Matrix multiplication memorisation

Question

So I'm writing an exam about matrices in a few weeks time, and I'd like to know if anybody has any tips about multiplying matrices.

Answer 1

There are a couple of different ways to think about matrix multiplication. Here is what I usually do.

To multiply two matrices $A$ and $B$, you should see $A$ as a stack of rows and $B$ as a sequence of columns. I.e. think in the following way:

$$ A = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\a_n \end{pmatrix} ~~~~ B = \begin{pmatrix} b_1 & b_2 & \cdots & b_k \end{pmatrix} $$

Note that $k$ can be different from $n$, but the $a$ and $b$-vectors need to have the same length for the multiplication to be valid.

Now you only have to remember how to do a dot product since your matrix product will be: $$AB = \begin{pmatrix} a_1 \cdot b_1 & a_1 \cdot b_2 & \cdots & a_1 \cdot b_k \\ a_2 \cdot b_1 & a_2 \cdot b_2 & \cdots & a_2 \cdot b_k \\ \vdots & \vdots & \vdots & \vdots \\ a_n \cdot b_1 & a_n \cdot b_2 & \cdots & a_n \cdot b_k \end{pmatrix} $$ and the rule to remember is that the element at row $i$ and column $j$ will be the dot product of $a_i$ and $b_j$. I usually then calculate $AB$ row by row, keeping the $a$-vector fixed and changing the $b$-vector as I progress over the column. Then I pick the next $a$, calculate the dot product with each $b$ over the columns, etc.

An example. Let $$A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} ~~~~ B = \begin{pmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \end{pmatrix} $$ so with the notation above (I don't suggest you do this when doing the calculations, this is just for clarity): $$\begin{align} a_1 &= \begin{pmatrix} 1 & 2 \end{pmatrix} \\ a_2 &= \begin{pmatrix} -1 & 1 \end{pmatrix} \end{align}$$ and $$\begin{align} b_1 &= \begin{pmatrix} 2 \\ 1 \end{pmatrix} \\ b_2 &= \begin{pmatrix} 3 \\ 1 \end{pmatrix} \\ b_3 &= \begin{pmatrix} -1 \\ 1 \end{pmatrix} \end{align}$$

So, the first row of $AB$ will be (note that we only use $a_1$ here): $$\begin{align} (AB)_{1,1} &= a_1 \cdot b_1 = 1 \cdot 2 + 2 \cdot 1 = 4 \\ (AB)_{1,2} &= a_1 \cdot b_2 = 1 \cdot 3 + 2 \cdot 1 = 5 \\ (AB)_{1,3} &= a_1 \cdot b_3 = 1 \cdot (-1) + 2 \cdot 1 = 1 \end{align}$$

Then we move on to the second row (here we only use $a_2$): $$\begin{align} (AB)_{2,1} &= a_2 \cdot b_1 = -1 \\ (AB)_{2,2} &= a_2 \cdot b_2 = -2 \\ (AB)_{2,3} &= a_2 \cdot b_3 = 2 \end{align}$$

I usually do these dot products in my head as I go along. As a more physical remainder of how to do things, I usually trace over the $a$-row I am using with the pencil (moving the pencil horizontally), then trace over the $b$-column I am using (moving the pencil vertically).

I like this method because you only need to keep one accumulating value in your head at a time. When you're done with a dot product, write it down and move to the next.

Also, if you think you have calculated the element at row $i$ and column $j$ of $AB$ wrong, it's easy to think in reverse: $(AB)_{i,j}$ is just the dot product of row $a_i$ and column $b_j$. Calculate it again and check the result.