Matrix Operations

Main idea

Multiplying two matrices means doing one transformation and then the other: $AB$ is "apply $B$ , then $A$ ." That's why the order matters, and why each entry of $AB$ comes from a row of $A$ dotted with a column of $B$ .

{\htmlData{sym=0}{(AB)_{ij}}} = \sum_k a_{ik}b_{kj}, \qquad AB \ne BA \text{ in general}, \qquad (AB)^T = B^T{\htmlData{sym=1}{A^T}}

What the symbols mean

$(AB)_{ij}$
Entry of the product
Row i of A times column j of B.
$A^T$
Transpose
Rows become columns.

Explanation

A matrix product is simply two transformations done one after the other, packed into a single matrix. It's called "multiplication" because it follows a lot of the same rules as multiplying numbers, though not all of them. It's good for anything that happens in stages, like a robot arm with several joints or a 3D scene that gets moved, then viewed, then projected. The formula $(AB)_{ij} = \sum_k a_{ik}b_{kj}$ looks dense, but it's just this: take row $i$ of $A$ (one row of numbers), take column $j$ of $B$ (one column of numbers), multiply them in pairs and add. That's one entry, and you repeat it for every spot. Bear with the bookkeeping; the idea is simple.

Sizes must match. An $m\times n$ times an $n\times p$ gives $m\times p$ : the inner sizes agree, the outer sizes survive. That makes sense once you see that a row of $A$ and a column of $B$ have to be the same length to be dotted together.

Columns view. You already know that $A\vec x$ sends a vector through the transformation $A$ . Column $j$ of $AB$ is just $A$ times column $j$ of $B$ . So $AB$ takes each column of $B$ and sends it through $A$ , which is why $AB$ means "do $B$ first."

Order matters. $AB$ and $BA$ are usually different. Think about real moves: rotating a book and then sliding its top over isn't the same as sliding first and then rotating. There's an animation further down that shows exactly this with a rotation and a shear.

A common mistake. It's tempting to think matrices behave like numbers, so $AB = 0$ must mean $A = 0$ or $B = 0$ , and $AB = AC$ lets you cancel $A$ . Neither is true. A nonzero $A$ can flatten everything $B$ produces down to zero, and once information is flattened, you can't cancel your way back to it.

Identity. $I$ has 1s on the diagonal and 0s elsewhere. It's the "do nothing" transformation, so $AI = IA = A$ , the same way multiplying a number by 1 leaves it alone.

See it

A vector v sweeps around a circle, and two matrices act on it: B, a 90° rotation, and A, a shear. Rotating then shearing (AB) and shearing then rotating (BA) send v to two different places, and as you can see, the two results never land on the same spot for any direction of v. Matrix multiplication depends on order, just like real moves do.

Worked examples

Example 1

Compute $\begin{bmatrix}1 & 2\\ 3 & 4\end{bmatrix}\begin{bmatrix}5 & 6\\ 7 & 8\end{bmatrix}$ .

1
Dot each row of the first matrix with each column of the second. Row 1 $(1, 2)$ gives $(5 + 14,\; 6 + 16)$ , and row 2 $(3, 4)$ gives $(15 + 28,\; 18 + 32)$ .

$\begin{bmatrix}19 & 22\\ 43 & 50\end{bmatrix}$

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Compute $\begin{bmatrix}1 & 2\\ 3 & 4\end{bmatrix}\begin{bmatrix}5 & 6\\ 7 & 8\end{bmatrix}$ , row by row.

Problem 2+5 XP

Now compute $\begin{bmatrix}5 & 6\\ 7 & 8\end{bmatrix}\begin{bmatrix}1 & 2\\ 3 & 4\end{bmatrix}$ , row by row.

Problem 3+5 XP

A $2\times3$ matrix times a $3\times4$ matrix has what size? Give rows, columns.

Problem 4+5 XP

Give the transpose of $\begin{bmatrix}1 & 2 & 3\\ 4 & 5 & 6\end{bmatrix}$ , row by row.

Problem 5+5 XP

For $A = \begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}$ , compute $A^2$ , row by row.

Problem 6+5 XP

A market stall sells apples at $\$ 0.50 $and bread at$ $3 $. On Monday it sold$ 10 $apples and$ 4 $loaves; on Tuesday$ 6 $apples and$ 8 $loaves. Write the sales as$ Q = \begin{bmatrix}10 & 4\ 6 & 8\end{bmatrix} $and the prices as$ \vec p = \binom{0.5}{3} $. Compute$ Q\vec p$, the revenue each day.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find every matrix $B = \begin{bmatrix}a & b\\ c & d\end{bmatrix}$ that commutes with $A = \begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}$ (so $AB = BA$ ), and give the one with $a = 2$ and $b = 3$ , row by row.

Where it's used

A robot arm's pose is a product of joint matrices, and a 3D scene's vertices go through model, view and projection matrices multiplied together, in that order.

Robot Arm KinematicsLin Alg · Robotics 3D TransformationsLin Alg · Game Development

The Matrix of a Linear Transformation The Inverse of a Matrix