We already saw in our introductory example how to multiply matrices, but it's complex enough to warrant another look. We multiplied the tables (matrices)

\begin{pmatrix}2 & 0 & 2\\ 1 & 3 & 0\\ 1 & 1 & 2\end{pmatrix}\times\begin{pmatrix}1 & 2\\ 0 & 1\\ 0 & 0\end{pmatrix}

and found

\begin{pmatrix}2 & 4\\ 1 & 5\\ 1 & 3\end{pmatrix}

How did we get this? Recall what the matrices represented:

The first has a row per candidate, and each column j represents how many voters had them as their j^{\th} choice.
The second matrix had a column for each scoring system:
- Column 1: 1 point for first place, 0 for second and third.
- Column 2: 2 points for winning, 1 for second place and 0 for third place.
The final matrix has rows representing candidates, and a column for each voting system.

This is important: when we multiplied a left matrix by a right matrix - L\times R=M

the rows of the output matched the rows of L
the columns of the output matched the columns of R

Why does this happen, and what does it mean for matrix multiplication for generally? When we do a matrix multiplication, we take the i^{\th} row of the first matrix L, and take the dot (scalar) product with the j^{\th} column of the second matrix R. This gives us the entry of the output matrix in the i^{\th} row and j^{\th} column.

<ul>
<li>Center: a 3×3 matrix multiplied by a 3×2 matrix equals a 3×2 matrix.
<ul>
<li>Left matrix: rows are (2, 0, 2), (1, 3, 0), (1, 1, 2). The first row is outlined in orange; the third row is outlined in blue.</li>
<li>Right matrix: columns are (1, 0, 0) and (2, 1, 0). The first column is outlined in orange; the second column is outlined in blue.</li>
<li>Resulting matrix shown as
( 2 4
1 5
1 3 )
with the entry 2 at the top-left colored/orange and the entry 3 at the bottom-right colored/blue.</li>
</ul>
</li>
<li>Top: an orange dot-product computation (2, 0, 2) · (1, 0, 0) = 2 with an orange arrow pointing down to the orange 2 in the result.</li>
<li>Bottom: a blue dot-product computation (1, 1, 2) · (2, 1, 0) = 3 with a blue arrow pointing up to the blue 3 in the result.</li>
<li>Color-coding links: orange highlights pair the first row (left) with the first column (right) and its resulting entry; blue highlights pair the third row with the second column and its resulting entry.</li>
</ul>

The diagram above only shows how 2 entries were calculated, but in actuality there are 6!

	Candidate A	Candidate B	Candidate C
Albert	1st	2nd	3rd
Britney	1st	2nd	3rd
Chaz	3rd	1st	2nd
Denise	3rd	2nd	1st

	1st	2nd	3rd
Candidate A
Candidate B
Candidate C

	1st	2nd	3rd
Candidate A	2	0	2
Candidate B	1	3	0
Candidate C	1	1	2

	Candidate A	Candidate B	Candidate C
Albert	1st	2nd	3rd
Britney	1st	2nd	3rd
Chaz	3rd	1st	2nd
Denise	3rd	2nd	1st

Foundations of Matrices

Foundations of Matrices