Matrix Addition, Subtraction and Multiplication

What a matrix is. A matrix is a rectangular array of numbers (its entries or elements) enclosed in brackets. Its order (or dimensions) is written $m \times n$, where $m$ is the number of rows and $n$ is the number of columns. Always read rows first, columns second — "$2 \times 3$" means $2$ rows and $3$ columns.

For example: $A = \begin{pmatrix} 1 & 4 & -2 \\ 0 & 3 & 5 \end{pmatrix} \text{ is } 2\times 3, \qquad B = \begin{pmatrix} 7 \\ -1 \\ 2 \end{pmatrix} \text{ is } 3\times 1.$

We label the entry in row $i$, column $j$ as $a_{ij}$. So in $A$ above, $a_{12} = 4$ and $a_{23} = 5$ (row 2, column 3).

Conformability — the single most important idea in this whole subtopic — is the rule that decides whether an operation is defined:

• Addition and subtraction: the two matrices must have the same order (same number of rows AND same number of columns).
• Multiplication $AB$: the number of columns of $A$ must equal the number of rows of $B$.

If matrices are not conformable for the operation you are attempting, the operation does not exist — you write "not defined", you do not force an answer.

Cambridge tip. Before doing ANY matrix calculation, write the order under each matrix (e.g. "$2\times 3$"). It takes two seconds and instantly tells you whether the operation is legal and what shape the answer will be.

Addition and subtraction are done element by element (component-wise). The two matrices must share the same order, and the result has that same order: $(A + B)_{ij} = a_{ij} + b_{ij}, \qquad (A - B)_{ij} = a_{ij} - b_{ij}.$

For example: $\begin{pmatrix} 1 & 4 \\ 0 & 3 \end{pmatrix} + \begin{pmatrix} 5 & -1 \\ 2 & 6 \end{pmatrix} = \begin{pmatrix} 1+5 & 4+(-1) \\ 0+2 & 3+6 \end{pmatrix} = \begin{pmatrix} 6 & 3 \\ 2 & 9 \end{pmatrix}.$

Scalar multiplication. To multiply a matrix by a single number (a scalar) $k$, multiply every entry by $k$: $kA = \begin{pmatrix} ka_{11} & ka_{12} \\ ka_{21} & ka_{22} \end{pmatrix}.$

So $3\begin{pmatrix} 1 & -2 \\ 4 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -6 \\ 12 & 0 \end{pmatrix}$.

Combining them. Scalar multiples slot straight into sums and differences. An expression such as $2A - 3B$ means "scale $A$ by $2$, scale $B$ by $3$, then subtract element by element" — and this is only defined when $A$ and $B$ have the same order.

Properties (easy ones). Addition is commutative ($A + B = B + A$) and associative ($(A+B)+C = A+(B+C)$), exactly like ordinary numbers. The zero matrix $\mathbf{0}$ (every entry $0$) is the additive identity: $A + \mathbf{0} = A$. These nice properties do not carry over to multiplication, as the next section shows.

Cambridge tip. Mark schemes credit each correct entry, so lay your working out in grid form and double-check signs — a single sign slip in one entry loses an accuracy mark even when the method is perfect.

Matrix multiplication is not element-by-element. Instead, the entry in row $i$, column $j$ of the product $AB$ is found by taking row $i$ of $A$ and column $j$ of $B$, multiplying matching terms and adding: $(AB)_{ij} = \sum_{k} a_{ik}\,b_{kj} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots$

This is exactly the dot product of a row with a column. For this to work, the row of $A$ and the column of $B$ must be the same length — i.e. the columns of $A$ must equal the rows of $B$.

The shape rule. If $A$ is $m\times n$ and $B$ is $n\times p$, then $AB$ is defined and has order $m\times p$: $\underbrace{(m\times n)}_{A}\;\underbrace{(n\times p)}_{B} \;\longrightarrow\; (m\times p).$ The inner numbers ($n$ and $n$) must match — they "cancel"; the outer numbers ($m$ and $p$) give the order of the answer. If the inner numbers differ, $AB$ is not defined.

A full $2\times 2$ example: $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} = \begin{pmatrix} 1\cdot 5 + 2\cdot 7 & 1\cdot 6 + 2\cdot 8 \\ 3\cdot 5 + 4\cdot 7 & 3\cdot 6 + 4\cdot 8 \end{pmatrix} = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix}.$

The same row-into-column rule works for $3\times 3$ products and for non-square products, e.g. a $(2\times 3)(3\times 2)$ gives a $2\times 2$ answer.

Cambridge tip. Write the answer's order first (from the outer dimensions), then fill the entries. If you cannot even state the order, the inner dimensions don't match and the product is "not defined".

Matrix multiplication obeys most of the algebra rules you know — but not commutativity. This is the big conceptual jump from numbers.

1. Non-commutative: $AB \neq BA$ in general. Order matters. Indeed $AB$ and $BA$ may even have different orders, and one may be defined while the other is not. A clean $2\times 2$ counterexample: $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}.$ $AB = \begin{pmatrix} 1\cdot1+2\cdot1 & 1\cdot0+2\cdot1 \\ 0\cdot1+1\cdot1 & 0\cdot0+1\cdot1 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix},$ $BA = \begin{pmatrix} 1\cdot1+0\cdot0 & 1\cdot2+0\cdot1 \\ 1\cdot1+1\cdot0 & 1\cdot2+1\cdot1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}.$ Since $\begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} \neq \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}$, we have $AB \neq BA$. Never assume you can swap the order of a matrix product.

2. Associative: $(AB)C = A(BC)$. You may regroup a product however you like, so writing $ABC$ is unambiguous. (You still may not reorder the factors.)

3. Distributive: $A(B + C) = AB + AC$ and $(B + C)A = BA + CA$. Multiplication spreads over addition — but because order matters, the two distributive laws are genuinely different (left-multiply vs right-multiply by $A$).

4. Scalars float freely. For a scalar $k$: $k(AB) = (kA)B = A(kB)$.

The identity matrix. The $n\times n$ identity $I$ (ones on the leading diagonal, zeros elsewhere) acts like the number $1$: $AI = IA = A$ whenever the product is defined. It is the one matrix that does commute with everything of the right order.

Cambridge tip. When a question says "show that $AB \neq BA$", you MUST compute both products in full and point to a differing entry. Asserting non-commutativity without the two grids scores nothing.