Matrix multiplication is composition

We know how to apply one matrix at a time, so we do exactly that, starting with the move that happens first.

By Day 4's column picture, $S\yellow{\mathbf x}$ is $1$ times the first column of $S$ plus $1$ times the second column:

Now the quarter turn acts on $(2, 1)$, so we use the entries $2$ and $1$ as weights on the columns of $R$:

The point ends at $(-1, 2)$. We wrote the whole trip as $R(S\yellow{\mathbf x})$, with $S$ closest to $\yellow{\mathbf x}$ because it acts first.

The columns of $M$ must be the places where $\green{\hat\imath}$ and $\red{\hat\jmath}$ end up after both moves, so we follow each arrow through $S$ and then $R$.

Each column of $M$ came from sending a column of $S$ through $R$. That pattern is the whole rule for multiplying matrices.

The columns of $AB$ are $A$ applied to the columns of $B$, so we compute two matrix–vector products.

The first column of $B$ is $\yellow{(1, 1)}$. Its entries are the weights on the columns of $A$:

The second column of $B$ is $\blue{(1, -1)}$:

Stacking the two results as columns gives $AB = \begin{bmatrix} 1 & 3 \\ 2 & 0 \end{bmatrix}$. Each column of the answer is a combination of the columns of $A$, and the matching column of $B$ supplies the weights.

The first column of $AB$ is $A\mathbf b_1$. Call the entries of $\mathbf b_1$ $p$ and $q$. Then we need

Matching entries gives $p + 2q = 5$ and $p - q = -1$. Subtracting the second equation from the first removes $p$ and leaves $3q = 6$, so $q = 2$. Then $p = -1 + q = 1$.

So $\mathbf b_1 = (1, 2)$. Check: $1(1, 1) + 2(2, -1) = (5, -1)$.

The entry in row 2, column 3 is the second entry of column 3, and column 3 of $AB$ is $A\mathbf b_3$ with $\mathbf b_3 = (-1, 3)$:

The entry we want is $15$. Now look at how the $15$ was made: $(-3)(-1) + (4)(3)$. The numbers $-3$ and $4$ are row 2 of $A$, and $-1$ and $3$ are column 3 of $B$. Nothing else was needed.

The turn acts first, so the product is $SR$ with $R$ on the right. Its columns are $S$ applied to the columns of $R$.

The first column of $R$ is $(0, 1)$, and $S(0, 1) = 0\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 1\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

The second column of $R$ is $(-1, 0)$, and $S(-1, 0) = -1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 0\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$.

So $SR = \begin{bmatrix} 1 & -1 \\ 1 & 0 \end{bmatrix}$, and $SR\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

In the other order, $RS = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$ sent $(1, 1)$ to $(-1, 2)$. The same two moves in a different order put the same point in a different place.

By the column rule, $MB$ is zero exactly when $M$ sends each column of $B$ to $\mathbf 0$. So we look for nonzero vectors $(p, q)$ with $M(p, q) = \mathbf 0$.

Both entries are zero when $p + 2q = 0$, because the second entry is twice the first. Taking $q = 1$ gives $p = -2$, and taking $q = -2$ gives $p = 4$. So one answer is

This worked because the columns of $M$ lie on one line, so a combination of them can cancel to zero.

Scale first, then subtract entry by entry. Tripling $A$ gives $\begin{bmatrix} 3 & -6 & 9 \\ 0 & 12 & -3 \end{bmatrix}$. Subtracting $B$ in each position:

Flipping $A$ turns its rows $(1, 2)$ and $(-3, 4)$ into columns, and flipping $B$ does the same with its two rows of length 3:

$B^T$ is $3 \times \underline{2}$ and $A^T$ is $\underline{2} \times 2$, so $B^TA^T$ is defined and is $3 \times 2$. Row by row with the row–column rule:

So $B^TA^T = \begin{bmatrix} 12 & 14 \\ 2 & 4 \\ 5 & 15 \end{bmatrix}$, which is exactly $AB$ flipped. The other order, $A^TB^T$, is $2 \times 2$ times $3 \times 2$, which is not even defined.