Matrix Multiplication

How to multiply matrices — rows times columns — and why this operation encodes function composition and systems of linear maps.

Matrix multiplication — (AB)ᵢⱼ = row i of A · col j of B

2×3×3×2=2×2— inner dims match ✓

A (2×3)

120314

B (3×2)

210312

C = AB (2×2)

271014

Hover a result cell to see the row·column dot product

Definition

To multiply matrices $A$ (size $m \times n$ ) and $B$ (size $n \times p$ ), the number of columns of $A$ must equal the number of rows of $B$ . The result $C = AB$ has size $m \times p$ .

The $(i,j)$ entry of $C$ is the dot product of row $i$ of $A$ with column $j$ of $B$ :

$C_{ij} = \sum_{k=1}^n A_{ik} B_{kj}$

Matrix multiplication is:

Associative: $(AB)C = A(BC)$
Distributive: $A(B+C) = AB + AC$
NOT commutative: $AB \neq BA$ in general

Key properties

Associative: $(AB)C = A(BC)$ — a chain of products can be grouped however you like
Distributive over addition: $A(B+C) = AB+AC$ and $(A+B)C = AC+BC$
Scalar compatibility: $c(AB) = (cA)B = A(cB)$ for any scalar $c$
Not commutative: $AB \neq BA$ in general — order matters
The identity matrix $I$ acts as a multiplicative identity: $AI = IA = A$

Common mistakes

Assuming $AB = BA$ : matrix multiplication only commutes in special cases (e.g. both diagonal, or one is the identity) — never assume it by default
Mismatched dimensions: $A_{m\times n}B_{p\times q}$ is only defined when $n = p$ — always check the inner dimensions match
$(AB)^2 \neq A^2B^2$ in general: $(AB)(AB) = ABAB$ , which simplifies to $A^2B^2$ only if $A$ and $B$ commute

2×2 multiplication

$\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}5&6\\7&8\end{pmatrix} = \begin{pmatrix}1\cdot5+2\cdot7 & 1\cdot6+2\cdot8\\3\cdot5+4\cdot7&3\cdot6+4\cdot8\end{pmatrix} = \begin{pmatrix}19&22\\43&50\end{pmatrix}$

Try it

Show that matrix multiplication is not commutative: find $2\times2$ matrices $A$ and $B$ with $AB \neq BA$ .

Solution

$A = \begin{pmatrix}1&0\\0&0\end{pmatrix}$ , $B = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ .

$AB = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ , $BA = \begin{pmatrix}0&0\\0&0\end{pmatrix}$ . So $AB \neq BA$ .

Related concepts

Needs first

Dot Product Matrices Matrix Operations

Determinants Linear Transformations Matrix Inverse

Uses this

Eigenvalues and Eigenvectors

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