are given by linear combinations of these three elements (further reading https://en.wikipedia.org/wiki/Basis_(linear_algebra))
So what's a row decomposition? Well then we just work with the elements
$$
\mathbf e_1^T=\begin{pmatrix}1&0&0\end{pmatrix}, \quad \mathbf e_2^T=\begin{pmatrix}0&1&0\end{pmatrix},\quad \mathbf e_3^T=\begin{pmatrix}0&0&1\end{pmatrix}.
$$
Let's just see this at work for the column vectors. Take the equation
\begin{equation}
\begin{pmatrix}1&0&1\\0&0&1\\1&2&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\2\\4\end{pmatrix}.
\end{equation}
Since the $\mathbf e_i$ form a basis, we can rewrite this as
\begin{equation}
(x+z)\mathbf e_1+z\mathbf e_2+(x+2y)\mathbf e_3=\mathbf e_1+2\mathbf e_2+4\mathbf e_3.
\end{equation}
The linear independence property of basis vectors means that the coefficients must match. This is how we get 
\begin{equation}\begin{aligned}x+z&=1,\\z&=2,\\x+2y&=4.\end{aligned}\end{equation}
To do a row decomposition, we can just work with the transposition of our equation
\begin{equation}
\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}1&0&1\\0&0&2\\1&1&0\end{pmatrix}=\begin{pmatrix}1&2&4\end{pmatrix}.
\end{equation}
Can you see how to decompose this into combinations of $\mathbf e_1^T,\mathbf e_2^T$ and $\mathbf e_3^T$?
I would like to add, for one, two, four (and eight sort of) dimensions we have the division algebras. This means this talking of reciprocal can be made rigorous. The dimensions correspond to the real numbers, the complex numbers, the quaternions and the octonions. However, in all other dimensions, the inverse of a direction is not a well-defined quantity. https://en.wikipedia.org/wiki/Division_algebra
