Multiplication

A matrix is a motion. So what is multiplying two of them?

Rotate space, then shear it. Two motions, one after the other. The result is still linear — grid lines stayed straight and evenly spaced through both steps — so some single matrix must describe the combined trip. Finding it is matrix multiplication.

Working one out

Let $A$ shear and $B$ rotate by 90 degrees:

$$A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$$

Apply $B$ first, then $A$. Use the only tool you need: track the basis vectors.

Where does $\hat{\imath}$ end up? $B$ sends it to $(0,1)$, its first column. Now feed $(0,1)$ into $A$, reading $A$ by columns: $0 \cdot (1,0) + 1 \cdot (1,1) = (1,1)$. So $\hat{\imath}$ finishes at $(1,1)$, and that is the first column of the answer.

Where does $\hat{\jmath}$ end up? $B$ sends it to $(-1,0)$. Feed that into $A$: $-1 \cdot (1,0) + 0 \cdot (1,1) = (-1,0)$. Second column.

$$AB = \begin{bmatrix} 1 & -1 \ 1 & 0 \end{bmatrix}$$

Every number you computed answered one question: where does this basis vector end up after both motions? That is all matrix multiplication has ever been.

Two rules that stop being arbitrary

Order matters. $AB \neq BA$ in general, because rotating then shearing is physically a different motion from shearing then rotating. Try both in the visual and watch them disagree. The notation reads right to left: in $AB$, the matrix $B$ acts first, because in $A(Bx)$ the vector meets $B$ first.

Grouping does not. $(AB)C = A(BC)$, and you need no algebra to believe it. Both sides describe the same three motions performed in the same order. How you mentally bracket them cannot change what physically happens.

Everything so far has lived in the plane. Time to see whether any of it depended on that.