The order (or dimension) of a matrix is expressed as:
$$m \times n \quad (\text{rows} \times \text{columns})$$
A matrix with $m$ rows and $n$ columns has order $m \times n$ and contains $m \times n$ entries in total.
$$B = \begin{pmatrix} 1 & 4 & 7 \ 2 & 5 & 8 \ 3 & 6 & 9 \ 0 & 1 & 2 \end{pmatrix} \quad \Rightarrow \quad \text{Order: } 4 \times 3$$
$B$ has 4 rows and 3 columns, so 12 entries in total.
The element in row $i$ and column $j$ of matrix $A$ is written $a_{ij}$.
$$A = \begin{pmatrix} 3 & 8 & 1 \ 6 & 2 & 9 \end{pmatrix}$$
| Context | What the order tells you |
|---|---|
| \$3 \times 1$ column matrix | 3 items in a single category |
| \$1 \times 4$ row matrix | 4 attributes of a single item |
| $n \times n$ square matrix | Same number of inputs and outputs |
The product $AB$ is only defined when the number of columns in $A$ equals the number of rows in $B$.
$$A: m \times \mathbf{k} \quad B: \mathbf{k} \times n \quad \Rightarrow \quad AB: m \times n$$
Given:
$$C = \begin{pmatrix} 5 & 0 \ -1 & 3 \ 2 & 7 \end{pmatrix}$$
Can $CD$ be formed if $D$ has order \$2 \times 4$?
Number of columns in $C = 2$ = number of rows in $D = 2$. Yes. The product $CD$ has order \$3 \times 4$.
STUDY HINT: A quick memory aid — “rows come before columns” in the order notation, just as you read left-to-right, top-to-bottom. The subscript $a_{ij}$ follows the same order: $i$ = row, $j$ = column.
VCAA FOCUS: Exam questions often provide two matrices and ask whether their product is defined, and if so, what order the result will be. Check column count of first = row count of second.
