My question arises from an excerpt from Chapter 1.4 of Gilbert Strang's "Linear Algebra and its Applications".
Strang illustrates a few points on the titular concept of this post using the following matrix $A$ and vector $x$.
$$A=\ \begin{bmatrix} 1 & 1 & 6 \\ 3 & 0 & 1 \\ 1 & 1 & 4 \\ \end{bmatrix} $$
$$x= \begin{bmatrix} 2 \\ 5 \\ 0 \\ \end{bmatrix} $$
He notes that the standard method, elementary of matrix multiplication, namely $Ax$ by rows, is a valid operation because it recreates the original system of linear equations arising from this problem, thus maintaining an equality.
This method creates this solution:
$$ \begin{bmatrix} 1 & 1 & 6 \\ \end{bmatrix} \begin{bmatrix} 2 \\ 5 \\ 0 \\ \end{bmatrix}\hat i + \begin{bmatrix} 3 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 2 \\ 5 \\ 0 \\ \end{bmatrix}\hat j + \begin{bmatrix} 1 & 1 & 4 \\ \end{bmatrix} \begin{bmatrix} 2 \\ 5 \\ 0 \\ \end{bmatrix}\hat k $$
Whilst the $Ax$ by columns, interpretation, which, I am guessing, has the $\hat i, \hat j$ and $\hat k$ components of $x$ scale the set of $\hat i, \hat j$ and $\hat k$ components of matrix $A$ produces:
$$2\left( \begin{array}{c} 1 \\ 3\\ 1 \end{array} \right)+ 5\left( \begin{array}{c} 1 \\ 0\\ 1 \end{array} \right)+ 0\left( \begin{array}{c} 6 \\ 3\\ 7 \end{array} \right).$$
Both should produce the vector $b= 7\hat i + 6\hat j + 7\hat k$, but I'm a bit confused as to how the $Ax$ by rows algorithm makes sense. For example, the row $$\begin{bmatrix} 1 & 1 & 6 \\ \end{bmatrix}$$ describes the $LHS$ of a plane equation in $R^3$. Why would treating this as a vector, and multiplying it by $x$ produce the $\hat i$ scalar of $b$? And why can we define a vector as a column or a row of a matrix? It seems odd that a $3x3$ matrix can have $2$ different sets of vectors yet still describe the same matrix.