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This question is not a duplicate of the original, in which user Shuchang proved the question. Presently I'm asking about further intuition or a picture, and no proofs please.

$1.$ Intuitively, in (Row), how and why can $\mathbf{A}$ left-multiply into the column vector form of $\mathbf{B}$ ?
In (Coln), how and why can $\mathbf{B}$ right-multiply the row vector form of $\mathbf{A}$ ?

$2.$ Would someone please provide a like picture? I'd like to perceive all this intuitively/naturally; I can follow the algebra in Shuchang Zhang's answer, but want to stop /brooking it.

For relief, I denote all row vectors with superscripts and column with subscripts. Define $\mathbf{A} = \left[\begin{matrix} \vec{a^1} \\ \vdots \\ \vec{a^i} \\ \vdots \\ \vec{a^m} \end{matrix}\right]_{m \times n} \& \quad \mathbf{B} = \left[\vec{b_1} \cdots \vec{b_i} \cdots \vec{b_p}\right]_{n \times p}$ ,
where each of the $m$ $\vec{a^i}$s has size $(1 \times n)$ and each of the $p$ $\vec{b_i}$s size $(n \times 1)$. Then:

$\mathbf{AB} = \mathbf{A}\left[\vec{b_1} \cdots \vec{b_i} \cdots \vec{b_p}\right]_{n \times p} = \left[\mathbf{A}\vec{b_1} \cdots \mathbf{A}\vec{b_i} \cdots \mathbf{A}\vec{b_p}\right]_{m \times p}. \tag{Row}$

$\text{ Also, } \qquad \mathbf{AB} = \left[\begin{matrix} \vec{a^1} \\ \vdots \\ \vec{a^i} \\ \vdots \\ \vec{a^m} \end{matrix}\right]\mathbf{B} = \left[\begin{matrix} \vec{a^1}\mathbf{B} \\ \vdots \\ \vec{a^i}\mathbf{B} \\ \vdots \\ \vec{a^m}\mathbf{B} \end{matrix}\right]_{m \times p}. \tag{Coln}$

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Have you thought about the usefulness of this "matrix multiplication" in compactly expressing a system of linear equations? –  hardmath May 26 at 20:18

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