# matrix multiplication by columns

Yo,

I need some help with understanding matrix multiplication by columns. Consider the two matrices:

$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 6 & 5 & 4 \\ 7 & 8 & 9 \end{array} \right) \left( \begin{array}{ccc} 3 & 2 & 1 \\ 4 & 5 & 6 \\ 9 & 8 & 7 \end{array} \right)$

So I'm familiar with the standard algorithm where element $AB_{ij}$ is found by multiplying the $i^{th}$ row of A with the $j^{th}$ column of B.

Apparently there is another way to multiply matrices where you work with whole columns of A to get the product AB. Does anyone know how to do that? If so, could you please provide a general algorithm? I've never heard of it and I can't find it anywhere.

You can find it in the starting lectures of Gilbert Strang. Anyway, here is just a teaser for you $$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 6 & 5 & 4 \\ 7 & 8 & 9 \end{array} \right)\begin{pmatrix}x\\y\\z \end{pmatrix} = \begin{pmatrix} 1\\6\\7 \end{pmatrix}x+ \begin{pmatrix} 2\\5\\8 \end{pmatrix}y+ \begin{pmatrix} 3\\4\\9 \end{pmatrix}z$$ How 'bout that?
$$\begin{pmatrix} x & y & z \end{pmatrix} \cdot \left( \begin{array}{ccc} 1 & 2 & 3 \\ 6 & 5 & 4 \\ 7 & 8 & 9 \end{array} \right) = x\cdot \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} + y\cdot \begin{pmatrix} 6 & 5 & 4 \end{pmatrix} + z\cdot \begin{pmatrix} 7 & 8 & 9 \end{pmatrix}.$$